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Unformatted text preview: College Algebra
√ Version 3 = 1.7320508075688772 . . .
by Carl Stitz, Ph.D.
Lakeland Community College Jeff Zeager, Ph.D.
Lorain County Community College Modified by
Joel Robbin and Mike Schroeder
University of Wisconsin, Madison
June 29, 2010 Table of Contents
Preface
0 Basic Algebra
0.1 The Laws of Algebra
0.2 Kinds of Numbers .
0.3 Exponents . . . . . .
0.4 Absolute Value . . .
0.5 Solving Equations .
0.6 Exercises . . . . . .
0.7 Answers . . . . . . . v .
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. 1
1
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14 1 Coordinates
1.1 The Cartesian Coordinate Plane
1.1.1 Distance in the Plane . .
1.1.2 Exercises . . . . . . . . .
1.1.3 Answers . . . . . . . . . .
1.2 Relations . . . . . . . . . . . . .
1.2.1 Exercises . . . . . . . . .
1.2.2 Answers . . . . . . . . . .
1.3 Graphs of Equations . . . . . . .
1.3.1 Exercises . . . . . . . . .
1.3.2 Answers . . . . . . . . . .
1.4 Three Interesting Curves . . . . .
1.4.1 Circles . . . . . . . . . . .
1.4.2 Parabolas . . . . . . . . .
1.4.3 Ellipses . . . . . . . . . .
1.4.4 Exercises . . . . . . . . .
1.4.5 Answers . . . . . . . . . . .
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. 2 Functions
77
2.1 Introduction to Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
2.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 iv Table of Contents 2.2 2.3 2.4 2.5 2.1.2 Answers . . . . .
Function Notation . . .
2.2.1 Exercises . . . .
2.2.2 Answers . . . . .
Function Arithmetic . .
2.3.1 Exercises . . . .
2.3.2 Answers . . . . .
Graphs of Functions . .
2.4.1 General Function
2.4.2 Exercises . . . .
2.4.3 Answers . . . . .
Transformations . . . .
2.5.1 Exercises . . . .
2.5.2 Answers . . . . . ......
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. 3 Linear and Quadratic Functions
3.1 Linear Functions . . . . . . . . . . .
3.1.1 Exercises . . . . . . . . . . .
3.1.2 Answers . . . . . . . . . . . .
3.2 Deﬁning Functions (Word Problems)
3.2.1 Exercises . . . . . . . . . . .
3.2.2 Answers . . . . . . . . . . . .
3.3 Quadratic Functions . . . . . . . . .
3.3.1 Exercises . . . . . . . . . . .
3.3.2 Answers . . . . . . . . . . . .
3.4 Inequalities . . . . . . . . . . . . . .
3.4.1 Exercises . . . . . . . . . . .
3.4.2 Answers . . . . . . . . . . . .
4 Polynomial Functions
4.1 Graphs of Polynomials . . . .
4.1.1 Exercises . . . . . . .
4.1.2 Answers . . . . . . . .
4.2 The Factor Theorem and The
4.2.1 Exercises . . . . . . .
4.2.2 Answers . . . . . . . . .
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Remainder Theorem
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............ 5 Rational Functions
5.1 Introduction to Rational Functions
5.1.1 Exercises . . . . . . . . . .
5.1.2 Answers . . . . . . . . . . .
5.2 Graphs of Rational Functions . . .
5.2.1 Exercises . . . . . . . . . . .
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. 295
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329 7 Exponential and Logarithmic Functions
7.1 Introduction to Exponential and Logarithmic Functions
7.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . .
7.1.2 Answers . . . . . . . . . . . . . . . . . . . . . . .
7.2 Properties of Logarithms . . . . . . . . . . . . . . . . . .
7.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . .
7.2.2 Answers . . . . . . . . . . . . . . . . . . . . . . .
7.3 Exponential Equations and Inequalities . . . . . . . . .
7.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . .
7.3.2 Answers . . . . . . . . . . . . . . . . . . . . . . .
7.4 Logarithmic Equations and Inequalities . . . . . . . . .
7.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . .
7.4.2 Answers . . . . . . . . . . . . . . . . . . . . . . .
7.5 Applications of Exponential and Logarithmic Functions
7.5.1 Applications of Exponential Functions . . . . . .
7.5.2 Applications of Logarithms . . . . . . . . . . . .
7.5.3 Exercises . . . . . . . . . . . . . . . . . . . . . .
7.5.4 Answers . . . . . . . . . . . . . . . . . . . . . . . .
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395
397 8 Systems of Equations
8.1 Systems of Linear Equations: Gaussian Elimination
8.1.1 Exercises . . . . . . . . . . . . . . . . . . . .
8.1.2 Answers . . . . . . . . . . . . . . . . . . . . .
8.2 Systems of Linear Equations: Augmented Matrices*
8.2.1 Exercises . . . . . . . . . . . . . . . . . . . .
8.2.2 Answers . . . . . . . . . . . . . . . . . . . . .
8.3 Determinants and Cramer’s Rule* . . . . . . . . . .
8.3.1 Deﬁnition and Properties of the Determinant
8.3.2 Cramer’s Rule . . . . . . . . . . . . . . . . .
8.3.3 Exercises . . . . . . . . . . . . . . . . . . . . .
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437 5.3 5.2.2 Answers . . . . . . . . . . . . .
Rational Inequalities and Applications
5.3.1 Exercises . . . . . . . . . . . .
5.3.2 Answers . . . . . . . . . . . . . 6 Further Topics in Functions
6.1 Function Composition . .
6.1.1 Exercises . . . . .
6.1.2 Answers . . . . . .
6.2 Inverse Functions . . . . .
6.2.1 Exercises . . . . .
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. vi Table of Contents 8.4 8.3.4 Answers . . . .
Systems of NonLinear
8.4.1 Exercises . . .
8.4.2 Answers . . . . ......
Equations
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and Inequalities
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. 488 Laws of Algebra Proved
The Laws of Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Analogy between Addition and Multiplication . . . . . . . . . . . . . . . . . .
Consequences of the Distributive Law . . . . . . . . . . . . . . . . . . . . . . . . . 491
. 491
. 491
. 496 9 Sequences and Series
9.1 Sequences . . . . . . . . . . . . . . . .
9.1.1 Exercises . . . . . . . . . . . .
9.1.2 Answers . . . . . . . . . . . . .
9.2 Series and Summation Notation . . . .
9.2.1 Exercises . . . . . . . . . . . .
9.2.2 Answers . . . . . . . . . . . . .
9.3 IRAs and Mortgages . . . . . . . . . .
9.3.1 Exercises . . . . . . . . . . . .
9.3.2 Answers . . . . . . . . . . . . .
9.4 Inﬁnite sums and Repeating Decimals*
9.4.1 Exercises . . . . . . . . . . . . .
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. 10 Complex Numbers and the Fundamental Theorem
10.1 Complex Numbers . . . . . . . . . . . . . . . . . . .
10.2 The Fundamental Theorem of Algebra . . . . . . . .
10.2.1 Exercises . . . . . . . . . . . . . . . . . . . .
10.2.2 Answers . . . . . . . . . . . . . . . . . . . . .
A The
A.1
A.2
A.3 of Algebra
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. Preface
This book is a modiﬁed version of the Open Source Precalculus Project initiated by Carl Stitz and
Jeﬀ Seager. The original version is available at
http://www.stitzzeager.com/Free_College_Algebra_Book_Download.html.
As indicated on that website you may go to
http://www.lulu.com/product/paperback/collegealgebra/11396948
to order a lowcost, royalty free printed version of the book from lulu.com. Neither author receives
royalties from lulu.com, and, in most cases, it is far cheaper to purchase the printed version from
lulu than to print out the entire book at home.
The version you are viewing was modiﬁed by Joel Robbin and Mike Schroeder for use in
Math 112 at the University of Wisconsin Madison. A companion workbook for the course is being
published by Kendall Hunt Publishing Co. 4050 Westmark Drive, Dubuque, IA 52002. Neither Joel
Robbin nor Mike Schroeder nor anyone else at the University of Wisconsin receives any royalties
from sales of the workbook to UW students.
The original version of this book contains the following acknowledgements:
The authors are indebted to the many people who support this project. From Lakeland Community College, we wish to thank the following people: Bill Previts, who
not only class tested the book but added an extraordinary amount of exercises to it;
Rich Basich and Ivana Gorgievska, who class tested and promoted the book; Don Anthan and Ken White, who designed the electric circuit applications used in the text;
Gwen Sevits, Assistant Bookstore Manager, for her patience and her eﬀorts to get the
book to the students in an eﬃcient and economical fashion; Jessica Novak, Marketing
and Communication Specialist, for her eﬀorts to promote the book; Corrie Bergeron,
Instructional Designer, for his enthusiasm and support of the text and accompanying
YouTube videos; Dr. Fred Law, Provost, and the Board of Trustees of Lakeland Community College for their strong support and deep commitment to the project. From
Lorain County Community College, we wish to thank: Irina Lomonosov for class testing
the book and generating accompanying PowerPoint slides; Jorge Gerszonowicz, Kathryn
Arocho, Heather Bubnick, and Florin Muscutariu for their unwavering support of the
project; Drs. Wendy Marley and Marcia Ballinger, Lorain CCC, for the Lorain CCC
vii viii Table of Contents
enrollment data used in the text. We would also like to extend a special thanks to
Chancellor Eric Fingerhut and the Ohio Board of Regents for their support and promotion of the project. Last, but certainly not least, we wish to thank Dimitri Moonen, our
dear friend from across the Atlantic, who took the time each week to email us typos
and other corrections. Chapter 0 Basic Algebra
0.1 The Laws of Algebra Terminology and Notation. In this section we review the notations used in algebra. Some
are peculiar to this book. For example the notation A := B indicates that the equality holds
by deﬁnition of the notations involved. Two other notations which will become important when
we solve equations are =⇒ and ⇐⇒ . The notation P =⇒ Q means that P implies Q i.e.
“If P , then Q”. For example, x = 2 =⇒ x2 = 4. (Note however that the converse statement
x2 = 4 =⇒ x = 2 is not always true since it might be that x = −2.) The notation P ⇐⇒ Q
means P =⇒ Q and Q =⇒ P , i.e. “P if and only if Q”. For example 3x − 6 = 0 ⇐⇒ x = 2.
The notations =⇒ and ⇐⇒ are explained more carefully in Section 0.5 below.
Implicit Multiplication. In mathematics the absence of an operation symbol usually indicates
multiplication: ab mean a × b. Sometimes a dot is used to indicate multiplication and in computer
languages an asterisk is often used.
ab := a · b := a ∗ b := a × b
Order of operations. Parentheses are used to indicate the order of doing the operations: in
evaluating an expression with parentheses the innermost matching pairs are evaluated ﬁrst as in
((1 + 2)2 + 5)2 = (32 + 5)2 = (9 + 5)2 = 142 = 196.
There are conventions which allow us not to write the parentheses. For example, multiplication is
done before addition
ab + c
means (ab) + c and not a(b + c),
and powers are done before multiplication:
ab2 c means a(b2 )c and not (ab)2 c. In the absence of other rules and parentheses, the left most operations are done ﬁrst.
a−b−c means (a − b) − c and not a − (b − c). 2 Basic Algebra The long fraction line indicates that the division is done last:
a+b
c means (a + b)/c and not a + (b/c). In writing fractions the length of the fraction line indicates which fraction is evaluated ﬁrst:
a
b
c means a/(b/c) and not (a/b)/c, a
b means (a/b)/c and not a/(b/c).
c
The length of the horizontal line in the radical sign indicates the order of evaluation:
√
√
a+b
means (a + b) and not ( a) + b.
√
√
a+b
means ( a) + b and not (a + b).
The Laws of Algebra. There are four fundamental operations which can be performed on
numbers.
1. Addition. The sum of a and b is denoted a + b.
2. Multiplication. The product of a and b is denoted ab.
3. Reversing the sign. The negative of a is denoted −a.
4. Inverting. The reciprocal of a (for a = 0) is denoted by a−1 or by 1
.
a These operations satisfy the following laws. Associative
Commutative
Identity
Inverse
Distributive a + (b + c) = (a + b) + c
a + (b + c) = (a + b) + c
a+0=0+a=a
a + (−a) = (−a) + a = 0
a(b + c) = ab + ac a(bc) = (ab)c
a(bc) = (ab)c
a·1=1·a=a
a · a−1 = a−1 · a = 1
(a + b)c = ac + bc The operations of subtraction and division are then deﬁned by a − b := a + (−b) a ÷ b := a
1
:= a · b−1 = a · .
b
b 0.1 The Laws of Algebra 3 All the rules of calculation that you learned in elementary school follow from the above fundamental laws. In particular, the Commutative and Associative Laws say that you can add a bunch
of numbers in any order and similarly you can multiply a bunch of numbers in any order. For
example, (A + B ) + (C + D) = (A + C ) + (B + D), (A · B ) · (C · D) = (A · C ) · (B · D). Because both addition and multiplication satisfy the commutative, associative, identity, and
inverse laws, there are other analogies: (i) −(−a) = a (a−1 )−1 = a (ii) −(a + b) = −a − b (ab)−1 = a−1 b−1 (iii) −(a − b) = b − a (iv) (a − b) + (c − d) = (a + c) − (b + d) ac
ac
·=
bd
bd (v) a − b = (a + c) − (b + c) ac
a
=
b
bc (vi) (a − b) − (c − d) = (a − b) + (d − c) a/b
ad
=·
c/d
bc a
b −1 = b
a These identities1 are proved in Appendix A.2 Here are some further identities which are proved
using the distributive law. 1 An identity is an equation which is true for all values of the variables which appear in it. 4 Basic Algebra (i) a · 0 = 0 (ii) −a = (−1)a (iii) a(−b) = −ab (iv) (−a)(−b) = ab (v) ac
ad + cb
+=
bd
bd (vi) (a + b)(c + d) = ab + ad + bc + bd (vii) (a + b)2 = a2 + 2ab + b2 (viii) (a + b)(a − b) = a2 − b2 These proved in Appendix A.3.
An important consequence of the fact that a · 0 = 0 · a = 0 is the following ZeroProduct
Property. It is used to solve equations. ZeroProduct Property
pq = 0 ⇐⇒ p = 0 or q = 0 (or both). Proof: If p = 0 (or q = 0) then pq = 0. Conversely, if p = 0, then q = p−1 pq = p−1 0 = 0.
Definition 0.1. For a natural number n and any number a the nth power of a is
an := a · a · a · · · a
n factors
The zeroth power is
a0 := 1
and negative powers are deﬁned by
a−n := 1
.
an The Laws of Exponents. The following laws are easy to understand when m and n are integers.
In Theorem 0.1 below we will learn that these laws also hold whenever a and b are positive real
numbers and m and n are any real numbers, not just integers. 0.1 The Laws of Algebra (i)
(ii)
(iii)
(iv)
(v) 5 am an = am+n
(am )n = amn
am
= am−n
an
(ab)m = am bm
a m am
=m
b
b Example 0.1.1. Simplify
Solution.
11
+
ab a2 a3 = (aa)(aaa) = a5
(a2 )3 = (aa)(aa)(aa) = a6
a2
1
= a−3 = 3
a5
a
(ab)2 = (ab)(ab) = (aa)(bb) = a2 b2
a2 a a
aa
a2
=·=
=2
b
bb
bb
b e.g.
e.g.
e.g.
e.g.
e.g. 11
+
ab
−1 = −1 1
=
11
+
ab ab
11
+
ab =
ab ab
ab
.
=
ab ab
b+a
+
a
b Three Column Calculations. An algebraic calculation often involves substituting expressions
for letters in general laws. To avoid making mistakes it is advisable to arrange the computation
neatly and use equal signs between quantities which you assert are equal. When you check your
work, ask yourself at each step what general principle you used and how you substituted into that
general expression. The following three column calculation illustrates this technique.2
step
12
+
3x x
2
+
3x x
x
6
+
=
3x 3x
x+6
=
3x = by
A
AC
=
B
CB
A
CA
=
B
CB
AB
A+B
+
=
C
C
C with
A = 1, B = 3, C = x
A = 2, B = x, C = 3
A = x, B = 6, C = 3x When in Doubt. If you are in doubt as to whether some general equation is true you can plug
in numbers: if the two sides of the equation are not equal the general equation is false.3 Thus, in
general,
(A + B )2 = A2 + B 2
2
We *don’t* expect this level of detail on exams; we *do* expect you to do calculations like this without making
mistakes.
3
But maybe not conversely! 6 Basic Algebra because when A = 2 and B = 3 we have (A + B )2 = (2 + 3)2 = 52 = 25 but A2 + B 2 = 22 + 32 =
4 + 9 = 13 and 25 = 13. The correct general law is
(A + B )2 = A2 + 2AB + B 2
(when A = 2 and B = 3 A2 + 2AB + B 2 = 4 + 12 + 9 = 25 = 52 = (2 + 3)2 ) and this shows that
(A + B )2 = A2 + B 2 =⇒ 2AB = 0 so that (A + B )2 = A2 + B 2 only when A = 0 or B = 0 (or
both). 0.2 Kinds of Numbers We distinguish the following diﬀerent kinds of numbers.
The natural numbers are 1, 2, 3 . . ..
The integers are . . . − 3, −2, −1, 0, 1, 2, 3 . . ..
The rational numbers are ratios of integers like 3/2, 14/99, −1/2.
The real numbers are numbers which have an inﬁnite decimal expansion like 3
= 1.5000 . . . ,
2 14
= 0.141414 . . . ,
99 √ 2 = 1.4142135623730951 . . . . The complex numbers are those numbers of form z = x + iy where x and y are real numbers
and i is a special new number called the imaginary unit which has the property that i2 = −1;
Every integer is a rational number (because n = n/1), every rational number is a real number (see
Remark 0.2 below), and every real number is a complex number (because x = x + 0i). A real
number which is not rational is called irrational.
New Numbers  New Solutions. Each kind of number enables us to solve equations that the
previous kind couldn’t solve:
The solution of the equation x + 5 = 3 is x = −2 which is an integer but not a natural
number.
The solution of the equation 5x = 3 is x = 3
5 which is a rational number but not an integer.
√
√
The equation x2 = 2 has two solutions x = 2. The number 2 is a real number but not a
rational number.
The equation x2 = 4 has two real solutions x = ±2 but the equation z 2 = −4 has no real
solutions because the square of a nonzero real number is always positive. However it does
have two complex solutions, namely z = ±2i. 0.2 Kinds of Numbers 7 We will not use complex numbers until Chapter 10 but may refer to them implicitly as in
The equation x2 = −4 has no (real) solution.
Rational Numbers  Repeating Decimals. It will be proved in Theorem 9.3 that a real
number is rational if and only if its decimal expansion eventually repeats periodically forever as in
the following examples:
1
= 0.3333 . . . ,
3 17
= 2.83333 . . . ,
6 7
= 1.250000 . . . ,
4 22
= 3.142857 142857 142857 . . . .
7 Unless the decimal expansion of a real number is eventually zero, as in 1 = 0.5000 . . ., any ﬁnite
2
part of the decimal expansion is close to, but not exactly equal to, the real number. For example
1.414 is close to the square root of two but not exactly equal:
√
(1.414)2 = 1.999396 = 2,
( 2)2 = 2.
If we compute the square root to more decimal places we get a better approximation, but it still
isn’t exactly correct:
(1.4142135623730951)2 = 2.00000000000000014481069235364401.
√
The square root of 2 is Irrational. Here is a proof that 2 is irrational. If it were rational
there would be integers m and n with
m2
= 2.
n
By canceling common factors we may assume that m and n have no common factors and hence
that they are not both even. Now m2 = 2n2 so m2 is even so m is even, say m = 2p. Then
4p2 = (2p)2 = m2 = 2n2 so 2p2 = n2 so n2 is even so n is even. This contradicts the fact m and n
are not both even.
The Number Line. The choice of two points (representing 0 and 1) on a line determines a
correspondence between the points of the line and the real numbers as indicated in the following
picture.
−5 −4 −3 −2 −1 0 1 2 3 4 5 The correspondence is called a coordinate system on the line. The line is called a number line.
When the point A corresponds to the number a we say that the number a is the coordinate of
the point A. The positive numbers are the real numbers on the same side of 0 as 1 and the
negative numbers are on the other side. We usually draw the number line as above so that it is 8 Basic Algebra horizontal and 1 is to the right of 0. We write say a is less than b and write a < b b is to the right
of a, i.e. when b − a is positive. it is equivalent to say that b is greater than a or a to the left of
b and to write b > a. The notation a ≤ b means that a is less than or equal to b i.e. either a < b
or else a = b. Similarly, b ≥ a means that b is greater than or equal to a i.e. either b > a or else
b = a. Thus when a < b, a number 4
c is between a and b ⇐⇒ a < c < b.
Sometimes we insert the word strictly for emphasis: a is strictly less than b means that a < b (not
just a ≤ b).
Order. The order relation just described is characterized by the following.
(Trichotomy) Every real number is either positive, negative, or zero (and no number satisﬁes two
of these conditions).
(Sum) The sum of two positive numbers is positive.
(Product) The product of two positive numbers is positive.
This characterization together with the notation explained in the previous paragraph implies the
following: (i) Either a < b, a = b, or a > b. (ii) If a < b and b < c, then a < c. (iii) If a < b, then a + c < b + c. (iv) If a < b and c > 0, then ac < bc. (v) If a < b and c < 0, then ac > bc.
1
1
If 0 < a < b, then 0 < < .
b
a (vi) Interval Notation. The open interval (a, b) is the set of all real numbers x such that a < x < b,
and the closed interval [a, b] is the set of all real numbers x such that a ≤ x ≤ b. Thus
x is in the set (a, b) ⇐⇒ a < x < b
and
x is in the set [a, b] ⇐⇒ a ≤ x ≤ b.
4 The notation ⇐⇒ is an abbreviation for “if and only if”. 0.3 Exponents 9 These notations are extended to include half open intervals and unbounded intervals as in
x is in the set (a, b] ⇐⇒ a < x ≤ b, x is in the set (a, ∞) ⇐⇒ a < x, x is in the set (−∞, a] ⇐⇒ x ≤ a, etc. The union symbol ∪ is used to denote a set consisting of more than one interval as in
x is in the set (a, b) ∪ (c, ∞) ⇐⇒ either a < x < b or else c < x.
The symbol ∞ is pronounced inﬁnity and is used to indicate that an interval is unbounded. It is
not a number so we never write (c, ∞].
√
Example 0.2.1. Which is bigger: π or 10? (Don’t use a calculator.)
Solution. π = 3.14 . . . < 3.15. and
3.152 = (3 + 0.15)2 = 32 + 2 × 3 × 0.15 + 0.152 = 9 + 0.90 + 0.0225 = 9.9225 < 10
√
so π < 3.15 < 10. 0.3 Exponents The proof of the following theorem requires a more careful deﬁnition of the set of real numbers
than we have given and is best left for more advanced courses.
Theorem 0.1. Suppose that a is a positive real number. Then there is one and only one way
to deﬁne ax for all real numbers x such that
(i) ax+y = ax · ay , a0 = 1, a1 = a, 1x = 1. (ii) If a > 1 and x < y then ax < ay .
(iii) If a < 1 and x < y then ax > ay .
With this deﬁnition, the laws of exponents in Paragraph 0.1 continue to hold when a and b
are positive real numbers and m and n are arbitrary real numbers. The number ax is positive
(when a is positive) regardless of the sign of x. In particular by property (v) in Paragraph 0.1 we have (ax )y = axy so (am/n )n = am and
(am )1/n = am/n . Hence for positive numbers a and b we have
b = am/n ⇐⇒ bn = am . 10 Basic Algebra When m = 1 and n is a natural number the number a1/n is called the nth root (square root if
n = 2 and cube root if n = 3) and is sometimes denoted
√
n a := a1/n . When n is absent, n = 2 is understood:
√ a := a1/2 . nth roots A number b is said to be an nth root of a iﬀ bn = a. When n is odd, every real number
√
a has exactly one (real) nth root and this is denoted by n a. When n is even, a positive real number
√
a has two (real) nth roots (and n a denotes the one which is positive) but a negative number has
no real nth roots. (In trigonometry it is proved that every nonzero complex number has exactly n
distinct complex nth roots.)
The equation b2 = 9 has two solutions, namely b = 3 and b = −3 and each is “a” square root of
9 but only b = 3 is “the” square root of 9. However −2 is the (only) real cube root of −8 because
(−2)3 = −8. The number −9 has no real square root (because b2 = (−b)2 > 0 if b = 0) but does
have two complex square roots (because (3i)2 = (−3i)2 = −9). For most of this book5 we only use
real numbers and we say that
√
a is undeﬁned when a < 0
and that
you can’t take the square root of a negative number.
√
√
Also a always denotes the nonnegative square root: thus (−3)2 = 32 = 9 but 9 = 3 and
√
9 = −3. 0.4 Absolute Value There are a few ways to describe what is meant by the absolute value x of a real number x. You
may have been taught that x is the distance from the real number x to the 0 on the number. So,
for example, 5 = 5 and  − 5 = 5, since each is 5 units from 0 on the number line.
distance is 5 units −5 −4 −3 −2 −1 distance is 5 units 0 1 2 3 4 5 √
Another way to √
deﬁne absolute value is by the equation x = x2 . Using this deﬁnition, we
√
have 5 = (5)2 = 25 = 5 and  − 5 = (−5)2 = 25 = 5. The long and short of both of these
procedures is that x takes negative real numbers and assigns them to their positive counterparts
5 More precisely until Chapter 10 0.4 Absolute Value 11 while it leaves positive numbers alone. This last description is the one we shall adopt, and is
summarized in the following deﬁnition.
Definition 0.2. The absolute value of a real number x, denoted x, is given by −x, if x < 0
x = x, if x ≥ 0 In Deﬁnition 0.2, we deﬁne x using a piecewisedeﬁned function. (See page 93 in Section 2.2.)
To check that this deﬁnition agrees with what we previously understood as absolute value, note
that since 5 ≥ 0, to ﬁnd 5 we use the rule x = x, so 5 = 5. Similarly, since −5 < 0, we use the
rule x = −x, so that  − 5 = −(−5) = 5. This is one of the times when it’s best to interpret the
expression ‘−x’ as ‘the opposite of x’ as opposed to ‘negative x.’ Before we embark on studying
absolute value functions, we remind ourselves of the properties of absolute value.
Theorem 0.2. Properties of Absolute Value: Let a, b, and x be real numbers and let n be
an integer.a Then
Product Rule: ab = ab
Power Rule: an  = an whenever an is deﬁned
Quotient Rule: a
a
=
, provided b = 0
b
b The Triangle Inequality: a + b ≤ a + b
x = 0 if and only if x = 0.
For c > 0, x = c if and only if x = c or x = −c.
For c < 0, x = c has no solution.
a Recall that this means n = 0, ±1, ±2, . . . . The proof of the Product and Quotient Rules in Theorem 0.2 boils down to checking four cases:
when both a and b are positive; when they are both negative; when one is positive and the other
is negative; when one or both are zero. For example, suppose we wish to show ab = ab. We
need to show this equation is true for all real numbers a and b. If a and b are both positive, then
so is ab. Hence, a = a, b = b, and ab = ab. Hence, the equation ab = ab is the same as 12 Basic Algebra ab = ab which is true. If both a and b are negative, then ab is positive. Hence, a = −a, b = −b,
and ab = ab. The equation ab = ab becomes ab = (−a)(−b), which is true. Suppose a is
positive and b is negative. Then ab is negative, and we have ab = −ab, a = a and b = −b.
The equation ab = ab reduces to −ab = a(−b) which is true. A symmetric argument shows the
equation ab = ab holds when a is negative and b is positive. Finally, if either a or b (or both)
are zero, then both sides of ab = ab are zero, and so the equation holds in this case, too. All
of this rhetoric has shown that the equation ab = ab holds true in all cases. The proof of the
Quotient Rule is very similar, with the exception that b = 0. The Power Rule can be shown by
repeated application of the Product Rule. The last three properties can be proved using Deﬁnition
0.2 and by looking at the cases when x ≥ 0, in which case x = x, or when x < 0, in which case
x = −x. For example, if c > 0, and x = c, then if x ≥ 0, we have x = x = c. If, on the other
hand, x < 0, then −x = x = c, so x = −c. The remaining properties are proved similarly and are
left as exercises. 0.5 Solving Equations Definition 0.3. A number a is called a solution of an equation containing the variable x if
the equation becomes a true statement when a is substituted for x. A solution of an equation is
sometimes also called a root of the equation. Two equations are said to be equivalent iﬀ they
have exactly the same solutions. We will sometimes use the symbol ⇐⇒ to indicate that two
equations are equivalent.
Usually two equations are equivalent because one can be obtained from the other by performing
an operation to both sides of the equation which can be reversed by another operation of the same
kind. For example, the equations 3x + 7 = 13 and x = 2 are equivalent because
3x + 7 = 13 ⇐⇒
⇐⇒ 3x = 6 (subtract 7 from both sides),
x = 2 (divide both sides by 3). The reasoning is reversible: we can go from x = 2 to 3x = 6 by multiplying both sides by by 3 and
from 3x = 6 to 3x + 7 = 13 by adding 7 to both sides.
We use the symbol =⇒ when we want to assert that one equation implies another but do not
want to assert the converse. The guiding principal here is
If an equation E results from an equation E by performing the same operation to both
sides, then E =⇒ E , i.e. every solution of E is a solution of E .
If the operation is not “reversible” as explained above, there is the possibility that the set of
solutions gets bigger in which case the new solutions are called extraneous solutions. (They do
not solve the original equation.) The simplest example of how an extraneous solution can arise is
x = 3 =⇒ x2 = 9 (square both sides) but the operation of squaring both sides is not reversible: it is incorrect to conclude that x2 = 9
implies that x = 3. What is correct is that x2 = 9 ⇐⇒ x = ±3, i.e. either x = 3 or else x = −3.
When solving an equation you may use operations which are not reversible provided that you 0.6 Exercises 13 Always check your answer!
In addition to catching mistakes, this will show you which – if any – of the solutions you found are
extraneous.
Here are two ways in which extraneous solutions can arise:
(i) Squaring both sides of an equation.
(ii) Multiplying both sides of of an equation by a quantity not known to be nonzero.
As an example of (i) consider the equation
√ 10 − x = −x − 2. Squaring both sides gives the quadratic equation 10 − x = x2 + 4x + 4 which has two solutions
√
√
x = −6 and x = 1. Now 10 − (−6) = −(−6) − 2 but 10 − 1 = −1 − 2 (Remember that means
the positive square root.) Thus x = −6 is the only solution of the original equation and x = 1 is
an extraneous solution.
As an example of (ii) consider
1
1
=2+
.
x−1
x−1
This equation has no solution: if it did we would subtract (x − 1)−1 from both sides and deduce
that 0 = 2 which is false. But if we multiply both sides by x − 1 we get 1 = 2(x − 1) + 1 which has
the (extraneous) solution x = 1. 0.6 Exercises 1. True or false?
(i) a+b
ab
=+?
c
cc (iii) (a + b)/c = (a/c) + (b/c) ?
(v) a+b
ab
=+?
c+d
cd (ii) c
cc
=+?
a+b
ab (iv) c/(a + b) = (c/a) + (c/b) ?
(vi) ab
a·b
=·?
c·d
cd 2. Factor:
(a) 2x3 − 12x2 + 6x = (b) 18a4 b2 − 30a3 b3 = (c) x2 − a2 = (d) 4x2 − 1
=
9 14 Basic Algebra
3. Simplify:
x3 − 9x
x2 + 6 x + 9
1
−2
a
(c) 1
−4
a2 x2 + 2 x − 3
x2 − 4
·2
x2 + 4 x + 4 x + 4 x − 5
2
1
+2
ab ab
(d)
3
4
− ab
a3 b (a) (b) 4. Use the properties of exponents to rewrite the given expression as a simple fraction with only
positive exponents.
(a) k −4 s 2
k 9 s−7 9 5. Reduce the fraction 0.7
1. s+t
to lowest terms.
sx4 + tx4 Answers
(ii) False.: (i) True.
1
.
A
2+3
5
(v) False:
=,
4+5
9 (iii) True: 1/A = 1
11
=+
2+3
23 (iv) False.
24
22
+=.
35
15 (vi) True. 2.
(a) 2x3 − 12x2 + 6x = l2x(x2 − 6x + 3)
(c) x2 − a2 = (x − a)(x + a) (b) 18a4 b2 − 30a3 b3 = 6a3 b2 (3a − 5b)
1
1
1
(d) 4x2 − = 2x −
2x +
9
3
3 3.
x3 − 9x
x(x − 3)
=
x2 + 6 x + 9
x+3
1
−2
a
a
(c)
=
1
1 + 2a
−4
a2 (a) 4. (a)
5. k −4 s 2
k 9 s−7 9 = s9
k 13 9 = s+t
s+t
1
=4
= 4.
4 + tx4
sx
x (s + t)
x x2 + 2 x − 3
x2 − 4
(x + 3)(x − 2)
·2
=
x2 + 4 x + 4 x + 4 x − 5
(x + 2)(x + 5)
1
2
+
a2 (b + 2)
ab ab2
(d)
=
4
3
b(3 − 4a2 )
−
3b
a
ab (b) s81
.
k 117 Chapter 1 Coordinates
1.1 The Cartesian Coordinate Plane In order to visualize the pure excitement that is Algebra, we need to unite Algebra and Geometry.
Simply put, we must ﬁnd a way to draw algebraic things. Let’s start with possibly the greatest
mathematical achievement of all time: the Cartesian Coordinate Plane.1 Imagine two real
number lines crossing at a right angle at 0 as below.
y
4
3
2
1 −4 −3 −2 −1 1 2 3 4 x −1
−2
−3
−4 The horizontal number line is usually called the xaxis while the vertical number line is usually
called the y axis.2 As with the usual number line, we imagine these axes extending oﬀ indeﬁnitely
in both directions. Having two number lines allows us to locate the position of points oﬀ of the
number lines as well as points on the lines themselves.
1
2 So named in honor of Ren´ Descartes.
e
The labels can vary depending on the context of application. 16 Coordinates For example, consider the point P below on the left. To use the numbers on the axes to label
this point, we imagine dropping a vertical line from the xaxis to P and extending a horizontal line
from the y axis to P . We then describe the point P using the ordered pair (2, −4). The ﬁrst
number in the ordered pair is called the abscissa or xcoordinate and the second is called the
ordinate or y coordinate.3 Taken together, the ordered pair (2, −4) comprise the Cartesian
coordinates of the point P . In practice, the distinction between a point and its coordinates is
blurred; for example, we often speak of ‘the point (2, −4).’ We can think of (2, −4) as instructions
on how to reach P from the origin by moving 2 units to the right and 4 units downwards. Notice
that the order in the ordered pair is important − if we wish to plot the point (−4, 2), we would
move to the left 4 units from the origin and then move upwards 2 units, as below on the right. y y 4 4 3 3 (−4, 2)
2
1 −4 −3 −2 2
1 −1 1 2 3 4 x −4 −3 −2 −1 1 −1
−2 3 4 x −2 −3 2 −1 −3 −4 P −4 P (2, −4) Example 1.1.1. Plot the following points: A(5, 8), B − 5 , 3 , C (−5.8, −3), D(4.5, −1), E (5, 0),
2
F (0, 5), G(−7, 0), H (0, −9), O(0, 0).4
Solution. To plot these points, we start at the origin and move to the right if the xcoordinate is
positive; to the left if it is negative. Next, we move up if the y coordinate is positive or down if it
is negative. If the xcoordinate is 0, we start at the origin and move along the y axis only. If the
y coordinate is 0 we move along the xaxis only.
3
Again, the names of the coordinates can vary depending on the context of the application. If, for example, the
horizontal axis represented time we might choose to call it the taxis. The ﬁrst number in the ordered pair would
then be the tcoordinate.
4
The letter O is almost always reserved for the origin. 1.1 The Cartesian Coordinate Plane 17
y
9
8 A(5, 8) 7
6
5 F (0, 5) 4
3
5
B −2, 3 G(−7, 0) 2
1 O(0, 0) −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 E (5, 0)
3 4 5 6 7 8 9 x −1 D(4.5, −1) −2
−3 C (−5.8, −3) −4
−5
−6
−7
−8
−9 H (0, −9) When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs
(x, y ) as x and y take values from the real numbers. Below is a summary of important facts about
Cartesian coordinates.
Important Facts about the Cartesian Coordinate Plane
(a, b) and (c, d) represent the same point in the plane if and only if a = c and b = d.
(x, y ) lies on the xaxis if and only if y = 0.
(x, y ) lies on the y axis if and only if x = 0.
The origin is the point (0, 0). It is the only point common to both axes. 18 Coordinates The axes divide the plane into four regions called quadrants. They are labeled with Roman
numerals and proceed counterclockwise around the plane: y
4 Quadrant II Quadrant I
3 x < 0, y > 0 x > 0, y > 0
2
1 −4 −3 −2 −1 1 2 3 4 x −1
−2 Quadrant III Quadrant IV
−3 x < 0, y < 0 x > 0, y < 0
−4 For example, (1, 2) lies in Quadrant I, (−1, 2) in Quadrant II, (−1, −2) in Quadrant III, and
(1, −2) in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically
refer to the point as lying on the positive or negative xaxis (if y = 0) or on the positive or negative
y axis (if x = 0). For example, (0, 4) lies on the positive y axis whereas (−117, 0) lies on the
negative xaxis. Such points do not belong to any of the four quadrants.
One of the most important concepts in all of mathematics is symmetry.There are many types
of symmetry in mathematics, but three of them can be discussed easily using Cartesian Coordinates. Definition 1.1. Two points (a, b) and (c, d) in the plane are said to be
symmetric about the xaxis if a = c and b = −d
symmetric about the y axis if a = −c and b = d
symmetric about the origin if a = −c and b = −d Schematically, 1.1 The Cartesian Coordinate Plane 19
y Q(−x, y ) P (x, y ) 0 x R(−x, −y ) S (x, −y ) In the above ﬁgure, P and S are symmetric about the xaxis, as are Q and R; P and Q are
symmetric about the y axis, as are R and S ; and P and R are symmetric about the origin, as are
Q and S .
Example 1.1.2. Let P be the point (−2, 3). Find the points which are symmetric to P about the:
1. xaxis 2. y axis 3. origin Check your answer by graphing.
Solution. The ﬁgure after Deﬁnition 1.1 gives us a good way to think about ﬁnding symmetric
points in terms of taking the opposites of the x and/or y coordinates of P (−2, 3).
1. To ﬁnd the point symmetric about the xaxis, we replace the y coordinate with its opposite
to get (−2, −3).
2. To ﬁnd the point symmetric about the y axis, we replace the xcoordinate with its opposite
to get (2, 3).
3. To ﬁnd the point symmetric about the origin, we replace the x and y coordinates with their
opposites to get (2, −3).
y
3 P (−2, 3) (2, 3) 2
1 −3 −2 −1
−1 1 2 3 −2
−3 (−2, −3) (2, −3) x 20 Coordinates One way to visualize the processes in the previous example is with the concept of reﬂections.
If we start with our point (−2, 3) and pretend the xaxis is a mirror, then the reﬂection of (−2, 3)
across the xaxis would lie at (−2, −3). If we pretend the y axis is a mirror, the reﬂection of (−2, 3)
across that axis would be (2, 3). If we reﬂect across the xaxis and then the y axis, we would go
from (−2, 3) to (−2, −3) then to (2, −3), and so we would end up at the point symmetric to (−2, 3)
about the origin. We summarize and generalize this process below. Reﬂections
To reﬂect a point (x, y ) about the:
xaxis, replace y with −y .
y axis, replace x with −x.
origin, replace x with −x and y with −y . 1.1.1 Distance in the Plane Another important concept in geometry is the notion of length. If we are going to unite Algebra
and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of
what distance in the plane means. Suppose we have two points, P (x1 , y1 ) and Q (x2 , y2 ) , in the
plane. By the distance d between P and Q, we mean the length of the line segment joining P with
Q. (Remember, given any two distinct points in the plane, there is a unique line containing both
points.) Our goal now is to create an algebraic formula to compute the distance between these two
points. Consider the generic situation below on the left.
Q (x2 , y2 ) Q (x2 , y2 ) d P (x1 , y1 ) d P (x1 , y1 ) (x2 , y1 ) With a little more imagination, we can envision a right triangle whose hypotenuse has length
d as drawn above on the right. From the latter ﬁgure, we see that the lengths of the legs of the
triangle are x2 − x1  and y2 − y1  so the Pythagorean Theorem gives us
x2 − x1 2 + y2 − y1 2 = d2 1.1 The Cartesian Coordinate Plane 21 (x2 − x1 )2 + (y2 − y1 )2 = d2
(Do you remember why we can replace the absolute value notation with parentheses?) By extracting
the square root of both sides of the second equation and using the fact that distance is never
negative, we get Equation 1.1. The Distance Formula: The distance d between the points P (x1 , y1 ) and
Q (x2 , y2 ) is:
d= (x2 − x1 )2 + (y2 − y1 )2 It is not always the case that the points P and Q lend themselves to constructing such a triangle.
If the points P and Q are arranged vertically or horizontally, or describe the exact same point, we
cannot use the above geometric argument to derive the distance formula. It is left to the reader to
verify Equation 1.1 for these cases.
Example 1.1.3. Find and simplify the distance between P (−2, 3) and Q(1, −3).
Solution.
(x2 − x1 )2 + (y2 − y1 )2 d= (1 − (−2))2 + (−3 − 3)2 =
√ 9 + 36
√
=35 = √
So, the distance is 3 5. Example 1.1.4. Find all of the points with xcoordinate 1 which are 4 units from the point (3, 2).
Solution. We shall soon see that the points we wish to ﬁnd are on the line x = 1, but for now
we’ll just view them as points of the form (1, y ). Visually, 22 Coordinates
y
3 (3, 2) 2
1 distance is 4 units
2 3 x −1 (1, y )
−2
−3 We require that the distance from (3, 2) to (1, y ) be 4. The Distance Formula, Equation 1.1, yields d= (x2 − x1 )2 + (y2 − y1 )2 4= (1 − 3)2 + (y − 2)2 4= 4 + (y − 2)2 42 = 4 + (y − 2)2 2 squaring both sides 16 = 4 + (y − 2)2
12 = (y − 2)2
(y − 2)2 = 12
√
y − 2 = ± 12
√
y − 2 = ±2 3
√
y = 2±2 3 extracting the square root √
√
We obtain two answers: (1, 2 + 2 3) and (1, 2 − 2 3). The reader is encouraged to think about
why there are two answers.
Related to ﬁnding the distance between two points is the problem of ﬁnding the midpoint of
the line segment connecting two points. Given two points, P (x1 , y1 ) and Q (x2 , y2 ), the midpoint,
M , of P and Q is deﬁned to be the point on the line segment connecting P and Q whose distance
from P is equal to its distance from Q. 1.1 The Cartesian Coordinate Plane 23
Q (x2 , y2 ) M
P (x1 , y1 )
If we think of reaching M by going ‘halfway over’ and ‘halfway up’ we get the following formula. Equation 1.2. The Midpoint Formula: The midpoint M of the line segment connecting
P (x1 , y1 ) and Q (x2 , y2 ) is:
x1 + x2 y1 + y2
M=
,
2
2 If we let d denote the distance between P and Q, we leave it as an exercise to show that the
distance between P and M is d/2 which is the same as the distance between M and Q. This suﬃces
to show that Equation 1.2 gives the coordinates of the midpoint.
Example 1.1.5. Find the midpoint of the line segment connecting P (−2, 3) and Q(1, −3).
Solution.
x1 + x2 y1 + y2
,
2
2
(−2) + 1 3 + (−3)
,
2
2 = 10
−,
22 =
The midpoint is = = M 1
− ,0
2 1
− ,0 .
2 An interesting application5 of the midpoint formula follows.
5 This is a key concept in the development of inverse functions. See Section 6.2 24 Coordinates Example 1.1.6. Prove that the points (a, b) and (b, a) are symmetric about the line y = x.
Solution. By ‘symmetric about the line y = x’, we mean that if a mirror were placed along
the line y = x, the points (a, b) and (b, a) would be mirror images of one another. (You should
compare and contrast this with the other types of symmetry presented back in Deﬁnition 1.1.)
Schematically,
(a, b) y=x
∗
(b, a) From the ﬁgure, we see that this problem amounts to showing that the midpoint of the line
segment connecting (a, b) and (b, a) lies on the line y = x. Applying Equation 1.2 yields = a+b b+a
,
2
2 = M a+b a+b
,
2
2 Since the x and y coordinates of this point are the same, we ﬁnd that the midpoint lies on the line
y = x, as required. 1.1 The Cartesian Coordinate Plane 1.1.2 25 Exercises √
1. Plot and label the points A(−3, −7), B (1.3, −2), C (π, 10), D(0, 8), E (−5.5, 0), F (−8, 4),
G(9.2, −7.8) and H (7, 5) in the Cartesian Coordinate Plane given below.
y
9
8
7
6
5
4
3
2
1 −9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 −1
−2
−3
−4
−5
−6
−7
−8
−9 2. For each point given in Exercise 1 above
Identify the quadrant or axis in/on which the point lies.
Find the point symmetric to the given point about the xaxis.
Find the point symmetric to the given point about the y axis.
Find the point symmetric to the given point about the origin. 6 7 8 9 x 26 Coordinates
3. For each of the following pairs of points, ﬁnd the distance d between them and ﬁnd the
midpoint M of the line segment connecting them.
(a) (1, 2), (−3, 5)
(b) (3, −10), (−1, 2)
1
3
(c)
,4 ,
, −1
2
2
7
23
,
,2
(d) − ,
32
3 11 19
24 6
, − ,−
.
,
55
5
5
√√
√
√
(f)
2, 3 , − 8, − 12
√√
√√
(g) 2 45, 12 ,
20, 27 .
(e) (h) (0, 0), (x, y ) 4. Find all of the points of the form (x, −1) which are 4 units from the point (3, 2).
5. Find all of the points on the y axis which are 5 units from the point (−5, 3).
6. Find all of the points on the xaxis which are 2 units from the point (−1, 1).
7. Find all of the points of the form (x, −x) which are 1 unit from the origin.
8. Let’s assume for a moment that we are standing at the origin and the positive y axis points
due North while the positive xaxis points due East. Our Sasquatchometer tells us that
Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates
of his position? How far away is he from us? If he runs 7 miles due East what would his new
position be?
9. Verify the Distance Formula 1.1 for the cases when:
(a) The points are arranged vertically. (Hint: Use P (a, y1 ) and Q(a, y2 ).)
(b) The points are arranged horizontally. (Hint: Use P (x1 , b) and Q(x2 , b).)
(c) The points are actually the same point. (You shouldn’t need a hint for this one.)
10. Verify the Midpoint Formula by showing the distance between P (x1 , y1 ) and M and the
distance between M and Q(x2 , y2 ) are both half of the distance between P and Q.
11. Show that the points A, B and C below are the vertices of a right triangle.
(a) A(−3, 2), B (−6, 4), and C (1, 8) (b) A(−3, 1), B (4, 0) and C (0, −3) 12. Find a point D(x, y ) such that the points A(−3, 1), B (4, 0), C (0, −3) and D are the corners
of a square. Justify your answer.
13. The world is not ﬂat.6 Thus the Cartesian Plane cannot possibly be the end of the story.
Discuss with your classmates how you would extend Cartesian Coordinates to represent the
three dimensional world. What would the Distance and Midpoint formulas look like, assuming
those concepts make sense at all?
6 There are those who disagree with this statement. Look them up on the Internet some time when you’re bored. 1.1 The Cartesian Coordinate Plane 1.1.3 27 Answers √
1. The required points A(−3, −7), B (1.3, −2), C (π, 10), D(0, 8), E (−5.5, 0), F (−8, 4),
G(9.2, −7.8), and H (7, 5) are plotted in the Cartesian Coordinate Plane below.
y
9
8 D(0, 8) 7
6 H (7, 5)
5
4 F (−8, 4) √
C (π, 10) 3
2
1 E (−5.5, 0)
−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 x −1
−2 B (1.3, −2)
−3
−4
−5
−6
−7 A(−3, −7) −8 G(9.2, −7.8) −9 2. (a) The
(b) The
point A(−3, −7) is
in Quadrant III
symmetric about xaxis with (−3, 7)
symmetric about y axis with (3, −7)
symmetric about origin with (3, 7)
point B (1.3, −2) is
in Quadrant IV symmetric about xaxis with (1.3, 2)
symmetric about y axis with (−1.3, −2)
symmetric about origin with (−1.3, 2)
√
(c) The point C (π, 10) is
in Quadrant I
√
symmetric about xaxis with (π, − 10)
√
symmetric about y axis with (−π, 10) 28 Coordinates
√ symmetric about origin with (−π, − 10) symmetric about xaxis with (−8, −4)
symmetric about y axis with (8, 4)
symmetric about origin with (8, −4) (d) The point D(0, 8) is
on the positive y axis
symmetric about xaxis with (0, −8)
symmetric about y axis with (0, 8)
symmetric about origin with (0, −8) (g) The point G(9.2, −7.8) is
(e) The point E (−5.5, 0) is
on the negative xaxis
symmetric about xaxis with (−5.5, 0)
symmetric about y axis with (5.5, 0)
symmetric about origin with (5.5, 0) (h) The point H (7, 5) is
(f) The point F (−8, 4) is
in Quadrant II 3. (a) d = 5, M = −1, 7
2 (e) d = √
(b) d = 4 10, M = (1, −4)
(c) d = √ 26, M = 1, in Quadrant IV
symmetric about xaxis with (9.2, 7.8)
symmetric about y axis with (−9.2, −7.8)
symmetric about origin with (−9.2, 7.8) in Quadrant I
symmetric about xaxis with (7, −5)
symmetric about y axis with (−7, 5)
symmetric about origin with (−7, −5)
√ 74, M = √
(f) d = 3 5, M = 3
2 (g) d = √ 83, M = √ 13 13
,−
.
10 10
√
√
2
3
−
,−
2
2
√
√53
.
4 5,
2 37
57
,M=
,
2
64
√
√
4. (3 + 7, −1), (3 − 7, −1) xy
(h) d = x2 + y 2 , M =
,
22
√
√
6. (−1 + 3, 0), (−1 − 3, 0) 5. (0, 3) 7. (d) d = √ √
2
2
2 ,− 2 √ √
2
2
2, 2 ,− 8. (−3, −4), 5 miles, (4, −4)
10. (a) The distance from A to B is
√
C is 65. Since √ 13, the distance from A to C is
√ 2 13 + √ 2 52 = √ √ 52, and the distance from B to 2 65 , we are guaranteed by the converse of the Pythagorean Theorem that the triangle is right. 1.2 Relations We now turn our attention to sets of points in the plane. Definition 1.2. A relation is a set of points in the plane. 1.2 Relations 29 Throughout this text we will see many diﬀerent ways to describe relations. In this section we
will focus our attention on describing relations graphically, by means of the list (or roster) method
and algebraically. Depending on the situation, one method may be easier or more convenient to
use than another. Consider the set of points below
y
4
3 (4, 3)
2 (−2, 1)
1 −4 −3 −2 −1 1 2 3 4 x −1
−2
−3 (0, −3) −4 These three points constitute a relation. Let us call this relation R. Above, we have a graphical
description of R. Although it is quite pleasing to the eye, it isn’t the most portable way to describe
R. The list (or roster) method of describing R simply lists all of the points which belong to R.
Hence, we write: R = {(−2, 1), (4, 3), (0, −3)}.1 The roster method can be extended to describe
inﬁnitely many points, as the next example illustrates.
Example 1.2.1. Graph the following relations.
1. A = {(0, 0), (−3, 1), (4, 2), (−3, 2)}
2. HLS1 = {(x, 3) : −2 ≤ x ≤ 4}
3. HLS2 = {(x, 3) : −2 ≤ x < 4}
4. V = {(3, y ) : y is a real number}
Solution.
1. To graph A, we simply plot all of the points which belong to A, as shown below on the left.
2. Don’t let the notation in this part fool you. The name of this relation is HLS1 , just like the
name of the relation in part 1 was R. The letters and numbers are just part of its name, just
1 We use ‘set braces’ {} to indicate that the points in the list all belong to the same set, in this case, R. 30 Coordinates
like the numbers and letters of the phrase ‘King George III’ were part of George’s name. The
next hurdle to overcome is the description of HLS1 itself − a variable and some seemingly
extraneous punctuation have found their way into our nice little roster notation! The way
to make sense of the construction {(x, 3) : −2 ≤ x ≤ 4} is to verbalize the set braces {}
as ‘the set of’ and the colon : as ‘such that’. In words, {(x, 3) : −2 ≤ x ≤ 4} is: ‘the set
of points (x, 3) such that −2 ≤ x ≤ 4.’ The purpose of the variable x in this case is to
describe inﬁnitely many points. All of these points have the same y coordinate, 3, but the
xcoordinate is allowed to vary between −2 and 4, inclusive. Some of the points which belong
to HLS1 include some friendly points like: (−2, 3), (−1, 3), (0, 3), (1, 3), (2, 3), (3, 3), and
√
5
(4, 3). However, HLS1 also contains the points (0.829, 3), − 6 , 3 , ( π, 3), and so on. It is
impossible to list all of these points, which is why the variable x is used. Plotting several
friendly representative points should convince you that HLS1 describes the horizontal line
segment from the point (−2, 3) up to and including the point (4, 3). y y 4
3 3 2 2 1
−4 −3 −2 −1 4 1
1 2 The graph of A 3 4 x −4 −3 −2 −1 1 2 3 4 x The graph of HLS1 3. HLS2 is hauntingly similar to HLS1 . In fact, the only diﬀerence between the two is that
instead of ‘−2 ≤ x ≤ 4’ we have ‘−2 ≤ x < 4’. This means that we still get a horizontal line
segment which includes (−2, 3) and extends to (4, 3), but does not include (4, 3) because of
the strict inequality x < 4. How do we denote this on our graph? It is a common mistake to
make the graph start at (−2, 3) end at (3, 3) as pictured below on the left. The problem with
this graph is that we are forgetting about the points like (3.1, 3), (3.5, 3), (3.9, 3), (3.99, 3),
and so forth. There is no real number that comes ‘immediately before’ 4, and so to describe
the set of points we want, we draw the horizontal line segment starting at (−2, 3) and draw
an ‘open circle’ at (4, 3) as depicted below on the right. 1.2 Relations 31
y y 4 4 3 3 2 2 1 1 −4 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 This is NOT the correct graph of HLS2 1 2 3 4 x The graph of HLS2 4. Our last example, V , describes the set of points (3, y ) such that y is a real number. All of
these points have an xcoordinate of 3, but the y coordinate is free to be whatever it wants
to be, without restriction. Plotting a few ‘friendly’ points of V should convince you that all
the points of V lie on a vertical line which crosses the xaxis at x = 3. Since there is no
restriction on the y coordinate, we put arrows on the end of the portion of the line we draw
to indicate it extends indeﬁnitely in both directions. The graph of V is below on the left.
y
4 y 3
2
−4 −3 −2 −1 1 2 3 4 x −1 1 −2
1 2 3 4 x −1 −3 −2 −4 −3 The graph of y = −2 −4 The graph of V The relation V in the previous example leads us to our ﬁnal way to describe relations: algebraically. We can simply describe the points in V as those points which satisfy the equation x = 3.
Most likely, you have seen equations like this before. Depending on the context, ‘x = 3’ could mean
we have solved an equation for x and arrived at the solution x = 3. In this case, however, ‘x = 3’
describes a set of points in the plane whose xcoordinate is 3. Similarly, the equation y = −2 in
this context corresponds to all points in the plane whose y coordinate is −2. Since there are no
restrictions on the xcoordinate listed, we would graph the relation y = −2 as the horizontal line
above on the right. In general, we have the following. 32 Coordinates Equations of Vertical and Horizontal Lines
The graph of the equation x = a is a vertical line through (a, 0).
The graph of the equation y = b is a horizontal line through (0, b). In the next section, and in many more after that, we shall explore the graphs of equations in
great detail.2 For now, we shall use our ﬁnal example to illustrate how relations can be used to
describe entire regions in the plane.
Example 1.2.2. Graph the relation: R = {(x, y ) : 1 < y ≤ 3}
Solution. The relation R consists of those points whose y coordinate only is restricted between 1
and 3 excluding 1, but including 3. The xcoordinate is free to be whatever we like. After plotting
some3 friendly elements of R, it should become clear that R consists of the region between the
horizontal lines y = 1 and y = 3. Since R requires that the y coordinates be greater than 1, but not
equal to 1, we dash the line y = 1 to indicate that those points do not belong to R. Graphically,
y
4
3
2
1 −4 −3 −2 −1 1 2 3 4 x The graph of R 1.2.1 Exercises 1. Graph the following relations.
(a) {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}
(b) {(−2, 2), (−2, −1), (3, 5), (3, −4)}
(c)
n, 4 − n2 : n = 0, ±1, ±2
6
(d)
k , k : k = ±1, ±2, ±3, ±4, ±5, ±6
2. Graph the following relations.
2
3 In fact, much of our time in College Algebra will be spent examining the graphs of equations.
The word ‘some’ is a relative term. It may take 5, 10, or 50 points until you see the pattern. 1.2 Relations 33 (a) {(x, −2) : x > −4} (e) {(x, y ) : y < 4} (b) {(2, y ) : y ≤ 5} (f) {(x, y ) : x ≤ 3, y < 2} (c) {(−2, y ) : −3 < y < 4} (g) {(x, y ) : x > 0, y < 4}
√
2
(h) {(x, y ) : − 2 ≤ x ≤ 3 , π < y ≤ 9 }
2 (d) {(x, y ) : x ≤ 3} 3. Describe the following relations using the roster method.
y y 4 3 3 2 2 1 1
−4 −3 −2 −1
−1 1 −4 −3 −2 −1
−1 x 1 2 3 x −2
−3 (a) The graph of relation A
(d) y The graph of relation D
y 3 5 2 4 1 3 −3 −2 −1
−1 1 2 3 2 x 1 −2
−1
−1 −3 (b) The graph of relation B 1 2 3 4 5 x (e) The graph of relation E
y y 2 4 1 3
2
1
−3 −2 −1 −4 −3 −2 −1
−1
−2
1 2 3 4. Graph the following lines. 2 3 4 −3 x (c) The graph of relation C 1 (f) The graph of relation F 5x 34 Coordinates
(a) x = −2 (b) y = 3 5. What is another name for the line x = 0? For y = 0?
6. Some relations are fairly easy to describe in words or with the roster method but are rather
diﬃcult, if not impossible, to graph. Discuss with your classmates how you might graph the
following relations. Please note that in the notation below we are using the ellipsis, . . . ,
to denote that the list does not end, but rather, continues to follow the established pattern
indeﬁnitely. For the ﬁrst two relations, give two examples of points which belong to the
relation and two points which do not belong to the relation.
(a) {(x, y ) : x is an odd integer, and y is an even integer.}
(b) {(x, 1) : x is an irrational number }
(c) {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4), (32, 5), . . .}
(d) {. . . , (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9), . . .} 1.2.2 Answers
(c) 1. (a) y y
9
8
7
6
5
4
3
2
1
−3−2−1 4
3
2
1
−2 −1 123 y
5
4
3
2
1 −2
−3
−4 2. 2 x (d)
y
6
5
4
3
2
1 (b) −2−11
− x 1 −6−5−4−3−2−11
−
123 x −2
−3
−4
−5
−6 123456 x 1.2 Relations 35
y (a)
y 4
3 −4 −3 −2 −1
−1 1 2 3 4 x
2
1 −3 −3 −2 −1 1 2 (f) (b) y y
5 3 4 2 3 1 2
1
1 −1 2 x 3 2 1 −1 3 x 1 2 3 −2 −2 −3 −3 (c) (g) y y 4
3
2
1 4
3
2 x −3 −2 −1 1
− 1 −2
−3
−1 (d) x (h) y y 3
7
2
6
1
5
−1 1 2 3 x 4
3 −2 2 −3 1 (e) −3 −2 −1 3. (a) A = {(−4, −1), (−3, 0), (−2, 1), (−1, 2), (0, 3), (1, 4)} 1 x 3 x 36 Coordinates
(b) B = {(x, y ) : x > −2}
(c) C = {(x, y ) : y ≥ 0}
(d) D = {(x, y ) : −3 < x ≤ 2}
(e) E = {(x, y ) : x ≥ 0,y ≥ 0}
(f) F = {(x, y ) : −4 < x < 5, −3 < y < 2}
y y 3
2 4. (a) 2 1
−3 −2 −1 3 1 x −3 −2 −1 The line x = −2 (b) x The line y = 3 5. The line x = 0 is the y axis and the line y = 0 is the xaxis. 1.3 Graphs of Equations In the previous section, we said that an equation in x and y determines a relation.1 In this section,
we begin to explore this topic in greater detail. The main idea of this section is The Fundamental Graphing Principle
The graph of an equation is the set of points which satisfy the equation. That is, a point (x, y )
is on the graph of an equation if and only if x and y satisfy the equation. Example 1.3.1. Determine if (2, −1) is on the graph of x2 + y 3 = 1.
Solution. To check, we substitute x = 2 and y = −1 into the equation and see if the equation is
satisﬁed
? (2)2 + (−1)3 = 1
3=1
Hence, (2, −1) is not on the graph of x2 + y 3 = 1.
We could spend hours randomly guessing and checking to see if points are on the graph of the
equation. A more systematic approach is outlined in the following example.
1 An inequalities in x and y also determines a relation. 1.3 Graphs of Equations 37 Example 1.3.2. Graph x2 + y 3 = 1.
Solution. To eﬃciently generate points on the graph of this equation, we ﬁrst solve for y
x2 + y 3 = 1 3 y 3 = 1 − x2
√
y 3 = 3 1 − x2
√
y = 3 1 − x2 We now substitute a value in for x, determine the corresponding value y , and plot the resulting
point, (x, y ). For example, for x = −3, we substitute
y= 3 1 − x2 = 3 1 − (−3)2 = √
3 −8 = −2, so the point (−3, −2) is on the graph. Continuing in this manner, we generate a table of points
which are on the graph of the equation. These points are then plotted in the plane as shown below. x
−3
−2 y y (x, y )
3 −2
(−3, −2)
√
√
− 3 3 (−2, − 3 3) 2
1 −1 0 (−1, 0) 0 1 (0, 1) 1
2 0
√
3 −3 (1, 0)
√
(2, − 3 3) 3 −2 (3, −2) −4 −3 −2 −1 1 2 3 4 x −1
−2
−3 Remember, these points constitute only a small sampling of the points on the graph of this
equation. To get a better idea of the shape of the graph, we could plot more points until we feel
comfortable ‘connecting the dots.’ Doing so would result in a curve similar to the one pictured
below. 38 Coordinates
y
3
2
1
−4 −3 −2 −1
−1 1 2 3 4 x −2
−3 Don’t worry if you don’t get all of the little bends and curves just right − Calculus is where the
art of precise graphing takes center stage. For now, we will settle with our naive ‘plug and plot’
approach to graphing. If you feel like all of this tedious computation and plotting is beneath you,
then you can reach for a graphing calculator, input the formula as shown above, and graph.2 Of all of the points on the graph of an equation, the places where the graph crosses the axes
hold special signiﬁcance. These are called the intercepts of the graph. Intercepts come in two
distinct varieties: xintercepts and y intercepts. They are deﬁned below. Definition 1.3. Suppose the graph of an equation is given.
A point at which a graph meets the y axis is called an y intercept of the graph. In our previous example the graph had two xintercepts, (−1, 0) and (1, 0), and one y intercept,
(0, 1). The graph of an equation can have any number of intercepts, including none at all! Since
xintercepts lie on the xaxis, we can ﬁnd them by setting y = 0 in the equation. Similarly, since
y intercepts lie on the y axis, we can ﬁnd them by setting x = 0 in the equation. Keep in mind,
intercepts are points and therefore must be written as ordered pairs. To summarize, 2 Remember: At UW we don’t allow calculators on exams. But using them intelligently outside of class can be a
great beneﬁt. 1.3 Graphs of Equations 39 Steps for ﬁnding the intercepts of the graph of an equation
Given an equation involving x and y :
the xintercepts always have the form (x, 0); to ﬁnd the xintercepts of the graph, set
y = 0 and solve for x.
y intercepts always have the form (0, y ); to ﬁnd the y intercepts of the graph, set x = 0
and solve for y . Another fact which you may have noticed about the graph in the previous example is that it
seems to be symmetric about the y axis. To actually prove this analytically, we assume (x, y ) is
a generic point on the graph of the equation. That is, we assume x2 + y 3 = 1. As we learned
in Section 1.1, the point symmetric to (x, y ) about the y axis is (−x, y ). To show the graph is
symmetric about the y axis, we need to show that (−x, y ) is on the graph whenever (x, y ) is. In
other words, we need to show (−x, y ) satisﬁes the equation x2 + y 3 = 1 whenever (x, y ) does.
Substituting gives
? (−x)2 + (y )3 = 1
x2 + y 3 = 1
When we substituted (−x, y ) into the equation x2 + y 3 = 1, we obtained the original equation back
when we simpliﬁed. This means (−x, y ) satisﬁes the equation and hence is on the graph. In this
way, we can check whether the graph of a given equation possesses any of the symmetries discussed
in Section 1.1. The results are summarized below. Steps for testing if the graph of an equation possesses symmetry
To test the graph of an equation for symmetry
About the y axis: Substitute (−x, y ) into the equation and simplify. If the result is
equivalent to the original equation, the graph is symmetric about the y axis.
About the xaxis: Substitute (x, −y ) into the equation and simplify. If the result is
equivalent to the original equation, the graph is symmetric about the xaxis.
About the origin: Substitute (−x, −y ) into the equation and simplify. If the result is
equivalent to the original equation, the graph is symmetric about the origin. 40 Coordinates Intercepts and symmetry are two tools which can help us sketch the graph of an equation
analytically, as evidenced in the next example.
Example 1.3.3. Find the x and y intercepts (if any) of the graph of (x − 2)2 + y 2 = 1. Test for
symmetry. Plot additional points as needed to complete the graph.
Solution. To look for xintercepts, we set y = 0 and solve:
(x − 2)2 + y 2 = 1
(x − 2)2 + 02 = 1
(x − 2)2 = 1
√
(x − 2)2 =
1 extract square roots x − 2 = ±1
x = 2±1
x = 3, 1
We get two answers for x which correspond to two xintercepts: (1, 0) and (3, 0). Turning our
attention to y intercepts, we set x = 0 and solve:
(x − 2)2 + y 2 = 1
(0 − 2)2 + y 2 = 1
4 + y2 = 1
y 2 = −3
Since there is no real number which squares to a negative number (Do you remember why?), we
are forced to conclude that the graph has no y intercepts.
Plotting the data we have so far, we get
y
2
1 −1
−2 (1, 0)
1 (3, 0)
2 3 4 x 1.3 Graphs of Equations 41 Moving along to symmetry, we can immediately dismiss the possibility that the graph is symmetric about the y axis or the origin. If the graph possessed either of these symmetries, then the
fact that (1, 0) is on the graph would mean (−1, 0) would have to be on the graph. (Why?) Since
(−1, 0) would be another xintercept (and we’ve found all of these), the graph can’t have y axis or
origin symmetry. The only symmetry left to test is symmetry about the xaxis. To that end, we
substitute (x, −y ) into the equation and simplify
(x − 2)2 + y 2 = 1
? (x − 2)2 + (−y )2 = 1
(x − 2)2 + y 2 = 1
Since we have obtained our original equation, we know the graph is symmetric about the xaxis.
This means we can cut our ‘plug and plot’ time in half: whatever happens below the xaxis is
reﬂected above the xaxis, and viceversa. Proceeding as we did in the previous example, we obtain
y
2
1 −1 1 2 3 4 x −2 A couple of remarks are in order. First, it is entirely possible to choose a value for x which does
not correspond to a point on the graph. For example, in the previous example, if we solve for y as
is our custom, we get:
y = ± 1 − (x − 2)2 .
Upon substituting x = 0 into the equation, we would obtain
√
√
y = ± 1 − (0 − 2)2 = ± 1 − 4 = ± −3,
which is not a real number. This means there are no points on the graph with an xcoordinate
of 0. When this happens, we move on and try another point. This is another drawback of the
‘plugandplot’ approach to graphing equations. Luckily, we will devote much of the remainder
of this book developing techniques which allow us to graph entire families of equations quickly.3
Second, it is instructive to show what would have happened had we tested the equation in the last
3 Without the use of a calculator, if you can believe it! 42 Coordinates example for symmetry about the y axis. Substituting (−x, y ) into the equation yields
(x − 2)2 + y 2 = 1
? (−x − 2)2 + y 2 = 1
? ((−1)(x + 2))2 + y 2 = 1
? (x + 2)2 + y 2 = 1.
This last equation does not appear to be equivalent to our original equation. However, to prove
it is not symmetric about the y axis, we need to ﬁnd a point (x, y ) on the graph whose reﬂection
(−x, y ) is not. Our xintercept (1, 0) ﬁts this bill nicely, since if we substitute (−1, 0) into the
equation we get
? (x − 2)2 + y 2 = 1
(−1 − 2)2 + 02 = 1
9 = 1.
This proves that (−1, 0) is not on the graph. 1.3.1 Exercises 1. For each equation given below
Find the x and y intercept(s) of the graph, if any exist.
Following the procedure in Example 1.3.2, create a table of sample points on the graph
of the equation.
Plot the sample points and create a rough sketch of the graph of the equation.
Test for symmetry. If the equation appears to fail any of the symmetry tests, ﬁnd a
point on the graph of the equation whose reﬂection fails to be on the graph as was done
at the end of Example 1.3.3 (a) y = x2 + 1
(b) y = x2 − 2x − 8 (c) y = x3 − x
(d) y = x3
4
√ − 3x x−2
√
(f) y = 2 x + 4 − 2 (e) y = (g) 3x − y = 7
(h) 3x − 2y = 10
(i) (x + 2)2 + y 2 = 16
(j) x2 − y 2 = 1
(k) 4y 2 − 9x2 = 36
(l) x3 y = −4 1.3 Graphs of Equations 43 2. The procedures which we have outlined in the Examples of this section and used in the exercises given above all rely on the fact that the equations were “wellbehaved”. Not everything
in Mathematics is quite so tame, as the following equations will show you. Discuss with your
classmates how you might approach graphing these equations. What diﬃculties arise when
trying to apply the various tests and procedures given in this section? For more information,
including pictures of the curves, each curve name is a link to its page at www.wikipedia.org.
For a much longer list of fascinating curves, click here.
(a) x3 + y 3 − 3xy = 0 Folium of Descartes
(b) x4 = x2 + y 2 Kampyle of Eudoxus 1.3.2 (c) y 2 = x3 + 3x2 Tschirnhausen cubic
(d) (x2 + y 2 )2 = x3 + y 3 Crooked egg Answers 1. (a) y = x2 + 1 (b) y = x2 − 2x − 8 The graph has no xintercepts xintercepts: (4, 0), (−2, 0) y intercept: (0, 1) y intercept: (0, −8) x y (x, y ) −2 5 (−2, 5) −1 2 (−1, 2) 0 1 (0, 1) 1 2 (1, 2) 2 5 (2, 5) y
5
4
3
2
1
−2 −1 1 2 x The graph is not symmetric about the
xaxis (e.g. (2, 5) is on the graph but
(2, −5) is not)
The graph is symmetric about the
y axis
The graph is not symmetric about the
origin (e.g. (2, 5) is on the graph but
(−2, −5) is not) 44 Coordinates x y (x, y ) −3 7 (−3, 7) −2 −2 0 (−2, 0) −1 0 (−1, 0) −5 (−1, −5) 0 0 (0, 0) 0 −8 (0, −8) 1 0 (1, 0) 1 −9 (1, −9) 2 6 (2, 6) −1 x y (x, y ) −6 (−2, −6) y 2 −8 (2, −8) 3 −5 (3, −5) 6
5
4
3 4 0 (4, 0) 2 5 7 (5, 7) −2 −1
−1 1
2 x −2 y −3 7
6
5
4
3
2
1
−3 2 1
−−
−2
−3
−4
−5
−6
−7
−8
−9 1 −4
−5
−6 12345 x The graph is not symmetric about the
xaxis (e.g. (−3, 7) is on the graph but
(−3, −7) is not) The graph is not symmetric about the
xaxis. (e.g. (2, 6) is on the graph but
(2, −6) is not)
The graph is not symmetric about the
y axis. (e.g. (2, 6) is on the graph but
(−2, 6) is not)
The graph is symmetric about the
origin. The graph is not symmetric about the
y axis (e.g. (3, 7) is on the graph but
(3, 7) is not)
The graph is not symmetric about the
origin (e.g. (−3, 7) is on the graph but
(3, −7) is not)
(c) y = x3 − x
xintercepts: (−1, 0), (0, 0), (1, 0)
y intercept: (0, 0) (d) y = x3
4 − 3x 1.3 Graphs of Equations 45 √
xintercepts: ±2 3, 0 x y (x, y ) 2 0 (2, 0) 3 1 (3, 1) 6 2 (6, 2) y intercept: (0, 0)
x y (x, y ) −4 −4 (−4, −4) −3 9
4 −2 4 −3, 9
4 11 (−2, 4) 3 (11, 3) y
3 11
4 −1, 11
4 0 0 (0, 0) 1 − 11
4 1, − 11
4 2 −4 (2, −4) 3 9
−4 9
3, − 4 4 4 (4, 4) −1 2
1
1 3 4 5 6 7 8 9 10 11 x The graph is not symmetric about the
xaxis (e.g. (3, 1) is on the graph but
(3, −1) is not)
The graph is not symmetric about the
y axis (e.g. (3, 1) is on the graph but
(−3, 1) is not) y The graph is not symmetric about the
origin (e.g. (3, 1) is on the graph but
(−3, −1) is not) 4
3
2
1
−4 −3 −2 −1
−1 2 1 2 3 4 x −2
−3
−4 The graph is not symmetric about the
xaxis (e.g. (−4, −4) is on the graph
but (−4, 4) is not)
The graph is not symmetric about the
y axis (e.g. (−4, −4) is on the graph
but (4, −4) is not)
The graph is symmetric about the
origin
√
(e) y = x − 2
xintercept: (2, 0)
The graph has no y intercepts 46 Coordinates
√
(f) y = 2 x + 4 − 2 x xintercept: (−3, 0) y (x, y ) −2 −13 (−2, −13) −1 −10 (−1, −10) y intercept: (0, 2)
x y (x, y )
0 −2
0 (−3, 0)
√
−2, 2 − 2
√
−2, 3 − 2 −4 (1, −4) 2 −1 (2, −1) 3 2 (3, 2) (−4, −2) −3 (0, −7) 1 −4 −7 −2
−1 √
2 2−2
√
2 3−2 y
3 0 2 1 (0, 2)
√
−2, 5 − 2 √
2 5−2
y 2
1
−2 −1
−1 1 2 3 x −2 3 −3 2 −4 1 −5 −4 −3 −2 −1
−1 1 2 x −6
−7 −2 −8 −3 −9
−10
−11 The graph is not symmetric about the
xaxis (e.g. (−4, −2) is on the graph
but (−4, 2) is not)
The graph is not symmetric about the
y axis (e.g. (−4, −2) is on the graph
but (4, −2) is not)
The graph is not symmetric about the
origin (e.g. (−4, −2) is on the graph
but (4, 2) is not)
(g) 3x − y = 7
Rewrite as: y = 3x − 7.
7
xintercept: ( 3 , 0) y intercept: (0, −7) −12
−13 The graph is not symmetric about the
xaxis (e.g. (3, 2) is on the graph but
(3, −2) is not)
The graph is not symmetric about the
y axis (e.g. (3, 2) is on the graph but
(−3, 2) is not)
The graph is not symmetric about the
origin (e.g. (3, 2) is on the graph but
(−3, −2) is not) 1.3 Graphs of Equations
(h) 3x − 2y = 10
Rewrite as: y =
xintercepts: 47 3x−10
2. 10
3 ,0 y intercept: (0, −5) x
−6
−4 (x, y ) 0
√
±2 3 (−6, 0)
√
−4, ±2 3
(−2, ±4)
√
0, ±2 3
(2, 0) y (x, y ) −2 −8 (−2, −8) 0 ±4
√
±2 3 −1 − 13
2 −1, − 13
2 2 0 0 −5 (0, −5) 1 7
−2 7
1, − 2 2 −2 (2, −2) x −2 y y
5
4
2
1
−7 −6 −5 −4 −3 −2 −1
−1 y 1 2 3 x −2 2 −3 1
−3 −2 −1
−1 3 −4
1 2 3 4 x −5 −2
−3
−4
−5
−6
−7
−8
−9 The graph is not symmetric about the
xaxis (e.g. (2, −2) is on the graph but
(2, 2) is not) The graph is symmetric about the
xaxis
The graph is not symmetric about the
y axis (e.g. (−6, 0) is on the graph but
(6, 0) is not)
The graph is not symmetric about the
origin (e.g. (−6, 0) is on the graph but
(6, 0) is not) The graph is not symmetric about the
y axis (e.g. (2, −2) is on the graph but
(−2, −2) is not)
The graph is not symmetric about the
origin (e.g. (2, −2) is on the graph but
(−2, 2) is not)
(i) (x + 2)2 + y 2 = 16
Rewrite as y = ± 16 − (x + 2)2 .
xintercepts: (−6, 0), (2, 0)
√
y intercepts: 0, ±2 3 (j) x2 − y 2 = 1
√
Rewrite as: y = ± x2 − 1. 48 Coordinates
xintercepts: (−1, 0), (1, 0) x The graph has no y intercepts
−4
x
−3
−2 y
(x, y )
√
√
± 8 (−3, ± 8)
√
√
± 3 (−2, ± 3) −2
0
2 −1 0 (−1, 0) 0
√
±3
√
±8 (1, 0)
√
(2, ± 3)
√
(3, ± 8)
4
1
2
3 y y
√
±3 5
√
±3 2 (x, y )
√
−4, ±3 5
√
−2, ±3 2 ±3
√
±3 2
√
±3 5 (0, ±3)
√
2, ±3 2
√
4, ±3 5 y
7
6
5
4
3
2
1 3
2 −4 −3 −2 −1
−1 1 1 2 3 4 x −2
−3 −2 −1
−1 1 2 3 x −3 −2 −4 −3 −5
−6
−7 The graph is symmetric about the
xaxis
The graph is symmetric about the
y axis
The graph is symmetric about the
origin
(k) 4y 2 − 9x2 = 36
√ Rewrite as: y = ± 9x2 +36
.
2 The graph has no xintercepts
y intercepts: (0, ±3) The graph is symmetric about the
xaxis
The graph is symmetric about the
y axis
The graph is symmetric about the
origin 1.4 Three Interesting Curves 49
y (l) x3 y = −4
Rewrite as: y = − 32 4
.
x3 The graph has no xintercepts
The graph has no y intercepts
4 x y (x, y ) −2 1
2 1
(−2, 2 ) −1 4 (−1, 4) −1
2 32 (− 1 , 32)
2 1
2 −32 ( 1 , −32)
2 1 −4 (1, −4) 2 −1
2 1
(2, − 2 ) −2 −1 −4 1 2 x −32 The graph is not symmetric about the
xaxis (e.g. (1, −4) is on the graph but
(1, 4) is not)
The graph is not symmetric about the
y axis (e.g. (1, −4) is on the graph but
(−1, −4) is not)
The graph is symmetric about the
origin 1.4
1.4.1 Three Interesting Curves
Circles Recall from geometry that a circle can be determined by ﬁxing a point (called the center) and a
positive number (called the radius) as follows. Definition 1.4. A circle with center (h, k ) and radius r > 0 is the set of all points (x, y ) in
the plane whose distance to (h, k ) is r. 50 Coordinates (x, y )
r
(h, k ) From the picture, we see that a point (x, y ) is on the circle if and only if its distance to (h, k )
is r. We express this relationship algebraically using the Distance Formula, Equation 1.1, as
r= (x − h)2 + (y − k )2 By squaring both sides of this equation, we get an equivalent equation (since r > 0) which gives us
the standard equation of a circle.
Equation 1.3. The Standard Equation of a Circle: The equation of a circle with center
(h, k ) and radius r > 0 is (x − h)2 + (y − k )2 = r2 . Example 1.4.1. Write the standard equation of the circle with center (−2, 3) and radius 5.
Solution. Here, (h, k ) = (−2, 3) and r = 5, so we get
(x − (−2))2 + (y − 3)2 = (5)2
(x + 2)2 + (y − 3)2 = 25
Example 1.4.2. Graph (x + 2)2 + (y − 1)2 = 4. Find the center and radius.
Solution. From the standard form of a circle, Equation 1.3, we have that x + 2 is x − h, so h = −2
and y − 1 is y − k so k = 1. This tells us that our center is (−2, 1). Furthermore, r2 = 4, so r = 2.
Thus we have a circle centered at (−2, 1) with a radius of 2. Graphing gives us
y
4
3
2
1
−4 −3 −2 −1
−1 1 x 1.4 Three Interesting Curves 51 If we were to expand the equation in the previous example and gather up like terms, instead of
the easily recognizable (x + 2)2 + (y − 1)2 = 4, we’d be contending with x2 + 4x + y 2 − 2y + 1 = 0.
If we’re given such an equation, we can complete the square in each of the variables to see if it ﬁts
the form given in Equation 1.3 by following the steps given below.
To Put a Circle into Standard Form
1. Group the same variables together on one side of the equation and put the constant on
the other side.
2. Complete the square on both variables as needed.
3. Divide both sides by the coeﬃcient of the squares. (For circles, they will be the same.) Example 1.4.3. Complete the square to ﬁnd the center and radius of 3x2 − 6x + 3y 2 + 4y − 4 = 0.
Solution. 3x2 − 6x + 3y 2 + 4y − 4 = 0
3x2 − 6x + 3y 2 + 4y = 4
4
3 x2 − 2x + 3 y 2 + y
3 add 4 to both sides =4 4
4
3 x2 − 2x + 1 + 3 y 2 + y +
3
9 = 4 + 3(1) + 3 2
3 2 3(x − 1)2 + 3 y + 2
3 2 (x − 1)2 + y + factor out leading coeﬃcients
4
9 complete the square in x, y = 25
3 factor = 25
9 divide both sides by 3 From Equation 1.3, we identify x − 1 as x − h, so h = 1, and y + 2 as y − k , so k = − 2 . Hence,
3
3
5
the center is (h, k ) = 1, − 2 . Furthermore, we see that r2 = 25 so the radius is r = 3 .
3
9
It is possible to obtain equations like (x − 3)2 + (y + 1)2 = 0 or (x − 3)2 + (y + 1)2 = −1, neither
of which describes a circle. (Do you see why not?) The reader is encouraged to think about what, if
any, points lie on the graphs of these two equations. The next example uses the Midpoint Formula,
Equation 1.2, in conjunction with the ideas presented so far in this section. 52 Coordinates Example 1.4.4. Write the standard equation of the circle which has (−1, 3) and (2, 4) as the
endpoints of a diameter.
Solution. We recall that a diameter of a circle is a line segment containing the center and two
points on the circle. Plotting the given data yields
y 4
3 r
(h, k) 2
1
−2 −1 1 2 3 x Since the given points are endpoints of a diameter, we know their midpoint (h, k ) is the center
of the circle. Equation 1.2 gives us
(h, k ) = x1 + x2 y1 + y2
,
2
2 = −1 + 2 3 + 4
,
2
2 = 17
,
22 The diameter of the circle is the distance between the given points, so we know that half of the
distance is the radius. Thus,
r= 1
2 (x2 − x1 )2 + (y2 − y1 )2 = 1
2 (2 − (−1))2 + (4 − 3)2 =
=
√
Finally, since 10
2 2 1√ 2
3 + 12
2
√
10
2 10
= , our answer becomes
4 1
x−
2 2 7
+ y−
2 2 = 10
4 We close this section with the most important4 circle in all of mathematics: the Unit Circle.
4 While this may seem like an opinion, it is indeed a fact. 1.4 Three Interesting Curves 53 Definition 1.5. The Unit Circle is the circle centered at (0, 0) with a radius of 1. The
standard equation of the Unit Circle is x2 + y 2 = 1. √ 3
.
Example 1.4.5. Find the points on the unit circle with y coordinate
2
√
3
Solution. We replace y with
in the equation x2 + y 2 = 1 to get
2
√2
3
2
= 1.
x+
2
√
√
13
1
1
13
From this we get
= 1 so
= so x = ± . The points are
,
and − ,
. These
4
4
2
22
22
√
3
and the unit circle x2 + y 2 = 1,
two points are the intersection of the horizontal line y =
2
3
x2 + 1.4.2 x2 Parabolas Definition 1.6. Let F be a point in the plane and D be a line not containing F . The set of
all points equidistant from F and D is called the parabola with focus F and directrix D. Schematically, we have the following. F V
D
Each dashed line from the point F to a point on the curve has the same length as the dashed
line from the point on the curve to the line D. The point suggestively labeled V is, as you should
expect, the vertex. The vertex is the point on the parabola closest to the focus.
We want to use only the distance deﬁnition of parabola to derive the equation of a parabola
and, if all is right with the universe, we should get an expression much like those studied in Section 54 Coordinates 3.3. Let p denote the directed5 distance from the vertex to the focus, which by deﬁnition is the
same as the distance from the vertex to the directrix. For simplicity, assume that the vertex is
(0, 0) and that the parabola opens upwards. Hence, the focus is (0, p) and the directrix is the line
y = −p. Our picture becomes
y (x, y )
(0, p)
x (0, 0)
y = −p (x, −p) From the deﬁnition of parabola, we know the distance from (0, p) to (x, y ) is the same as the
distance from (x, −p) to (x, y ). Using the Distance Formula, Equation 1.1, we get (x − 0)2 + (y − p)2 =
x2 + (y − p)2 = (x − x)2 + (y − (−p))2
(y + p)2 x2 + (y − p)2 = (y + p)2 square both sides x2 + y 2 − 2py + p2 = y 2 + 2py + p2 expand quantities x2 = 4py gather like terms 2 Solving for y yields y = xp , which is a quadratic function of the form found in Equation 3.4
4
with a = 41p and vertex (0, 0).
We know from previous experience that if the coeﬃcient of x2 is negative, the parabola opens
2
downwards. In the equation y = xp this happens when p < 0. In our formulation, we say that p is
4
a ‘directed distance’ from the vertex to the focus: if p > 0, the focus is above the vertex; if p < 0,
the focus is below the vertex. The focal length of a parabola is p.
What if we choose to place the vertex at an arbitrary point (h, k )? We can either use transformations (vertical and horizontal shifts from Section 2.5) or rederive the equation from Deﬁnition
1.6 to arrive at the following.
5 We’ll talk more about what ‘directed’ means later. 1.4 Three Interesting Curves 55 Equation 1.4. The Standard Equation of a Verticala Parabola: The equation of a
(vertical) parabola with vertex (h, k ) and focal length p is
(x − h)2 = 4p(y − k )
If p > 0, the parabola opens upwards; if p < 0, it opens downwards.
a That is, a parabola which opens either upwards or downwards. Notice that in the standard equation of the parabola above, only one of the variables, x, is
squared. This is a quick way to distinguish an equation of a parabola from that of a circle because
in the equation of a circle, both variables are squared.
Example 1.4.6. Graph (x + 1)2 = −8(y − 3). Find the vertex, focus, and directrix.
Solution. We recognize this as the form given in Equation 1.4. Here, x − h is x + 1 so h = −1,
and y − k is y − 3 so k = 3. Hence, the vertex is (−1, 3). We also see that 4p = −8 so p = −2. Since
p < 0, the focus will be below the vertex and the parabola will open downwards. The distance from
the vertex to the focus is p = 2, which means the focus is 2 units below the vertex. If we start at
(−1, 3) and move down 2 units, we arrive at the focus (−1, 1). The directrix, then, is 2 units above
the vertex and if we move 2 units up from (−1, 3), we’d be on the horizontal line y = 5.
y
5
4
3
2
1
−6 −5 −4 −3 −2 −1
−1 1 2 3 4 x −2 Of all of the information requested in the previous example, only the vertex is part of the graph
of the parabola. So in order to get a sense of the actual shape of the graph, we need some more
information. While we could plot a few points randomly, a more useful measure of how wide a
parabola opens is the length of the parabola’s latus rectum.6 The latus rectum of a parabola
is the line segment parallel to the directrix which contains the focus. The endpoints of the latus
rectum are, then, two points on ‘opposite’ sides of the parabola. Graphically, we have the following.
6 No, I’m not making this up. 56 Coordinates the latus rectum
F
V
D
It turns out7 that the length of the latus rectum is 4p, which, in light of Equation 1.4, is easy
to ﬁnd. In our last example, for instance, when graphing (x + 1)2 = −8(y − 3), we can use the fact
that the length of the latus rectum is  − 8 = 8, which means the parabola is 8 units wide at the
focus, to help generate a more accurate graph by plotting points 4 units to the left and right of the
focus.
Example 1.4.7. Find the standard form of the parabola with focus (2, 1) and directrix y = −4.
Solution. Sketching the data yields,
y
1
x
−1
−1
−2 1 2 3
The vertex lies on this vertical line
midway between the focus and the directrix −3 From the diagram, we see the parabola opens upwards. (Take a moment to think about it if you
don’t see that immediately.) Hence, the vertex lies below the focus and has an xcoordinate of 2.
To ﬁnd the y coordinate, we note that the distance from the focus to the directrix is 1 − (−4) = 5,
which means the vertex lies 5/2 units (halfway) below the focus. Starting at (2, 1) and moving
down 5/2 units leaves us at (2, −3/2), which is our vertex. Since the parabola opens upwards, we
know p is positive. Thus p = 5/2. Plugging all of this data into Equation 1.4 give us
(x − 2)2 = 4 5
2 (x − 2)2 = 10 y + 7 Consider this an exercise to show what follows. y− −
3
2 3
2 1.4 Three Interesting Curves 57 If we interchange the roles of x and y , we can produce ‘horizontal’ parabolas: parabolas which
open to the left or to the right. The directrices8 of such animals would be vertical lines and the
focus would either lie to the left or to the right of the vertex. A typical ‘horizontal’ parabola is
sketched below.
D V F Equation 1.5. The Standard Equation of a Horizontal Parabola: The equation of a
(horizontal) parabola with vertex (h, k ) and focal length p is
(y − k )2 = 4p(x − h)
If p > 0, the parabola opens to the right; if p < 0, it opens to the left. Example 1.4.8. Graph (y − 2)2 = 12(x + 1). Find the vertex, focus, and directrix.
Solution. We recognize this as the form given in Equation 1.5. Here, x − h is x + 1 so h = −1,
and y − k is y − 2 so k = 2. Hence, the vertex is (−1, 2). We also see that 4p = 12 so p = 3.
Since p > 0, the focus will be the right of the vertex and the parabola will open to the right. The
distance from the vertex to the focus is p = 3, which means the focus is 3 units to the right. If
we start at (−1, 2) and move right 3 units, we arrive at the focus (2, 2). The directrix, then, is 3
units to the left of the vertex and if we move left 3 units from (−1, 2), we’d be on the vertical line
x = −4. Since the length of the latus rectum is 4p = 12, the parabola is 12 units wide at the
focus, and thus there are points 6 units above and below the focus on the parabola.
8 plural of ‘directrix’ 58 Coordinates
y
8
7
6
5
4
3
2
1
−5 −4 −3 −2 −1
−1 1 2 3 x −2
−3
−4 As with circles, not all parabolas will come to us in the forms in Equations 1.4 or 1.5. If we
encounter an equation with two variables in which exactly one variable is squared, we can attempt
to put the equation into a standard form using the following steps. To Put a Parabola into Standard Form
1. Group the variable which is squared on one side of the equation and put the nonsquared
variable and the constant on the other side.
2. Complete the square if necessary and divide by the coeﬃcient of the perfect square.
3. Factor out the coeﬃcient of the nonsquared variable from it and the constant. Example 1.4.9. Consider the equation y 2 + 4y + 8x = 4. Put this equation into standard form
and graph the parabola. Find the vertex, focus, and directrix.
Solution. We need to get a perfect square (in this case, using y ) on the lefthand side of the
equation and factor out the coeﬃcient of the nonsquared variable (in this case, the x) on the
other. 1.4 Three Interesting Curves 59 y 2 + 4y + 8x = 4
y 2 + 4y = −8x + 4
y 2 + 4y + 4 = −8x + 4 + 4 complete the square in y only
(y + 2)2 = −8x + 8 factor (y + 2)2 = −8(x − 1) Now that the equation is in the form given in Equation 1.5, we see that x − h is x − 1 so h = 1,
and y − k is y + 2 so k = −2. Hence, the vertex is (1, −2). We also see that 4p = −8 so that
p = −2. Since p < 0, the focus will be the left of the vertex and the parabola will open to the left.
The distance from the vertex to the focus is p = 2, which means the focus is 2 units to the left
of 1, so if we start at (1, −2) and move left 2 units, we arrive at the focus (−1, −2). The directrix,
then, is 2 units to the right of the vertex, so if we move right 2 units from (1, −2), we’d be on the
vertical line x = 3. Since the length of the latus rectum is 4p is 8, the parabola is 8 units wide at
the focus, so there are points 4 units above and below the focus on the parabola. y
2
1
−2 −1
−1 1 2 x −2
−3
−4
−5
−6 In studying quadratic functions, we have seen parabolas used to model physical phenomena
such as the trajectories of projectiles. Other applications of the parabola concern its ‘reﬂective
property’ which necessitates knowing about the focus of a parabola. For example, many satellite
dishes are formed in the shape of a paraboloid of revolution as depicted below. 60 Coordinates Every cross section through the vertex of the paraboloid is a parabola with the same focus.
To see why this is important, imagine the dashed lines below as electromagnetic waves heading
towards a parabolic dish. It turns out that the waves reﬂect oﬀ the parabola and concentrate at
the focus which then becomes the optimal place for the receiver. If, on the other hand, we imagine
the dashed lines as emanating from the focus, we see that the waves are reﬂected oﬀ the parabola
in a coherent fashion as in the case in a ﬂashlight. Here, the bulb is placed at the focus and the
light rays are reﬂected oﬀ a parabolic mirror to give directional light. F Example 1.4.10. A satellite dish is to be constructed in the shape of a paraboloid of revolution.
If the receiver placed at the focus is located 2 ft above the vertex of the dish, and the dish is to be
12 feet wide, how deep will the dish be?
Solution. One way to approach this problem is to determine the equation of the parabola suggested to us by this data. For simplicity, we’ll assume the vertex is (0, 0) and the parabola opens
upwards. Our standard form for such a parabola is x2 = 4py . Since the focus is 2 units above the
vertex, we know p = 2, so we have x2 = 8y . Visually, 1.4 Three Interesting Curves 61
y
(6, y )
12 units wide ? 2 −6 6 x Since the parabola is 12 feet wide, we know the edge is 6 feet from the vertex. To ﬁnd the
depth, we are looking for the y value when x = 6. Substituting x = 6 into the equation of the
parabola yields 62 = 8y or y = 36/8 = 9/2 = 4.5. Hence, the dish will be 9/2 or 4.5 feet deep. 1.4.3 Ellipses In the deﬁnition of a circle, Deﬁnition 1.4, we ﬁxed a point called the center and considered all
of the points which were a ﬁxed distance r from that one point. For our next conic section, the
ellipse, we ﬁx two distinct points and a distance d to use in our deﬁnition.
Definition 1.7. Given two distinct points F1 and F2 in the plane and a ﬁxed distance d, an
ellipse is the set of all points (x, y ) in the plane such that the sum of the distance from F1 to
(x, y ) and the distance from F2 to (x, y ) is d. The points F1 and F2 are called the focia of the
ellipse.
a the plural of ‘focus’ (x, y )
d1
F1 d2
F2 d1 + d2 = d for all (x, y ) on the ellipse 62 Coordinates We may imagine taking a length of string and anchoring it to two points on a piece of paper.
The curve traced out by taking a pencil and moving it so the string is always taut is an ellipse. Minor Axis The center of the ellipse is the midpoint of the line segment connecting the two foci. The
major axis of the ellipse is the line segment connecting two opposite ends of the ellipse which
also contains the center and foci. The minor axis of the ellipse is the line segment connecting
two opposite ends of the ellipse which contains the center but is perpendicular to the major axis.
The vertices of an ellipse are the points of the ellipse which lie on the major axis. Notice that the
center is also the midpoint of the major axis, hence it is the midpoint of the vertices. In pictures
we have, Major Axis V1 F1 C F2 V2 An ellipse with center C ; foci F1 , F2 ; and vertices V1 , V2 Note that the major axis is the longer of the two axes through the center, and likewise, the
minor axis is the shorter of the two. In order to derive the standard equation of an ellipse, we
assume that the ellipse has its center at (0, 0), its major axis along the xaxis, and has foci (c, 0)
and (−c, 0) and vertices (−a, 0) and (a, 0). We will label the y intercepts of the ellipse as (0, b) and
(0, −b) (We assume a, b, and c are all positive numbers.) Schematically, 1.4 Three Interesting Curves 63
y (0, b) (x, y ) x (−a, 0) (−c, 0) (c, 0) (a, 0) (0, −b) Note that since (a, 0) is on the ellipse, it must satisfy the conditions of Deﬁnition 1.7. That
is, the distance from (−c, 0) to (a, 0) plus the distance from (c, 0) to (a, 0) must equal the ﬁxed
distance d. Since all of these points lie on the xaxis, we get
distance from (−c, 0) to (a, 0) + distance from (c, 0) to (a, 0) = d
(a + c) + (a − c) = d
2a = d
In other words, the ﬁxed distance d mentioned in the deﬁnition of the ellipse is none other than
the length of the major axis. We now use that fact (0, b) is on the ellipse, along with the fact that
d = 2a to get
distance from (−c, 0) to (0, b) + distance from (c, 0) to (0, b) = 2a
(0 − (−c))2 + (b − 0)2 + (0 − c)2 + (b − 0)2
√
√
b2 + c2 + b2 + c2
√
2 b2 + c2
√
b2 + c2 = 2a
= 2a
= 2a
=a From this, we get a2 = b2 + c2 , or b2 = a2 − c2 , which will prove useful later. Now consider a
point (x, y ) on the ellipse. Applying Deﬁnition 1.7, we get 64 Coordinates distance from (−c, 0) to (x, y ) + distance from (c, 0) to (x, y ) = 2a
(x − (−c))2 + (y − 0)2 + (x − c)2 + (y − 0)2 = 2a (x + c)2 + y 2 + (x − c)2 + y 2 = 2a In order to make sense of this situation, we need to do some rearranging, squaring, and more
rearranging.9
(x + c)2 + y 2 + (x − c)2 + y 2 = 2a
(x + c)2 + y 2 = 2a −
(x + c)2 + y 2 2 = (x − c)2 + y 2 2a − (x − c)2 + y 2 2 (x + c)2 + y 2 = 4a2 − 4a (x − c)2 + y 2 + (x − c)2 + y 2
4a (x − c)2 + y 2 = 4a2 + (x − c)2 − (x + c)2 4a (x − c)2 + y 2 = 4a2 − 4cx a
a (x − c)2 + y 2 = a2 − cx (x − c)2 + y 2 a2 (x − c)2 + y 2 2 = a2 − cx 2 = a4 − 2a2 cx + c2 x2 a2 x2 − 2a2 cx + a2 c2 + a2 y 2 = a4 − 2a2 cx + c2 x2
a2 x2 − c2 x2 + a2 y 2 = a4 − a2 c2
a2 − c2 x2 + a2 y 2 = a2 a2 − c2
We are nearly ﬁnished. Recall that b2 = a2 − c2 so that
a2 − c2 x2 + a2 y 2 = a2 a2 − c2
b2 x2 + a2 y 2 = a2 b2
x2 y 2
+2 =1
a2
b
As with parabolas, ellipses have a reﬂective property. If we imagine the dashed lines below as
sound waves, then the waves emanating from one focus reﬂect oﬀ the top of the ellipse and head
9 In other words, tons and tons of Intermediate Algebra. Stay sharp, this is not for the faint of heart. 1.4 Three Interesting Curves 65 towards the other focus. Such geometry is exploited in the construction of socalled ‘Whispering
Galleries’. If a person whispers at one focus, a person standing at the other focus will hear the ﬁrst
person as if they were standing right next to them. F1 F2 Example 1.4.11. Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded
whispering gallery. Recall that a whispering gallery is a room which, in cross section, is half of an
ellipse. If the room is 40 feet high at the center and 100 feet wide at the ﬂoor, how far from the
outer wall should each of them stand so that they will be positioned at the foci of the ellipse?
Solution. Graphing the data yields
y 40 units tall
x 100 units wide
It’s most convenient to imagine this ellipse centered at (0, 0). Since the ellipse is 100 units wide
and 40 units tall, we get a = 50 and b = 40. Hence, our ellipse has the equation
x2
y2
+ 2 = 1.
502 40
√
√
We’re looking for the foci, and we get c = 502 − 402 = 900 = 30, so that the foci are 30 units
from the center. That means they are 50 − 30 = 20 units from the vertices. Hence, Jason and
Jamie should stand 20 feet from opposite ends of the gallery. 66 Coordinates 1.4.4 Exercises 1. Find the standard equation of the circle given the center and radius and sketch its graph.
(a) Center (−1, −5), radius 10
(b) Center (4, −2), radius 3
7
(c) Center −3, 13 , radius 1
2 √
(d) Center (π, e2 ), radius 3 91
√
(e) Center −e, 2 , radius π 2. Complete the square in order to put the equation into standard form. Identify the center and
the radius or explain why the equation does not represent a circle.
(a) x2 − 4x + y 2 + 10y = −25
(b) −2x2 − 36x − 2y 2 − 112 = 0
(c) x2 + y 2 + 8x − 10y − 1 = 0 (d) x2 + y 2 + 5x − y − 1 = 0
(e) 4x2 + 4y 2 − 24y + 36 = 0
(f) x2 + x + y 2 − 6 y = 1
5 3. Find the standard equation of the circle which satisﬁes the following criteria:
(a) center (3, 5), passes through (−1, −2) (c) endpoints of a diameter: (3, 6) and (−1, 4) (b) center (3, 6), passes through (−1, 4) (d) endpoints of a diameter:
√ 4. Verify the following points lie on the Unit Circle: (±1, 0), (0, ±1), ±
√ and ± 1
2, 4 √
2
, ± 22
2 , 3
2 , −1 , ±1, ±
2 √ 3
2 3
1
2 , ±2 5. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot
tall platform making its overall height is 136 feet.10 Find an equation for the wheel assuming
that its center lies on the y axis.
6. Sketch the graph of the given parabola. Find the vertex, focus and directrix. Include the
endpoints of the latus rectum in your sketch.
(a) (y − 2)2 = −12(x + 3)
(b) (y + 4)2 = 4x
(c) (x − 3)2 = 16y
(d) x + 72
3 =2 y+ 5
2 (e) (x − 1)2 = 4(y + 3)
(f) (x + 2)2 = −20(y − 5)
(g) (y − 4)2 − 18(x − 2)
(h) y + 32
2 = −7 x + 9
2 7. Put the equation into standard form and identify the vertex, focus and directrix. 10 Source: Cedar Point’s webpage. 1.4 Three Interesting Curves 67 (a) y 2 − 10y − 27x + 133 = 0
(b) 25x2 + 20x + 5y − 1 = 0
(c) x2 + 2x − 8y + 49 = 0 (d) 2y 2 + 4y + x − 8 = 0
(e) x2 − 10x + 12y + 1 = 0
(f) 3y 2 − 27y + 4x + 211 = 0
4 8. Find an equation for the parabola which ﬁts the given criteria.
(a) Vertex (7, 0), focus (0, 0)
(b) Vertex (−8, −9), Both (0, 0), (−16, 0) are points on the curve
(c) Focus (10, 1), directrix x = 5
(d) The endpoints of latus rectum are (−2, −7) and (−4, −7)
9. Graph the ellipse. Find the center, the lines which contain the major and minor axes, the
vertices, the foci and the eccentricity.
x2
y2
+
=1
169 25
(x − 2)2 (y + 3)2
(b)
+
=1
4
9
(x + 5)2 (y − 4)2
(c)
+
=1
16
1
(x − 1)2 (y − 3)2
(d)
+
=1
10
11 (e) (x − 1)2 (y + 3)2
+
=1
9
4 (f) (x + 2)2 (y − 5)2
+
=1
16
20 (g) (a) (x − 4)2 (y − 2)2
+
=1
8
18 10. Put the equation in standard form. Find the center, the lines which contain the major and
minor axes, the vertices, the foci and the eccentricity.
(a) 12x2 + 3y 2 − 30y + 39 = 0 (d) 9x2 + 4y 2 − 4y − 8 = 0 (b) 5x2 + 18y 2 − 30x + 72y + 27 = 0 (e) 9x2 + 25y 2 − 54x − 50y − 119 = 0 (c) x2 − 2x + 2y 2 − 12y + 3 = 0 (f) 6x2 + 5y 2 − 24x + 20y + 14 = 0 11. Find the standard form of the equation of the ellipse which has the given properties.
(a) Center (3, 7), Vertex (3, 2), Focus (3, 3)
(b) All points on the ellipse are in Quadrant IV except (0, −9) and (8, 0)11
√ (c) Foci (0, ±4), Point on curve 2, 5 3 5
(d) Vertex (−10, 5), Focus (−2, 5), Eccentricity
11 1
2 One might also say that the ellipse is “tangent to the axes” at those two points. 68 Coordinates 12. The Earth’s orbit around the sun is an ellipse with the sun at one focus and eccentricity
e ≈ 0.0167. The length of the semimajor axis (that is, half of the major axis) is deﬁned
to be 1 astronomical unit (AU). The vertices of the elliptical orbit are given special names:
‘aphelion’ is the vertex farther from the sun, and ‘perihelion’ is the vertex closest to the sun.
Find the distance in AU between the sun and aphelion and the distance in AU between the
sun and perihelion.
13. With the help of your classmates, research whispering galleries and other ways ellipses have
been used in architecture and design.
14. With the help of your classmates, research “extracorporeal shockwave lithotripsy”. It uses
the reﬂective property of the ellipsoid to dissolve kidney stones. 1.4 Three Interesting Curves 1.4.5 69 Answers
y 2 2 (d) (x − π )2 + y − e2 1. (a) (x + 1)2 + (y + 5)2 = 100 = 91 3 y 5
e2 + −11 −1 9 √
3 91 x e2 −5 √
e2 − 3 91
−15 x (b) (x − 4)2 + (y + 2)2 = 9 √
π − 3 91 y (e) (x + e)2 + y − 1
1 4 7 √
π + 3 91 π √ x 2 2 = π2 y −2 √
2+π −5 √
2 x (c) (x + 3)2 + y − 72
13 = −e − π 1
4 −e −e + π
√
2−π y 27
26
7
13 7
−2 −3 1
26
5
−2 2. (a) (x − 2)2 + (y + 5)2 = 4
Center (2, −5), radius r = 2
(b) (x + 9)2 + y 2 = 25
Center (−9, 0), radius r = 5
(c) (x + 4)2 + (y − 5)2 = 42 √
Center (−4, 5), radius r = 42 x (d) x + 52
2 + y− 12
2 = 30
4 5
Center − 2 , 1 , radius r =
2
(e) x2 + (y − 3)2 = 0
This is not a circle.
2
2
(f) x + 1 + y − 3 = 161
2
5
100
1
Center − 2 , 3 , radius r =
5 √ 30
2 √ 161
10 70 Coordinates
3. (a) (x − 3)2 + (y − 5)2 = 65
(b) (x − 3)2 + (y − 6)2 = 20 (c) (x − 1)2 + (y − 5)2 = 5
(d) (x − 1)2 + y − 32
2 = 13
2 5 6 5. x2 + (y − 72)2 = 4096
y 6. (a) (y − 2)2 = −12(x + 3) 8 Vertex (−3, 2)
Focus (−6, 2)
Directrix x = 0
Endpoints of latus rectum (−6, 8), (−6, −4) 7
6
5
4
3
2
1 x −7 −6 −5 −4 −3 −2 −1
−1
−2
−3
−4 y (b) (y + 4)2 = 4x
Vertex (0, −4)
Focus (1, −4)
Directrix x = −1
Endpoints of latus rectum (1, −2), (1, −6) −1
−1 1 2 3 x 4 −2
−3
−4
−5
−6
−7
−8 y (c) (x − 3)2 = −16y 4 Vertex (3, 0)
Focus (3, −4)
Directrix y = 4
Endpoints of latus rectum (−5, −4), (11, −4) 3
2
1
−5 −4 −3 −2 −1
−1 1 2 3 4 7 8 9 10 11 x −2
−3
−4 (d) x + 72
3 =2 y+
Vertex − 7 , − 5
3
2
Focus − 7 , −2
3
Directrix y = −3 5
2 Endpoints of latus rectum − 10 , −2 , − 4 , −2
3
3 1.4 Three Interesting Curves 71 y
2
1 −5 −4 −3 −2 −1 x
−1
−2
−3 y (e) (x − 1)2 = 4(y + 3)
Vertex (1, −3)
Focus (1, −2)
Directrix y = −4
Endpoints of latus rectum (3, −2), (−1, −2) −3 −2 −1 1 2 3 x 4 −1
−2
−3
−4 y (f) (x + 2)2 = −20(y − 5) 10 Vertex (−2, 5)
Focus (−2, 0)
Directrix y = 10
Endpoints of latus rectum (−12, 0), (8, 0) 9
8
7
6
5
4
3
2
1
−12 −10 −8 −6 −4 −2 4 6 8 x y (g) (y − 4)2 − 18(x − 2)
Vertex (2, 4)
Focus 123 , 4
Directrix x = − 5
2
Endpoints of latus rectum − 13 , −5 ,
2 2 13
11
9 13
2 , 13 7
5
3
1
−1 1
−
−3
−5 1 2 3 4 5 6 7 x 72 Coordinates
y 32
2 = −7 x + 9
2
9
Vertex − 2 , − 3
2
Focus − 25 , − 3
4
2
Directrix x = − 11
4
Endpoints of latus rectum − 25 , 2 , − 25 , −5
4
4 (h) y + 2
1 x −5 −4 −3 −2 −1
−1
−2
−3
−4
−5 7. (a) (y − 5)2 = 27(x − 4)
Vertex (4, 5)
Focus 443 , 5
Directrix x = − 11
4 1
(d) (y + 1)2 = − 2 (x − 10)
Vertex (10, −1)
Focus 789 , −1
Directrix x = 81
8 2 (e) (x − 5)2 = −12(y − 2)
Vertex (5, 2)
Focus (5, −1)
Directrix y = 5 (c) (x + 1)2 = 8(y − 6)
Vertex (−1, 6)
Focus (−1, 8)
Directrix y = 4 (f) y − 9 = − 4 (x − 2)
2
3
9
Vertex 2, 2
Focus 5 , 9
32
7
Directrix x = 3 1
(b) x + 2 = − 5 (y − 1)
5
2
Vertex − 5 , 1
2 19
Focus − 5 , 20
Directrix y = 21
20 8. (a) y 2 = −28(x − 7)
(b) (x + 8)2 = 64 (y + 9)
9
(c) (y − 1)2 = 10 x − 15
2
x2
y2
+
=1
9. (a)
169 25
Center (0, 0)
Major axis along y = 0
Minor axis along x = 0
Vertices (13, 0), (−13, 0)
Foci (12, 0), (−12, 0)
12
e=
13 (b) (x − 2)2 (y + 3)2
+
=1
4
9
Center (2, −3) 2 (d) (x − 1)2 = 6 y + 17 or
2
(x − 1)2 = −6 y + 11
2 y
5
4
3
2
1
−13 −1
−1 1 −2
−3
−4
−5 Major axis along x = 2
Minor axis along y = −3
Vertices (2, 0), (2, −6) 13 x 1.4 Three Interesting Curves
Foci √ , −3 +
(2
5
e=
3 √ 5), (2, −3 − 73
√ y 5) 1 2 3 4 x −1
−2
−3
−4
−5
−6 (x + 5)2 (y − 4)2
(c)
+
=1
16
1
Center (−5, 4)
Major axis along y = 4
Minor axis along x = −5
Vertices (−9, 4), (−1, 4) √
√
Foci √ 5 + 15, 4), (−5 − 15, 4)
(−
15
e=
4 (d) y
4
3
2
1
−9 −8 −7 −6 −5 −4 −3 −2 −1 y (x − 1)2 (y − 3)2
+
=1
10
11
Center (1, 3)
Major axis along x = 1
Minor axis along y = 3
√
√
Vertices (1, 3 + 11), (1, 3 − 11)
Foci √ , 2), (1, 4)
(1
11
e=
11 6
5
4
3
2
1
−2 −1 (e) (x − 1)2 (y + 3)2
+
=1
9
4
Center (1, −3)
Major axis along y = −3
Minor axis along x = 1
Vertices (4, −3), (−2, −3)
√
√
Foci √ + 5, −3), (1 − 5, −3)
(1
5
e=
3 1 2 3 4 1 2 3 4 x y
−2 −1
−1
−2
−3
−4
−5 x x 74 Coordinates (f) y (x + 2)2 (y − 5)2
+
=1
16
20
Center (−2, 5)
Major axis along x = −2
Minor axis along y√ 5
=
√
Vertices (−2, 5 + 2 5), (−2, 5 − 2 5)
Foci √ 2, 7), (−2, 3)
(−
5
e=
5 10
9
8
7
6
5
4
3
2
1
−6 −5 −4 −3 −2 −1 (g) 1 2 x y (x − 4)2 (y − 2)2
+
=1
8
18
Center (4, 2)
Major axis along x = 4
Minor axis along√ = 2
y
√
Vertices (4, 2 + 3 2), (4, 2 − 3 2)
√
√
Foci √ , 2 + 10), (4, 2 − 10)
(4
5
e=
3 7
6
5
4
3
2
1
1 2 3 4 5 6 7 x −1
−2
−3 10. (a) x2 (y − 5)2
+
=1
3
12
Center (0, 5)
Major axis along x = 0
Minor axis along√ = 5
y
√
Vertices (0, 5 − 2 3), (0, 5 + 2 3)
Foci √ , 2), (0, 8)
(0
e = 23 (b) (x − 3)2 (y + 2)2
+
=1
18
5
Center (3, −2)
Major axis along y = −2
Minor axis along x = 3
√
√
Vertices (3 − 3 2, −2), (3 + 3 2, −2)
√
√
Foci √ − 13, −2), (3 + 13, −2)
(3
26
e= 6 1.4 Three Interesting Curves (c) (x − 1)2 (y − 3)2
+
=1
16
8
Center (1, 3)
Major Axis along y = 3
Minor Axis along x = 1
Vertices (5,√ (−3, 3) √
3),
Foci √ + 2 2, 3), (1 − 2 2, 3)
(1
e = 22 75 (e) (x − 3)2 (y − 1)2
+
=1
25
9
Center (3, 1)
Major Axis along y = 1
Minor Axis along x = 3
Vertices (8, 1), (−2, 1)
Foci (7, 1), (−1, 1)
e= 4
5 2 (d) x2 4 y − 1
2
+
=1
1
9
Center 0, 1
2
Major Axis along x = 0 (the y axis)
1
Minor Axis along y = 2
Vertices (0,√ (0, −1) √
2),
1+ 5
Foci 0, 2
, 0, 1−2 5 (f) √ e= 5
3 (x − 3)2
9
(x − 8)2
(b)
64
y2
x2
+
(c)
9
25
(x − 6)2
(d)
256 11. (a) (y − 7)2
=1
25
(y + 9)2
+
=1
81
+ =1
+ (y − 5)2
=1
192 12. Distance from the sun to aphelion ≈ 1.0167 AU.
Distance from the sun to perihelion ≈ 0.9833 AU. (x − 2)2 (y + 2)2
+
=1
5
6
Center (2, −2)
Major Axis along x = 2
Minor Axis along √ = −2
y
√
Vertices 2, −2 + 6 , (2, −2 − 6)
Foci √ , −1), (2, −3)
(2
e = 66 76 Coordinates Chapter 2 Functions
2.1 Introduction to Functions One of the core concepts in College Algebra is the function. There are many ways to describe a
function and we begin by deﬁning a function as a special kind of relation. Definition 2.1. A relation in which each xcoordinate is matched with only one y coordinate
is said to describe y as a function of x. Example 2.1.1. Which of the following relations describe y as a function of x?
1. R1 = {(−2, 1), (1, 3), (1, 4), (3, −1)}
2. R2 = {(−2, 1), (1, 3), (2, 3), (3, −1)}
Solution. A quick scan of the points in R1 reveals that the xcoordinate 1 is matched with
two diﬀerent y coordinates: namely 3 and 4. Hence in R1 , y is not a function of x. On the
other hand, every xcoordinate in R2 occurs only once which means each xcoordinate has only one
corresponding y coordinate. So, R2 does represent y as a function of x.
Note that in the previous example, the relation R2 contained two diﬀerent points with the same
y coordinates, namely (1, 3) and (2, 3). Remember, in order to say y is a function of x, we just
need to ensure the same xcoordinate isn’t used in more than one point.1
To see what the function concept means geometrically, we graph R1 and R2 in the plane.
1 We will have occasion later in the text to concern ourselves with the concept of x being a function of y . In this
case, R1 represents x as a function of y ; R2 does not. 78 Functions
y y 4 4 3 3 2 2 1 1 −2 −1
−1 1 2 x 3 −2 −1
−1 The graph of R1 1 2 3 x The graph of R2 The fact that the xcoordinate 1 is matched with two diﬀerent y coordinates in R1 presents
itself graphically as the points (1, 3) and (1, 4) lying on the same vertical line, x = 1. If we turn
our attention to the graph of R2 , we see that no two points of the relation lie on the same vertical
line. We can generalize this idea as follows Theorem 2.1. The Vertical Line Test: A set of points in the plane represents y as a
function of x if and only if no two points lie on the same vertical line. It is worth taking some time to meditate on the Vertical Line Test; it will check to see how well
you understand the concept of ‘function’ as well as the concept of ‘graph’.
Example 2.1.2. Use the Vertical Line Test to determine which of the following relations describes
y as a function of x.
y y 4 4 3 3 2 2 1 1 1 2 3 −1 The graph of R x −1 1 x −1 The graph of S Solution. Looking at the graph of R, we can easily imagine a vertical line crossing the graph
more than once. Hence, R does not represent y as a function of x. However, in the graph of S ,
every vertical line crosses the graph at most once, and so S does represent y as a function of x. 2.1 Introduction to Functions 79 In the previous test, we say that the graph of the relation R fails the Vertical Line Test, whereas
the graph of S passes the Vertical Line Test. Note that in the graph of R there are inﬁnitely many
vertical lines which cross the graph more than once. However, to fail the Vertical Line Test, all you
need is one vertical line that ﬁts the bill, as the next example illustrates.
Example 2.1.3. Use the Vertical Line Test to determine which of the following relations describes
y as a function of x.
y y 4 4 3 3 2 2 1 1 −1 1 −1 x −1 1 x −1 The graph of S1 The graph of S2 Solution. Both S1 and S2 are slight modiﬁcations to the relation S in the previous example whose
graph we determined passed the Vertical Line Test. In both S1 and S2 , it is the addition of the
point (1, 2) which threatens to cause trouble. In S1 , there is a point on the curve with xcoordinate
1 just below (1, 2), which means that both (1, 2) and this point on the curve lie on the vertical line
x = 1. (See the picture below.) Hence, the graph of S1 fails the Vertical Line Test, so y is not a
function of x here. However, in S2 notice that the point with xcoordinate 1 on the curve has been
omitted, leaving an ‘open circle’ there. Hence, the vertical line x = 1 crosses the graph of S2 only
at the point (1, 2). Indeed, any vertical line will cross the graph at most once, so we have that the
graph of S2 passes the Vertical Line Test. Thus it describes y as a function of x.
y
4
3
2
1 −1 x
−1 S1 and the line x = 1 80 Functions Suppose a relation F describes y as a function of x. The sets of x and y coordinates are given
special names. Definition 2.2. Suppose F is a relation which describes y as a function of x.
The set of the xcoordinates of the points in F is called the domain of F .
The set of the y coordinates of the points in F is called the range of F . We demonstrate ﬁnding the domain and range of functions given to us either graphically or via
the roster method in the following example.
Example 2.1.4. Find the domain and range of the following functions
1. F = {(−3, 2), (0, 1), (4, 2), (5, 2)}
2. G is the function graphed below:
y
4
3
2
1 −1 1 x −1 The graph of G
Solution. The domain of F is the set of the xcoordinates of the points in F : {−3, 0, 4, 5} and
the range of F is the set of the y coordinates: {1, 2}.2
To determine the domain and range of G, we need to determine which x and y values occur as
coordinates of points on the given graph. To ﬁnd the domain, it may be helpful to imagine collapsing
the curve to the xaxis and determining the portion of the xaxis that gets covered. This is called
2 When listing numbers in a set, we list each number only once, in increasing order. 2.1 Introduction to Functions 81 projecting the curve to the xaxis. Before we start projecting, we need to pay attention to two
subtle notations on the graph: the arrowhead on the lower left corner of the graph indicates that the
graph continues to curve downwards to the left forever more; and the open circle at (1, 3) indicates
that the point (1, 3) isn’t on the graph, but all points on the curve leading up to that point are on
the curve.
y
y
4 4 3 3 2 project down 2 1 −1 1 1 x −1 −1 1 x −1 project up
The graph of G The graph of G We see from the ﬁgure that if we project the graph of G to the xaxis, we get all real numbers
less than 1. Using interval notation, we write the domain of G is (−∞, 1). To determine the range
of G, we project the curve to the y axis as follows:
y
y
4
3 3 2 project right 4 2 1 −1 project left
1 −1 The graph of G x 1 −1 1 x −1 The graph of G Note that even though there is an open circle at (1, 3), we still include the y value of 3 in our
range, since the point (−1, 3) is on the graph of G. We see that the range of G is all real numbers
less than or equal to 4, or, in interval notation: (−∞, 4]. 82 Functions All functions are relations, but not all relations are functions. Thus the equations which described the relations in Section 1.2 may or may not describe y as a function of x. The algebraic
representation of functions is possibly the most important way to view them so we need a process
for determining whether or not an equation of a relation represents a function. (We delay the
discussion of ﬁnding the domain of a function given algebraically until Section 2.2.)
Example 2.1.5. Determine which equations represent y as a function of x:
1. x3 + y 2 = 1
2. x2 + y 3 = 1
3. x2 y = 1 − 3y Solution. For each of these equations, we solve for y and determine whether each choice of x will
determine only one corresponding value of y .
1.
x3 + y 2 = 1
y 2 = 1 − x3
√
y2 =
1 − x3
extract square roots
√
y = ± 1 − x3
√
If we substitute x = 0 into our equation for y , we get: y = ± 1 − 03 = ±1, so that (0, 1)
and (0, −1) are on the graph of this equation. Hence, this equation does not represent y as
a function of x.
2.
x2 + y 3 = 1 3 y 3 = 1 − x2
√
y 3 = 3 1 − x2
√
y = 3 1 − x2 For every choice of x, the equation y =
equation describes y as a function of x. √
3 1 − x2 returns only one value of y . Hence, this 2.1 Introduction to Functions 83 3.
x2 y = 1 − 3y
x2 y + 3y = 1
y x2 + 3 =1 y= x2 factor
1
+3 For each choice of x, there is only one value for y , so this equation describes y as a function
of x.
Of course, we could always use our graphing calculator to verify our responses to the previous
example. For example, if we wanted to verify that the ﬁrst equation does not represent y as a
function of x, we could enter the equation for y into the calculator as indicated below and graph.
Note that we need to enter both solutions – the positive and the negative square root – for y . The
resulting graph clearly fails the Vertical Line Test, so does not represent y as a function of x. 2.1.1 Exercises 1. Determine which of the following relations represent y as a function of x. Find the domain
and range of those relations which are functions.
(a) {(−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}
(b) {(−3, 0), (1, 6), (2, −3), (4, 2), (−5, 6), (4, −9), (6, 2)}
(c) {(−3, 0), (−7, 6), (5, 5), (6, 4), (4, 9), (3, 0)}
(d) {(1, 2), (4, 4), (9, 6), (16, 8), (25, 10), (36, 12), . . .}
(e) {(x, y ) : x is an odd integer, and y is an even integer}
(f) {(x, 1) : x is an irrational number}
(g) {(1, 0), (2, 1), (4, 2), (8, 3), (16, 4), (32, 5), . . . }
(h) {. . . , (−3, 9), (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4), (3, 9), . . . }
(i) {(−2, y ) : −3 < y < 4}
(j) {(x, 3) : −2 ≤ x < 4} 84 Functions
2. Determine which of the following relations represent y as a function of x. Find the domain
and range of those relations which are functions.
y y
4 2 3 1 2
−4 −3 −2 −1 1 2 3 4 5 x −1 1 −2
−4 −3 −2 −1 1 x
−3 −1 (b) (a) y
3 y
5 2 4 1 3
−3 −2 −1 1 2 x 3 −1 2 −2 1 −3
−2 −1 1 x 2 (d) (c) y
4 y
3 3 2 2 1 1 1 2 3 4 5 6 7 8 9 x −4 −3 −2 −1 (e) 1 2 3 2 3 x (f)
y
4 y
4 3 3 2 2 1 1
−5 −4 −3 −2 −1 1
−1 −4 −3 −2 −1
−1 (g) 1 x
−2 (h) 4 x 2.1 Introduction to Functions 85 y
y 9
8
7
6
5
4
3
2
1
−3 2 −1
− −1 5
4
3
2
1 123 1 −5 −4 −3 −2 −1
−1 x 2 3 4 5 x 1 2 3 4 5 6 −2 −2
−3
−4
−5 −3
−4
−5 (i) (j)
y y 5 5 4 4 3 3 2 2 1 1 −5 −4 −3 −2 −1
−1 1 2 3 4 5 x −1
−1 −2 −2 −3 −3 −4 −4 −5 x −5 (k) (l) 3. Determine which of the following equations represent y as a function of x.
(a) y = x3 − x
√
(b) y = x − 2
(c) x3 y = −4
(d) x2 − y 2 = 1
x
(e) y = 2
x −9
(f) x = −6
(g) x = y2 +4 (h) y = x2 + 4
(i) x2 + y 2 = 4
√
(j) y = 4 − x2
(k) x2 − y 2 = 4
(l) x3 + y 3 = 4
(m) 2x + 3y = 4
(n) 2xy = 4 4. Explain why the height h of a Sasquatch is a function of its age N in years. Given that a
Sasquatch is 2 feet tall at birth, experiences growth spurts at ages 3, 23 and 57, and lives to
be about 150 years old with a maximum height of 9 feet, sketch a rough graph of the height
function.
5. Explain why the population P of Sasquatch in a given area is a function of time t. What
would be the range of this function?
6. Explain why the relation between your classmates and their email addresses may not be a
function. What about phone numbers and Social Security Numbers? 86 Functions
7. The process given in Example 2.1.5 for determining whether an equation of a relation represents y as a function of x breaks down if we cannot solve the equation for y in terms of x.
However, that does not prevent us from proving that an equation which fails to represent y
as a function of x actually fails to do so. What we really need is two points with the same
xcoordinate and diﬀerent y coordinates which both satisfy the equation so that the graph
of the relation would fail the Vertical Line Test 2.1. Discuss with your classmates how you
might ﬁnd such points for the relations given below.
(a) x3 + y 3 − 3xy = 0
(b) x4 = x2 + y 2 2.1.2 (c) y 2 = x3 + 3x2
(d) (x2 + y 2 )2 = x3 + y 3 Answers 1. (a) Function, domain = {−3, −2, −1, 0, 1, 2, 3}, range = {0, 1, 4, 9}.
(b) Not a function.
(c) Function, domain = {−7, −3, 3, 4, 5, 6}, range = {0, 4, 5, 6, 9}
(d) Function, domain = {1, 4, 9, 16, 25, 36, . . .} = {x : x is a perfect square},
range = {2, 4, 6, 8, 10, 12, . . .} = {y : y is a positive even integer}
(e) Not a function
(f) Function, domain = {x : x is irrational}, range = {1}.
(g) Function, domain = {x : x = 2n for some whole number n}, range = {y : y is any whole number},
(h) Function, domain = {x : x is any integer}, range = {y : y = n2 for some integer n}.
(i) Not a function.
(j) Function, domain = [−2, 4), range = {3}.
2. (a) Function, domain = {−4, −3, −2, −1, 0, 1}, range = {−1, 0, 1, 2, 3, 4}
(b) Not a function
(c) Function, domain = (−∞, ∞), range = [1, ∞)
(d) Not a function
(e) Function, domain = [2, ∞), range = [0, ∞)
(f) Function, domain = (−∞, ∞), range = (0, 4]
(g) Not a function
(h) Function, domain = [−5, −3) ∪ (−3, 3), range = (−2, −1) ∪ [0, 4)
(i) Function, domain = [−2, ∞), range = [−3, ∞)
(j) Not a function
(k) Function, domain = [−5, 4), range = [−4, 4)
(l) Function, domain = [0, 3) ∪ (3, 6], range = (−4, −1] ∪ [0, 4]
3. (a) Function (e) Function (b) Function (f) Not a function (c) Function (g) Not a function (d) Not a function (h) Function 2.2 Function Notation 87 (i) Not a function (l) Function (j) Function
(k) Not a function 2.2 (m) Function
(n) Function Function Notation In Deﬁnition 2.1, we described a function as a special kind of relation – one in which each xcoordinate is matched with only one y coordinate. In this section, we focus more on the process
by which the x is matched with the y . If we think of the domain of a function as a set of inputs
and the range as a set of outputs, we can think of a function f as a process by which each input
x is matched with only one output y . Since the output is completely determined by the input x
and the process f , we symbolize the output with function notation: ‘f (x)’, read ‘f of x.’ In this
case, the parentheses here do not indicate multiplication, as they do elsewhere in algebra. This
could cause confusion if the context is not clear. In other words, f (x) is the output which results
by applying the process f to the input x. This relationship is typically visualized using a diagram
similar to the one below. f x
Domain
(Inputs) y = f (x)
Range
(Outputs) The value of y is completely dependent on the choice of x. For this reason, x is often called the
independent variable, or argument of f , whereas y is often called the dependent variable.
As we shall see, the process of a function f is usually described using an algebraic formula.
For example, suppose a function f takes a real number and performs the following two steps, in
sequence
1. multiply by 3
2. add 4 88 Functions If we choose 5 as our input, in step 1 we multiply by 3 to get (5)(3) = 15. In step 2, we add 4 to
our result from step 1 which yields 15 + 4 = 19. Using function notation, we would write f (5) = 19
to indicate that the result of applying the process f to the input 5 gives the output 19. In general,
if we use x for the input, applying step 1 produces 3x. Following with step 2 produces 3x + 4 as
our ﬁnal output. Hence for an input x, we get the output f (x) = 3x + 4. Notice that to check our
formula for the case x = 5, we replace the occurrence of x in the formula for f (x) with 5 to get
f (5) = 3(5) + 4 = 15 + 4 = 19, as required.
Example 2.2.1. Suppose a function g is described by applying the following steps, in sequence
1. add 4
2. multiply by 3
Determine g (5) and ﬁnd an expression for g (x).
Solution. Starting with 5, step 1 gives 5 + 4 = 9. Continuing with step 2, we get (3)(9) = 27. To
ﬁnd a formula for g (x), we start with our input x. Step 1 produces x + 4. We now wish to multiply
this entire quantity by 3, so we use a parentheses: 3(x + 4) = 3x + 12. Hence, g (x) = 3x + 12. We
can check our formula by replacing x with 5 to get g (5) = 3(5) + 12 = 15 + 12 = 27 .
Most of the functions we will encounter in College Algebra will be described using formulas like
the ones we developed for f (x) and g (x) above. Evaluating formulas using this function notation
is a key skill for success in this and many other math courses.
Example 2.2.2. For f (x) = −x2 + 3x + 4, ﬁnd and simplify
1. f (−1), f (0), f (2)
2. f (2x), 2f (x)
3. f (x + 2), f (x) + 2, f (x) + f (2)
Solution.
1. To ﬁnd f (−1), we replace every occurrence of x in the expression f (x) with −1
f (−1) = −(−1)2 + 3(−1) + 4
= −(1) + (−3) + 4
=0
Similarly, f (0) = −(0)2 + 3(0) + 4 = 4, and f (2) = −(2)2 + 3(2) + 4 = −4 + 6 + 4 = 6. 2.2 Function Notation 89 2. To ﬁnd f (2x), we replace every occurrence of x with the quantity 2x
f (2x) = −(2x)2 + 3(2x) + 4
= −(4x2 ) + (6x) + 4
= −4x2 + 6x + 4
The expression 2f (x) means we multiply the expression f (x) by 2
2f (x) = 2 −x2 + 3x + 4
= −2x2 + 6x + 8
Note the diﬀerence between the answers for f (2x) and 2f (x). For f (2x), we are multiplying
the input by 2; for 2f (x), we are multiplying the output by 2. As we see, we get entirely
diﬀerent results. Also note the practice of using parentheses when substituting one algebraic
expression into another; we highly recommend this practice as it will reduce careless errors.
3. To ﬁnd f (x + 2), we replace every occurrence of x with the quantity x + 2
f (x + 2) = −(x + 2)2 + 3(x + 2) + 4
= − x2 + 4x + 4 + (3x + 6) + 4
= −x2 − 4x − 4 + 3x + 6 + 4
= −x2 − x + 6
To ﬁnd f (x) + 2, we add 2 to the expression for f (x)
f (x) + 2 = −x2 + 3x + 4 + 2 = −x2 + 3x + 6
Once again, we see there is a dramatic diﬀerence between modifying the input and modifying
the output. Finally, in f (x) + f (2) we are adding the value f (2) to the expression f (x).
From our work above, we see f (2) = 6 so that
f (x) + f (2) = −x2 + 3x + 4 + 6 = −x2 + 3x + 10
Notice that f (x + 2), f (x) + 2 and f (x) + f (2) are three diﬀerent expressions. Even though
function notation uses parentheses, as does multiplication, there is no general ‘distributive
property’ of function notation. 90 Functions
Suppose we wish to ﬁnd r(3) for r(x) = 2x
. Substitution gives
x2 − 9 r(3) = 2(3)
6
=,
(3)2 − 9
0 which is undeﬁned. The number 3 is not an allowable input to the function r; in other words, 3 is
not in the domain of r. Which other real numbers are forbidden in this formula? We think back
to arithmetic. The reason r(3) is undeﬁned is because substitution results in a division by 0. To
determine which other numbers result in such a transgression, we set the denominator equal to 0
and solve
x2 − 9 = 0
x2 = 9
√
√
x2 =
9 extract square roots
x = ±3
As long as we substitute numbers other than 3 and −3, the expression r(x) is a real number.
Hence, we write our domain in interval notation as (−∞, −3) ∪ (−3, 3) ∪ (3, ∞). When a formula
for a function is given, we assume (unless the contrary is explicitly stated) that the domain of
the function is the set of all real numbers for which the formula makes arithmetic sense when the
number is substituted into the formula. This set of numbers is often called the implied domain1
of the function. At this stage, there are only two mathematical sins we need to avoid: division by
0 and extracting even roots of negative numbers. The following example illustrates these concepts.
Example 2.2.3. Find the domain2 of the following functions.
1. f (x) = 2 4x
1−
x−3
√
2. g (x) = 4 − 3x
√
3. h(x) = 5 4 − 3x 4. r(x) = 4
√
6− x+3 5. I (x) = 3x2
x Solution.
1. In the expression for f , there are two denominators. We need to make sure neither of them is
0. To that end, we set each denominator equal to 0 and solve. For the ‘small’ denominator,
1
2 or, ‘implicit domain’
The word ‘implied’ is, well, implied. 2.2 Function Notation 91 we get x − 3 = 0 or x = 3. For the ‘large’ denominator 1− 4x
x−3 =0 1= 4x
x−3 (1)(x − 3) = 4x
$
(x − $
$$ 3) clear denominators
x $3
$−$ x − 3 = 4x
−3 = 3x
−1 = x
So we get two real numbers which make denominators 0, namely x = −1 and x = 3. Our
domain is all real numbers except −1 and 3: (−∞, −1) ∪ (−1, 3) ∪ (3, ∞).
2. The potential disaster for g is if the radicand3 is negative. To avoid this, we set 4 − 3x ≥ 0
4 − 3x ≥ 0
4 ≥ 3x
4
≥x
3
Hence, as long as x ≤ 4 , the expression 4 − 3x ≥ 0, and the formula g (x) returns a real
3
4
number. Our domain is −∞, 3 .
3. The formula for h(x) is hauntingly close to that of g (x) with one key diﬀerence – whereas
the expression for g (x) includes an even indexed root (namely a square root), the formula
for h(x) involves an odd indexed root (the ﬁfth root.) Since odd roots of real numbers (even
negative real numbers) are real numbers, there is no restriction on the inputs to h. Hence,
the domain is (−∞, ∞).
4. To ﬁnd the domain of r, we notice that we have two potentially hazardous issues: not only
do we have a denominator, we have a square root in that denominator. To satisfy the square
3 The ‘radicand’ is the expression ‘inside’ the radical. 92 Functions
root, we set the radicand x + 3 ≥ 0 so x ≥ −3. Setting the denominator equal to zero gives
6− √ x+3 = 0
√
x+3
6=
√
62 =
x+3 2 36 = x + 3
33 = x
Since we squared both sides in the course of solving this equation, we need to check our
√
√
answer. Sure enough, when x = 33, 6 − x + 3 = 6 − 36 = 0, and so x = 33 will cause
problems in the denominator. At last we can ﬁnd the domain of r: we need x ≥ −3, but
x = 33. Our ﬁnal answer is [−3, 33) ∪ (33, ∞).
2 x
5. It’s tempting to simplify I (x) = 3x = 3x, and, since there are no longer any denominators,
claim that there are no longer any restrictions. However, in simplifying I (x), we are assuming
0
x = 0, since 0 is undeﬁned.4 Proceeding as before, we ﬁnd the domain of I to be all real
numbers except 0: (−∞, 0) ∪ (0, ∞). It is worth reiterating the importance of ﬁnding the domain of a function before simplifying,
as evidenced by the function I in the previous example. Even though the formula I (x) simpliﬁes
to 3x, it would be inaccurate to write I (x) = 3x without adding the stipulation that x = 0. To
understand why this is so important imagine writing a computer program to evaluate I (x). The
computer is told to input x, square it, multiply by 3 and divide by x. If the input is zero, the
program will generate an error because division by zero is undeﬁned. Most computer programs are
not sophisticated enough to do the preliminary simpliﬁcation.
To ﬁnd the domain of a function deﬁned by an expression, do so before simplifying!
Our next example shows how a function can be used to model realworld phenomena.
Example 2.2.4. The height h in feet of a model rocket above the ground t seconds after lift oﬀ is
given by −5t2 + 100t, if 0 ≤ t ≤ 20
h(t) = 0, if t > 20
Find and interpret h(10) and h(60).
4 More precisely, the fraction
such creatures. 0
0 is an ‘indeterminant form’. Much time will be spent in Calculus wrestling with 2.2 Function Notation 93 Solution. There are a few qualities of h which may be oﬀputting. The ﬁrst is that, unlike
previous examples, the independent variable is t, not x. In this context, t is chosen because it
represents time. The second is that the function is broken up into two rules: one formula for values
of t between 0 and 20 inclusive, and another for values of t greater than 20. To ﬁnd h(10), we ﬁrst
notice that 10 is between 0 and 20 so we use the ﬁrst formula listed: h(t) = −5t2 + 100t. Hence,
h(10) = −5(10)2 + 100(10) = 500. In terms of the model rocket, this means that 10 seconds after
lift oﬀ, the model rocket is 500 feet above the ground. To ﬁnd h(60), we note that 60 is greater
than 20, so we use the rule h(t) = 0. This function returns a value of 0 regardless of what value is
substituted in for t, so h(60) = 0. This means that 60 seconds after lift oﬀ, the rocket is 0 feet above
the ground; in other words, a minute after lift oﬀ, the rocket has already returned to earth.
The type of function in the previous example is called a piecewisedeﬁned function, or ‘piecewise’ function for short. Many realworld phenomena (e.g. postal rates,5 income tax formulas6 )
are modeled by such functions. Also note that the domain of h in the above example is restricted
to t ≥ 0. For example, h(−3) is not deﬁned because t = −3 doesn’t satisfy any of the conditions
in any of the function’s pieces. There is no inherent arithmetic reason which prevents us from
calculating, say, −5(−3)2 + 100(−3), it’s just that in this applied setting, t = −3 is meaningless.
In this case, we say h has an applied domainof [0, ∞) 2.2.1 Exercises 1. Suppose f is a function that takes a real number x and performs the following three steps in
the order given: (1) square root; (2) subtract 13; (3) make the quantity the denominator of
a fraction with numerator 4. Find an expression for f (x) and ﬁnd its domain.
2. Suppose g is a function that takes a real number x and performs the following three steps in
the order given: (1) subtract 13; (2) square root; (3) make the quantity the denominator of
a fraction with numerator 4. Find an expression for g (x) and ﬁnd its domain.
3. Suppose h is a function that takes a real number x and performs the following three steps in
the order given: (1) square root; (2) make the quantity the denominator of a fraction with
numerator 4; (3) subtract 13. Find an expression for h(x) and ﬁnd its domain.
4. Suppose k is a function that takes a real number x and performs the following three steps
in the order given: (1) make the quantity the denominator of a fraction with numerator 4;
(2) square root; (3) subtract 13. Find an expression for k (x) and ﬁnd its domain.
5. For f (x) = x2 − 3x + 2, ﬁnd and simplify the following:
(a) f (3) (d) f (4x) (g) f (x − 4) (b) f (−1) (e) 4f (x) (h) f (x) − 4 (f) f (−x) (i) f x2 (c) f
5
6 3
2 See the United States Postal Service website http://www.usps.com/prices/firstclassmailprices.htm
See the Internal Revenue Service’s website http://www.irs.gov/pub/irspdf/i1040tt.pdf 94 Functions
6. Repeat Exercise 5 above for f (x) = 2
x3 7. Let f (x) = 3x2 + 3x − 2. Find and simplify the following:
2
a
f (a)
2 (a) f (2) (d) 2f (a) (g) f (b) f (−2) (e) f (a + 2) (c) f (2a) (f) f (a) + f (2) (h)
(i) f (a + h) 8. Let f (x) = x ≤ −3 x + 5,
√ 9 − x2 , −3 < x ≤ 3 −x + 5, x>3 (a) f (−4) (c) f (3) (e) f (−3.001) (b) f (−3) (d) f (3.001) (f) f (2) 9. Let f (x) = x2 x ≤ −1 1 − x2 if −1 < x ≤ 1 Compute the following function values. x √ if if x>1 (a) f (4) (d) f (0) (b) f (−3) (e) f (−1) (c) f (1) (f) f (−0.999) 10. Find the (implied) domain of the function.
(a) f (x) = x4 − 13x3 + 56x2 − 19
(b) f (x) = x2 +4 x+4
x2 − 36
√
(d) f (x) = 6x − 2
(c) f (x) = (g) f (x) = √
3 (i)
(j) 6
6x − 2 (k) 6x − 2 (l) (e) f (x) = √
(f) f (x) = (h) 6
√
4 − 6x − 2 (m) √
6x − 2
f (x) = 2
x − 36
√
3
6x − 2
f (x) = 2
x + 36
t
s(t) =
t−8
√
r
Q(r) =
r−8
θ
b(θ) = √
θ−8
y
α(y ) = 3
y−8 2.2 Function Notation
√ 95 x−7+
1
(o) g (v ) =
1
4− 2
v (n) A(x) = √ 9−x (p) u(w) = w−8
√
5− w 11. The population of Sasquatch in Portage County can be modeled by the function P (t) =
150t
, where t = 0 represents the year 1803. What is the applied domain of P ? What range
t + 15
“makes sense” for this function? What does P (0) represent? What does P (205) represent?
12. Recall that the set of integers is the set of numbers Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}.7 The
greatest integer of x, x , is deﬁned to be the largest integer k with k ≤ x.
(a) Find 0.785 , 117 , −2.001 , and π + 6
(b) Discuss with your classmates how x may be described as a piecewise deﬁned function.
HINT: There are inﬁnitely many pieces!
(c) Is a + b = a + b always true? What if a or b is an integer? Test some values, make
a conjecture, and explain your result.
13. Graph the following function:8 If your income is over but not over your 2009 tax is of the amount over $0 $13,620 4.60% $0 $13,620 $27,250 $626.52 + 6.15% $13,620 $27,250 $204,370 $1,464.77 + 6.50% $27,250 $204,370 $300,000 $12,977.57 + 6.75% $204,370 $19,432.60 + 7.75% $300,000 $300,000 14. Graph the following function:9
7 The use of the letter Z for the integers is ostensibly because the German word zahlen means ‘to count.’
http://www.revenue.wi.gov/faqs/pcs/taxrates.html
9
http://www.irs.gov/pub/irspdf/i1040tt.pdf
8 96 Functions If your taxable income is over but not over your tax is of the amount over $0 $16,700.00 . . . 10% $0 16,700 67,900.00 1,670 + 15% 16,700 67,900 137,050 .50 9,350 + 25% 67,900 137,050 208,850.50 26,637 + 28% 137,050 208,850 372,950.50 46,741 + 33% 208,850 372,950 ... 100,895 + 35% 372,950 15. We have through our examples tried to convince you that, in general, f (a + b) = f (a) +
f (b). It has been our experience that students refuse to believe us so we’ll try again with a
diﬀerent approach. With the help of your classmates, ﬁnd a function f for which the following
properties are always true.
(a) f (0) = f (−1 + 1) = f (−1) + f (1)
(b) f (5) = f (2 + 3) = f (2) + f (3)
(c) f (−6) = f (0 − 6) = f (0) − f (6)
(d) f (a + b) = f (a) + f (b) regardless of what two numbers we give you for a and b.
How many functions did you ﬁnd that failed to satisfy the conditions above? Did f (x) = x2
√
1
work? What about f (x) = x or f (x) = 3x + 7 or f (x) = ? Did you ﬁnd an attribute
x
common to those functions that did succeed? You should have, because there is only one
extremely special family of functions that actually works here. Thus we return to our previous
statement, in general, f (a + b) = f (a) + f (b). 2.2.2 Answers 4
1. f (x) = √
x − 13
Domain: [0, 169) ∪ (169, ∞)
4
x − 13
Domain: (13, ∞) 2. g (x) = √ 5. (a) 2
(b) 6
(c) − 1
4 4
3. h(x) = √ − 13
x
Domain: (0, ∞)
4
− 13
x
Domain: (0, ∞) 4. k (x) = (d) 16x2 − 12x + 2
(e) 4x2 − 12x + 8
(f) x2 + 3x + 2 (g) x2 − 11x + 30
(h) x2 − 3x − 2
(i) x4 − 3x2 + 2 2.2 Function Notation
6. (a)
(b)
(c)
(d)
(e)
7. (a)
(b)
(c)
(d)
(e) 97 2
27
−2
16
27
1
32x3
8
x3 2
x3
2
2
(g)
=3
3
2 + 48x − 64
(x − 4)
x − 12x
2
2 − 4x3
(h) 3 − 4 =
x
x3
2
(i) 6
x 16
4
12a2 + 6a − 2
6a2 + 6a − 4
3a2 + 15a + 14 (f) 3a2 + 3a + 14 (f) − 8. (a) f (−4) = 1
(b) f (−3) = 2 (g)
(h)
(i) 12
6
+ a −2
a2
3a2
3a
2 + 2 −1
3a2 + 6ah + 3h2 (e) f (−3.001) = 1.999 (c) f (3) = 0
(d) f (3.001) = 1.999 9. (a) f (4) = 4 + 3 a + 3h − 2 (f) f (2) = √ 5 (d) f (0) = 1 (b) f (−3) = 9 (e) f (−1) = 1 (c) f (1) = 0 (f) f (−0.999) ≈ 0.0447101778 10. (a) (−∞, ∞) (i) (−∞, ∞) (b) (−∞, ∞) (j) (−∞, 8) ∪ (8, ∞) (c) (−∞, −6) ∪ (−6, 6) ∪ (6, ∞) (k) [0, 8) ∪ (8, ∞) (d)
(e) 1
3, ∞
1
3, ∞ (l) (8, ∞)
(m) (−∞, 8) ∪ (8, ∞) (f) (−∞, ∞)
(g)
(h) 1
3, 3
1
3, 6 (n) [7, 9] ∪ (3, ∞) 1
(o) −∞, − 2 ∪ − 1 , 0 ∪ 0, 1 ∪
2
2 ∪ (6, ∞) 1
2, ∞ (p) [0, 25) ∪ (25, ∞) 11. The applied domain of P is [0, ∞). The range is some subset of the natural numbers because
we cannot have fractional Sasquatch. This was a bit of a trick question and we’ll address the
notion of mathematical modeling more thoroughly in later chapters. P (0) = 0 means that
there were no Sasquatch in Portage County in 1803. P (205) ≈ 139.77 would mean there were
139 or 140 Sasquatch in Portage County in 2008.
12. (a) 0.785 = 0, 117 = 117, −2.001 = −3, and π + 6 = 9 98 Functions 2.3 Function Arithmetic In the previous section we used the newly deﬁned function notation to make sense of expressions
such as ‘f (x) + 2’ and ‘2f (x)’ for a given function f . It would seem natural, then, that functions
should have their own arithmetic which is consistent with the arithmetic of real numbers. The
following deﬁnitions allow us to add, subtract, multiply and divide functions using the arithmetic
we already know for real numbers. Function Arithmetic
Suppose f and g are functions and x is an element common to the domains of f and g .
The sum of f and g , denoted f + g , is the function deﬁned by the formula: (f + g )(x) = f (x) + g (x)
The diﬀerence of f and g , denoted f − g , is the function deﬁned by the formula: (f − g )(x) = f (x) − g (x)
The product of f and g , denoted f g , is the function deﬁned by the formula: (f g )(x) = f (x)g (x)
The quotient of f and g , denoted f
, is the function deﬁned by the formula:
g
f
g (x) = f (x)
,
g (x) provided g (x) = 0. In other words, to add two functions, we add their outputs; to subtract two functions, we
subtract their outputs, and so on.
1
Example 2.3.1. Let f (x) = 6x2 − 2x and g (x) = 3 − . Find and simplify expressions for for the
x
following functions. In addition, ﬁnd the domain of each of these functions.
1. (f + g )(x) 3. (f g )(x) 2. (g − f )(x) 4. g
f (x) 2.3 Function Arithmetic 99 Solution.
1. (f + g )(x) is deﬁned to be f (x) + g (x). To that end, we get (f + g )(x) = f (x) + g (x) = 6x2 − 2x + 3 − = 6 x2 − 2x + 3 − 1
x 1
x = 6x3 2x2 3x 1
−
+
−
x
x
x
x = 6x3 − 2x2 + 3x − 1
x get common denominators To ﬁnd the domain of (f + g ) we do so before we simplify, that is, at the step
6x2 − 2x + 3 − 1
x We see x = 0, but everything else is ﬁne. Hence, the domain is (−∞, 0) ∪ (0, ∞).
2. (g − f )(x) is deﬁned to be g (x) − f (x). To that end, we get (g − f )(x) = g (x) − f (x) = 3− = 3− 1
x − 6x2 − 2x 1
− 6x2 + 2x
x = 3x 1 6x3 2x2
−−
+
x
x
x
x = −6x3 + 2x2 + 3x − 1
x get common denominators 100 Functions
Looking at the expression for (g − f ) before we simpliﬁed
3− 1
x − 6x2 − 2x we see, as before, x = 0 is the only restriction. The domain is (−∞, 0) ∪ (0, ∞).
3. (f g )(x) is deﬁned to be f (x)g (x). Substituting yields (f g )(x) = f (x)g (x)
1
x = 6x2 − 2x 3− = 6x2 − 2x 3x − 1
x = 2x(3x − 1)
1 3x − 1
x factor = 2&(3x − 1)
x
1 3x − 1
x
& cancel = 2(3x − 1)2
= 2 9x2 − 6x + 1
= 18x2 − 12x + 2
To determine the domain, we check the step just after we substituted
6x2 − 2x 3− 1
x which gives us, as before, the domain: (−∞, 0) ∪ (0, ∞).
4. g
f (x) is deﬁned to be g (x)
. Thus we have
f (x) 2.3 Function Arithmetic g
f (x) 101 = g (x)
f (x)
1
x
6x2 − 2x
3− = 1
x ·x
6x2 − 2x x
3− = simplify complex fractions 1
x
x
(6x2 − 2x) x
3− = = 3x − 1
(6x2 − 2x) x = 3x − 1
2x2 (3x − 1) factor = X
$1
(3x $$
$$− 1)
$
2x2$$− $
(3x $ 1) cancel = 1
2x2 To ﬁnd the domain, we consider the ﬁrst step after substitution:
1
x
6x2 − 2x
3− 1
To avoid division by zero in the ‘little’ fraction, x , we need x = 0. For the ‘big’ fraction we
2 − 2x = 0 and solve: 2x(3x − 1) = 0 and get x = 0, 1 . Thus we must exclude x = 1 as
set 6x
3
3
well, resulting in a domain of (−∞, 0) ∪ 0, 1 ∪ 1 , ∞ .
3
3 We close this section with concept of the diﬀerence quotient of a function. It is a critical
tool for Calculus and also a great way to practice function notation.1
1 You may need to brush up on your Intermediate Algebra skills, as well. 102 Functions Definition 2.3. Given a function, f , the diﬀerence quotient of f is the expression:
f (x + h) − f (x)
h Example 2.3.2. Find and simplify the diﬀerence quotients for the following functions
1. f (x) = x2 − x − 2 2. g (x) = 3
2x + 1 Solution.
1. To ﬁnd f (x + h), we replace every occurrence of x in the formula f (x) = x2 − x − 2 with the
quantity (x + h) to get
f (x + h) = (x + h)2 − (x + h) − 2
= x2 + 2xh + h2 − x − h − 2.
So the diﬀerence quotient is f (x + h) − f (x)
h = x2 + 2xh + h2 − x − h − 2 − x2 − x − 2
h = x2 + 2xh + h2 − x − h − 2 − x2 + x + 2
h = 2xh + h2 − h
h = h (2x + h − 1)
h = h
(2x + h − 1) = 2x + h − 1. h
factor cancel 2.3 Function Arithmetic 103 2. To ﬁnd g (x + h), we replace every occurrence of x in the formula g (x) = 3
with the
2x + 1 quantity (x + h)
g (x + h) = 3
2(x + h) + 1 = 3
,
2x + 2 h + 1 which yields g (x + h) − g (x)
h = = 3
3
−
2x + 2 h + 1 2x + 1
h
3
3
−
2x + 2h + 1 2x + 1 · (2x + 2h + 1)(2x + 1)
h
(2x + 2h + 1)(2x + 1) = 3(2x + 1) − 3(2x + 2h + 1)
h(2x + 2h + 1)(2x + 1) = 6x + 3 − 6x − 6h − 3
h(2x + 2h + 1)(2x + 1) = −6h
h(2x + 2h + 1)(2x + 1) = −6
h
h
(2x + 2h + 1)(2x + 1) = −6
.
(2x + 2h + 1)(2x + 1) For reasons which will become clear in Calculus, we do not expand the denominator. 2.3.1 Exercises 1. Let f (x) = √
1
x, g (x) = x + 10 and h(x) = .
x (a) Compute the following function values. 104 Functions
i. (f + g )(4) ii. (g − h)(7) iii. (f h)(25) iv. h
g (3) (b) Find the domain of the following functions then simplify their expressions.
i. (f + g )(x)
ii. (g − h)(x) iii. (f h)(x)
h
(x)
iv.
g v. g
(x)
h vi. (h − f )(x) √
2. Let f (x) = 3 x − 1, g (x) = 2x2 − 3x − 2 and h(x) = 3
.
2−x (a) Compute the following function values.
i. (f + g )(4) ii. (g − h)(1) iii. (f h)(0) iv. h
g (−1) (b) Find the domain of the following functions then simplify their expressions.
i. (f − g )(x) 3. Let f (x) = ii. (gh)(x) iii. √
6x − 2, g (x) = x2 − 36, and h(x) = f
g (x) iv. f
h (x) 1
.
x−4 (a) Compute the following function values.
i. (f + g )(3)
ii. (g − h)(8) f
(4)
g
iv. (f h)(8) iii. v. (g + h)(−4)
h
(−12)
vi.
g (b) Find the domain of the following functions and simplify their expressions.
i. (f + g )(x)
ii. (g − h)(x) f
(x)
g
iv. (f h)(x) iii. 4. Find and simplify the diﬀerence quotient
(a) f (x) = 2x − 5
(b) f (x) = −3x + 5
(c) f (x) = 6
(d) f (x) = 3x2 − x
(e) f (x) = −x2 + 2x − 1 v. (g + h)(x)
h
vi.
(x)
g f (x + h) − f (x)
for the following functions.
h
(f) f (x) = x3 + 1
2
(g) f (x) =
x
3
(h) f (x) =
1−x
x
(i) f (x) =
x−9 2.3 Function Arithmetic 105 √
(j) f (x) = x 2
(k) f (x) = mx + b where m = 0 2.3.2 (l) f (x) = ax2 + bx + c where a = 0 Answers 1. (a)
(b) i. (f + g )(4) = 16
i. (f + g )(x) = √ ii. (g −h)(7) = 118
7 x + x + 10 iii. (f h)(25) =
h
g iv. Domain: [0, ∞) Domain: (0, ∞) (b) i. (f + g )(4) = 23 iv. h
g 1
39 (3) = 1
x(x + 10) Domain: (−∞, −10) ∪ (−10, 0) ∪ (0, ∞)
g
v.
(x) = x(x + 10)
h
Domain: (−∞, 0) ∪ (0, ∞)
1√
vi. (h − f )(x) = − x
x
Domain: (0, ∞) 1
x
Domain: (−∞, 0) ∪ (0, ∞)
1
iii. (f h)(x) = √
x
ii. (g − h)(x) = x + 10 − 2. (a) (x) = 1
5 ii. (g − h)(1) = −6 √
i. (f − g )(x) = −2x2 + 3x + 3 x + 1
Domain: [0, ∞)
ii. (gh)(x) = −6x − 3 iii. (f h)(0) = −
f
g iii. (x) = 3
2 iv. h
g (−1) = √
3 x−1
2x2 − 3x − 2 Domain: [0, 2) ∪ (2, ∞)
√
√
f
(x) = −x x + 1 x + 2 x −
iv.
3
h Domain: (−∞, 2) ∪ (2, ∞) 1
3 2
3 Domain: [0, 2) ∪ (2, ∞)
3. (a) ii. (g − h)(8) =
(b) 111
4 i. (f + g )(x) = x2 − 36 +
Domain: iii. √ 6x − 2 1
,∞
3 1
x−4
Domain: (−∞, 4) ∪ (4, ∞)
√
f
6x − 2
iii.
(x) = 2
g
x − 36
ii. (g − h)(x) = x2 − 36 − 2 f
g √ 22
(4) = −
20
√
46
iv. (f h)(8) =
4 i. (f + g )(3) = −23 161
8
1
(−12) = −
1728 v. (g + h)(−4) = −
vi. h
g 1
, 6 ∪ (6, ∞)
3
√
6x − 2
iv. (f h)(x) =
x−4
1
Domain:
, 4 ∪ (4, ∞)
3
Domain: v. (g + h)(x) = x2 − 36 + 1
x−4 Rationalize the numerator. It won’t look ‘simpliﬁed’ per se, but work through until you can cancel the ‘h’. 106 Functions
Domain: (−∞, 4) ∪ (4, ∞)
1
h
(x) =
vi.
g
(x − 4) (x2 − 36) 3
(1 − x − h)(1 − x)
−9
(i)
(x − 9)(x + h − 9)
1
(j) √
√
x+h+ x
(k) m 4. (a)
(b)
(c)
(d)
(e)
(f) 2
−3
0
6x + 3h − 1
−2x − h + 2
3x2 + 3xh + h2
2
(g) −
x(x + h) 2.4 Domain:
(−∞, −6) ∪ (−6, 4) ∪ (4, 6) ∪ (6, ∞) (h) (l) 2ax + ah + b Graphs of Functions In Section 2.1 we deﬁned a function as a special type of relation; one in which each xcoordinate
was matched with only one y coordinate. We spent most of our time in that section looking at
functions graphically because they were, after all, just sets of points in the plane. Then in Section
2.2 we described a function as a process and deﬁned the notation necessary to work with functions
algebraically. So now it’s time to look at functions graphically again, only this time we’ll do so with
the notation deﬁned in Section 2.2. We start with what should not be a surprising connection. The Fundamental Graphing Principle for Functions
The graph of a function f is the set of points which satisfy the equation y = f (x). That is, the
point (x, y ) is on the graph of f if and only if y = f (x). Example 2.4.1. Graph f (x) = x2 − x − 6.
Solution. To graph f , we graph the equation y = f (x). To this end, we use the techniques
outlined in Section 1.3. Speciﬁcally, we check for intercepts, test for symmetry, and plot additional
points as needed. To ﬁnd the xintercepts, we set y = 0. Since y = f (x), this means f (x) = 0.
f (x) = x2 − x − 6
0 = x2 − x − 6
0 = (x − 3)(x + 2) factor
x − 3 = 0 or x + 2 = 0
x = −2, 3 2.4 Graphs of Functions 107 So we get (−2, 0) and (3, 0) as xintercepts. To ﬁnd the y intercept, we set x = 0. Using
function notation, this is the same as ﬁnding f (0) and f (0) = 02 − 0 − 6 = −6. Thus the y intercept
is (0, −6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses
none of the three symmetries discussed thus far. (You should verify this.) We can make a table
analogous to the ones we made in Section 1.3, plot the points and connect the dots in a somewhat
pleasing fashion to get the graph below on the right. x
−3 f (x) (x, f (x))
6 (−3, 6) y
7
6 −2 0 (−2, 0) 5
4
3 −1 −4 (−1, −4) 2
1 0 −6 (0, −6) 1 −6 (1, −6) 2 −4 (2, −4) −3 −2 −1
−1 1 2 3 4 x −2
−3
−4
−5
−6 3 0 (3, 0) 4 6 (4, 6) Graphing piecewisedeﬁned functions is a bit more of a challenge. 4 − x2 if x < 1
Example 2.4.2. Graph: f (x) = x − 3, if x ≥ 1
Solution. We proceed as before: ﬁnding intercepts, testing for symmetry and then plotting
additional points as needed. To ﬁnd the xintercepts, as before, we set f (x) = 0. The twist is that
we have two formulas for f (x). For x < 1, we use the formula f (x) = 4 − x2 . Setting f (x) = 0
gives 0 = 4 − x2 , so that x = ±2. However, of these two answers, only x = −2 ﬁts in the domain
x < 1 for this piece. This means the only xintercept for the x < 1 region of the xaxis is (−2, 0).
For x ≥ 1, f (x) = x − 3. Setting f (x) = 0 gives 0 = x − 3, or x = 3. Since x = 3 satisﬁes the
inequality x ≥ 1, we get (3, 0) as another xintercept. Next, we seek the y intercept. Notice that
x = 0 falls in the domain x < 1. Thus f (0) = 4 − 02 = 4 yields the y intercept (0, 4). As far
as symmetry is concerned, you can check that the equation y = 4 − x2 is symmetric about the
y axis; unfortunately, this equation (and its symmetry) is valid only for x < 1. You can also verify 108 Functions y = x − 3 possesses none of the symmetries discussed in the Section 1.3. When plotting additional
points, it is important to keep in mind the restrictions on x for each piece of the function. The
sticking point for this function is x = 1, since this is where the equations change. When x = 1, we
use the formula f (x) = x − 3, so the point on the graph (1, f (1)) is (1, −2). However, for all values
less than 1, we use the formula f (x) = 4 − x2 . As we have discussed earlier in Section 1.2, there is
no real number which immediately precedes x = 1 on the number line. Thus for the values x = 0.9,
x = 0.99, x = 0.999, and so on, we ﬁnd the corresponding y values using the formula f (x) = 4 − x2 .
Making a table as before, we see that as the x values sneak up to x = 1 in this fashion, the f (x)
values inch closer and closer1 to 4 − 12 = 3. To indicate this graphically, we use an open circle at
the point (1, 3). Putting all of this information together and plotting additional points, we get
y
4
3 x f (x) (x, f (x)) 0.9 3.19 (0.9, 3.19) 0.99 ≈ 3.02 (0.99, 3.02) 2
1
−3 −2 −1
−1 1 2 3 x −2 0.999 ≈ 3.002 (0.999, 3.002) −3
−4 In the previous two examples, the xcoordinates of the xintercepts of the graph of y = f (x)
were found by solving f (x) = 0. For this reason, they are called the zeros of f . Definition 2.4. The zeros of a function f are the solutions to the equation f (x) = 0. In other
words, x is a zero of f if and only if (x, 0) is an xintercept of the graph of y = f (x). Of the three symmetries discussed in Section 1.3, only two are of signiﬁcance to functions:
symmetry about the y axis and symmetry about the origin.2 Recall that we can test whether the
graph of an equation is symmetric about the y axis by replacing x with −x and checking to see
if an equivalent equation results. If we are graphing the equation y = f (x), substituting −x for
x results in the equation y = f (−x). In order for this equation to be equivalent to the original
equation y = f (x) we need f (−x) = f (x). In a similar fashion, we recall that to test an equation’s
graph for symmetry about the origin, we replace x and y with −x and −y , respectively. Doing
1
2 We’ve just stepped into Calculus here!
Why are we so dismissive about symmetry about the xaxis for graphs of functions? 2.4 Graphs of Functions 109 this substitution in the equation y = f (x) results in −y = f (−x). Solving the latter equation for
y gives y = −f (−x). In order for this equation to be equivalent to the original equation y = f (x)
we need −f (−x) = f (x), or, equivalently, f (−x) = −f (x). These results are summarized below. Steps for testing if the graph of a function possesses symmetry
The graph of a function f is symmetric:
About the y axis if and only if f (−x) = f (x) for all x in the domain of f .
About the origin if and only if f (−x) = −f (x) for all x in the domain of f . For reasons which won’t become clear until we study polynomials, we call a function even if
its graph is symmetric about the y axis or odd if its graph is symmetric about the origin. Apart
from a very specialized family of functions which are both even and odd,3 functions fall into one of
three distinct categories: even, odd, or neither even nor odd.
Example 2.4.3. Analytically determine if the following functions are even, odd, or neither even
nor odd. Verify your result with a graphing calculator.
5
2 − x2
5x
2. g (x) =
2 − x2
5x
3. h(x) =
2 − x3 5x
2x − x3
x
5. j (x) = x2 −
−1
100 1. f (x) = 4. i(x) = Solution. The ﬁrst step in all of these problems is to replace x with −x and simplify.
1.
f (x) = 5
2 − x2 f (−x) = 5
2 − (−x)2 f (−x) = 5
2 − x2 f (−x) = f (x)
Hence, f is even. The graphing calculator furnishes the following:
3 Any ideas? 110 Functions This suggests4 the graph of f is symmetric about the y axis, as expected.
2.
g (x) = 5x
2 − x2 g (−x) = 5(−x)
2 (−x)2 g (−x) = −5x
2 − x2 It doesn’t appear that g (−x) is equivalent to g (x). To prove this, we check with an x value.
After some trial and error, we see that g (1) = 5 whereas g (−1) = −5. This proves that g is
not even, but it doesn’t rule out the possibility that g is odd. (Why not?) To check if g is
odd, we compare g (−x) with −g (x) −g (x) = −
= 5x
2 − x2 −5x
2 − x2 −g (x) = g (−x)
Hence, g is odd. Graphically, 4 ‘Suggests’ is about the extent of what a graphing calculator can do. 2.4 Graphs of Functions 111 The calculator indicates the graph of g is symmetric about the origin, as expected.
3.
h(x) = 5x
2 − x3 h(−x) = 5(−x)
2 − (−x)3 h(−x) = −5x
2 + x3 Once again, h(−x) doesn’t appear to be equivalent to h(x). We check with an x value, for
example, h(1) = 5 but h(−1) = − 5 . This proves that h is not even and it also shows h is not
3
odd. (Why?) Graphically, The graph of h appears to be neither symmetric about the y axis nor the origin.
4.
i(x) = 5x
2x − x3 i(−x) = 5(−x)
2(−x) − (−x)3 i(−x) = −5x
−2x + x3 The expression i(−x) doesn’t appear to be equivalent to i(x). However, after checking some
x values, for example x = 1 yields i(1) = 5 and i(−1) = 5, it appears that i(−x) does, in
fact, equal i(x). However, while this suggests i is even, it doesn’t prove it. (It does, however,
prove i is not odd.) To prove i(−x) = i(x), we need to manipulate our expressions for i(x)
and i(−x) and show they are equivalent. A clue as to how to proceed is in the numerators:
in the formula for i(x), the numerator is 5x and in i(−x) the numerator is −5x. To rewrite
i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value
of the fraction the same, we need to multiply the denominator by −1 as well. Thus 112 Functions i(x) = 5x
2x − x3 = (−1)5x
(−1) (2x − x3 ) = −5x
−2x + x3 Hence, i(x) = i(−x), so i is even. The calculator supports our conclusion. 5.
j (x) = x2 − x
−1
100 j (−x) = (−x)2 −
j (−x) = x2 + −x
−1
100 x
−1
100 The expression for j (−x) doesn’t seem to be equivalent to j (x), so we check using x = 1 to
1
1
get j (1) = − 100 and j (−1) = 100 . This rules out j being even. However, it doesn’t rule out
j being odd. Examining −j (x) gives j (x) = x2 − x
−1
100 −j (x) = − x2 −
−j (x) = −x2 + x
−1
100 x
+1
100 The expression −j (x) doesn’t seem to match j (−x) either. Testing x = 2 gives j (2) =
and j (−2) = 151 , so j is not odd, either. The calculator gives:
50 149
50 2.4 Graphs of Functions 113 The calculator suggests that the graph of j is symmetric about the y axis which would imply
that j is even. However, we have proven that is not the case.
There are two lessons to be learned from the last example. The ﬁrst is that sampling function
values at particular x values is not enough to prove that a function is even or odd – despite the
fact that j (−1) = −j (1), j turned out not to be odd. Secondly, while the calculator may suggest
mathematical truths, it is the algebra which proves mathematical truths.5 2.4.1 General Function Behavior The last topic we wish to address in this section is general function behavior. As you shall see in
the next several chapters, each family of functions has its own unique attributes and we will study
them all in great detail. The purpose of this section’s discussion, then, is to lay the foundation for
that further study by investigating aspects of function behavior which apply to all functions. To
start, we will examine the concepts of increasing, decreasing, and constant. Before deﬁning
the concepts algebraically, it is instructive to ﬁrst look at them graphically. Consider the graph of
the function f given on the next page.
Reading from left to right, the graph ‘starts’ at the point (−4, −3) and ‘ends’ at the point
(6, 5.5). If we imagine walking from left to right on the graph, between (−4, −3) and (−2, 4.5), we
are walking ‘uphill’; then between (−2, 4.5) and (3, −8), we are walking ‘downhill’; and between
(3, −8) and (4, −6), we are walking ‘uphill’ once more. From (4, −6) to (5, −6), we ‘level oﬀ’, and
then resume walking ‘uphill’ from (5, −6) to (6, 5.5). In other words, for the x values between −4
and −2 (inclusive), the y coordinates on the graph are getting larger, or increasing, as we move
from left to right. Since y = f (x), the y values on the graph are the function values, and we say that
the function f is increasing on the interval [−4, −2]. Analogously, we say that f is decreasing
on the interval [−2, 3] increasing once more on the interval [3, 4], constant on [4, 5], and ﬁnally
increasing once again on [5, 6]. It is extremely important to notice that the behavior (increasing,
decreasing or constant) occurs on an interval on the xaxis. When we say that the function f is
increasing on [−4, −2] we do not mention the actual y values that f attains along the way. Thus,
we report where the behavior occurs, not to what extent the behavior occurs.6 Also notice that
we do not say that a function is increasing, decreasing or constant at a single x value. In fact,
5
6 Or, in other words, don’t rely too heavily on the machine!
The notions of how quickly or how slowly a function increases or decreases are explored in Calculus. 114 Functions we would run into serious trouble in our previous example if we tried to do so because x = −2
is contained in an interval on which f was increasing and one on which it is decreasing. (There’s
more on this issue and many others in the exercises.)
y
7 (−2, 4.5) (6, 5.5) 6
5
4
3
2
1 −4 −3 −2 −1
−1 1 2 3 4 5 6 7 x −2
−3 (−4, −3) −4
−5 (4, −6) −6 (5, −6) −7
−8
−9 (3, −8) The graph of y = f (x)
We’re now ready for the more formal algebraic deﬁnitions of what it means for a function to be
increasing, decreasing or constant. Definition 2.5. Suppose f is a function deﬁned on an interval I . We say f is:
increasing on I if and only if f (a) < f (b) for all real numbers a, b in I with a < b.
decreasing on I if and only if f (a) > f (b) for all real numbers a, b in I with a < b.
constant on I if and only if f (a) = f (b) for all real numbers a, b in I . It is worth taking some time to see that the algebraic descriptions of increasing, decreasing, and
constant as stated in Deﬁnition 2.5 agree with our graphical descriptions given earlier. You should
look back through the examples and exercise sets in previous sections where graphs were given to 2.4 Graphs of Functions 115 see if you can determine the intervals on which the functions are increasing, decreasing or constant.
Can you ﬁnd an example of a function for which none of the concepts in Deﬁnition 2.5 apply?
Now let’s turn our attention to a few of the points on the graph. Clearly the point (−2, 4.5)
does not have the largest y value of all of the points on the graph of f − indeed that honor goes
to (6, 5.5) − but (−2, 4.5) should get some sort of consolation prize for being ‘the top of the hill’
between x = −4 and x = 3. We say that the function f has a local maximum7 at the point
(−2, 4.5), because the y coordinate 4.5 is the largest y value (hence, function value) on the curve
‘near’8 x = −2. Similarly, we say that the function f has a local minimum9 at the point (3, −8),
since the y coordinate −8 is the smallest function value near x = 3. Although it is tempting to
say that local extrema10 occur when the function changes from increasing to decreasing or vice
versa, it is not a precise enough way to deﬁne the concepts for the needs of Calculus. At the risk of
being pedantic, we will present the traditional deﬁnitions and thoroughly vet the pathologies they
induce in the exercises. We have one last observation to make before we proceed to the algebraic
deﬁnitions and look at a fairly tame, yet helpful, example.
If we look at the entire graph, we see the largest y value (hence the largest function value) is
5.5 at x = 6. In this case, we say the maximum11 of f is 5.5; similarly, the minimum12 of f is
−8. We formalize these concepts in the following deﬁnitions. Definition 2.6. Suppose f is a function with f (a) = b.
We say f has a local maximum at the point (a, b) if and only if there is an open interval
I containing a for which f (a) ≥ f (x) for all x in I diﬀerent than a. The value f (a) = b
is called ‘a local maximum value of f ’ in this case.
We say f has a local minimum at the point (a, b) if and only if there is an open interval
I containing a for which f (a) ≤ f (x) for all x in I diﬀerent than a. The value f (a) = b
is called ‘a local minimum value of f ’ in this case.
The value b is called the maximum of f if b ≥ f (x) for all x in the domain of f .
The value b is called the minimum of f if b ≤ f (x) for all x in the domain of f . It’s important to note that not every function will have all of these features. Indeed, it is
possible to have a function with no local or absolute extrema at all! (Any ideas of what such a
7 Also called ‘relative maximum’.
We will make this more precise in a moment.
9
Also called a ‘relative minimum’.
10
‘Maxima’ is the plural of ‘maximum’ and ‘mimima’ is the plural of ‘minimum’. ‘Extrema’ is the plural of
‘extremum’ which combines maximum and minimum.
11
Sometimes called the ‘absolute’ or ‘global’ maximum.
12
Again, ‘absolute’ or ‘global’ minimum can be used.
8 116 Functions function’s graph would have to look like?) We shall see in the exercises examples of functions which
have one or two, but not all, of these features, some that have instances of each type of extremum
and some functions that seem to defy common sense. In all cases, though, we shall adhere to the
algebraic deﬁnitions above as we explore the wonderful diversity of graphs that functions provide
to us.
Here is the ‘tame’ example which was promised earlier. It summarizes all of the concepts
presented in this section as well as some from previous sections so you should spend some time
thinking deeply about it before proceeding to the exercises.
Example 2.4.4. Given the graph of y = f (x) below, answer all of the following questions.
y
4 (0, 3)
3
2
1 (−2, 0)
−4 −3 −2 (2, 0)
−1 1 2 3 4 x −1
−2
−3 (−4, −3) (4, −3)
−4 1. Find the domain of f .
2. Find the range of f . 9. List the intervals on which f is increasing.
10. List the intervals on which f is decreasing. 3. Determine f (2).
4. List the xintercepts, if any exist. 11. List the local maximums, if any exist. 5. List the y intercepts, if any exist. 12. List the local minimums, if any exist. 6. Find the zeros of f . 13. Find the maximum, if it exists. 7. Solve f (x) < 0.
8. Determine the number of solutions to the
equation f (x) = 1.
Solution. 14. Find the minimum, if it exists.
15. Does f appear to be even, odd, or neither? 2.4 Graphs of Functions 117 1. To ﬁnd the domain of f , we proceed as in Section 2.1. By projecting the graph to the xaxis,
we see the portion of the xaxis which corresponds to a point on the graph is everything from
−4 to 4, inclusive. Hence, the domain is [−4, 4].
2. To ﬁnd the range, we project the graph to the y axis. We see that the y values from −3 to
3, inclusive, constitute the range of f . Hence, our answer is [−3, 3].
3. Since the graph of f is the graph of the equation y = f (x), f (2) is the y coordinate of the
point which corresponds to x = 2. Since the point (2, 0) is on the graph, we have f (2) = 0.
4. The xintercepts are the points on the graph with y coordinate 0, namely (−2, 0) and (2, 0).
5. The y intercept is the point on the graph with xcoordinate 0, namely (0, 3).
6. The zeros of f are the xcoordinates of the xintercepts of the graph of y = f (x) which are
x = −2, 2.
7. To solve f (x) < 0, we look for the x values of the points on the graph where the y coordinate
is less than 0. Graphically, we are looking where the graph is below the xaxis. This happens
for the x values from −4 to −2 and again from 2 to 4. So our answer is [−4, −2) ∪ (2, 4].
8. To ﬁnd where f (x) = 1, we look for points on the graph where the y coordinate is 1. Even
though these points aren’t speciﬁed, we see that the curve has two points with a y value of
1, as seen in the graph below. That means there are two solutions to f (x) = 1.
y
4
3
2
1 −4 −3 −2 −1 1 2 3 4 x −1
−2
−3
−4 9. As we move from left to right, the graph rises from (−4, −3) to (0, 3). This means f is
increasing on the interval [−4, 0]. (Remember, the answer here is an interval on the xaxis.) 118 Functions 10. As we move from left to right, the graph falls from (0, 3) to (4, −3). This means f is decreasing
on the interval [0, 4]. (Remember, the answer here is an interval on the xaxis.)
11. The function has its only local maximum at (0, 3).
12. There are no local minimums. Why don’t (−4, −3) and (4, −3) count? Let’s consider the
point (−4, −3) for a moment. Recall that, in the deﬁnition of local minimum, there needs to
be an open interval I which contains x = −4 such that f (−4) < f (x) for all x in I diﬀerent
from −4. But if we put an open interval around x = −4 a portion of that interval will lie
outside of the domain of f . Because we are unable to fulﬁll the requirements of the deﬁnition
for a local minimum, we cannot claim that f has one at (−4, −3). The point (4, −3) fails for
the same reason – no open interval around x = 4 stays within the domain of f .
13. The maximum value of f is the largest y coordinate which is 3.
14. The minimum value of f is the smallest y coordinate which is −3.
15. The graph appears to be symmetric about the y axis. This suggests13 that f is even.
.
With few exceptions, we will not develop techniques in College Algebra which allow us to
determine the intervals on which a function is increasing, decreasing or constant or to ﬁnd the local
maximums and local minimums analytically; this is the business of Calculus.14 When we have need
to ﬁnd such beasts, we will resort to the calculator. Most graphing calculators have ‘Minimum’
and ‘Maximum’ features which can be used to approximate these values, as demonstrated below.
15x
. Use a graphing calculator to approximate the intervals on
+3
which f is increasing and those on which it is decreasing. Approximate all extrema.
Example 2.4.5. Let f (x) = x2 Solution. Entering this function into the calculator gives Using the Minimum and Maximum features, we get
13
14 but does not prove
Although, truth be told, there is only one step of Calculus involved, followed by several pages of algebra. 2.4 Graphs of Functions 119 To two decimal places, f appears to have its only local minimum at (−1.73, −4.33) and its only
local maximum at (1, 73, 4.33). Given the symmetry about the origin suggested by the graph, the
relation between these points shouldn’t be too surprising. The function appears to be increasing on
[−1.73, 1.73] and decreasing on (−∞, −1.73] ∪ [1.73, ∞). This makes −4.33 the (absolute) minimum
and 4.33 the (absolute) maximum.
Example 2.4.6. Find the points on the graph of y = (x − 3)2 which are closest to the origin. Round
your answers to two decimal places.
Solution. Suppose a point (x, y ) is on the graph of y = (x − 3)2 . Its distance to the origin, (0, 0),
is given by d= (x − 0)2 + (y − 0)2 = x2 + y 2 = x2 + [(x − 3)2 ]2 = x2 + (x − 3)4 Since y = (x − 3)2 Given a value for x, the formula d = x2 + (x − 3)4 is the distance from (0, 0) to the point
(x, y ) on the curve y = (x − 3)2 . What we have deﬁned, then, is a function d(x) which we wish
to minimize over all values of x. To accomplish this task analytically would require Calculus so as
we’ve mentioned before, we can use a graphing calculator to ﬁnd an approximate solution. Using
the calculator, we enter the function d(x) as shown below and graph. 120 Functions Using the Minimum feature, we see above on the right that the (absolute) minimum occurs near
x = 2. Rounding to two decimal places, we get that the minimum distance occurs when x = 2.00.
To ﬁnd the y value on the parabola associated with x = 2.00, we substitute 2.00 into the equation
to get y = (x − 3)2 = (2.00 − 3)2 = 1.00. So, our ﬁnal answer is (2.00, 1.00).15 (What does the y
value listed on the calculator screen mean in this problem?) 2.4.2 Exercises 1. Sketch the graphs of the following functions. State the domain of the function, identify any
intercepts and test for symmetry.
(a) f (x) = x−2
3 (b) f (x) = √
5−x (c) f (x) = √
3 x (d) f (x) = x2 1
+1 2. Analytically determine if the following functions are even, odd or neither.
(a) f (x) = 7x
(b) f (x) = 7x + 2
(c) f (x) = 1
x3 (d)
(e)
(f)
(g) (h) f (x) = x4 +x3 +x2 +x+1
√
(i) f (x) = 5 − x f (x) = 4
f (x) = 0
f (x) = x6 − x4 + x2 + 9
f (x) = −x5 − x3 + x (j) f (x) = x2 − x − 6 3. Given the graph of y = f (x) below, answer all of the following questions.
y
5
4
3
2
1
−5 −4 −3 −2 −1
−1 1 2 3 4 5 x −2
−3
−4
−5 15
It seems silly to list a ﬁnal answer as (2.00, 1.00). Indeed, Calculus conﬁrms that the exact answer to this
problem is, in fact, (2, 1). As you are well aware by now, the author is a pedant, and as such, uses the decimal places
to remind the reader that any result garnered from a calculator in this fashion is an approximation, and should be
treated as such. 2.4 Graphs of Functions
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h) Find the domain of f .
Find the range of f .
Determine f (−2).
List the xintercepts, if any exist.
List the y intercepts, if any exist.
Find the zeros of f .
Solve f (x) ≥ 0.
Determine the number of solutions to the
equation f (x) = 2. 121
(i) List the intervals where f is increasing.
(j) List the intervals where f is decreasing.
(k) List the local maximums, if any exist.
(l) List the local minimums, if any exist.
(m) Find the maximum, if it exists.
(n) Find the minimum, if it exists.
(o) Is f even, odd, or neither? 4. Use your graphing calculator to approximate the local and absolute extrema of the following
functions. Approximate the intervals on which the function is increasing and those on which
it is decreasing. Round your answers to two decimal places.
√
(a) f (x) = x4 − 3x3 − 24x2 + 28x + 48
(c) f (x) = 9 − x2
√
(b) f (x) = x2/3 (x − 4)
(d) f (x) = x 9 − x2
5. Sketch the graphs of the following piecewisedeﬁned functions. −2x − 4 if x < 0 x2 if x ≤ −2 (a) f (x) = (c) f (x) =
3x if x ≥ 0 3 − x if −2 < x < 2 4 if x ≥ 2 1 x if −6 < x < −1 √
(d) f (x) = x + 4 if −4 ≤ x < 5 x if −1 < x < 1 (b) f (x) =
√ √ x − 1 if x ≥ 5 x if 1 < x < 9
6. Let f (x) = x , the greatest integer function deﬁned in Exercise 12 in Section 2.2.
(a) Graph y = f (x). Be careful to correctly describe the behavior of the graph near the
integers.
(b) Is f even, odd, or neither? Explain.
(c) Discuss with your classmates which points on the graph are local minimums, local maximums or both. Is f ever increasing? Decreasing? Constant?
7. Use your graphing calculator to show that the following functions do not have any extrema,
neither local nor absolute. 122 Functions
(a) f (x) = x3 + x − 12 (b) f (x) = −5x + 2 8. In Exercise 11 in Section 2.2, we saw that the population of Sasquatch in Portage County
150t
could be modeled by the function P (t) =
, where t = 0 represents the year 1803. Use
t + 15
your graphing calculator to analyze the general function behavior of P . Will there ever be a
time when 200 Sasquatch roam Portage County?
9. One of the most important aspects of the Cartesian Coordinate Plane is its ability to put
Algebra into geometric terms and Geometry into algebraic terms. We’ve spent most of this
chapter looking at this very phenomenon and now you should spend some time with your
classmates reviewing what we’ve done. What major results do we have that tie Algebra and
Geometry together? What concepts from Geometry have we not yet described algebraically?
What topics from Intermediate Algebra have we not yet discussed geometrically?
10. It’s now time to “thoroughly vet the pathologies induced” by the precise deﬁnitions of local
maximum and local minimum. We’ll do this by providing you and your classmates a series
of exercises to discuss. You will need to refer back to Deﬁnition 2.5 (Increasing, Decreasing
and Constant) and Deﬁnition 2.6 (Maximum and Minimum) during the discussion.
(a) Consider the graph of the function f given below.
y
3
2
1
−2 −1
−1 1 2 x −2
−3 i.
ii.
iii.
iv. Show that f has a local maximum but not a local minimum at the point (−1, 1).
Show that f has a local minimum but not a local maximum at the point (1, 1).
Show that f has a local maximum AND a local minimum at the point (0, 1).
Show that f is constant on the interval [−1, 1] and thus has both a local maximum
AND a local minimum at every point (x, f (x)) where −1 < x < 1. (b) Using Example 2.4.4 as a guide, show that the function g whose graph is given below
does not have a local maximum at (−3, 5) nor does it have a local minimum at (3, −3).
Find its extrema, both local and absolute. What’s unique about the point (0, −4) on
this graph? Also ﬁnd the intervals on which g is increasing and those on which g is
decreasing. 2.4 Graphs of Functions 123
y
5
4
3
2
1
−3 −2 −1
−1 1 2 3 x −2
−3
−4 (c) We said earlier in the section that it is not good enough to say local extrema exist
where a function changes from increasing to decreasing or vice versa. As a previous
exercise showed, we could have local extrema when a function is constant so now we
need to examine some functions whose graphs do indeed change direction. Consider the
functions graphed below. Notice that all four of them change direction at an open circle
on the graph. Examine each for local extrema. What is the eﬀect of placing the “dot”
on the y axis above or below the open circle? What could you say if no function value
was assigned to x = 0?
y y 4 4 3 3 2 2 1 1 −2 −1
−1 1 2 x Function I
y i. −2 −1
−1 iii. 5 3 y 3 1 2 −2 −1
−1 1 2 1 x
−2 −1 Function II Answers x 4 2 2.4.3 2 Function III 4 ii. 1 iv. 1 2 Function IV x 124
1. (a) Functions 2.4 Graphs of Functions 125
y x−2
3
Domain: (−∞, ∞)
xintercept: (2, 0)
y intercept: 0, − 2
3
No symmetry
f (x) = 1 −1
−1 √
(b) f (x) = 5 − x
Domain: (−∞, 5]
xintercept: (5, 0)
√
y intercept: (0, 5)
No symmetry 4 3 x 1 2 2 y
3
2
1 −4 −3 −2 −1 √
(c) f (x) = 3 x
Domain: (−∞, ∞)
xintercept: (0, 0)
y intercept: (0, 0)
Symmetry about the origin 3 4 5 y
1
−8 −7 −6 −5 −4 −3 −2 −1
−1 1 2 3 4 5 6 7 8 x −2 y x2 1
−2 −1 1 2 x (f) f (x) = x6 − x4 + x2 + 9 is even 2. (a) f (x) = 7x is odd
(b) f (x) = 7x + 2 is neither
1
(c) f (x) = 3 is odd
x
(d) f (x) = 4 is even
(e) f (x) = 0 is even and odd (g) f (x) = −x5 − x3 + x is odd
(h) f (x) = x4 + x3 + x2 + x + 1 is neither
√
(i) f (x) = 5 − x is neither
(j) f (x) = x2 − x − 6 is neither
(f) −4, −1, 1 (k) (−3, 4), (2, 3) (b) [−5, 4] (g) [−4, −1], [1, 3] (l) (0, −1) (c) f (−2) = 2 (h) 4 (m) 4 (d) (−4, 0), (−1, 0), (1, 0) (i) [−5, −3], [0, 2] (n) −5 (e) (0, −1) (j) [−3, 0], [2, 3] (o) Neither 3. (a) [−5, 3] x 2 1
+1
Domain: (−∞, ∞)
No xintercepts
y intercept: (0, 1)
Symmetry about the y axis (d) f (x) = 1 126 Functions
(c) Absolute maximum f (0) = 3
Absolute minimum f (±3) = 0
Local maximum at (0, 3)
No local minimum
Increasing on [−3, 0]
Decreasing on [0, 3] 4. (a) No absolute maximum
Absolute minimum f (4.55) ≈ −175.46
Local minimum at (−2.84, −91.32)
Local maximum at (0.54, 55.73)
Local minimum at (4.55, −175.46)
Increasing on [−2.84, 0.54], [4.55, ∞)
Decreasing on (−∞, −2.84], [0.54, 4.55]
(b) No absolute maximum
No absolute minimum
Local maximum at (0, 0)
Local minimum at (1.60, −3.28)
Increasing on (−∞, 0], [1.60, ∞)
Decreasing on [0, 1.60] (d) Absolute maximum f (2.12) ≈ 4.50
Absolute minimum f (−2.12) ≈ −4.50
Local maximum (2.12, 4.50)
Local minimum (−2.12, −4.50)
Increasing on [−2.12, 2.12]
Decreasing on [−3, −2.12], [2.12, 3] 5. (a) (c)
y y 3 6 2 5 1 4 −2 −1
−1 3
1 x
2 −2 1 −3
−2 −1 1 2 x 3 −4 (d) (b) y y
3 3 2
2 1 1
−6 −5 −4 −3 −2 −1
−1
−4 −3 −2 −1 6. (a) 1 2 3 4 5 6 7 x 1 2 3 4 5 6 7 8 9 x 2.5 Transformations 127
.
.
. y
6
5
4
3
2
1
−6 −5 −4 −3 −2 −1 1 2 3 4 5 6 x −2
−3
−4
−5
−6 .
.
. The graph of f (x) = x .
(b) Note that f (1.1) = 1, but f (−1.1) = −2, and so f is neither even nor odd. 2.5 Transformations In this section, we study how the graphs of functions change, or transform, when certain specialized
modiﬁcations are made to their formulas. The transformations we will study fall into three broad
categories: shifts, reﬂections, and scalings, and we will present them in that order. Suppose the
graph below is the complete graph of f .
y
(5, 5)
5
4 (2, 3)
3 (4, 3)
2 (0, 1)
1 2 3 4 5 x y = f (x) The Fundamental Graphing Principle for Functions says that for a point (a, b) to be on the
graph, f (a) = b. In particular, we know f (0) = 1, f (2) = 3, f (4) = 3 and f (5) = 5. Suppose
we wanted to graph the function deﬁned by the formula g (x) = f (x) + 2. Let’s take a minute to
remind ourselves of what g is doing. We start with an input x to the function f and we obtain the
output f (x). The function g takes the output f (x) and adds 2 to it. In order to graph g , we need 128 Functions to graph the points (x, g (x)). How are we to ﬁnd the values for g (x) without a formula for f (x)?
The answer is that we don’t need a formula for f (x), we just need the values of f (x). The values
of f (x) are the y values on the graph of y = f (x). For example, using the points indicated on the
graph of f , we can make the following table.
x (x, f (x)) f (x) g (x) = f (x) + 2 (x, g (x)) 0 (0, 1) 1 3 (0, 3) 2 (2, 3) 3 5 (2, 5) 4 (4, 3) 3 5 (4, 5) 5 (5, 5) 5 7 (5, 7) In general, if (a, b) is on the graph of y = f (x), then f (a) = b, so g (a) = f (a) + 2 = b + 2.
Hence, (a, b + 2) is on the graph of g . In other words, to obtain the graph of g , we add 2 to the
y coordinate of each point on the graph of f . Geometrically, adding 2 to the y coordinate of a point
moves the point 2 units above its previous location. Adding 2 to every y coordinate on a graph
en masse is usually described as ‘shifting the graph up 2 units’. Notice that the graph retains the
same basic shape as before, it is just 2 units above its original location. In other words, we connect
the four points we moved in the same manner in which they were connected before. We have the
results sidebyside below.
y y
(5, 7) 7 7 6 6 (5, 5) (2, 5) 5 5 4 4 (4, 5)
(2, 3)
(0, 3) 3 (4, 3)
2 2 (0, 1) 1 1 2 3 4 5 x shift up 2 units 1 2 3 4 5 x −− − − − −→
−−−−−−
y = f (x) add 2 to each y coordinate y = g (x) = f (x) + 2 You’ll note that the domain of f and the domain of g are the same, namely [0, 5], but that the
range of f is [1, 5] while the range of g is [3, 7]. In general, shifting a function vertically like this
will leave the domain unchanged, but could very well aﬀect the range. You can easily imagine what
would happen if we wanted to graph the function j (x) = f (x) − 2. Instead of adding 2 to each of
the y coordinates on the graph of f , we’d be subtracting 2. Geometrically, we would be moving
the graph down 2 units. We leave it to the reader to verify that the domain of j is the same as f ,
but the range of j is [−1, 3]. What we have discussed is generalized in the following theorem. 2.5 Transformations 129 Theorem 2.2. Vertical Shifts. Suppose f is a function and k is a positive number.
To graph y = f (x) + k , shift the graph of y = f (x) up k units by adding k to the
y coordinates of the points on the graph of f .
To graph y = f (x) − k , shift the graph of y = f (x) down k units by subtracting k from
the y coordinates of the points on the graph of f . The key to understanding Theorem 2.2 and, indeed, all of the theorems in this section comes
from an understanding of the Fundamental Graphing Principle for Functions. If (a, b) is on the
graph of f , then f (a) = b. Substituting x = a into the equation y = f (x) + k gives y = f (a) + k =
b + k . Hence, (a, b + k ) is on the graph of y = f (x) + k , and we have the result. In the language
of ‘inputs’ and ‘outputs’, Theorem 2.2 can be paraphrased as “Adding to, or subtracting from, the
output of a function causes the graph to shift up or down, respectively”. So what happens if we
add to or subtract from the input of the function?
Keeping with the graph of y = f (x) above, suppose we wanted to graph g (x) = f (x + 2). In
other words, we are looking to see what happens when we add 2 to the input of the function.1 Let’s
try to generate a table of values of g based on those we know for f . We quickly ﬁnd that we run
into some diﬃculties.
x (x, f (x)) f (x) g (x) = f (x + 2) (x, g (x)) 0 (0, 1) 1 f (0 + 2) = f (2) = 3 (0, 3) 2 (2, 3) 3 f (2 + 2) = f (4) = 3 (2, 3) 4 (4, 3) 3 f (4 + 2) = f (6) =? 5 (5, 5) 5 f (5 + 2) = f (7) =? When we substitute x = 4 into the formula g (x) = f (x + 2), we are asked to ﬁnd f (4 + 2) = f (6)
which doesn’t exist because the domain of f is only [0, 5]. The same thing happens when we attempt
to ﬁnd g (5). What we need here is a new strategy. We know, for instance, f (0) = 1. To determine
the corresponding point on the graph of g , we need to ﬁgure out what value of x we must substitute
into g (x) = f (x + 2) so that the quantity x + 2, works out to be 0. Solving x + 2 = 0 gives x = −2,
and g (−2) = f ((−2) + 2) = f (0) = 1 so (−2, 1) on the graph of g . To use the fact f (2) = 3, we set
x + 2 = 2 to get x = 0. Substituting gives g (0) = f (0 + 2) = f (2) = 3. Continuing in this fashion,
we get
1 We have spent a lot of time in this text showing you that f (x + 2) and f (x) + 2 are, in general, wildly diﬀerent
algebraic animals. We will see momentarily that the geometry is also dramatically diﬀerent. 130 Functions x x+2 g (x) = f (x + 2) (x, g (x)) −2 0 g (−2) = f (0) = 1 (−2, 1) 0 2 g (0) = f (2) = 3 (0, 3) 2 4 g (2) = f (4) = 3 (2, 3) 3 5 g (3) = f (5) = 5 (3, 5) In summary, the points (0, 1), (2, 3), (4, 3) and (5, 5) on the graph of y = f (x) give rise to
the points (−2, 1), (0, 3), (2, 3) and (3, 5) on the graph of y = g (x), respectively. In general, if
(a, b) is on the graph of y = f (x), then f (a) = b. Solving x + 2 = a gives x = a − 2 so that
g (a − 2) = f ((a − 2) + 2) = f (a) = b. As such, (a − 2, b) is on the graph of y = g (x). The point
(a − 2, b) is exactly 2 units to the left of the point (a, b) so the graph of y = g (x) is obtained by
shifting the graph y = f (x) to the left 2 units, as pictured below.
y y
(5, 5) (3, 5) 5 5 4 4 (2, 3)
(0, 3) 3 (4, 3) (2, 3) 2 2 (−2, 1) (0, 1)
−2 −1 1 2 3 4 5 x shift left 2 units 1 −2 −1 1 2 3 4 5 x −− − − − −→
−−−−−−
y = f (x) subtract 2 from each xcoordinate y = g (x) = f (x + 2) Note that while the ranges of f and g are the same, the domain of g is [−2, 3] whereas the domain
of f is [0, 5]. In general, when we shift the graph horizontally, the range will remain the same, but
the domain could change. If we set out to graph j (x) = f (x − 2), we would ﬁnd ourselves adding
2 to all of the x values of the points on the graph of y = f (x) to eﬀect a shift to the right 2 units.
Generalizing, we have the following result.
Theorem 2.3. Horizontal Shifts. Suppose f is a function and h is a positive number.
To graph y = f (x + h), shift the graph of y = f (x) left h units by subtracting h from
the xcoordinates of the points on the graph of f .
To graph y = f (x − h), shift the graph of y = f (x) right h units by adding h to the
xcoordinates of the points on the graph of f . 2.5 Transformations 131 In other words, Theorem 2.3 says adding to or subtracting from the input to a function amounts
to shifting the graph left or right, respectively. Theorems 2.2 and 2.3 present a theme which will run
common throughout the section: changes to the outputs from a function aﬀect the y coordinates
of the graph, resulting in some kind of vertical change; changes to the inputs to a function aﬀect
the xcoordinates of the graph, resulting in some kind of horizontal change.
Example 2.5.1.
√
x. Plot at least three points.
√
2. Use your graph in 1 to graph g (x) = x − 1.
√
3. Use your graph in 1 to graph j (x) = x − 1.
√
4. Use your graph in 1 to graph m(x) = x + 3 − 2.
1. Graph f (x) = Solution.
1. Owing to the square root, the domain of f is x ≥ 0, or [0, ∞). We choose perfect squares to
build our table and graph below. From the graph we verify the domain of f is [0, ∞) and the
range of f is also [0, ∞).
x y f (x) (x, f (x)) (4, 2)
2 0 0 (1, 1) (0, 0) 1 (0, 0) 1 1 4 (1, 1) 2 1 2 3 y = f ( x) = (4, 2) √ 4 x x 2. The domain of g is the same as the domain of f , since the only condition on both functions
is that x ≥ 0. If we compare the formula for g (x) with f (x), we see that g (x) = f (x) − 1.
In other words, we have subtracted 1 from the output of the function f . By Theorem 2.2,
we know that in order to graph g , we shift the graph of f down one unit by subtracting
1 from each of the y coordinates of the points on the graph of f . Applying this to the three
points we have speciﬁed on the graph, we move (0, 0) to (0, −1), (1, 1) to (1, 0), and (4, 2) to
(4, 1). The rest of the points follow suit, and we connect them with the same basic shape as
before. We conﬁrm the domain of g is [0, ∞) and ﬁnd the range of g to be [−1, ∞).
y y
(4, 2) 2 2 (1, 1) (4, 1) 1 1 (1, 0) (0, 0)
1 2 3 4 x 1 2 3 4 (0, −1)
shift down 1 unit y = f (x) = √ −− − − − −→
−−−−−−
x subtract 1 from each y coordinate y = g ( x) = √ x−1 x 132 Functions 3. Solving x − 1 ≥ 0 gives x ≥ 1, so the domain of j is [1, ∞). To graph j , we note that
j (x) = f (x − 1). In other words, we are subtracting 1 from the input of f . According to
Theorem 2.3, this induces a shift to the right of the graph of f . We add 1 to the xcoordinates
of the points on the graph of f and get the result below. The graph reaﬃrms the domain of
j is [1, ∞) and tells us that the range is [0, ∞).
y y (5, 2) (4, 2)
2 2 (2, 1) (1, 1)
1 1 (0, 0)
1 2 3 y = f (x) = 4 √ 5 x shift right 1 unit (1, 0) 2 3 −− − − − −→
−−−−−−
x add 1 to each xcoordinate y = j (x) = 4 √ x 5 x−1 4. To ﬁnd the domain of m, we solve x + 3 ≥ 0 and get [−3, ∞). Comparing the formulas of
f (x) and m(x), we have m(x) = f (x + 3) − 2. We have 3 being added to an input, indicating
a horizontal shift, and 2 being subtracted from an output, indicating a vertical shift. We
leave it to the reader to verify that, in this particular case, the order in which we perform
these transformations is immaterial; we will arrive at the same graph regardless as to which
transformation we apply ﬁrst.2 We follow the convention ‘inputs ﬁrst’,3 and to that end we
ﬁrst tackle the horizontal shift. Letting m1 (x) = f (x + 3) denote this intermediate step,
Theorem 2.3 tells us that the graph of y = m1 (x) is the graph of f shifted to the left 3 units.
Hence, we subtract 3 from each of the xcoordinates of the points on the graph of f . y
(1, 2) y
(4, 2)
2 (−2, 1) (1, 1)
(−3, 0) (0, 0)
−3 −2 −1
−1 2
1 1
1 2 3 4 −3 −2 −1
−1 x 1 2 3 4 x −2 −2 shift left 3 units y = f (x) = √ −− − − − −→
−−−−−−
x subtract 3 from each xcoordinate y = m1 (x) = f (x + 3) = √
x+3 Since m(x) = f (x + 3) − 2 and f (x + 3) = m1 (x), we have m(x) = m1 (x) − 2. We can apply
Theorem 2.2 and obtain the graph of m by subtracting 2 from the y coordinates of each of
the points on the graph of m1 (x). The graph veriﬁes that the domain of m is [−3, ∞) and
we ﬁnd the range of m is [−2, ∞).
2
We shall see in the next example that order is generally important when applying more than one transformation
to a graph.
3
We could equally have chosen the convention ‘outputs ﬁrst’. 2.5 Transformations 133 y y
(1, 2)
2 1 (−2, 1) 2 1 (1, 0) (−3, 0)
−3 −2 −1
−1 1 2 3 4 x −3 −2 −1
−1 2 3 4 x −2 −2 shift down 2 units y = m1 (x) = f (x + 3) = 1 (−2, −1) √ −− − − − −→
−−−−−−
x+3 subtract 2 from each y coordinate (−3, −2)
y = m(x) = m1 (x) − 2 = √ x+3−2 Keep in mind that we can check our answer to any of these kinds of problems by showing that
any of the points we’ve moved lie on the graph of our ﬁnal answer. For example, we can check that
√
(−3, −2) is on the graph of m, by computing m(−3) = (−3) + 3 − 2 = 0 − 2 = −2 We now turn our attention to reﬂections. We know from Section 1.1 that to reﬂect a point (x, y )
across the xaxis, we replace y with −y . If (x, y ) is on the graph of f , then y = f (x), so replacing y
with −y is the same as replacing f (x) with −f (x). Hence, the graph of y = −f (x) is the graph of f
reﬂected across the xaxis. Similarly, the graph of y = f (−x) is the graph of f reﬂected across the
y axis. Returning to inputs and outputs, multiplying the output from a function by −1 reﬂects its
graph across the xaxis, while multiplying the input to a function by −1 reﬂects the graph across
the y axis.4 Theorem 2.4. Reﬂections. Suppose f is a function.
To graph y = −f (x), reﬂect the graph of y = f (x) across the xaxis by multiplying the
y coordinates of the points on the graph of f by −1.
To graph y = f (−x), reﬂect the graph of y = f (x) across the y axis by multiplying the
xcoordinates of the points on the graph of f by −1. Applying Theroem 2.4 to the graph of y = f (x) given at the beginning of the section, we can
graph y = −f (x) by reﬂecting the graph of f about the xaxis
4
The expressions −f (x) and f (−x) should look familiar  they are the quantities we used in Section 2.4 to test if
a function was even, odd, or neither. The interested reader is invited to explore the role of reﬂections and symmetry
of functions. What happens if you reﬂect an even function across the y axis? What happens if you reﬂect an odd
function across the y axis? What about the xaxis? 134 Functions y y
(5, 5) 5 5 4 4 (2, 3)
3 3 (4, 3)
2 2 (0, 1) 1
1 2 3 4 x 5 1 −1
−2 3 4 5 x −2 −3 2 (0, −1) −3 (4, −3)
(2, −3)
−4 −4 −5 −5 (5, −5) reﬂect across xaxis −− − − − −→
−−−−−− multiply each y coordinate by −1 y = f ( x) y = −f (x) By reﬂecting the graph of f across the y axis, we obtain the graph of y = f (−x).
y y
(−5, 5) (5, 5)
5 5 4 (−2, 3) (2, 3) 3 3 (−4, 3) (4, 3) 2 2 (0, 1)
−5 −4 −3 −2 −1 4 (0, 1)
1 2 3 4 5 x reﬂect across y axis −− − − − −→
−−−−−− y = f (x) multiply each xcoordinate by −1 −5 −4 −3 −2 −1 1 2 3 4 5 x y = f (−x) With the addition of reﬂections, it is now more important than ever to consider the order of
transformations, as the next example illustrates.
√
Example 2.5.2. Let f (x) = x. Use the graph of f from Example 2.5.1 to graph the following
functions below. Also, state their domains and ranges.
√
−x
√
2. j (x) = 3 − x
√
3. m(x) = 3 − x
1. g (x) = Solution.
√
1. The mere sight of −x usually causes alarm, if not panic. When we discussed domains
in Section 2.2, we clearly banished negatives from the radicals of even roots. However, we
must remember that x is a variable, and as such, the quantity −x isn’t always negative. For 2.5 Transformations 135 √
example, if x = −4, −x = 4, thus −x = −(−4) = 2 is perfectly welldeﬁned. To ﬁnd the
domain analytically, we set −x ≥ 0 which gives x ≤ 0, so that the domain of g is (−∞, 0].
Since g (x) = f (−x), Theorem 2.4 tells us the graph of g is the reﬂection of the graph of f
across the y axis. We can accomplish this by multiplying each xcoordinate on the graph
of f by −1, so that the points (0, 0), (1, 1), and (4, 2) move to (0, 0), (−1, 1), and (−4, 2),
respectively. Graphically, we see that the domain of g is (−∞, 0] and the range of g is the
same as the range of f , namely [0, ∞).
y y
(4, 2)
2 2
(−4, 2) (1, 1) 1 1
(−1, 1) (0, 0)
−4 −3 −2 −1 1 y = f (x) = 2 √ 3 x 4 reﬂect across y axis −− − − − −→
−−−−−− multiply each xcoordinate by −1 x (0, 0) −4 −3 −2 −1 1 2 y = g (x) = f (−x) = √ 3 4 x −x √
2. To determine the domain of j (x) = 3 − x, we solve 3 − x ≥ 0 and get x ≤ 3, or (−∞, 3].
To determine which transformations we need to apply to the graph of f to obtain the graph
√
√
of j , we rewrite j (x) = −x + 3 = f (−x + 3). Comparing this formula with f (x) = x, we
see that not only are we multiplying the input x by −1, which results in a reﬂection across
the y axis, but also we are adding 3, which indicates a horizontal shift to the left. Does it
matter in which order we do the transformations? If so, which order is the correct order?
Let’s consider the point (4, 2) on the graph of f . We refer to the discussion leading up to
Theorem 2.3. We know f (4) = 2 and wish to ﬁnd the point on y = j (x) = f (−x + 3) which
corresponds to (4, 2). We set −x + 3 = 4 and solve. Our ﬁrst step is to subtract 3 from both
sides to get −x = 1. Subtracting 3 from the xcoordinate 4 is shifting the point (4, 2) to
the left. From −x = 1, we then multiply5 both sides by −1 to get x = −1. Multiplying the
xcoordinate by −1 corresponds to reﬂecting the point about the y axis. Hence, we perform
the horizontal shift ﬁrst, then follow it with the reﬂection about the y axis. Starting with
√
f (x) = x, we let j1 (x) be the intermediate function which shifts the graph of f 3 units to
the left, j1 (x) = f (x + 3).
y y (1, 2) (4, 2)
2 2 (−2, 1) (1, 1) 1 1
(−3, 0) (0, 0)
−4 −3 −2 −1 1 y = f (x) = 2 √ 3 4 x shift left 3 units −4 −3 −2 −1 1 2 −− − − − −→
−−−−−−
x subtract 3 from each xcoordinate y = j1 (x) = f (x + 3) = 3 √ 4 x x+3 To obtain the function j , we reﬂect the graph of j1 about y axis. Theorem 2.4 tells us we
√
have j (x) = j1 (−x). Putting it all together, we have j (x) = j1 (−x) = f (−x + 3) = −x + 3,
5 Or divide  it amounts to the same thing. 136 Functions
which is what we want.6 From the graph, we conﬁrm the domain of j is (−∞, 3] and we get
the range is [0, ∞). y y (1, 2)
2 2 (−2, 1) (2, 1)
(−1, 2) 1 (3, 0) (−3, 0)
−4 −3 −2 −1 1 y = j1 (x) = √ 2 3 4 x reﬂect across y axis −− − − − −→
−−−−−− multiply each xcoordinate by −1 x+3 −4 −3 −2 −1 1 2 y = j (x) = j1 (−x) = 3 √ 4 x −x + 3 √
3. The domain of m works out to be the domain of f , [0, ∞). Rewriting m(x) = − x + 3, we
see m(x) = −f (x) + 3. Since we are multiplying the output of f by −1 and then adding
3, we once again have two transformations to deal with: a reﬂection across the xaxis and
a vertical shift. To determine the correct order in which to apply the transformations, we
imagine trying to determine the point on the graph of m which corresponds to (4, 2) on the
graph of f . Since in the formula for m(x), the input to f is just x, we substitute to ﬁnd
m(4) = −f (4) + 3 = −2 + 3 = 1. Hence, (4, 1) is the corresponding point on the graph of
m. If we closely examine the arithmetic, we see that we ﬁrst multiply f (4) by −1, which
corresponds to the reﬂection across the xaxis, and then we add 3, which corresponds to
the vertical shift. If we deﬁne an intermediate function m1 (x) = −f (x) to take care of the
reﬂection, we get
y y 3 3 (4, 2)
2 2 (1, 1)
1 1 (0, 0)
1 2 3 4 x (0, 0) −1 1 2 3 4 x −1 −2 −2 reﬂect across xaxis y = f (x) = √ −− − − − −→
−−−−−− x multiply each y coordinate by −1 (1, −1) (4, −2)
√
y = m1 (x) = −f (x) = − x To shift the graph of m1 up 3 units, we set m(x) = m1 (x) + 3. Since m1 (x) = −f (x), when
√
we put it all together, we get m(x) = m1 (x) + 3 = −f (x) + 3 = − x + 3. We see from the
graph that the range of m is (−∞, 3].
6
If we had done the reﬂection ﬁrst, then j1 (x) = f (−x). Following this by a shift left would give us j (x) =
√
j1 (x + 3) = f (−(x + 3)) = f (−x − 3) = −x − 3 which isn’t what we want. However, if we did the reﬂection ﬁrst
and followed it by a shift to the right 3 units, we would have arrived at the function j (x). We leave it to the reader
to verify the details. 2.5 Transformations 137 y y 3 (0, 3) 2 2 1 1 (1, 2)
(4, 1) 1 (0, 0) 2 3 x 4 1 −1
−2 2 3 4 x −1 (1, −1) shift up 3 units (4, −2) −2 −− − − − −→
−−−−−−
add 3 to each y coordinate √
y = m(x) = m1 (x) + 3 = − x + 3 √
y = m1 (x) = − x We now turn our attention to our last class of transformations, scalings. Suppose we wish to
graph the function g (x) = 2f (x) where f (x) is the function whose graph is given at the beginning
of the section. From its graph, we can build a table of values for g as before. y
(5, 5)
5 x 4 (x, f (x)) f (x) g (x) = 2f (x) (x, g (x)) (2, 3)
3 0 (0, 1) 1 2 (0, 2) 2 (4, 3) (2, 3) 3 6 (2, 6) 4 (4, 3) 3 6 (4, 6) 5 (5, 5) 5 10 (5, 10) 2 (0, 1)
1 2 3 y = f (x) 4 5 x In general, if (a, b) is on the graph of f , then f (a) = b so that g (a) = 2f (a) = 2b puts (a, 2b)
on the graph of g . In other words, to obtain the graph of g , we multiply all of the y coordinates of
the points on the graph of f by 2. Multiplying all of the y coordinates of all of the points on the
graph of f by 2 causes what is known as a ‘vertical scaling7 by a factor of 2’, and the results are
7 Also called a ‘vertical stretch’, ‘vertical expansion’ or ‘vertical dilation’ by a factor of 2. 138 Functions given below. y y
(5, 10) 10 10 9 9 8 8 7 7 6 6 (2, 6)
(5, 5) (4, 6)
5 5 4 4 (2, 3)
3 3 (4, 3)
2 (0, 2) (0, 1) 1 1 2 3 4 5 x 1 vertical scaling by a factor of 2 2 3 4 5 x −− − − − − − − − −→
−−−−−−−−−−
y = f (x) y = 2f (x) multiply each y coordinate by 2 If we wish to graph y = 1 f (x), we multiply the all of the y coordinates of the points on the
2
graph of f by 1 . This creates a ‘vertical scaling8 by a factor of 1 ’ as seen below.
2
2 y y
(5, 5)
5 5 4 4 (2, 3) ` 5´
5, 2 3 3 (4, 3) ` 3´
2, 2 2 2 (0, 1) `
1 2 3 y = f (x) 4 5 x vertical scaling by a factor of 1
2 0, 1
2 ` 3´
4, 2 1
´
1 −− − − − − − − − −→
−−−−−−−−−−
multiply each y coordinate by 1
2 y= 2 3 1
f (x)
2 These results are generalized in the following theorem. 8 Also called ‘vertical shrink,’‘vertical compression’ or ‘vertical contraction’ by a factor of 2. 4 5 x 2.5 Transformations 139 Theorem 2.5. Vertical Scalings. Suppose f is a function and a > 0. To graph y = af (x),
multiply all of the y coordinates of the points on the graph of f by a. We say the graph of
f has been vertically scaled by a factor of a.
If a > 1, we say the graph of f has undergone a vertical stretch (expansion, dilation) by
a factor of a.
If 0 < a < 1, we say the graph of f has undergone a vertical shrink (compression,
1
contraction) by a factor of a . A few remarks about Theorem 2.5 are in order. First, a note about the verbiage. To the
authors, the words ‘stretch’, ‘expansion’, and ‘dilation’ all indicate something getting bigger. Hence,
‘stretched by a factor of 2’ makes sense if we are scaling something by multiplying it by 2. Similarly,
we believe words like ‘shrink’, ‘compression’ and ‘contraction’ all indicate something getting smaller,
so if we scale something by a factor of 1 , we would say it ‘shrinks by a factor of 2’  not ‘shrinks by
2
a factor of 1 .’ This is why we have written the descriptions ‘stretch by a factor of a’ and ‘shrink by
2
1
a factor of a ’ in the statement of the theorem. Second, in terms of inputs and outputs, Theorem 2.5
says multiplying the outputs from a function by positive number a causes the graph to be vertically
scaled by a factor of a. It is natural to ask what would happen if we multiply the inputs of a
function by a positive number. This leads us to our last transformation of the section.
Referring to the graph of f given at the beginning of this section, suppose we want to graph
g (x) = f (2x). In other words, we are looking to see what eﬀect multiplying the inputs to f by 2
has on its graph. If we attempt to build a table directly, we quickly run into the same problem we
had in our discussion leading up to Theorem 2.3, as seen in the table on the left below. We solve
this problem in the same way we solved this problem before. For example, if we want to determine
the point on g which corresponds to the point (2, 3) on the graph of f , we set 2x = 2 so that x = 1.
Substituting x = 1 into g (x), we obtain g (1) = f (2 · 1) = f (2) = 3, so that (1, 3) is on the graph of
g . Continuing in this fashion, we obtain the table on the lower right.
x (x, f (x)) f (x) g (x) = f (2x) (x, g (x)) x 2x g (x) = f (2x) (x, g (x)) 0 (0, 1) 1 f (2 · 0) = f (0) = 1 (0, 1) 0 0 g (0) = f (0) = 1 (0, 0) 2 (2, 3) 3 f (2 · 2) = f (4) = 3 (2, 3) 1 2 g (1) = f (2) = 3 (1, 3) 4 (4, 3) 3 f (2 · 4) = f (8) =? 2 4 g (2) = f (4) = 3 (2, 3) 5 (5, 5) 5 f (2 · 5) = f (10) =? 5
2 5 g 5
2 = f (5) = 5 5
2, 5 In general, if (a, b) is on the graph of f , then f (a) = b. Hence g a = f 2 · a = f (a) = b
2
2
so that a , b is on the graph of g . In other words, to graph g we divide the xcoordinates of the
2 140 Functions points on the graph of f by 2. This results in a horizontal scaling9 by a factor of 1 .
2
y y `5 (5, 5)
5 2 5 4 ´
,5 4 (2, 3) (1, 3) 3 3 (4, 3) (2, 3) 2 2 (0, 1) (0, 1)
1 2 3 4 5 x y = f (x) horizontal scaling by a factor of 1
2 1 2 3 4 x 5 −− − − − − − − − −→
−−−−−−−−−−
multiply each xcoordinate by 1
2 y = g (x) = f (2x) If, on the other hand, we wish to graph y = f 1 x , we end up multiplying the xcoordinates
2
of the points on the graph of f by 2 which results in a horizontal scaling10 by a factor of 2, as
demonstrated below.
y y (10, 5) (5, 5) 5 5 4 4 (4, 3) (2, 3) 3 3 (8, 3) (4, 3)
2 2 (0, 1) (0, 1)
1 2 3 4 5 6 7 8 9 10 x horizontal scaling by a factor of 2 −− − − − − − − − −→
−−−−−−−−−−
y = f (x) multiply each xcoordinate by 2 1 2 3 4 5 6 7 8 9 10 x `1 ´
y = g (x) = f 2 x We have the following theorem.
Theorem 2.6. Horizontal Scalings. Suppose f is a function and b > 0. To graph y = f (bx),
divide all of the xcoordinates of the points on the graph of f by b. We say the graph of f
has been horizontally scaled by a factor of 1 .
b
If 0 < b < 1, we say the graph of f has undergone a horizontal stretch (expansion, dilation)
by a factor of 1 .
b
If b > 1, we say the graph of f has undergone a horizontal shrink (compression, contraction) by a factor of b. Theorem 2.6 tells us that if we multiply the input to a function by b, the resulting graph is
scaled horizontally by a factor of 1 since the xvalues are divided by b to produce corresponding
b
9
10 Also called ‘horizontal shrink,’‘horizontal compression’ or ‘horizontal contraction’ by a factor of 2.
Also called ‘horizontal stretch,’‘horizontal expansion’ or ‘horizontal dilation’ by a factor of 2. 2.5 Transformations 141 points on the graph of f (bx). The next example explores how vertical and horizontal scalings
sometimes interact with each other and with the other transformations introduced in this section.
√
Example 2.5.3. Let f (x) = x. Use the graph of f from Example 2.5.1 to graph the following
functions below. Also, state their domains and ranges.
√
1. g (x) = 3 x
√
2. j (x) = 9x
x+3
2 3. m(x) = 1 −
Solution. 1. First we note that the domain of g is [0, ∞) for the usual reason. Next, we have g (x) = 3f (x)
so by Theorem 2.5, we obtain the graph of g by multiplying all of the y coordinates of the
points on the graph of f by 3. The result is a vertical scaling of the graph of f by a factor of
3. We ﬁnd the range of g is also [0, ∞).
y y
(4, 6) 6 6 5 5 4 4 3 3 (4, 2) (1, 3) 2 2 (1, 1)
1 1 (0, 0) (0, 0)
1 2 3 y = f (x) = √ 4 x 1 vertical scale by a factor of 3 2 3 −− − − − − − − − − −
− − − − − − − − − −→
x multiply each y coordinate by 3 x 4 √ y = g (x) = 3f (x) = 3 x 2. To determine the domain of j , we solve 9x ≥ 0 to ﬁnd x ≥ 0. Our domain is once again
[0, ∞). We recognize j (x) = f (9x) and by Theorem 2.6, we obtain the graph of j by dividing
the xcoordinates of the points on the graph of f by 9. From the graph, we see the range of
j is also [0, ∞).
y y
(4, 2)
2 2 `4
9 (1, 1)
1 1 `1
9 ´
,2
´
,1 (0, 0) (0, 0)
1 2 y = f (x) = 3 √ 4 x horizontal scale by a factor of 1 1
9 2 3 −− − − − − − − − − −
− − − − − − − − − −→
x multiply each xcoordinate by 1
9 y = j (x) = f (9x) = 4 √ 9x x 142 Functions 3. Solving x+3
2 ≥ 0 gives x ≥ −3, so the domain of m is [−3, ∞). To take advantage of what we know of transformations, we rewrite m(x) = − 1
2x + 3
2 + 1, or m(x) = −f 1
2x + 3
2
1
2x + 1. Focusing on the inputs ﬁrst, we note that the input to f in the formula for m(x) is
+ 3.
2
Multiplying the x by 1 corresponds to a horizontal stretch by a factor of 2, and adding the
2
3
3
2 corresponds to a shift to the left by 2 . As before, we resolve which to perform ﬁrst by
thinking about how we would ﬁnd the point on m corresponding to a point on f , in this case,
1
(4, 2). To use f (4) = 2, we solve 2 x + 3 = 4. Our ﬁrst step is to subtract the 3 (the horizontal
2
2
1
5
shift) to obtain 2 x = 2 . Next, we multiply by 2 (the horizontal stretch) and obtain x = 5.
We deﬁne two intermediate functions to handle ﬁrst the shift, then the stretch. In accordance
with Theorem 2.3, m1 (x) = f x + 3
2 = x+ 3
2 will shift the graph of f to the left
y y
(4, 2) `5
2 2 2 (1, 1) 3
2 units. ´
,2 `1´
−2,1 1 (0, 0)
−3 −2 −1 1 2 3 4 5 −3 −2 −1 ´
`3
−2,0 x −1 1 2 3 4 x 5 −2 −2
3
shift left 2 units y = f (x) = √ −− − − − −→
−−−−−− subtract 3 from each xcoordinate
2 x 1
Next, m2 (x) = m1 2 x = 1 x +
2
graph of m1 by a factor of 2. y `5
2 2 3
2 ´q
`
3
y = m1 (x) = f x + 2 = x + 3
2 will, according to Theorem 2.6, horizontally stretch the y
´
,2 (5, 2)
2 (−1, 1) `1´
−2,1
−3 −2 −1 ´
`3
−2,0 1 2 3 4 5 −2 −1 x (−3, 0) −2 1 2 3 4 5 x −1
−2 horizontal scale by a factor of 2 y = m1 (x) = q x+ 3
2 −− − − − − − − −→
−−−−−−−−−
multiply each xcoordinate by 2 y = m2 (x) = m1 `1 ´ q1
x=
x+
2
2 3
2 1
We now examine what’s happening to the outputs. From m(x) = −f 2 x + 3 + 1, we see the
2
output from f is being multiplied by −1 (a reﬂection about the xaxis) and then a 1 is added
(a vertical shift up 1). As before, we can determine the correct order by looking at how the
point (4, 2) is moved. We have already determined that to make use of the equation f (4) = 2,
1
we need to substitute x = 5. We get m(5) = −f 2 (5) + 3 + 1 = −f (4) + 1 = −2 + 1 = −1.
2
We see that f (4) (the output from f ) is ﬁrst multiplied by −1 then the 1 is added meaning we 2.5 Transformations 143 ﬁrst reﬂect the graph about the xaxis then shift up 1. Theorem 2.4 tells us m3 (x) = −m2 (x)
will handle the reﬂection. y y
(5, 2)
2 2 (−1, 1) (−3, 0) −2 −1 (−3, 0) 1 2 3 4 −2 −1 x 5 1 −1 1 2 3 4 x 5 (−1, −1)
−2 −2 (5, −2)
reﬂect across xaxis y = m2 (x) = q 1
x
2 + −− − − − −→
−−−−−− 3
2 multiply each y coordinate by −1 y = m3 (x) = −m2 (x) = − q 1
x
2 + 3
2 Finally, to handle the vertical shift, Theorem 2.2 gives m(x) = m3 (x) + 1, and we see that
the range of m is (−∞, 1]. y y
(−3, 1) 2 (−3, 0) 2 1 (−1, 0) −2 −1 1 2 3 4 −2 −1 x 5 1 (−1, −1) 2 3 4 5 x (5, −1)
−2 −2 (5, −2)
shift up 1 unit y = m3 (x) = −m2 (x) = − q 1
x
2 + 3
2 −− − − − −→
−−−−−−
add 1 to each y coordinate q
y = m(x) = m3 (x) + 1 = − 1 x +
2 3
2 +1 Some comments about Example 2.5.3 are in order. First, recalling the properties of radicals
from Intermediate Algebra, we√
know that the functions g and j are the same, since j and g have
√√
√
the same domains and j (x) = 9x = 9 x = 3 x = g (x). (We invite the reader to verify that
the all of the points we plotted on the graph of g lie on the graph of j and viceversa.) Hence, for
√
f (x) = x, a vertical stretch by a factor of 3 and a horizontal shrink by a factor of 9 result in
the same transformation. While this kind of phenomenon is not universal, it happens commonly
enough with some of the families of functions studied in College Algebra that it is worthy of note.
Secondly, to graph the function m, we applied a series of four transformations. While it would have
been easier on the authors to simply inform the reader of which steps to take, we have strived to
explain why the order in which the transformations were applied made sense. We generalize the
procedure in the theorem below. 144 Functions Theorem 2.7. Transformations. Suppose f is a function. To graph
g (x) = Af (Bx + H ) + K
1. Subtract H from each of the xcoordinates of the points on the graph of f . This results
in a horizontal shift to the left if H > 0 or right if H < 0.
2. Divide the xcoordinates of the points on the graph obtained in Step 1 by B . This results
in a horizontal scaling, but may also include a reﬂection about the y axis if B < 0.
3. Multiply the y coordinates of the points on the graph obtained in Step 2 by A. This
results in a vertical scaling, but may also include a reﬂection about the xaxis if A < 0.
4. Add K to each of the y coordinates of the points on the graph obtained in Step 3. This
results in a vertical shift up if K > 0 or down if K < 0. Theorem 2.7 can be established by generalizing the techniques developed in this section. Suppose (a, b) is on the graph of f . Then f (a) = b, and to make good use of this fact, we set
Bx + H = a and solve. We ﬁrst subtract the H (causing the horizontal shift) and then divide by
1
B . If B is a positive number, this induces only a horizontal scaling by a factor of B . If B < 0,
then we have a factor of −1 in play, and dividing by it induces a reﬂection about the y axis. So
we have x = a−H as the input to g which corresponds to the input x = a to f . We now evaluate
B
g a−H = Af B · a−H + H + K = Af (a) + K = Ab + K . We notice that the output from f is
B
B
ﬁrst multiplied by A. As with the constant B , if A > 0, this induces only a vertical scaling. If
A < 0, then the −1 induces a reﬂection across the xaxis. Finally, we add K to the result, which is
our vertical shift. A less precise, but more intuitive way to paraphrase Theorem 2.7 is to think of
the quantity Bx + H is the ‘inside’ of the function f . What’s happening inside f aﬀects the inputs
or xcoordinates of the points on the graph of f . To ﬁnd the xcoordinates of the corresponding
points on g , we undo what has been done to x in the same way we would solve an equation. What’s
happening to the output can be thought of as things happening ‘outside’ the function, f . Things
happening outside aﬀect the outputs or y coordinates of the points on the graph of f . Here, we
follow the usual order of operations agreement: we ﬁrst multiply by A then add K to ﬁnd the
corresponding y coordinates on the graph of g . Example 2.5.4. Below is the complete graph of y = f (x). Use it to graph g (x) = 4−3f (1−2x)
.
2 2.5 Transformations 145
y
(0, 3)
3
2
1 (−2, 0)
−4 −3 (2, 0) −2 −1 1 2 3 4 x −1
−2
−3 (−4, −3) (4, −3) Solution. We use Theorem 2.7 to track the ﬁve ‘key points’ (−4, −3), (−2, 0), (0, 3), (2, 0) and
(4, −3) indicated on the graph of f to their new locations. We ﬁrst rewrite g (x) in the form
3
presented in Theorem 2.7, g (x) = − 2 f (−2x + 1) + 2. We set −2x + 1 equal to the xcoordinates of
the key points and solve. For example, solving −2x + 1 = −4, we ﬁrst subtract 1 to get −2x = −5
5
then divide by −2 to get x = 2 . Subtracting the 1 is a horizontal shift to the left 1 unit. Dividing by
−2 can be thought of as a two step process: dividing by 2 which compresses the graph horizontally
by a factor of 2 followed by dividing (multiplying) by −1 which causes a reﬂection across the y axis.
We summarize the results in the table below.
(a, f (a)) a −2x + 1 = a x (−4, −3) −4 −2x + 1 = −4 x= 5
2 (−2, 0) −2 −2x + 1 = −2 x= 3
2 x= 1
2 (0, 3) 0 −2x + 1 = 0 (2, 0) 2 −2x + 1 = 2 x = − 1
2 (4, −3) 4 −2x + 1 = 4 x = − 3
2 Next, we take each of the x values and substitute them into g (x) = − 3 f (−2x + 1) + 2 to get
2
5
the corresponding y values. Substituting x = 2 , and using the fact that f (−4) = −3, we get
g 5
2 3
=− f
2 −2 5
2 3
3
9
13
+ 1 + 2 = − f (−4) + 2 = − (−3) + 2 = + 2 =
2
2
2
2 3
We see the output from f is ﬁrst multiplied by − 2 . Thinking of this as a two step process,
3
multiplying by 2 then by −1, we see we have a vertical stretch by a factor of 3 followed by a
2
reﬂection across the xaxis. Adding 2 results in a vertical shift up 2 units. Continuing in this
manner, we get the table below. 146 Functions x g (x) (x, g (x)) 5
2 13
2 5 13
2, 2 3
2 2 3
2, 2 1
2 −5
2 1
5
2, −2 −1
2 2 −1, 2
2 −3
2 13
2 − 3 , 13
22 To graph g , we plot each of the points in the table above and connect them in the same order
and fashion as the points to which they correspond. Plotting f and g sidebyside gives
` y −3,
2 13
2 ´ 6 ´ 4 (0, 3) 3 3
`1´
−2,2 2
1 −4 −3 −2 −1
−1 `3
2 ´
,2 (2, 0)
1 2 3 4 x −4 −3 −2 −1
−1 −2
(−4, −3) 13
2 , 5 4 (−2, 0) 2 6 5 `5 y −3 2 `1 ,−5
2 3 4 x −2 −3 1 −4 (4, −3) 2 ´ −4 The reader is strongly encouraged11 to graph the series of functions which shows the gradual
transformation of the graph of f into the graph of g . We have outlined the sequence of transformations in the above exposition; all that remains is to plot all ﬁve intermediate stages.
Our last example turns the tables and asks for the formula of a function given a desired sequence
of transformations. If nothing else, it is a good review of function notation.
Example 2.5.5. Let f (x) = x2 . Find and simplify the formula of the function g (x) whose graph
is the result of f undergoing the following sequence of transformations. Check your answer using
a graphing calculator.
1. Vertical shift up 2 units
2. Reﬂection across the xaxis
11 You really should do this once in your life. 2.5 Transformations 147 3. Horizontal shift right 1 unit
4. Horizontal stretch by a factor of 2
Solution. We build up to a formula for g (x) using intermediate functions as we’ve seen in previous
examples. We let g1 take care of our ﬁrst step. Theorem 2.2 tells us g1 (x) = f (x)+2 = x2 +2. Next,
we reﬂect the graph of g1 about the xaxis using Theorem 2.4: g2 (x) = −g1 (x) = − x2 + 2 =
−x2 − 2. We shift the graph to the right 1 unit, according to Theorem 2.3, by setting g3 (x) =
g2 (x − 1) = −(x − 1)2 − 2 = −x2 + 2x − 3. Finally, we induce a horizontal stretch by a factor of 2
2
1
1
1
using Theorem 2.6 to get g (x) = g3 2 x = − 2 x + 2 2 x − 3 which yields g (x) = − 1 x2 + x − 3.
4
We use the calculator to graph the stages below to conﬁrm our result. shift up 2 units −− − − − −→
−−−−−−
add 2 to each y coordinate y = f (x) = x2 y = g1 (x) = f (x) + 2 = x2 + 2 reﬂect across xaxis −− − − − −→
−−−−−− multiply each y coordinate by −1 y = g1 (x) = x2 + 2 y = g2 (x) = −g1 (x) = −x2 − 2 shift right 1 unit −− − − − −→
−−−−−−
add 1 to each xcoordinate y = g2 (x) = −x2 − 2 y = g3 (x) = g2 (x − 1) = −x2 + 2x − 3 148 Functions horizontal stretch by a factor of 2 −− − − − − − − −→
−−−−−−−−−
multiply each xcoordinate by 2 y = g3 (x) = −x2 + 2x − 3 y = g (x) = g3 1
2x = − 1 x2 + x − 3
4 We have kept the viewing window the same in all of the graphs above. This had the undesirable
consequence of making the last graph look ‘incomplete’ in that we cannot see the original shape
of f (x) = x2 . Altering the viewing window results in a more complete graph of the transformed
function as seen below. y = g (x)
This example brings our ﬁrst chapter to a close. In the chapters which lie ahead, be on the
lookout for the concepts developed here to resurface as we study diﬀerent families of functions. 2.5 Transformations 2.5.1 149 Exercises 1. The complete graph of y = f (x) is given below. Use it to graph the following functions.
y
(0, 4) 4
3
2
1
−4 −3
−1
(−2, 0) −1
−2 1 3 4 x (2, 0)
(4, −2) −3
−4 The graph of y = f (x)
(a) y = f (x) − 1 (d) y = f (2x) (g) y = f (x + 1) − 1 (b) y = f (x + 1) (e) y = −f (x) (h) y = 1 − f (x) (f) y = f (−x) (i) y = 1 f (x + 1) − 1
2 (c) y = 1
2 f (x) 2. The complete graph of y = S (x) is given below. Use it to graph the following functions.
y
(1, 3)
3
2
1
(−2, 0)
−2 (0, 0) −1 1 (2, 0) x −1
−2
−3
(−1, −3) The graph of y = S (x)
(a) y = S (x + 1) 1
(c) y = 2 S (−x + 1) (b) y = S (−x + 1) (d) y = 1 S (−x + 1) + 1
2 150 Functions 3. The complete graph of y = f (x) is given below. Use it to graph the following functions.
y
3 (0, 3) 2
1 −3 −2 −1 (−3, 0) 1
−1 (c) j (x) = f x − 3 x (3, 0) (g) d(x) = −2f (x) (a) g (x) = f (x) + 3
(b) h(x) = f (x) − 2 1
2
2
3 2
3x
− 1 f (3x)
4 (h) k (x) = f
(i) m(x) = (d) a(x) = f (x + 4) (j) n(x) = 4f (x − 3) − 6 (e) b(x) = f (x + 1) − 1 (k) p(x) = 4 + f (1 − 2x) (f) c(x) = 3
5 f (x) (l) q (x) = − 1 f
2 x+4
2 −3 √
4. The graph of y = f (x) = 3 x is given below on the left and the graph of y = g (x) is given
on the right. Find a formula for g based on transformations of the graph of f . Check your
answer by conﬁrming that the points shown on the graph of g satisfy the equation y = g (x).
y y 5 5 4 4 3 3 2 2 1 1 −11 10 9 −8 −7 −6 −5 −4 −3 −2 −1
−−
−1 1 2 3 4 5 6 7 8 x −11 10 9 −8 −7 −6 −5 −4 −3 −2 −1
−−
−1 −2 y= 4 5 6 7 8 x −4 −5 3 −3 −4 2 −2 −3 1 −5 √
3
x y = g (x) 5. For many common functions, the properties of algebra make a horizontal scaling the same
as a vertical scaling by (possibly) a diﬀerent factor. For example, we stated earlier that
√
√
9x = 3 x. With the help of your classmates, ﬁnd the equivalent vertical scaling produced
√
2
by the horizontal scalings y = (2x)3 , y = 5x, y = 3 27x and y = 1 x . What about
2
√
2
y = (−2x)3 , y =  − 5x, y = 3 −27x and y = − 1 x ?
2 2.5 Transformations 151 6. We mentioned earlier in the section that, in general, the order in which transformations are
applied matters, yet in our ﬁrst example with two transformations the order did not matter.
(You could perform the shift to the left followed by the shift down or you could shift down
and then left to achieve the same result.) With the help of your classmates, determine the
situations in which order does matter and those in which it does not.
7. What happens if you reﬂect an even function across the y axis?
8. What happens if you reﬂect an odd function across the y axis?
9. What happens if you reﬂect an even function across the xaxis?
10. What happens if you reﬂect an odd function across the xaxis?
11. How would you describe symmetry about the origin in terms of reﬂections?
12. As we saw in Example 2.5.5, the viewing window on the graphing calculator aﬀects how we see
the transformations done to a graph. Using two diﬀerent calculators, ﬁnd viewing windows
so that f (x) = x2 on the one calculator looks like g (x) = 3x2 on the other. 2.5.2 Answers
y 1. (a) y = f (x) − 1
(−1, 4) 4 y
3
4
3 2
(0, 3) 1 2
1
−4 −3 −2 −1
12
−1
(−2, −1)
(2, −1)
−2
−3
−4 (b) y = f (x + 1) 3 4 x −4 −3 −2 −1
−1
(−3, 0)
−2
−3
−4 (4, −3) (c) y = 1 f (x)
2 12
(1, 0) 3 (3, −2) 4 x 152 Functions
y (f) y = f (−x) 4 y 3
2 4
(0, 2) 3 1 2 −4 −3
−1
(−2, 0) −1
−2 1 3 4 1 x (2, 0) (4, −1) −4 −3 −3
(−4, −2) −4 −1
(−2, 0) −1
−2
−3
−4 (d) y = f (2x)
y
4 (0, 4) 3
2
1 (1, 0) −4 −3 −2 (−1, 0) 2 −2 3 4 x (2, −2) −3
−4 (e) y = −f (x)
y
4
3 (4, 2) 2
(−2, 0) 1 −4 −3 −2 −1
−1 (2, 0)
1 2 −2
−3
−4 (0, 4) (0, −4) 3 4 x 1 3
(2, 0) 4 x 2.5 Transformations 153
(i) y = 1 f (x + 1) − 1
2 (g) y = f (x + 1) − 1
y y 4 4 3 (−1, 3) 3 2 2 1 (−1, 1) −4 −3 −2 −1
−1 1 234
(1, −1) x −2 (−3, −1) −3 1 −4 −3 −2 −1
−1
(−3, −1)
−2 1 234
(1, −1)
(3, −2) −3 (3, −3) −4 −4 (h) y = 1 − f (x)
y
4 (4, 3) 3
2 (−2, 1) (2, 1) 1
−4 −3 −2 −1
−1 1 2 3 4 x −2
(0, −3) −3
−4 (b) y = S (−x + 1) 2. (a) y = S (x + 1)
y
3 y (0, 3) 3 2 −3 2 1
(−3, 0) (0, 3) 1 (−1, 0)
−2 (1, 0) −1 x −1
−2 1 2 −2 −3
(−2, −3) (−1, 0)
−1 −3 (1, 0) (2, −3) (3, 0)
3 x x 154 Functions
(c) y = 1 S (−x + 1)
2 (d) y = 1 S (−x + 1) + 1
2 y y 2 3
` 3´
0, 2 ` 5´
0, 2 1 2
(1, 0)
1 (−1, 0)
−1 (3, 0) 2 (1, 1)
x 3 −1
` −2 3
2, − 2 ´ 1 x 3
` −1 2, − 1
2 ´ (d) a(x) = f (x + 4) 3. (a) g (x) = f (x) + 3
y 6 (3, 1) 1
(−1, 1) y (−4, 3) (0, 6) 3 5 2 4 1 3
(−3, 3) −7 −6 −5 −4 −3 −2 −1
(−7, 0)
(−1, 0) (3, 3) 2
1 (e) b(x) = f (x + 1) − 1 −3 −2 −1 1 2 3 y x (−1, 2) −1 2 (b) h(x) = f (x) −
y
3 x 1 1
2
−4 −3 −2 −1 “
”
5
0, 2 1 2 x −1
(−4, −1) (2, −1) 2 (f) c(x) = 3 f (x)
5 1 2
“ −3 −” −1
−3, − 1
−1
2 1 y
“ 2“ 3 ”x
3, − 1
2 2 0, 9
5 ” 1 −3 −2 −1
(−3, 0)
−1 (c) j (x) = f x − 2
3
y“ 2,3
3 ” 3
2
1 −“ −2 ” 1
3
−
−7,0
−1
3 1 2 3
“ 11 , 0
3 x
” 1 2 3
(3, 0) x 2.5 Transformations 155 (g) d(x) = −2f (x) (j) n(x) = 4f (x − 3) − 6 y y (−3, 0) (3, 0) (3, 6)
6 −3 −2 −1 1 2 x 3 −1 5 −2 4 −3 3 −4 2 −5 1 −6
1 (0, −6) 2 3 4 5 6 x −1
−2
−3
−4
−5
−6 (h) k (x) = f 2
3x (6, −6) (0, −6) y (k) p(x) = 4 + f (1 − 2x) = f (−2x + 1) + 4 (0, 3)
3 y “ 2 7 1 1,7
2 ” 6 “ −4 ” −3 −2 −1
−9,0
−1
2 1 2 3 4
“ 5 x
” 9,0
2 (i) m(x) = − 1 f (3x)
4 4
(−1, 4)
3 y (2, 4) 2
1 (−1, 0) (1, 0) −1 1 x −1 1 2 x −1
−1 `
´
3
0, − 4 1
(l) q (x) = − 2 f x+4
2 − 3 = −1f
2 −10 9 −8 −7 −6 −5 −4 −3 −2 −1
−
−1
−2 (−10, −3) −3
−4
“
”
−4, − 9
2 √
√
4. g (x) = −2 3 x + 3 − 1 or g (x) = 2 3 −x − 3 − 1 1
2x y 1 2 x (2, −3) +2 3 156 Functions Chapter 3 Linear and Quadratic Functions
3.1 Linear Functions We now begin the study of families of functions. Our ﬁrst family, linear functions, are old friends as
we shall soon see. Recall from Geometry that two distinct points in the plane determine a unique
line containing those points, as indicated below.
P (x0 , y0 )
Q (x1 , y1 ) To give a sense of the ‘steepness’ of the line, we recall we can compute the slope of the line
using the formula below.
Equation 3.1. The slope m of the line containing the points P (x0 , y0 ) and Q (x1 , y1 ) is:
m= y1 − y0
,
x1 − x0 provided x1 = x0 . A couple of notes about Equation 3.1 are in order. First, don’t ask why we use the letter ‘m’
to represent slope. There are many explanations out there, but apparently no one really knows
for sure.1 Secondly, the stipulation x1 = x0 ensures that we aren’t trying to divide by zero. The
reader is invited to pause to think about what is happening geometrically; the anxious reader can
skip along to the next example.
1 See www.mathforum.org or www.mathworld.wolfram.com for discussions on this topic. 158 Linear and Quadratic Functions Example 3.1.1. Find the slope of the line containing the following pairs of points, if it exists. Plot
each pair of points and the line containing them.
1. P (0, 0), Q(2, 4) 4. P (−3, 2), Q(4, 2) 2. P (−1, 2), Q(3, 4) 5. P (2, 3), Q(2, −1) 3. P (−2, 3), Q(2, −3) 6. P (2, 3), Q(2.1, −1) Solution. In each of these examples, we apply the slope formula, Equation 3.1.
y
4
Q
3 1. m= 4−0
4
= =2
2−0
2 2
1 P1 2 3 x 4 y
4
Q 2. m= 4−2
2
1
==
3 − (−1)
4
2 3
2
P
1 −1 3 2 x 1 1 2 y
4
P 3
2
1 3. −3 − 3
−6
3
m=
=
=−
2 − (−2)
4
2 −3 −2 −1 3 x −1
−2
Q −3
−4 y
3 4. m= 2−2
0
= =0
4 − (−3)
7 2
Q P
1 −4 −3 −2 −1 1 2 3 4 x 3.1 Linear Functions 159
y
3 P 2
1 5. −4
−1 − 3
=
, which is undeﬁned
m=
2−2
0 1 x 2 −1 Q −2
−3 y
3 P 2
1 6. −4
−1 − 3
=
= −40
m=
2.1 − 2
0.1 1
−1 x 2
Q −2
−3 A few comments about Example 3.1.1 are in order. First, for reasons which will be made clear
soon, if the slope is positive then the resulting line is said to be increasing. If it is negative, we
say the line is decreasing. A slope of 0 results in a horizontal line which we say is constant, and
an undeﬁned slope results in a vertical line.2 Second, the larger the slope is in absolute value, the
steeper the line. You may recall from Intermediate Algebra that slope can be described as the
ratio ‘ rise ’. For example, in the second part of Example 3.1.1, we found the slope to be 1 . We can
run
2
interpret this as a rise of 1 unit upward for every 2 units to the right we travel along the line, as
shown below.
y
4
‘up 1’
3
‘over 2’
2
1 −1
2 1 2 3 x Some authors use the unfortunate moniker ‘no slope’ when a slope is undeﬁned. It’s easy to confuse the notions
of ‘no slope’ with ‘slope of 0’. For this reason, we will describe slopes of vertical lines as ‘undeﬁned’. 160 Linear and Quadratic Functions Using more formal notation, given points (x0 , y0 ) and (x1 , y1 ), we use the Greek letter delta ‘∆’
to write ∆y = y1 − y0 and ∆x = x1 − x0 . In most scientiﬁc circles, the symbol ∆ means ‘change
in’. Hence, we may write
∆y
m=
,
∆x
which describes the slope as the rate of change of y with respect to x. Rates of change abound
in the ‘real world,’ as the next example illustrates.
Example 3.1.2. At 6 AM, it is 24◦ F; at 10 AM, it is 32◦ F.
1. Find the slope of the line containing the points (6, 24) and (10, 32).
2. Interpret your answer to the ﬁrst part in terms of temperature and time.
3. Predict the temperature at noon.
Solution.
1. For the slope, we have m = 32−24
10−6 = 8
4 = 2. 2. Since the values in the numerator correspond to the temperatures in ◦ F, and the values in
2◦ F
2
,
the denominator correspond to time in hours, we can interpret the slope as 2 = =
1
1 hour
◦ F per hour. Since the slope is positive, we know this corresponds to an increasing line.
or 2
Hence, the temperature is increasing at a rate of 2◦ F per hour.
3. Noon is two hours after 10 AM. Assuming a temperature increase of 2◦ F per hour, in two
hours the temperature should rise 4◦ F. Since the temperature at 10 AM is 32◦ F, we would
expect the temperature at noon to be 32 + 4 = 36◦ F.
Now it may well happen that in the previous scenario, at noon the temperature is only 33◦ F.
This doesn’t mean our calculations are incorrect. Rather, it means that the temperature change
throughout the day isn’t a constant 2◦ F per hour. Mathematics is often used to describe, or model,
real world phenomena. Mathematical models are just that: models. The predictions we get out
of the models may be mathematically accurate, but may not resemble what happens in the real
world.
In Section 1.2, we discussed the equations of vertical and horizontal lines. Using the concept
of slope, we can develop equations for the other varieties of lines. Suppose a line has a slope of m
and contains the point (x0 , y0 ). Suppose (x, y ) is another point on the line, as indicated below.
(x, y ) (x0 , y0 ) 3.1 Linear Functions 161 We have
m= y − y0
x − x0 m ( x − x 0 ) = y − y0
y − y0 = m (x − x0 )
y = m (x − x0 ) + y0 .
We have just derived the pointslope form of a line.3
Equation 3.2. The pointslope form of the line with slope m containing the point (x0 , y0 )
is the equation y = m (x − x0 ) + y0 . Example 3.1.3. Write the equation of the line containing the points (−1, 3) and (2, 1).
Solution. In order to use Equation 3.2 we need to ﬁnd the slope of the line in question. So we
∆y
1
use Equation 3.1 to get m = ∆x = 2−−31) = − 2 . We are spoiled for choice for a point (x0 , y0 ).
3
(−
We’ll use (−1, 3) and leave it to the reader to check that using (2, 1) results in the same equation.
Substituting into the pointslope form of the line, we get y = m ( x − x 0 ) + y0
2
y = − (x − (−1)) + 3
3
2
2
y = − x− +3
3
3
2
7
y = − x+ .
3
3
We can check our answer by showing that both (−1, 3) and (2, 1) are on the graph of y = − 2 x + 7
3
3
algebraically, as we did in Section 1.3. In simplifying the equation of the line in the previous example, we produced another form of a
line, the slopeintercept form. This is the familiar y = mx + b form you have probably seen in
3 We can also understand this equation in terms of applying transformations to the function I (x) = x. See the
exercises. 162 Linear and Quadratic Functions Intermediate Algebra. The ‘intercept’ in ‘slopeintercept’ comes from the fact that if we set x = 0,
we get y = b. In other words, the y intercept of the line y = mx + b is (0, b).
Equation 3.3. The slopeintercept form of the line with slope m and y intercept (0, b) is
the equation y = mx + b. Note that if we have slope m = 0, we get the equation y = b which matches our formula for
a horizontal line given in Section 1.2. The formula given in Equation 3.3 can be used to describe
all lines except vertical lines. All lines except vertical lines are functions (why?) and so we have
ﬁnally reached a good point to introduce linear functions.
Definition 3.1. A linear function is a function of the form
f (x) = mx + b,
where m and b are real numbers with m = 0. The domain of a linear function is (−∞, ∞). For the case m = 0, we get f (x) = b. These are given their own classiﬁcation.
Definition 3.2. A constant function is a function of the form
f (x) = b,
where b is real number. The domain of a constant function is (−∞, ∞). Recall that to graph a function, f , we graph the equation y = f (x). Hence, the graph of a
linear function is a line with slope m and y intercept (0, b); the graph of a constant function is a
horizontal line (with slope m = 0) and a y intercept of (0, b). Now think back to Section 2.4.1,
speciﬁcally Deﬁnition 2.5 concerning increasing, decreasing and constant functions. A line with
positive slope was called an increasing line because a linear function with m > 0 is an increasing
function. Similarly, a line with a negative slope was called a decreasing line because a linear function
with m < 0 is a decreasing function. And horizontal lines were called constant because, well, we
hope you’ve already made the connection.
Example 3.1.4. Graph the following functions. Identify the slope and y intercept. 3.1 Linear Functions 163 1. f (x) = 3 3. f (x) = 3 − 2x
4 2. f (x) = 3x − 1 4. f (x) = x2 − 4
x−2 Solution.
1. To graph f (x) = 3, we graph y = 3. This is a horizontal line (m = 0) through (0, 3).
2. The graph of f (x) = 3x − 1 is the graph of the line y = 3x − 1. Comparison of this equation
with Equation 3.3 yields m = 3 and b = −1. Hence, our slope is 3 and our y intercept is
(0, −1). To get another point on the line, we can plot (1, f (1)) = (1, 2).
y
4 y 3
2 4 1
3
−2 −1
−1 2 1 2 x −2
1 −3
−4 −3 −2 −1 1 2 3 x f (x) = 3x − 1 f (x) = 3 3. At ﬁrst glance, the function f (x) = 3−2x does not ﬁt the form in Deﬁnition 3.1 but after some
4
x
3
rearranging we get f (x) = 3−2x = 3 − 24 = − 1 x + 4 . We identify m = − 1 and b = 3 . Hence,
4
4
2
2
4
1
our graph is a line with a slope of − 2 and a y intercept of 0, 3 . Plotting an additional
4
point, we can choose (1, f (1)) to get 1, 1 .
4 4. If we simplify the expression for f , we get
f (x) = $
x2 − 4 $$$ x + 2)
(x − 2)(
=
= x + 2.
$
x−2
(x − 2)
$$$ If we were to state f (x) = x + 2, we would be committing a sin of omission. Remember, to
ﬁnd the domain of a function, we do so before we simplify! In this case, f has big problems
when x = 2, and as such, the domain of f is (−∞, 2) ∪ (2, ∞). To indicate this, we write
f (x) = x + 2, x = 2. So, except at x = 2, we graph the line y = x + 2. The slope m = 1
and the y intercept is (0, 2). A second point on the graph is (1, f (1)) = (1, 3). Since our
function f is not deﬁned at x = 2, we put an open circle at the point that would be on the
line y = x + 2 when x = 2, namely (2, 4). 164 Linear and Quadratic Functions
y
4 y 3 2 2 1 −3 −2 −1 1
1 2 3 x −1 1 2 3 x x2 −4
f (x) =
x−2 3 − 2x
f (x) =
4 The last two functions in the previous example showcase some of the diﬃculty in deﬁning a
linear function using the phrase ‘of the form’ as in Deﬁnition 3.1, since some algebraic manipulations
may be needed to rewrite a given function to match ‘the form.’ Keep in mind that the domains of
2−
linear and constant functions are all real numbers, (−∞, ∞), and so while f (x) = x −24 simpliﬁed
x
to a formula f (x) = x + 2, f is not considered a linear function since its domain excludes x = 2.
However, we would consider
2x2 + 2
f (x) = 2
x +1
to be a constant function since its domain is all real numbers (why?) and
2$$+ $
x2 $ 1
2x2 + 2
=
$ =2
x2 + 1
x2 $$
$$+ 1
$ f (x) = The following example uses linear functions to model some basic economic relationships.
Example 3.1.5. The cost, C , in dollars, to produce x PortaBoy game systems for a local retailer
is given by C (x) = 80x + 150 for x ≥ 0.
1. Find and interpret C (10).
2. How many PortaBoys can be produced for $15,000?
3. Explain the signiﬁcance of the restriction on the domain, x ≥ 0.
4. Find and interpret C (0).
5. Find and interpret the slope of the graph of y = C (x).
Solution.
1. To ﬁnd C (10), we replace every occurrence of x with 10 in the formula for C (x) to get
C (10) = 80(10) + 150 = 950. Since x represents the number of PortaBoys produced, and
C (x) represents the cost, in dollars, C (10) = 950 means it costs $950 to produce 10 PortaBoys
for the local retailer. 3.1 Linear Functions 165 2. To ﬁnd how many PortaBoys can be produced for $15,000, we set the cost, C (x), equal to
15000, and solve for x
C (x) = 15000
80x + 150 = 15000
80x = 14850
14850
x=
= 185.625
80
Since we can only produce a whole number amount of PortaBoys, we can produce 185
PortaBoys for $15,000.
3. The restriction x ≥ 0 is the applied domain, as discussed in Section 2.2. In this context,
x represents the number of PortaBoys produced. It makes no sense to produce a negative
quantity of game systems.4
4. To ﬁnd C (0), we replace every occurrence of x with 0 in the formula for C (x) to get C (0) =
80(0) + 150 = 150. This means it costs $150 to produce 0 PortaBoys. The $150 is often called
the ﬁxed or startup cost of this venture. (What might contribute to this cost?)
5. If we were to graph y = C (x), we would be graphing the portion of the line y = 80x + 150
for x ≥ 0. We recognize the slope, m = 80. Like any slope, we can interpret this as a rate of
change. In this case, C (x) is the cost in dollars, while x measures the number of PortaBoys
so
∆y
∆C
80
$80
m=
=
= 80 =
=
.
∆x
∆x
1
1 PortaBoy
In other words, the cost is increasing at a rate of $80 per PortaBoy produced. This is often
called the variable cost for this venture.
The next example asks us to ﬁnd a linear function to model a related economic problem.
Example 3.1.6. The local retailer in Example 3.1.5 has determined that the number of PortaBoy
game systems sold in a week, x, is related to the price of each system, p, in dollars. When the
price was $220, 20 game systems were sold in a week. When the systems went on sale the following
week, 40 systems were sold at $190 a piece.
1. Find a linear function which ﬁts this data. Use the weekly sales, x, as the independent
variable and the price p, as the dependent variable.
2. Find a suitable applied domain.
4 Actually, it makes no sense to produce a fractional part of a game system, either, as we saw in the previous part
of this example. This absurdity, however, seems quite forgivable in some textbooks but not to us. 166 Linear and Quadratic Functions 3. Interpret the slope.
4. If the retailer wants to sell 150 PortaBoys next week, what should the price be?
5. What would the weekly sales be if the price were set at $150 per system?
Solution.
1. We recall from Section 2.2 the meaning of ‘independent’ and ‘dependent’ variable. Since x
is to be the independent variable, and p the dependent variable, we treat x as the input
variable and p as the output variable. Hence, we are looking for a function of the form
p(x) = mx + b. To determine m and b, we use the fact that 20 PortaBoys were sold during
the week the price was 220 dollars and 40 units were sold when the price was 190 dollars.
Using function notation, these two facts can be translated as p(20) = 220 and p(40) = 190.
Since m represents the rate of change of p with respect to x, we have
m= ∆p
190 − 220
−30
=
=
= −1.5.
∆x
40 − 20
20 We now have determined p(x) = −1.5x + b. To determine b, we can use our given data again.
Using p(20) = 220, we substitute x = 20 into p(x) = 1.5x + b and set the result equal to 220:
−1.5(20) + b = 220. Solving, we get b = 250. Hence, we get p(x) = −1.5x + 250. We can
check our formula by computing p(20) and p(40) to see if we get 220 and 190, respectively.
Incidentally, this equation is sometimes called the pricedemand5 equation for this venture.
2. To determine the applied domain, we look at the physical constraints of the problem. Certainly, we can’t sell a negative number of PortaBoys, so x ≥ 0. However, we also note that the
slope of this linear function is negative, and as such, the price is decreasing as more units are
sold. Another constraint, then, is that the price, p(x) ≥ 0. Solving −1.5x + 250 ≥ 0 results
500
in −1.5x ≥ −250 or x ≤
= 166.6. Since x represents the number of PortaBoys sold in a
3
week, we round down to 166. As a result, a reasonable applied domain for p is [0, 166].
3. The slope m = −1.5, once again, represents the rate of change of the price of a system with
respect to weekly sales of PortaBoys. Since the slope is negative, we have that the price
is decreasing at a rate of $1.50 per PortaBoy sold. (Said diﬀerently, you can sell one more
PortaBoy for every $1.50 drop in price.)
4. To determine the price which will move 150 PortaBoys, we ﬁnd p(150) = −1.5(150)+250 = 25.
That is, the price would have to be $25.
5. If the price of a PortaBoy were set at $150, we have p(x) = 150, or, −1.5x+250 = 150. Solving,
we get −1.5x = −100 or x = 66.6. This means you would be able to sell 66 PortaBoys a week
if the price were $150 per system.
5 Or simply the demand equation 3.1 Linear Functions 167 Not all realworld phenomena can be modeled using linear functions. Nevertheless, it is possible
to use the concept of slope to help analyze nonlinear functions using the following:
Definition 3.3. Let f be a function deﬁned on the interval [a, b]. The average rate of change
of f over [a, b] is deﬁned as:
∆f
f (b) − f (a)
=
∆x
b−a Geometrically, if we have the graph of y = f (x), the average rate of change over [a, b] is the
slope of the line which connects (a, f (a)) and (b, f (b)). This is called the secant line through these
points. For that reason, some textbooks use the notation msec for the average rate of change of a
function. Note that for a linear function m = msec , or in other words, its rate of change over an
interval is the same as its average rate of change.
y = f (x) (b, f (b)) (a, f (a)) The graph of y = f (x) and its secant line through (a, f (a)) and (b, f (b))
The interested reader may question the adjective ‘average’ in the phrase ‘average rate of change.’
In the ﬁgure above, we can see that the function changes wildly on [a, b], yet the slope of the secant
line only captures a snapshot of the action at a and b. This situation is entirely analogous to the
average speed on a trip. Suppose it takes you 2 hours to travel 100 miles. Your average speed is
100 miles
= 50 miles per hour. However, it is entirely possible that at the start of your journey, you
2 hours
traveled 25 miles per hour, then sped up to 65 miles per hour, and so forth. The average rate of
change is akin to your average speed on the trip. Your speedometer measures your speed at any
one instant along the trip, your instantaneous rates of change, and this is one of the central
themes of Calculus.6
When interpreting rates of change, we interpret them the same way we did slopes. In the
context of functions, it may be helpful to think of the average rate of change as:
6 Here we go again... 168 Linear and Quadratic Functions change in outputs
change in inputs
Example 3.1.7. The revenue of selling x units at a price p per unit is given by the formula R = xp.
Suppose we are in the scenario of Examples 3.1.5 and 3.1.6.
1. Find and simplify an expression for the weekly revenue R as a function of weekly sales, x.
2. Find and interpret the average rate of change of R over the interval [0, 50].
3. Find and interpret the average rate of change of R as x changes from 50 to 100 and compare
that to your result in part 2.
4. Find and interpret the average rate of change of weekly revenue as weekly sales increase from
100 PortaBoys to 150 PortaBoys.
Solution.
1. Since R = xp, we substitute p(x) = −1.5x + 250 from Example 3.1.6 to get
R(x) = x(−1.5x + 250) = −1.5x2 + 250x
2. Using Deﬁnition 3.3, we get the average rate of change is
∆R
R(50) − R(0)
8750 − 0
=
=
= 175.
∆x
50 − 0
50 − 0
Interpreting this slope as we have in similar situations, we conclude that for every additional
PortaBoy sold during a given week, the weekly revenue increases $175.
3. The wording of this part is slightly diﬀerent than that in Deﬁnition 3.3, but its meaning is to
ﬁnd the average rate of change of R over the interval [50, 100]. To ﬁnd this rate of change,
we compute
∆R
R(100) − R(50)
10000 − 8750
=
=
= 25.
∆x
100 − 50
50
In other words, for each additional PortaBoy sold, the revenue increases by $25. Note while
the revenue is still increasing by selling more game systems, we aren’t getting as much of an
increase as we did in part 2 of this example. (Can you think of why this would happen?)
4. Translating the English to the mathematics, we are being asked to ﬁnd the average rate of
change of R over the interval [100, 150]. We ﬁnd
R(150) − R(100)
3750 − 10000
∆R
=
=
= −125.
∆x
150 − 100
50
This means that we are losing $125 dollars of weekly revenue for each additional PortaBoy
sold. (Can you think why this is possible?) 3.1 Linear Functions 169 We close this section with a new look at diﬀerence quotients, ﬁrst introduced in Section 2.2. If
we wish to compute the average rate of change of a function f over the interval [x, x + h], then we
would have
∆f
f (x + h) − f (x)
f (x + h) − f (x)
=
=
∆x
(x + h) − x
h
As we have indicated, the rate of change of a function (average or otherwise) is of great importance in Calculus.7 7 So, we are not torturing you with these for nothing. 170 3.1.1 Linear and Quadratic Functions Exercises 1. Find both the pointslope form and the slopeintercept form of the line with the given slope
which passes through the given point.
√√
(a) m = 1 , P (−1, 4)
(c) m = −5, P ( 3, 2 3)
7
√
(b) m = − 2, P (0, −3)
(d) m = 678, P (−1, −12)
2. Find the slopeintercept form of the line which passes through the given points.
(a) P (0, 0), Q(−3, 5) (c) P (5, 0), Q(0, −8) (b) P (−1, −2), Q(3, −2) (d) P (3, −5), Q(7, 4) 3. Water freezes at 0◦ Celsius and 32◦ Fahrenheit and it boils at 100◦ C and 212◦ F.
(a) Find a linear function F that expresses temperature in the Fahrenheit scale in terms of
degrees Celsius. Use this function to convert 20◦ C into Fahrenheit.
(b) Find a linear function C that expresses temperature in the Celsius scale in terms of
degrees Fahrenheit. Use this function to convert 110◦ F into Celsius.
(c) Is there a temperature n such that F (n) = C (n)?
4. A salesperson is paid $200 per week plus 5% commission on her weekly sales of x dollars.
Find a linear function that represents her total weekly pay in terms of x. What must her
weekly sales be in order for her to earn $475.00 for the week?
5. Find all of the points on the line y = 2x + 1 which are 4 units from the point (−1, 3).
6. Economic forces beyond anyone’s control have changed the cost function for PortaBoys to
C (x) = 105x + 175. Rework Example 3.1.5 with this new cost function.
7. In response to the economic forces in the exercise above, the local retailer sets the selling
price of a PortaBoy at $250. Remarkably, 30 units were sold each week. When the systems
went on sale for $220, 40 units per week were sold. Rework Examples 3.1.6 and 3.1.7 with
this new data. What diﬃculties do you encounter?
8. Legend has it that a bull Sasquatch in rut will howl approximately 9 times per hour when it is
40◦ F outside and only 5 times per hour if it’s 70◦ F . Assuming that the number of howls per
hour, N , can be represented by a linear function of temperature Fahrenheit, ﬁnd the number
of howls per hour he’ll make when it’s only 20◦ F outside. What is the applied domain of this
function? Why?
9. (Parallel Lines) Recall from Intermediate Algebra that parallel lines have the same slope.
(Please note that two vertical lines are also parallel to one another even though they have
an undeﬁned slope.) In the exercises below, you are given a line and a point which is not on
that line. Find the line parallel to the given line which passes through the given point. 3.1 Linear Functions 171
(b) y = −6x + 5, P (3, 2) (a) y = 3x + 2, P (0, 0) 10. (Perpendicular Lines) Recall from Intermediate Algebra that two nonvertical lines are perpendicular if and only if they have negative reciprocal slopes. That is to say, if one line has
slope m1 and the other has slope m2 then m1 · m2 = −1. (You will be guided through a proof
of this result in the next exercise.) Please note that a horizontal line is perpendicular to a
vertical line and vice versa, so we assume m1 = 0 and m2 = 0. In the exercises below, you are
given a line and a point which is not on that line. Find the line perpendicular to the given
line which passes through the given point.
1
(a) y = 3 x + 2, P (0, 0) (b) y = −6x + 5, P (3, 2) 11. We shall now prove that y = m1 x + b1 is perpendicular to y = m2 x + b2 if and only if
m1 · m2 = −1. To make our lives easier we shall assume that m1 > 0 and m2 < 0. We can
also “move” the lines so that their point of intersection is the origin without messing things
up, so we’ll assume b1 = b2 = 0. (Take a moment with your classmates to discuss why this is
okay.) Graphing the lines and plotting the points O(0, 0) , P (1, m1 ) and Q(1, m2 ) gives us
the following set up.
y P O x Q The line y = m1 x will be perpendicular to the line y = m2 x if and only if OP Q is a right
triangle. Let d1 be the distance from O to P , let d2 be the distance from O to Q and let d3
be the distance from P to Q. Use the Pythagorean Theorem to show that OP Q is a right
triangle if and only if m1 · m2 = −1 by showing d2 + d2 = d2 if and only if m1 · m2 = −1.
1
2
3 172 Linear and Quadratic Functions 12. The function deﬁned by I (x) = x is called the Identity Function.
(a) Discuss with your classmates why this name makes sense.
(b) Show that the pointslope form of a line (Equation 3.2) can be obtained from I using a
sequence of the transformations deﬁned in Section 2.5.
13. Compute the average rate of change of the given function over the speciﬁed interval.
(a) f (x) = x3 , [−1, 2]
1
(b) f (x) = , [1, 5]
x
√
(c) f (x) = x, [0, 16] (d) f (x) = x2 , [−3, 3]
x+4
(e) f (x) =
, [5, 7]
x−3
(f) f (x) = 3x2 + 2x − 7, [−4, 2] 14. Compute the average rate of change of the given function over the interval [x, x + h]. Here
we assume [x, x + h] is in the domain of each function.
(a) f (x) = x3
(b) f (x) =
15. Explain: 1
x x+4
x−3
(d) f (x) = 3x2 + 2x − 7
(c) f (x) = 3.1 Linear Functions 3.1.2
1. Answers 173 174 Linear and Quadratic Functions
(a) y − 4 = 1 (x + 1)
7
1
y = 7 x + 29
7
√
(b) y + 3 = − 2(x − 0)
√
y = − 2x − 3 5
2. (a) y = − 3 x √
√
(c) y − 2 3 = −5(x − 3)
√
y = −5x + 7 3
(d) y + 12 = 678(x + 1)
y = 678x + 666
8
(c) y = 5 x − 8 (d) y = 9 x −
4 (b) y = −2 47
4 9
3. (a) F (C ) = 5 C + 32 (b) C (F ) = 5 F −
9 160
9 (c) F (−40) = −40 = C (−40).
4. W (x) = 200 + .05x, She must make $5500 in weekly sales.
5. (−1, −1) and 11 27
5, 5 2
8. N (T ) = − 15 T + 43
3 Having a negative number of howls makes no sense and since N (107.5) = 0 we can put an
upper bound of 107.5◦ on the domain. The lower bound is trickier because there’s nothing
other than common sense to go on. As it gets colder, he howls more often. At some point
it will either be so cold that he freezes to death or he’s howling nonstop. So we’re going to
say that he can withstand temperatures no lower than −60◦ so that the applied domain is
[−60, 107.5].
9. (a) y = 3x
10. (a) y = −3x
23 − (−1)3
=3
2 − (−1)
1
−1
1
(b) 5 1 = −
5−1 √ 5
√
16 − 0
1
(c)
=
16 − 0
4 13. (a) 14. (a) 3x2 + 3xh + h2
−1
(b)
x(x + h) (b) y = −6x + 20
1
(b) y = 6 x + (d)
(e)
(f) 3
2 32 − (−3)2
=0
3 − (−3)
7+4
7−3 5+4
− 5−3
7
=−
7−5
8 (3(2)2 + 2(2) − 7) − (3(−4)2 + 2(−4) − 7)
= −4
2 − (−4) −7
(x − 3)(x + h − 3)
(d) 6x + 3h + 2
(c) 3.2 Defining Functions (Word Problems) 3.2 175 Defining Functions (Word Problems) Example 3.2.1. Suppose you have two numbers which add to 12.
(i) If one of the numbers is 2, what is the sum of the squares of the two numbers?
Solution: If one of the numbers is 2, then the other number must be 12 − 2 = 10. Each of their
squares would be 22 = 4 and 102 = 100. So the sum of their squares would be 22 + 102 , which is
104.
(ii) If one of the numbers is x, then ﬁnd a function f (x) for the sum of the squares of the two
numbers.
Solution: If one of the numbers is x, then the other number must be 12 − x. Each of their squares
would be x2 and (12 − x)2 . So the sum of their squares would be f (x) = x2 + (12 − x)2 .
Example 3.2.2. The sum of two numbers is 12. Find a function f (x) which computes the sum of
the cubes of the two numbers, where x is one of the two numbers.
Solution: If x is one of two numbers which add to 12, then the other number must be 12 − x. So
their cubes would be x3 and (12 − x)3 , making
f (x) = x3 + (12 − x)3 .
To ﬁnd a function it helps to remember the following facts:
The distance between (x1 , y1 ) and (x2 , y2 ) is (x2 − x1 )2 + (y2 − y1 )2 . The midpoint between (x1 , y1 ) and (x2 , y2 ) is x1 + x2 y1 + y2
,
.
2
2 A rectangle with dimensions l and w
has area w l·w and perimeter 2l + 2 w . √ A right triangle with dimensions b and h
has area h 1
2 bh and hypothenuse √ b2 + h2 b b2 + h2 176 Linear and Quadratic Functions A circle with radius r has area
and circumference πr2 r 2πr A rectangular box with dimensions l, w, and h
has volume l·w·h and surface area 2lw + 2lh + 2hw . The surface area with no top is h w lw + 2lh + 2wh . r
Example 3.2.3. A square of side s is inscribed inside a circle of radius r. s r
s (i) Find a formula for the diameter of the circle in terms of r.
Solution: The diameter of a circle is twice the length of the radius. So d = 2 · r.
(i) Find a formula for the length of the side of the square in terms of the diameter, then the radius.
Solution: Using the Pythagorean theorem, we ﬁnd that
d2 = s2 + s2 = 2s2
We can solve for the side length to get
d
s2 =
√ s2 = d2
2 √
d2
d2
d
= √ =√
2
2
2 s s d
√
2
√
√
(Note that s2 = s and d2 = d because s and d represent distances and distances are always
positive) Now, we can ﬁnd the side length in terms of the radius:
s= √
√
√
d
(2r)
(2r)
2
2r 2
s= √ = √ = √ ·√ =
= r 2.
2
2
2
2
2 3.2 Defining Functions (Word Problems) 177 (iv) Find a function P (r) which computes the perimeter of the inscribed square as a function of
the radius r.
Solution: The perimeter of a square is P = 4s, and using what we found s to be above, we get
√
√
P (r) = 4s = 4(r 2) = 4 2 · r.
(v) Find a function A(r) which computes the area of the inscribed square as a function of the
radius r.
Solution: The area of a square is A = s2 , and using what we found s to be above, we get
√
√
A(s) = s2 = (r 2)2 = r2 ( 2)2 = 2r2
√
Example 3.2.4. Let f (x) = x.
(i) Find the function d(a), which computes the distance from the point (1, 0) to the point on the
graph of y = f (x) whose xcoordinate is a.
Solution: To ﬁnd this function, we need to convert this statement into a mathematical expression,
so the distance from the point (1, 0) to the point
d(a) = on the graph of y = f (x) whose xcoordinate is a.
To ﬁnd the distance between two points, we need to know the coordinates of the two points, then
we can use the distance formula. We know one point is (1, 0). The other point is on the graph
√
y = f (x) = x and has xcoordinate equalling a. Then the y coordinate for that point would be
√
√
y = f (a) = a, making the second point be (a, a).
y √
(a, a) (1, 0) a x Now, our function is
√
d(a) = the distance from the point (1, 0) to the point (a, a).
We can use the distance formula to express this distance and get that
d(a) = (1 − a)2 + (0 − √ a)2 = (1 − a)2 + a = 1 − 2a + a2 + a = 1 − a + a2 . (ii) Find a function A(a) which computes the area of a triangle whose vertices are (0, 0), (a, 0),
and (a, b), where (a, b) is on the graph of y = f (x). 178 Linear and Quadratic Functions
y √
(a, a) (1, 0) a x Solution: To compute the area of a triangle, we must use the formula
1
A = (base) × (height)
2
Using the picture, we can determine that
base = a, height = b
So the formula for the area is 1
A = ab.
2
Now, we need to reduce this formula to a single variable a. So we must ﬁnd a relationship between
a and b (an equation) which we can solve for b and substitute. As in part (i) the point (a, b) lies
√
√
on the graph of y = f (x) = x, it must be so that b = f (a) = a. This is our relationship. Now
√
we can substitute ( b) in for b in our area equation to ﬁnd area completely in terms of a:
1√
1√
A(a) = a( a) = a a.
2
2
Example 3.2.5. You own land along side a river, which you intend to fence. You have 400 feet of
fence which you intend to fence three sides of a rectangular plot, allowing the river to compose the
ﬁnal side. (i) If the side parallel to the river is to be 100 feet, ﬁnd the area of the resulting plot of land.
Solution: If we use all of the fencing, then we would use
100 feet along the side parallel to the river. Of the remaining
300 feet (400 − 100) we can split it in half and put 150 feet
( 400−100 ) on each side. This would make the enclosed area
2
be
A = 100 × 150 = 15000 150 100 3.2 Defining Functions (Word Problems) 179 (ii) Let x be the length of fence parallel to the river. Find the function A(x) which computes the
area of the enclosed rectangular area.
Solution: To ﬁnd the area of this rectangular
region, we start with the equation for the area of
a rectangle. Let x and y be the dimensions of the
rectangle as depicted. Then y x
A = x · y.
To make A be a function of x, we need to be able to compute y in terms of x. Notice in the example
before, to ﬁgure out the value of y , we ﬁrst ﬁgured out how the fence was being allocated. There
are two lengths of y and a length of x which would be needing fence, and if we use all of the fence,
this means that
x + 2 · y = 400
If we solve for y , we ﬁnd that
2y = 400 − x
y= 1
2 (400 − x) Using this computed value for y in terms of x, we can substitute in the equation for area to get
A = A(x) = x · 1
(400 − x)
2 1
1
= x · (200 − x) = 200x − x2 .
2
2 Example 3.2.6. A piece of wire which is 20cm long is to be cut in to two smaller wires.
(i) If one length is 8cm (and thus the other is 12cm) and each is bent into the shape of a square,
what would be the enclosed area?
Solution: One needs to recognize that the cut
length of wire becomes the perimeter of the resulting shape. If the perimeter of the ﬁrst square
is 8cm, then the side length would be one fourth
of 8cm, or 2cm. So the area of the ﬁrst square
would be 22 = 4cm2 . The second square will
have perimeter 12cm, which means its side length
would be 3cm, so its area would be 32 = 9cm2 .
So the total area enclosed in both squares would
be 9 + 4 = 13cm2 . 2cm
3cm (ii) Let x be one of the cut wire lengths. Find the function A(x) which computes the enclosed area
if each length of wire is shaped into a square. 180 Linear and Quadratic Functions Solution: If one of our cut lengths is x, then
the other must be 20 − x. If the piece of length x
is shaped into a square, that means the square’s
perimeter is x. So its side length is x . Thus its
4
enclosed area would be
x
4 2 = x2
.
16 The other piece is of length 20 − x, meaning the
square into which it is shaped will have a side
−
length of 204 x = 5 − x . So the enclosed area of
4
2
this square would be 5 − 1 x . This means that
4
the total enclosed area by both squares (A(x))
would be
A(x) = x2
1
+ 5− x
16
4 1
4x 1
4 (20 − x) 2 . (iii) Let x be one of the cut wire lengths. Find the function B (x) which computes the enclosed
area if each length of wire is shaped into a circle.
Solution: Here, if the wire of length x is shaped
into a circle, then its circumference is x. So its
radius for the ﬁrst circle can be determined by
x
. This means the area of this
2πr = x, so r =
2π
x2
circle would be π
. Similarly, the area of
2π
20 − x 2
.
the second circle is computed to be π
2π
So the area function B (x) would be
B (x) = π x
2π 2 +π 20 − x
2π 2 = x
2π 20−x
2π x2 + (20 − x)2
.
4π (Try to simplify to this!)
Example 3.2.7. An opentop box is created by taking a 8 × 12 in2 piece of cardboard and cutting
squares from the corners, then folding up the sides. 3.2 Defining Functions (Word Problems) 181 (i) If you cut out 2 × 2 squares from each corner, what would be the volume of the resulting box?
Solution: If we cut out squares of side length 2, then the lengths which become the length and
width of the base become 8 − 2(2) = 4in and 12 − 2(2) = 8in. The resulting height of the box is
the same as the cut length, 2in. So the volume of this box would be
Volume = 2 × 4 × 8 = 64in3 .
2
2
4 4
8 8
(ii) Let x be the length of the side of a cutout square. Find the function V (x) which computes
the resulting volume of the box.
Solution: To compute the volume of this box, we need to know the length, width, and height
in terms of x. As noted above, the height would be precisely x. The way we computed our other
dimensions above is to subtract twice x from each of the original dimensions, so our length would
be 12 − 2x and the width would be 8 − 2x. Therefore, the volume of this box would be
V (x) = x(12 − 2x)(8 − 2x).
x
x
8 − 2x 8 − 2x
12 − 2x 12 − 2x 3.2.1 Exercises 3.2.2 Answers 182 3.3 Linear and Quadratic Functions Quadratic Functions You learned about quadratic equations in Intermediate Algebra. In this section, we review those
equations in the context of our next family of functions: the quadratic functions.
Definition 3.4. A quadratic function is a function of the form
f (x) = ax2 + bx + c,
where a, b, and c are real numbers with a = 0. The domain of a quadratic function is (−∞, ∞). Example 3.3.1. Graph each of the following quadratic functions. Find the zeros of each function
and the x and y intercepts of each graph, if any exist. From the graph, determine the domain and
range of each function, list the intervals on which the function is increasing, decreasing, or constant
and ﬁnd the relative and absolute extrema, if they exist.
1. f (x) = x2 − 4x + 3. 2. g (x) = −2(x − 3)2 + 1. Solution.
1. To ﬁnd the zeros of f , we set f (x) = 0 and solve the equation x2 − 4x + 3 = 0. Factoring
gives us (x − 3)(x − 1) = 0 so that x = 3 or x = 1. The xintercepts are then (1, 0) and (3, 0).
To ﬁnd the y intercept, we set x = 0 and ﬁnd that y = f (0) = 3. Hence, the y intercept is
(0, 3). Plotting additional points, we get
y
8
7
6
5
4
3
2
1 −1 1 2 3 4 5 x −1 f (x) = x2 − 4x + 3
From the graph, we see the domain is (−∞, ∞) and the range is [−1, ∞). The function f
is increasing on [2, ∞) and decreasing on (−∞, 2]. A relative minimum occurs at the point
(2, −1) and the value −1 is both the relative and absolute minimum of f . 3.3 Quadratic Functions 183 2. Note that the formula for g (x) doesn’t match the form given in Deﬁnition 3.4. However, if we
took the time to expand g (x) = −2(x − 3)2 + 1, we would get g (x) = −2x2 + 12x − 17 which
does match with Deﬁnition 3.4. When we ﬁnd the zeros of g , we can use either formula, since
both are equivalent. Using the formula which was given to us, we get g (x) = 0
−2(x − 3)2 + 1 = 0
−2(x − 3)2 = −1
(x − 3)2 = 1
2 divide by −2
1
2 x−3 = ± extract square roots √
x−3 = ± 2
2
√ 2
2
√
6± 2
2 rationalize the denominator x = 3±
x= √ get a common denominator √ Hence, we have two xintercepts: 6+2 2 , 0 and 6−2 2 , 0 . (The inquisitive reader may
wonder what we would have done had we chosen to set the expanded form of g (x) equal to
zero. Since −2x2 + 12x − 17 does not factor nicely, we would have had to resort to other
methods, which are reviewed later in this section, to solve −2x2 + 12x − 17 = 0.) To ﬁnd
the y intercept, we set x = 0 and get g (0) = −17. Our y intercept is then (0, −17). Plotting
some additional points, we get 184 Linear and Quadratic Functions
y
1
−1 −1
−2
−3
−4
−5
−6
−7
−8
−9
−10
−11
−12
−13
−14
−15
−16
−17 1 2 3 4 5 x g (x) = −2(x − 3)2 + 1 The domain of g is (−∞, ∞) and the range is (−∞, 1]. The function g is increasing on (−∞, 3]
and decreasing on [3, ∞). The relative maximum occurs at the point (3, 1) with 1 being both the
relative and absolute maximum value of g .
Hopefully the previous examples have reminded you of some of the basic characteristics of the
graphs of quadratic equations. First and foremost, the graph of y = ax2 + bx + c where a, b, and
c are real numbers with a = 0 is called a parabola. If the coeﬃcient of x2 , a, is positive, the
parabola opens upwards; if a is negative, it opens downwards, as illustrated below.1 vertex vertex
a>0 a<0
Graphs of y = ax2 + bx + c. The point at which the relative minimum (if a > 0) or relative maximum (if a < 0) occurs is
called the vertex of the parabola. Note that each of the parabolas above is symmetric about the
dashed vertical line which contains its vertex. This line is called the axis of symmetry of the
parabola. As you may recall, there are two ways to quickly ﬁnd the vertex of a parabola, depending
on which form we are given. The results are summarized below.
1 We will justify the role of a in the behavior of the parabola later in the section. 3.3 Quadratic Functions 185 Equation 3.4. Vertex Formulas for Quadratic Functions: Suppose a, b, c, h, and k are
real numbers with a = 0.
If f (x) = a(x − h)2 + k , the vertex of the graph of y = f (x) is the point (h, k ).
If f (x) = ax2 + bx + c, the vertex of the graph of y = f (x) is the point − b
,f
2a − b
2a . Example 3.3.2. Use Equation 3.4 to ﬁnd the vertex of the graphs in Example 3.3.1.
Solution.
1. The formula f (x) = x2 − 4x + 3 is in the form f (x) = ax2 + bx + c. We identify a = 1, b = −4,
and c = 3, so that
−4
b
=−
= 2,
−
2a
2(1)
and
f − b
2a = f (2) = −1, so the vertex is (2, −1) as previously stated.
2. We see that the formula g (x) = −2(x − 3)2 + 1 is in the form g (x) = a(x − h)2 + k . We identify
a = −2, x − h as x − 3 (so h = 3), and k = 1 and get the vertex (3, 1), as required.
The formula f (x) = a(x − h)2 + k , a = 0 in Equation 3.4 is sometimes called the standard
form of a quadratic function; the formula f (x) = ax2 + bx + c, a = 0 is sometimes called the
general form of a quadratic function.
To see why the formulas in Equation 3.4 produce the vertex, let us ﬁrst consider a quadratic
function in standard form. If we consider the graph of the equation y = a(x − h)2 + k we see
that when x = h, we get y = k , so (h, k ) is on the graph. If x = h, then x − h = 0 and so
(x − h)2 is a positive number. If a > 0, then a(x − h)2 is positive, and so y = a(x − h)2 + k is
always a number larger than k . That means that when a > 0, (h, k ) is the lowest point on the
graph and thus the parabola must open upwards, making (h, k ) the vertex. A similar argument
shows that if a < 0, (h, k ) is the highest point on the graph, so the parabola opens downwards,
and (h, k ) is also the vertex in this case. Alternatively, we can apply the machinery in Section 2.5.
The vertex of the parabola y = x2 is easily seen to be the origin, (0, 0). We leave it to the reader
to convince oneself that if we apply any of the transformations in Section 2.5 (shifts, reﬂections,
and/or scalings) to y = x2 , the vertex of the resulting parabola will always be the point the graph
corresponding to (0, 0). To obtain the formula f (x) = a(x − h)2 + k , we start with g (x) = x2
and ﬁrst deﬁne g1 (x) = ag (x) = ax2 . This is results in a vertical scaling and/or reﬂection.2 Since
2 Just a scaling if a > 0. If a < 0, there is a reﬂection involved. 186 Linear and Quadratic Functions we multiply the output by a, we multiply the y coordinates on the graph of g by a, so the point
(0, 0) remains (0, 0) and remains the vertex. Next, we deﬁne g2 (x) = g1 (x − h) = a(x − h)2 . This
induces a horizontal shift right or left h units3 moves the vertex, in either case, to (h, 0). Finally,
f (x) = g2 (x) + k = a(x − h)2 + k which eﬀects a vertical shift up or down k units4 resulting in the
vertex moving from (h, 0) to (h, k ).
To verify the vertex formula for a quadratic function in general form, we complete the square
to convert the general form into the standard form.5 f (x) = ax2 + bx + c
b
= a x2 + x + c
a
b
b2
= a x2 + x + 2
a
4a
= a x+ b
2a 2 + +c−a 4ac − b2
4a b2
4a2 complete the square factor; get a common denominator Comparing this last expression with the standard form, we identify (x − h) as x + 2ba so that
b
2
2
h = − . Instead of memorizing the value k = 4ac−b , we see that f − 2ba = 4ac−b . As such,
4a
4a
2a
we have derived the vertex formula for the general form as well. Note that the value a plays the
exact same role in both the standard and general equations of a quadratic function − it is the
coeﬃcient of x2 in each. No matter what the form, if a > 0, the parabola opens upwards; if a < 0,
the parabola opens downwards.
Now that we have the completed square form of the general form of a quadratic function, it is
time to remind ourselves of the quadratic formula. In a function context, it gives us a means to
ﬁnd the zeros of a quadratic function in general form.
Equation 3.5. The Quadratic Formula: If a, b, c are real numbers with a = 0, then the
solutions to ax2 + bx + c = 0 are
√
−b ± b2 − 4ac
x=
.
2a 3 Right if h > 0, left if h < 0.
Up if k > 0, down if k < 0
5
Actually, we could also take the standard form, f (x) = a(x − h)2 + k, expand it, and compare the coeﬃcients of
it and the general form to deduce the result. However, we will have another use for the completed square form of the
general form of a quadratic, so we’ll proceed with the conversion.
4 3.3 Quadratic Functions 187 Assuming the conditions of Equation 3.5, the solutions to ax2 + bx + c = 0 are precisely the
zeros of f (x) = ax2 + bx + c. We have shown an equivalent formula for f is
2 b
f (x) = a x +
2a + 4ac − b2
.
4a Hence, an equation equivalent to ax2 + bx + c = 0 is
a x+ b
2a 2 + 4ac − b2
= 0.
4a Solving gives a x+ b
2a 2 + 4ac − b2
4a a x+ x+ 2 b
2a 1
b
a x+
a
2a
b
2a x+ =0 =−
2 4ac − b2
4a
b2 − 4ac
4a = 1
a = b2 − 4ac
4a2 2 b
2a b
x+
2a b2 − 4ac
4a2 =±
√
=± b2 − 4ac
2a b
x=− ±
2a
x= extract square roots −b ± √ b2 − 4ac
2a √ b2 − 4ac
2a In our discussions of domain, we were warned against having negative numbers underneath the
√
square root. Given that b2 − 4ac is part of the Quadratic Formula, we will need to pay special
attention to the radicand b2 − 4ac. It turns out that the quantity b2 − 4ac plays a critical role in 188 Linear and Quadratic Functions determining the nature of the solutions to a quadratic equation. It is given a special name and is
discussed below.
Definition 3.5. If a, b, c are real numbers with a = 0, then the discriminant of the quadratic
equation ax2 + bx + c = 0 is the quantity b2 − 4ac. Theorem 3.1. Discriminant Trichotomy: Let a, b, and c be real numbers with a = 0.
If b2 − 4ac < 0, the equation ax2 + bx + c = 0 has no real solutions.
If b2 − 4ac = 0, the equation ax2 + bx + c = 0 has exactly one real solution.
If b2 − 4ac > 0, the equation ax2 + bx + c = 0 has exactly two real solutions. The proof of Theorem 3.1 stems from the position of the discriminant in the quadratic equation,
and is left as a good mental exercise for the reader. The next example exploits the fruits of all of
our labor in this section thus far.
Example 3.3.3. The proﬁt function for a product is deﬁned by the equation Proﬁt = Revenue −
Cost, or P (x) = R(x) − C (x). Recall from Example 3.1.7 that the weekly revenue, in dollars, made
by selling x PortaBoy Game Systems is given by R(x) = −1.5x2 + 250x. The cost, in dollars, to
produce x PortaBoy Game Systems is given in Example 3.1.5 as C (x) = 80x + 150, x ≥ 0.
1. Determine the weekly proﬁt function, P (x).
2. Graph y = P (x). Include the x and y intercepts as well as the vertex and axis of symmetry.
3. Interpret the zeros of P .
4. Interpret the vertex of the graph of y = P (x).
5. Recall the weekly pricedemand equation for PortaBoys is: p(x) = −1.5x + 250, where p(x)
is the price per PortaBoy, in dollars, and x is the weekly sales. What should the price per
system be in order to maximize proﬁt?
Solution.
1. To ﬁnd the proﬁt function P (x), we subtract
P (x) = R(x) − C (x) = −1.5x2 + 250x − (80x + 150) = −1.5x2 + 170x − 150. 3.3 Quadratic Functions 189 2. To ﬁnd the xintercepts, we set P (x) = 0 and solve −1.5x2 + 170x − 150 = 0. The mere
thought of trying to factor the left hand side of this equation could do serious psychological
damage, so we resort to the quadratic formula, Equation 3.5. Identifying a = −1.5, b = 170,
and c = −150, we obtain x=
= = = −b ± √ b2 − 4ac
2a −170 ± 1702 − 4(−1.5)(−150)
2(−1.5) √
−170 ± 28000
−3
√
170 ± 20 70
3
√ √ We get two xintercepts: 170−20 70 , 0 and 170+20 70 , 0 . To ﬁnd the y intercept, we set
3
3
x = 0 and ﬁnd y = P (0) = −150 for a y intercept of (0, −150). To ﬁnd the vertex, we use
the fact that P (x) = −1.5x2 + 170x − 150 is in the general form of a quadratic function and
170
appeal to Equation 3.4. Substituting a = −1.5 and b = 170, we get x = − 2(−1.5) = 170 .
3
70
To ﬁnd the y coordinate of the vertex, we compute P 13 = 14000 and ﬁnd our vertex is
3
170 14000
, 3 . The axis of symmetry is the vertical line passing through the vertex so it is the
3
line x = 170 . To sketch a reasonable graph, we approximate the xintercepts, (0.89, 0) and
3
(112.44, 0), and the vertex, (56.67, 4666.67). (Note that in order to get the xintercepts and
the vertex to show up in the same picture, we had to scale the xaxis diﬀerently than the
y axis. This results in the lefthand xintercept and the y intercept being uncomfortably close
to each other and to the origin in the picture.)
y
4000
3000
2000
1000 10 20 30 40 50 60 70 80 90 100 110 120 x 3. The zeros of P are the solutions to P (x) = 0, which we have found to be approximately
0.89 and 112.44. Since P represents the weekly proﬁt, P (x) = 0 means the weekly proﬁt
is $0. Sometimes, these values of x are called the ‘breakeven’ points of the proﬁt function,
since these are places where the revenue equals the cost; in other words we gave sold enough 190 Linear and Quadratic Functions
product to recover the cost spent to make the product. More importantly, we see from the
graph that as long as x is between 0.89 and 112.44, the graph y = P (x) is above the xaxis,
meaning y = P (x) > 0 there. This means that for these values of x, a proﬁt is being made.
Since x represents the weekly sales of PortaBoy Game Systems, we round the zeros to positive
integers and have that as long as 1, but no more than 112 game systems are sold weekly, the
retailer will make a proﬁt. 4. From the graph, we see the maximum value of P occurs at the vertex, which is approximately
(56.67, 4666.67). As above, x represents the weekly sales of PortaBoy systems, so we can’t
sell 56.67 game systems. Comparing P (56) = 4666 and P (57) = 4666.5, we conclude we will
make a maximum proﬁt of $4666.50 if we sell 57 game systems.
5. In the previous part, we found we need to sell 57 PortaBoys per week to maximize proﬁt.
To ﬁnd the price per PortaBoy, we substitute x = 57 into the pricedemand function to get
p(57) = −1.5(57) + 250 = 164.5. The price should be set at $164.50.
We conclude this section with a more complicated absolute value function.
Example 3.3.4. Graph f (x) = x2 − x − 6.
Solution. Using the deﬁnition of absolute value, Deﬁnition 0.2, we have − x2 − x − 6 , if x2 − x − 6 < 0
f (x) = x2 − x − 6, if x2 − x − 6 ≥ 0
The trouble is that we have yet to develop any analytic techniques to solve nonlinear inequalities
such as x2 − x − 6 < 0. You won’t have to wait long; this is one of the main topics of Section 3.4.
Nevertheless, we can attack this problem graphically. To that end, we graph y = g (x) = x2 − x − 6
using the intercepts and the vertex. To ﬁnd the xintercepts, we solve x2 − x − 6 = 0. Factoring
gives (x − 3)(x + 2) = 0 so x = −2 or x = 3. Hence, (−2, 0) and (3, 0) are xintercepts. The y −1
intercept is found by setting x = 0, (0, −6). To ﬁnd the vertex, we ﬁnd x = − 2ba = − 2(1) = 1 , and
2
2 y = 1 − 1 − 6 = − 25 = −6.25. Plotting, we get the parabola seen below on the left. To obtain
2
2
4
points on the graph of y = f (x) = x2 − x − 6, we can take points on the graph of g (x) = x2 − x − 6
and apply the absolute value to each of the y values on the parabola. We see from the graph of g
that for x ≤ −2 or x ≥ 3, the y values on the parabola are greater than or equal to zero (since the
graph is on or above the xaxis), so the absolute value leaves this portion of the graph alone. For
x between −2 and 3, however, the y values on the parabola are negative. For example, the point
(0, −6) on y = x2 − x − 6 would result in the point (0,  − 6) = (0, −(−6)) = (0, 6) on the graph of
f (x) = x2 − x − 6. Proceeding in this manner for all points with xcoordinates between −2 and
3 results in the graph seen above on the right. 3.3 Quadratic Functions 191 y
7 y 7 6 6 5 5 4 4 3 3 2 2 1 1 −3 −2 −1
−1 1 2 3 x −3 −2 −1
−1 −2 −5 −6 x −4 −5 3 −3 −4 2 −2 −3 1 −6 y = g (x) = x2 − x − 6 y = f (x) = x2 − x − 6 If we take a step back and look at the graphs of g and f in the last example, we notice that to
obtain the graph of f from the graph of g , we reﬂect a portion of the graph of g about the xaxis.
We can see this analytically by substituting g (x) = x2 − x − 6 into the formula for f (x) and calling
to mind Theorem 2.4 from Section 2.5. −g (x), if g (x) < 0
f (x) = g (x), if g (x) ≥ 0
The function f is deﬁned so that when g (x) is negative (i.e., when its graph is below the xaxis),
the graph of f is its refection across the xaxis. This is a general template to graph functions
of the form f (x) = g (x). From this perspective, the graph of f (x) = x can be obtained by
reﬂection the portion of the line g (x) = x which is below the xaxis back above the xaxis creating
the characteristic ‘∨’ shape. 192 3.3.1 Linear and Quadratic Functions Exercises 1. Graph each of the following quadratic functions. Find the x and y intercepts of each graph,
if any exist. If it is given in the general form, convert it into standard form. Find the
domain and range of each function and list the intervals on which the function is increasing
or decreasing. Identify the vertex and the axis of symmetry and determine whether the vertex
yields a relative and absolute maximum or minimum.
(a) f (x) = x2 + 2 (e) f (x) = 2x2 − 4x − 1 (b) f (x) = −(x + 2)2 (f) f (x) = −3x2 + 4x − 7 (c) f (x) = x2 − 2x − 8 (g) f (x) = −3x2 + 5x + 4
1
(h) 6 f (x) = x2 −
x−1
100 (d) f (x) = −2(x + 1)2 + 4
2. Graph f (x) = 1 − x2  3. Find all of the points on the line y = 1 − x which are 2 units from (1, −1)
4. With the help of your classmates, show that if a quadratic function f (x) = ax2 + bx + c has
two real zeros then the xcoordinate of the vertex is the midpoint of the zeros.
5. Assuming no air resistance or forces other than the Earth’s gravity, the height above the
ground at time t of a falling object is given by s(t) = −4.9t2 + v0 t + s0 where h is in meters, t
is in seconds, v0 is the object’s initial velocity in meters per second and s0 is its initial position
in meters.
(a) What is the applied domain of this function?
(b) Discuss with your classmates what each of v0 > 0, v0 = 0 and v0 < 0 would mean.
(c) Come up with a scenario in which s0 < 0.
(d) Let’s say a slingshot is used to shoot a marble straight up from the ground (s0 = 0) with
an initial velocity of 15 meters per second. What is the marble’s maximum height above
the ground? At what time will it hit the ground?
(e) Now shoot the marble from the top of a tower which is 25 meters tall. When does it hit
the ground?
(f) What would the height function be if instead of shooting the marble up oﬀ of the tower,
you were to shoot it straight DOWN from the top of the tower?
6. The International Silver Strings Submarine Band holds a bake sale each year to fund their
trip to the National Sasquatch Convention. It has been determined that the cost in dollars
of baking x cookies is C (x) = 0.1x + 25 and that the demand function for their cookies is
p = 10 − .01x. How many cookies should they bake in order to maximize their proﬁt?
6 We have already seen the graph of this function. It was used as an example in Section 2.4 to show how the
graphing calculator can be misleading. 3.3 Quadratic Functions 193 7. The two towers of a suspension bridge are 400 feet apart. The parabolic cable7 attached to
the tops of the towers is 10 feet above the point on the bridge deck that is midway between
the towers. If the towers are 100 feet tall, ﬁnd the height of the cable directly above a point
of the bridge deck that is 50 feet to the right of the lefthand tower.
8. What is the largest rectangular area one can enclose with 14 inches of string?
9. Solve the following quadratic equations for the indicated variable.
(a) x2 − 10y 2 = 0 for x
(b) y 2 − 4y = x2 − 4 for x (e) y 2 − 3y = 4x for y (c) x2 − mx = 1 for x 7 (d) −gt2 + v0 t + s0 = 0 for t (Assume g = 0.)
(f) y 2 − 4y = x2 − 4 for y The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power
line does not form a parabola. We shall see in Exercise ?? in Section 7.5 what shape a free hanging cable makes. 194 3.3.2 Linear and Quadratic Functions Answers 1. (a) f (x) = x2 + 2
No xintercepts
y intercept (0, 2)
Domain: (−∞, ∞)
Range: [2, ∞)
Decreasing on (−∞, 0]
Increasing on [0, ∞)
Vertex (0, 2) is a minimum
Axis of symmetry x = 0 (b) f (x) = −(x + 2)2
xintercept (−2, 0)
y intercept (0, −4)
Domain: (−∞, ∞)
Range: (−∞, 0]
Increasing on (−∞, −2]
Decreasing on [−2, ∞)
Vertex (−2, 0) is a maximum
Axis of symmetry x = −2
(c) f (x) = x2 − 2x − 8 = (x − 1)2 − 9
xintercepts (−2, 0) and (4, 0)
y intercept (0, −8)
Domain: (−∞, ∞)
Range: [−9, ∞)
Decreasing on (−∞, 1]
Increasing on [1, ∞)
Vertex (1, −9) is a minimum
Axis of symmetry x = 1 (d) f (x) = −2(x + 1)2 √ 4
+
√
xintercepts (−1 − 2, 0) and (−1 + 2, 0)
y intercept (0, 2)
Domain: (−∞, ∞)
Range: (−∞, 4])
Increasing on (−∞, −1]
Decreasing on [−1, ∞)
Vertex (−1, 4) is a maximum
Axis of symmetry x = −1 y
10
9
8
7
6
5
4
3
2
1
−2 −1 1 x 2 y
x −4 −3 −2 −1
−1
−2
−3
−4
−5
−6
−7
−8 y
2
1
−2 −1
−1 1 2 3 x 4 −2
−3
−4
−5
−6
−7
−8
−9 y
4
3
2
1
−3 −2 −1 −1
−2
−3
−4 1 x 3.3 Quadratic Functions 195 (e) f (x) = 2x2 − 4x − 1 = 2(x − 1)2 − 3
√
√
xintercepts −1 − 26 , 0 and −1 + 26 , 0
y intercept (0, −1)
Domain: (−∞, ∞)
Range: [−3, ∞)
Increasing on [1, ∞)
Decreasing on (−∞, 1]
Vertex (1, −3) is a minimum
Axis of symmetry x = 1 (f) f (x) = −3x2 + 4x − 7 = −3(x − 2 )2 −
3
No xintercepts
y intercept (0, −7)
Domain: (−∞, ∞)
Range: (−∞, − 17 ]
3
2
Increasing on (−∞, 3 ]
Decreasing on [ 2 , ∞)
3
Vertex ( 2 , − 17 ) is a maximum
3
3
Axis of symmetry x = 2
3 y
4
3
2
1 −1
−1 1 2 x 3 −2
−3 y 17
3 1 −1 x 2 −2
−3
−4
−5
−6
−7
−8
−9
−10
−11
−12
−13
−14 2 (g) f (x) = −3x2 + 5x + 4 = −3 x − 5 +
√6
√
xintercepts 5−6 73 , 0 and 5+6 73 , 0
y intercept (0, 4)
Domain: (−∞, ∞)
73
Range: −∞, 12
5
Increasing on −∞, 6
Decreasing on 5 , ∞
6
Vertex 5 , 73 is a minimum
6 12
Axis of symmetry x = 5
6 73
12 y
6
5
4
3
2
1 −1
−1
−2
−3 1 2 3 x 196 Linear and Quadratic Functions
(h) f (x) = x2 − 1
100 x − 1 =
√
1+ 40001
200 x− 40001
12
200 √ − 40000
1− 40001
200 xintercepts
and
y intercept (0, −1)
Domain: (−∞, ∞)
40001
Range: − 40000 , ∞
1
Decreasing on −∞, 200
1
Increasing on 200 , ∞
40001
1
Vertex 200 , − 40000 is a minimum8
1
Axis of symmetry x = 200
2. y = 1 − x2  7
6
5
4
3
2
1
−2 −1 3. y y
8 1 2 x √
√
3 − 7 −1 + 7
,
,
2
2 √
√
3 + 7 −1 − 7
,
2
2 7
6
5
4
3
2
1
−2 −1 1 2 x 5. (a) The applied domain is [0, ∞).
(d) The height function is this case is s(t) = −4.9t2 + 15t. The vertex of this parabola
is approximately (1.53, 11.48) so the maximum height reached by the marble is 11.48
meters. It hits the ground again when t ≈ 3.06 seconds.
(e) The revised height function is s(t) = −4.9t2 + 15t + 25 which has zeros at t ≈ −1.20 and
t ≈ 4.26. We ignore the negative value and claim that the marble will hit the ground
after 4.26 seconds.
(f) Shooting down means the initial velocity is negative so the height functions becomes
s(t) = −4.9t2 − 15t + 25.
6. 495 cookies
7. Make the vertex of the parabola (0, 10) so that the point on the top of the lefthand tower
where the cable connects is (−200, 100) and the point on the top of the righthand tower is
9
(200, 100). Then the parabola is given by p(x) = 4000 x2 + 10. Standing 50 feet to the right of
the lefthand tower means you’re standing at x = −150 and p(−150) = 60.625. So the cable
is 60.625 feet above the bridge deck there.
8. The largest rectangle has area 12.25in2 .
8 You’ll need to use your calculator to zoom in far enough to see that the vertex is not the y intercept. 3.3 Quadratic Functions
√
9. (a) x = ±y 10
(b) x = ±(y − 2)
√
m ± m2 + 4
(c) x =
2 197 (d) t =
(e) y = v0 ±
3± √ (f) y = 2 ± x 2
v0 + 4gs0
2g 16x + 9
2 198 3.4 Linear and Quadratic Functions Inequalities In this section, not only do we develop techniques for solving various classes of inequalities analytically, we also look at them graphically. The next example motivates the core ideas.
Example 3.4.1. Let f (x) = 2x − 1 and g (x) = 5.
1. Solve f (x) = g (x).
2. Solve f (x) < g (x).
3. Solve f (x) > g (x).
4. Graph y = f (x) and y = g (x) on the same set of axes and interpret your solutions to parts 1
through 3 above.
Solution.
1. To solve f (x) = g (x), we replace f (x) with 2x − 1 and g (x) with 5 to get 2x − 1 = 5. Solving
for x, we get x = 3.
2. The inequality f (x) < g (x) is equivalent to 2x − 1 < 5. Solving gives x < 3 or (−∞, 3).
3. To ﬁnd where f (x) > g (x), we solve 2x − 1 > 5. We get x > 3, or (3, ∞).
4. To graph y = f (x), we graph y = 2x − 1, which is a line with a y intercept of (0, −1) and a
slope of 2. The graph of y = g (x) is y = 5 which is a horizontal line through (0, 5).
y
8
7
6 y = g ( x) 5
4
3
2 y = f (x) 1 1 2 3 4 x −1 To see the connection between the graph and the algebra, we recall the Fundamental Graphing
Principle for Functions in Section 2.4: the point (a, b) is on the graph of f if and only if
f (a) = b. In other words, a generic point on the graph of y = f (x) is (x, f (x)), and a generic 3.4 Inequalities 199 point on the graph of y = g (x) is (x, g (x)). When we seek solutions to f (x) = g (x), we are
looking for values x whose y values on the graphs of f and g are the same. In part 1, we found
x = 3 is the solution to f (x) = g (x). Sure enough, f (3) = 5 and g (3) = 5 so that the point
(3, 5) is on both graphs. We say the graphs of f and g intersect at (3, 5). In part 2, we set
f (x) < g (x) and solved to ﬁnd x < 3. For x < 3, the point (x, f (x)) is below (x, g (x)) since
the y values on the graph of f are less than the y values on the graph of g there. Analogously,
in part 3, we solved f (x) > g (x) and found x > 3. For x > 3, note that the graph of f is
above the graph of g , since the y values on the graph of f are greater than the y values on
the graph of g for those values of x.
y y 8 8 7 7 6 y = f (x) 6 y = g (x) 5 5 4 4 3 3 2 y = g (x) 2 y = f (x) 1 1 1 2 3 −1 f (x) < g (x) 4 x 1 2 3 4 x −1 f (x) > g (x) The preceding example demonstrates the following, which is a consequence of the Fundamental
Graphing Principle for Functions. Graphical Interpretation of Equations and Inequalities
Suppose f and g are functions.
The solutions to f (x) = g (x) are precisely the x values where the graphs of y = f (x) and
y = g (x) intersect.
The solutions to f (x) < g (x) are precisely the x values where the graph of y = f (x) is
below the graph of y = g (x).
The solutions to f (x) > g (x) are precisely the x values where the graph of y = f (x)
above the graph of y = g (x). 200 Linear and Quadratic Functions The next example turns the tables and furnishes the graphs of two functions and asks for
solutions to equations and inequalities.
Example 3.4.2. The graphs of f and g are below. The graph of y = f (x) resembles the upside
down ∨ shape of an absolute value function while the graph of y = g (x) resembles a parabola. Use
these graphs to answer the following questions.
y
4
y = g (x) 3
(−1, 2) (1, 2) 2
1 −2 −1 1 2 x −1
y = f (x) 1. Solve f (x) = g (x). 3. Solve f (x) ≥ g (x). 2. Solve f (x) < g (x). Solution.
1. To solve f (x) = g (x), we look for where the graphs of f and g intersect. These appear to be
at the points (−1, 2) and (1, 2), so our solutions to f (x) = g (x) are x = −1 and x = 1.
2. To solve f (x) < g (x), we look for where the graph of f is below the graph of g . This appears
to happen for the x values less than −1 and greater than 1. Our solution is (−∞, −1) ∪ (1, ∞).
3. To solve f (x) ≥ g (x), we look for solutions to f (x) = g (x) as well as f (x) > g (x). We solved
the former equation and found x = ±1. To solve f (x) > g (x), we look for where the graph
of f is above the graph of g . This appears to happen between x = −1 and x = 1, on the
interval (−1, 1). Hence, our solution to f (x) ≥ g (x) is [−1, 1].
y y 4 4
y = g (x) 3
(−1, 2) (−1, 2) (1, 2) 2 −1 (1, 2) 2 1
−2 y = g (x) 3 1
1 2 x −1 −2 −1 1 y = f (x)
f (x) < g (x) 2 x −1
y = f (x)
f (x) ≥ g (x) 3.4 Inequalities 201 We now turn our attention to solving inequalities involving the absolute value. We have the
following theorem from Intermediate Algebra to help us.
Theorem 3.2. Inequalities Involving the Absolute Value: Let c be a real number.
For c > 0, x < c is equivalent to −c < x < c.
For c ≤ 0, x < c has no solution.
For c ≥ 0, x > c is equivalent to x < −c or x > c.
For c < 0, x > c is true for all real numbers. We could argue Theorem 3.2 using cases. However, in light of what we have developed in this
section, we can understand these statements graphically. For instance, if c > 0, the graph of y = c
is a horizontal line which lies above the xaxis through (0, c). To solve x < c, we are looking for
the x values where the graph of y = x is below the graph of y = c. We know the graphs intersect
when x = c, and this happens when x = c or x = −c. Graphing, we get
y
(−c, c) −c (c, c) c x We see the graph of y = x is below y = c for x between −c and c, and hence we get x < c is
equivalent to −c < x < c. The other properties in Theorem 3.2 can be shown similarly.
Example 3.4.3. Solve the following inequalities analytically; check your answers graphically.
1. x − 1 ≥ 3 3. 2 < x − 1 ≤ 5 2. 4 − 32x + 1 > −2 4. x + 1 ≥ x+4
2 Solution.
1. To solve x − 1 ≥ 3, we seek solutions to x − 1 > 3 as well as solutions to x − 1 = 3.
From Theorem 3.2, x − 1 > 3 is equivalent to x − 1 < −3 or x − 1 > 3. From Theorem 0.2,
x − 1 = 3 is equivalent to x − 1 = −3 or x − 1 = 3. Combining these equations with the
inequalities, we solve x − 1 ≤ −3 or x − 1 ≥ 3. Our answer is x ≤ −2 or x ≥ 4, which, in
interval notation is (−∞, −2] ∪ [4, ∞). Graphically, we have 202 Linear and Quadratic Functions
y
4
3
2 −4 −3 −2 −1 1 2 3 4 5 x We see the graph of y = x − 1 (the ∨) is above the horizontal line y = 3 for x < −2 and
x > 4, and, hence, this is where x − 1 > 3. The two graphs intersect when x = −2 and
x = 4, and so we have graphical conﬁrmation of our analytic solution.
2. To solve 4 − 32x + 1 > −2 analytically, we ﬁrst isolate the absolute value before applying
Theorem 3.2. To that end, we get −32x + 1 > −6 or 2x + 1 < 2. Rewriting, we now have
3
−2 < 2x + 1 < 2 so that − 3 < x < 1 . In interval notation, we write − 2 , 1 . Graphically we
2
2
2
1
see the graph of y = 4 − 32x + 1 is above y = −2 for x values between − 3 and 2 .
2
y
4
3
2
1 −2 −1 1 2 x −1
−2
−3
−4 3. Rewriting the compound inequality 2 < x − 1 ≤ 5 as ‘2 < x − 1 and x − 1 ≤ 5’ allows
us to solve each piece using Theorem 3.2. The ﬁrst inequality, 2 < x − 1 can be rewritten
as x − 1 > 2 and so x − 1 < −2 or x − 1 > 2. We get x < −1 or x > 3. Our solution
to the ﬁrst inequality is then (−∞, −1) ∪ (3, ∞). For x − 1 ≤ 5, we combine results in
Theorems 0.2 and 3.2 to get −5 ≤ x − 1 ≤ 5 so that −4 ≤ x ≤ 6, or [−4, 6]. Our solution to
2 < x − 1 ≤ 5 is comprised of values of x which satisfy both parts of the inequality, and so we
take what’s called the ‘set theoretic intersection’ of (−∞, −1) ∪ (3, ∞) with [−4, 6] to obtain
[−4, −1) ∪ (3, 6]. Graphically, we see the graph of y = x − 1 is ‘between’ the horizontal lines
y = 2 and y = 5 for x values between −4 and −1 as well as those between 3 and 6. Including
the x values where y = x − 1 and y = 5 intersect, we get 3.4 Inequalities 203
y
8
7
6
5
4
3
2 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 x We need to exercise some special caution when solving x + 1 ≥ x+4 . When variables are both
2
inside and outside of the absolute value, it’s usually best to refer to the deﬁnition of absolute
value, Deﬁnition 0.2, to remove the absolute values and proceed from there. To that end, we have
x + 1 = −(x + 1) if x < −1 and x + 1 = x + 1 if x ≥ −1. We break the inequality into cases,
the ﬁrst case being when x < −1. For these values of x, our inequality becomes −(x + 1) ≥ x+4 .
2
Solving, we get −2x − 2 ≥ x + 4, so that −3x ≥ 6, which means x ≤ −2. Since all of these solutions
fall into the category x < −1, we keep them all. For the second case, we assume x ≥ −1. Our
inequality becomes x + 1 ≥ x+4 , which gives 2x + 2 ≥ x + 4 or x ≥ 2. Since all of these values
2
of x are greater than or equal to −1, we accept all of these solutions as well. Our ﬁnal answer is
(−∞, −2] ∪ [2, ∞).
y
4
3
2 −4 −3 −2 −1 1 2 3 4 x We now turn our attention to quadratic inequalities. In the last example of Section 3.3, we
needed to determine the solution to x2 − x − 6 < 0. We will now revisit this problem using some of
the techniques developed in this section not only to reinforce our solution in Section 3.3, but to also
help formulate a general analytic procedure for solving all quadratic inequalities. If we consider
f (x) = x2 − x − 6 and g (x) = 0, then solving x2 − x − 6 < 0 corresponds graphically to ﬁnding
the values of x for which the graph of y = f (x) = x2 − x − 6 (the parabola) is below the graph of
y = g (x) = 0 (the xaxis.) We’ve provided the graph again for reference. 204 Linear and Quadratic Functions
y
6
5
4
3
2
1
−3 −2 −1
−1 1 2 3 x −2
−3
−4
−5
−6 y = x2 − x − 6 We can see that the graph of f does dip below the xaxis between its two xintercepts. The
zeros of f are x = −2 and x = 3 in this case and they divide the domain (the xaxis) into three
intervals: (−∞, −2), (−2, 3), and (3, ∞). For every number in (−∞, −2), the graph of f is above
the xaxis; in other words, f (x) > 0 for all x in (−∞, −2). Similarly, f (x) < 0 for all x in (−2, 3),
and f (x) > 0 for all x in (3, ∞). We can schematically represent this with the sign diagram
below. (+) 0 (−) 0 (+)
−2 3 Here, the (+) above a portion of the number line indicates f (x) > 0 for those values of x; the
(−) indicates f (x) < 0 there. The numbers labeled on the number line are the zeros of f , so we
place 0 above them. We see at once that the solution to f (x) < 0 is (−2, 3).
Our next goal is to establish a procedure by which we can generate the sign diagram without
graphing the function. An important property1 of quadratic functions is that if the function is
positive at one point and negative at another, the function must have at least one zero in between.
Graphically, this means that a parabola can’t be above the xaxis at one point and below the xaxis
at another point without crossing the xaxis. This allows us to determine the sign of all of the
function values on a given interval by testing the function at just one value in the interval. This
gives us the following. 1 We will give this property a name in Chapter 4 and revisit this concept then. 3.4 Inequalities 205 Steps for Solving a Quadratic Inequality
1. Rewrite the inequality, if necessary, as a quadratic function f (x) on one side of the inequality and 0 on the other.
2. Find the zeros of f and place them on the number line with the number 0 above them.
3. Choose a real number, called a test value, in each of the intervals determined in step 2.
4. Determine the sign of f (x) for each test value in step 3, and write that sign above the
corresponding interval.
5. Choose the intervals which correspond to the correct sign to solve the inequality. Example 3.4.4. Solve the following inequalities analytically using sign diagrams. Verify your
answer graphically.
1. 2x2 ≤ 3 − x 3. 9x2 + 4 ≤ 12x 2. x2 > 2x + 1 4. 2x − x2 ≥ x − 1 − 1 Solution.
1. To solve 2x2 ≤ 3 − x, we ﬁrst get 0 on one side of the inequality which yields 2x2 + x − 3 ≤ 0.
We ﬁnd the zeros of f (x) = 2x2 + x − 3 by solving 2x2 + x − 3 = 0 for x. Factoring gives
(2x + 3)(x − 1) = 0, so x = − 3 ,or x = 1. We place these values on the number line with 0
2
above them and choose test values in the intervals −∞, − 3 , − 3 , 1 , and (1, ∞). For the
2
2
interval −∞, − 3 , we choose2 x = −2; for − 3 , 1 , we pick x = 0; and for (1, ∞), x = 2.
2
2
Evaluating the function at the three test values gives us f (−2) = 3 > 0 (so we place (+)
3
above −∞, − 3 ; f (0) = −3 < 0 (so (−) goes above the interval − 2 , 1 ); and, f (2) = 7
2
(which means (+) is placed above (1, ∞)). Since we are solving 2x2 + x − 3 ≤ 0, we look for
solutions to 2x2 + x − 3 < 0 as well as solutions for 2x2 + x − 3 = 0. For 2x2 + x − 3 < 0, we
3
need the intervals which we have a (−). Checking the sign diagram, we see this is − 2 , 1 .
We know 2x2 + x − 3 = 0 when x = − 3 and x = 1, so or ﬁnal answer is − 3 , 1 . To check our
2
2
solution graphically, we refer to the original inequality, 2x2 ≤ 3 − x. We let g (x) = 2x2 and
h(x) = 3 − x. We are looking for the x values where the graph of g is below that of h (the
solution to g (x) < h(x)) as well as the two graphs intersect (the solutions to g (x) = h(x).)
The graphs of g and h are given on the right with the sign chart on the left.
2 We have to choose something in each interval. If you don’t like our choices, please feel free to choose diﬀerent
numbers. You’ll get the same sign chart. 206 Linear and Quadratic Functions
y
7
6
5 (+) 0 (−) 0 (+)
3
−2
−2 4 1
0 3 2 2 −2 −1 1 2 x 2. Once again, we rewrite x2 > 2x + 1 as x2 − 2x − 1 > 0 and we identify f (x) = x2 − 2x − 1.
When we go to ﬁnd the zeros of f , we ﬁnd, to our chagrin, that the quadratic x2 − 2x − 1
doesn’t factor nicely. Hence, we resort to the quadratic formula to solve x2 − 2x − 1 = 0, and
√
arrive at x = 1 ± 2. As before, these zeros divide the number line into three pieces. To
√
√
help us decide on test values, we approximate 1 − 2 ≈ −0.4 and 1 + 2 ≈ 2.4. We choose
x = −1, x = 0, and x = 3 as our test values and ﬁnd f (−1) = 2, which is (+); f (0) = −1
which is (−); and f (3) = 2 which is (+) again. Our solution to x2 − 2x − 1 > 0 is where
√
√
we have (+), so, in interval notation −∞, 1 − 2 ∪ 1 + 2, ∞ . To check the inequality
x2 > 2x + 1 graphically, we set g (x) = x2 and h(x) = 2x + 1. We are looking for the x values
where the graph of g is above the graph of h. As before we present the graphs on the right
and the sign chart on the left.
y
8
7
6
5 (−)
(+)
0
0
√
√
1− 2
1+ 2
−1
0
3 (+) 4
3
2 −1 1 2 3 x 3. To solve 9x2 +4 ≤ 12x, as before, we solve 9x2 − 12x +4 ≤ 0. Setting f (x) = 9x2 − 12x +4 = 0,
we ﬁnd the only one zero of f , x = 2 . This one x value divides the number line into two
3
intervals, from which we choose x = 0 and x = 1 as test values. We ﬁnd f (0) = 4 > 0 and
f (1) = 1 > 0. Since we are looking for solutions to 9x2 − 12x + 4 ≤ 0, we are looking for 3.4 Inequalities 207 x values where 9x2 − 12x + 4 < 0 as well as where 9x2 − 12x + 4 = 0. Looking at our sign
diagram, there are no places where 9x2 − 12x + 4 < 0 (there are no (−)), so our solution
2
is only x = 3 (where 9x2 − 12x + 4 = 0). We write this as 2 . Graphically, we solve
3
9x2 + 4 ≤ 12x by graphing g (x) = 9x2 + 4 and h(x) = 12x. We are looking for the x values
where the graph of g is below the graph of h (for 9x2 + 4 < 12x) and where the two graphs
2
intersect (9x2 + 4 = 12x). We see the line and the parabola touch at 3 , 8 , but the parabola
3
is always above the line otherwise.
y
13
12
11
10 (+)
0 0
2
3 9 (+) 8
7
6 1 5
4
3
2
1
−1 1 x 4. To solve our last inequality, 2x − x2 ≥ x − 1 − 1, we rewrite the absolute value using cases.
For x < 1, x − 1 = −(x − 1) = 1 − x, so we get 2x − x2 ≥ 1 − x − 1, or x2 − 3x ≤ 0.
Finding the zeros of f (x) = x2 − 3x, we get x = 0 and x = 3. However, we are only concerned
with the portion of the number line where x < 1, so the only zero that we concern ourselves
with is x = 0. This divides the interval x < 1 into two intervals: (−∞, 0) and (0, 1). We
1
1
choose x = −1 and x = 2 as our test values. We ﬁnd f (−1) = 4 and f 2 = − 5 . Solving
4
2 − 3x ≤ 0 for x < 1 gives us [0, 1). Next, we turn our attention to the case x ≥ 1. Here,
x
x − 1 = x − 1, so our original inequality becomes 2x − x2 ≥ x − 1 − 1, or x2 − x − 2 ≤ 0.
Setting g (x) = x2 − x − 2, we ﬁnd the zeros of g to be x = −1 and x = 2. Of these, only
x = 2 lies in the region x ≥ 1, so we ignore x = −1. Our test intervals are now [1, 2) and
(2, ∞). We choose x = 1 and x = 3 as our test values and ﬁnd g (1) = −2 and g (3) = 4. To
solve g (x) ≤ 0, we have [1, 2).
(+) 0 (−)
1 0
−1 (−) 1
2 0 (+) 2
1 3 Combining these into one sign diagram, we get our solution is [0, 2]. Graphically, to check
2x − x2 ≥ x − 1 − 1, we set h(x) = 2x − x2 and i(x) = x − 1 − 1 and look for the x values
3 In this case, we say the line y = 12x is tangent to y = 9x2 + 4 at
functions is the stuﬀ of legends, I mean, Calculus. `2
3 ´
, 8 . Finding tangent lines to arbitrary 208 Linear and Quadratic Functions
where the graph of h is above the the graph of i (the solution of h(x) > i(x)) as well as the
xcoordinates of the intersection points of both graphs (where h(x) = i(x)). The combined
sign chart is given on the left and the graphs are on the right.
y
1 (+) 0 (−) 0 (+)
0
−1 0 −1 1 2 3 x −1 2
3 −2 It is quite possible to encounter inequalities where the analytical methods developed so far will
fail us. In this case, we resort to using the graphing calculator to approximate the solution, as the
next example illustrates.
Example 3.4.5. Suppose the revenue R, in thousands of dollars, from producing and selling x
hundred LCD TVs is given by R(x) = −5x3 + 35x2 + 155x for x ≥ 0, while the cost, in thousands
of dollars, to produce x hundred LCD TVs is given by C (x) = 200x + 25 for x ≥ 0. How many
TVs, to the nearest TV, should be produced to make a proﬁt?
Solution. Recall that proﬁt = revenue − cost. If we let P denote the proﬁt, in thousands of
dollars, which results from producing and selling x hundred TVs then
P (x) = R(x) − C (x) = −5x3 + 35x2 + 155x − (200x + 25) = −5x3 + 35x2 − 45x − 25,
where x ≥ 0. If we want to make a proﬁt, then we need to solve P (x) > 0; in other words,
−5x3 + 35x2 − 45x − 25 > 0. We have yet to discuss how to go about ﬁnding the zeros of P , let
alone making a sign diagram for such an animal,4 as such we resort to the graphing calculator.
After ﬁnding a suitable window, we get We are looking for the x values for which P (x) > 0, that is, where the graph of P is above the
xaxis. We make use of the ‘Zero’ command and ﬁnd two xintercepts.
4 The procedure, as we shall see in Chapter 4 is identical to what we have developed here. 3.4 Inequalities
209 We remember that x denotes the number of TVs in hundreds, so if we are to ﬁnd our solution
using the calculator, we need our answer to two decimal places. The zero5 2.414 . . . corresponds to
241.4 . . . TVs. Since we can’t make a fractional part of a TV, we round this up to 242 TVs.7 The
other zero seems dead on at 5, which corresponds to 500 TVs. Hence to make a proﬁt, we should
produce (and sell) between 242 and 499 TVs, inclusive.
Our last example in the section demonstrates how inequalities can be used to describe regions
in the plane, as we saw earlier in Section 1.2.
Example 3.4.6. Sketch the following relations.
1. R = {(x, y ) : y > x}.
2. S = {(x, y ) : y ≤ 2 − x2 }.
3. T = {(x, y ) : x < y ≤ 2 − x2 }.
Solution.
1. The relation R consists of all points (x, y ) whose y coordinate is greater than x. If we graph
y = x, then we want all of the points in the plane above the points on the graph. Dotting
the graph of y = x as we have done before to indicate the points on the graph itself are not
in the relation, we get the shaded region below on the left.
2. For a point to be in S , its y coordinate must be less than or equal to the y coordinate on the
parabola y = 2 − x2 . This is the set of all points below or on the parabola y = 2 − x2 .
y y 2
1
−2 2
1 −1 1
−1 The graph of R 2x −2 −1 1 2x −1 The graph of S 5
Note the y coordinates of the points here aren’t registered as 0. They are expressed in Scientiﬁc Notation. For
instance, 1E − 11 corresponds to 0.00000000001, which is pretty close in the calculator’s eyes6 to 0.
6
but not a Mathematician’s
7
Notice that P (241) < 0 and P (242) > 0 so we need to round up to 242 in order to make a proﬁt. 210 Linear and Quadratic Functions 3. Finally, the relation T takes the points whose y coordinates satisfy both the conditions in
R and S . So we shade the region between y = x and y = 2 − x2 , keeping those points
on the parabola, but not the points on y = x. To get an accurate graph, we need to ﬁnd
where these two graphs intersect, so we set x = 2 − x2 . Proceeding as before, breaking this
equation into cases, we get x = −1, 1. Graphing yields
y
2
1 −2 −1 1
−1 The graph of T 2 x 3.4 Inequalities 3.4.1 211 Exercises 1. Solve the inequality. Express your answer in interval form.
(a) 3x − 5 ≤ 4 (j) x2 + 4 ≤ 4x (b) 7x + 2 > 10 (k) x2 + 1 < 0 (c) 1 < 2x − 9 ≤ 3 (l) 3x2 ≤ 11x + 4 (d)  − 2x + 1 ≥ x + 5 (m) x > x2 (e) x + 3 ≥ 6x + 9 (n) 2x2 − 4x − 1 > 0 (f) x2 + 2x − 3 ≥ 0 (o) 5x + 4 ≤ 3x2 (g) 16x2 + 8x + 1 > 0 (p) 2 ≤ x2 − 9 < 9 (h) x2 + 9 < 6x (q) x2 ≤ 4x − 3 (i) 9x2 + 16 ≥ 24x (r) x2 + x + 1 ≥ 0 2. Prove the second, third and fourth parts of Theorem 3.2.
3. If a slingshot is used to shoot a marble straight up into the air from 2 meters above the
ground with an initial velocity of 30 meters per second, for what values of time t will the
marble be over 35 meters above the ground? (Refer to Exercise 5 in Section 3.3 for assistance
if needed.) Round your answers to two decimal places.
4. What temperature values in degrees Celsius are equivalent to the temperature range 50◦ F to
95◦ F ? (Refer to Exercise 3 in Section 3.1 for assistance if needed.)
5. The surface area S of a cube with edge length x is given by S (x) = 6x2 for x > 0. Suppose the
cubes your company manufactures are supposed to have a surface area of exactly 42 square
centimeters, but the machines you own are old and cannot always make a cube with the
precise surface area desired. Write an inequality using absolute value that says the surface
area of a given cube is no more than 3 square centimeters away (high or low) from the target
of 42 square centimeters. Solve the inequality and express your answer in interval form.
6. Sketch the following relations.
(a) R = {(x, y ) : y ≤ x − 1}
(b) R = (x, y ) : y > x2 + 1
(c) R = {(x, y ) : −1 < y ≤ 2x + 1} (d) R = (x, y ) : x2 ≤ y < x + 2
(e) R = {(x, y ) : x − 4 < y < 2 − x}
(f) R = (x, y ) : x2 < y ≤ 4x − 3 7. Suppose f is a function, L is a real number and ε is a positive number. Discuss with your
classmates what the inequality f (x) − L < ε means algebraically and graphically.8
8 Understanding this type of inequality is really important in Calculus. 212 3.4.2 Linear and Quadratic Functions Answers 1. (a)
(b) 1
3, 3 (k) No solution −∞, − 12
7 ∪ 8
7, ∞ (l) − 1 , 4
3 (c) [3, 4) ∪ (5, 6] (m) (0, 1) 4
(d) −∞, − 3 ∪ [6, ∞) √ (e) − 12 , − 6
7
5 (n) (f) (−∞, −3] ∪ [1, ∞) (o) −∞, 1 − 6
2 √ 6
2 ,∞ ∪ 1+ √ 1
(g) −∞, − 4 ∪ − 1 , ∞
4 (p) (h) No solution √
5+ 73
,∞
6
“√
i h√
” “ √ i h√
√
√”
−3 2, − 11 ∪ − 7, 0 ∪ 0, 7 ∪
11, 3 2 −∞, 5−6 73 ∪
√ (i) (−∞, ∞) (q) −2 − (j) {2} √
7 ∪ [1, 3] (r) (−∞, ∞) 7, −2 + 3. 1.44 < t < 4.68
4. From our previous work C (F ) = 5 (F − 32) so 50 ≤ F ≤ 95 becomes 10 ≤ C ≤ 35.
9
5. The surface area could go as low as 39cm2 or as high as 45cm2 so we have 39 ≤ S (x) ≤ 45.
Using absolute value and the fact that S (x) = 6x2 we get 6x2 − 42 ≤ 3. Solving this inequality
13
15
,
.
yields
2
2
6. (a) (b)
y y 3 4 2 3 1
−2 −1 2
1 2 3 1 x −1
−2
−3 −2 −1 1 2 x 3.4 Inequalities 213 (c) (e)
y y
5 4 4 3 3 2 2 1 1
−2 −1 1 2 3 x 1 2 −1
−2 −1 1 x 2 −2
−3 (d) −4 y
4 (f) 3
2 20 1 15
10 −1 1 2 x 5
−4 −3 −2 −1 3 214 Linear and Quadratic Functions Chapter 4 Polynomial Functions
4.1 Graphs of Polynomials Three of the families of functions studied thus far: constant, linear and quadratic, belong to a much
larger group of functions called polynomials. We begin our formal study of general polynomials
with a deﬁnition and some examples.
Definition 4.1. A polynomial function is a function of the form:
f (x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 ,
where a0 , a1 . . . . an are real numbers and n ≥ 1 is a natural number.a The domain of a
polynomial function is (−∞, ∞).
a Recall this means n is a ‘counting number’ n = 1, 2, 3, . . . There are several things about Deﬁnition 4.1 that may be oﬀputting or downright frightening.
The best thing to do is look at an example. Consider f (x) = 4x5 − 3x2 +2x − 5. Is this a polynomial
function? We can rewrite the formula for f as f (x) = 4x5 + 0x4 + 0x3 + (−3)x2 + 2x + (−5).
Comparing this with Deﬁnition 4.1, we identify n = 5, a5 = 4, a4 = 0, a3 = 0, a2 = −3, a1 = 2,
and a0 = −5. In other words, a5 is the coeﬃcient of x5 , a4 is the coeﬃcient of x4 , and so forth; the
subscript on the a’s merely indicates to which power of x the coeﬃcient belongs. The business of
restricting n to be a natural number lets us focus on wellbehaved algebraic animals.1
Example 4.1.1. Determine if the following functions are polynomials. Explain your reasoning. 1 Enjoy this while it lasts. Before we’re through with the book, you’ll have been exposed to the most terrible of
algebraic beasts. We will tame them all, in time. 216 1. g (x) = 4. f (x) = Polynomial Functions
4 + x3
x
√
3 2. p(x) = 4x + x3
x 3. q (x) = 5. h(x) = x x 4x + x3
x2 + 4 6. z (x) = 0 Solution. 3 +4
1. We note directly that the domain of g (x) = x x is x = 0. By deﬁnition, a polynomial has
all real numbers as its domain. Hence, g can’t be a polynomial. 3 2. Even though p(x) = x +4x simpliﬁes to p(x) = x2 + 4, which certainly looks like the form
x
given in Deﬁnition 4.1, the domain of p, which, as you may recall, we determine before we
simplify, excludes 0. Alas, p is not a polynomial function for the same reason g isn’t. 3. After what happened with p in the previous part, you may be a little shy about simplifying
3
q (x) = x 2+4x to q (x) = x, which certainly ﬁts Deﬁnition 4.1. If we look at the domain of
x +4
q before we simpliﬁed, we see that it is, indeed, all real numbers. A function which can be
written in the form of Deﬁnition 4.1 whose domain is all real numbers is, in fact, a polynomial. 4. We can rewrite f (x) =
polynomial. √
3 1 x as f (x) = x 3 . Since 1
3 is not a natural number, f is not a 5. The function h(x) = x isn’t a polynomial, since it can’t be written as a combination of
powers of x (even though it can be written as a piecewise function involving polynomials.)
As we shall see in this section, graphs of polynomials possess a quality2 that the graph of h
does not. 6. There’s nothing in Deﬁnition 4.1 which prevents all the coeﬃcients an , etc., from being 0.
Hence, z (x) = 0, is an honesttogoodness polynomial. 2 One which really relies on Calculus to verify. 4.1 Graphs of Polynomials 217 Definition 4.2. Suppose f is a polynomial function.
Given f (x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 with an = 0, we say – The natural number n is called the degree of the polynomial f .
– The term an xn is called the leading term of the polynomial f .
– The real number an is called the leading coeﬃcient of the polynomial f .
– The real number a0 is called the constant term of the polynomial f .
If f (x) = a0 , and a0 = 0, we say f has degree 0.
If f (x) = 0, we say f has no degree.a a Some authors say f (x) = 0 has degree −∞ for reasons not even we will go into. The reader may well wonder why we have chosen to separate oﬀ constant functions from the
other polynomials in Deﬁnition 4.2. Why not just lump them all together and, instead of forcing
n to be a natural number, n = 1, 2, . . ., let n be a whole number, n = 0, 1, 2, . . .. We could unify
all the cases, since, after all, isn’t a0 x0 = a0 ? The answer is ‘yes, as long as x = 0.’ The function
f (x) = 3 and g (x) = 3x0 are diﬀerent, because their domains are diﬀerent. The number f (0) = 3 is
deﬁned, whereas g (0) = 3(0)0 is not.3 Indeed, much of the theory we will develop in this chapter
doesn’t include the constant functions, so we might as well treat them as outsiders from the start.
One good thing that comes from Deﬁnition 4.2 is that we can now think of linear functions as
degree 1 (or ‘ﬁrst degree’) polynomial functions and quadratic functions as degree 2 (or ‘second
degree’) polynomial functions.
Example 4.1.2. Find the degree, leading term, leading coeﬃcient and constant term of the following
polynomial functions.
4−x
5 1. f (x) = 4x5 − 3x2 + 2x − 5 3. h(x) = 2. g (x) = 12x + x3 4. p(x) = (2x − 1)3 (x − 2)(3x + 2) Solution.
1. There are no surprises with f (x) = 4x5 − 3x2 + 2x − 5. It is written in the form of Deﬁnition
4.2, and we see the degree is 5, the leading term is 4x5 , the leading coeﬃcient is 4 and the
constant term is −5.
3 Technically, 00 is an indeterminant form, which is a special case of being undeﬁned. The authors realize this is
beyond pedantry, but we wouldn’t mention it if we didn’t feel it was neccessary. 218 Polynomial Functions 2. The form given in Deﬁnition 4.2 has the highest power of x ﬁrst. To that end, we rewrite
g (x) = 12x + x3 = x3 + 12x, and see the degree of g is 3, the leading term is x3 , the leading
coeﬃcient is 1 and the constant term is 0.
3. We need to rewrite the formula for h so that it resembles the form given in Deﬁnition 4.2:
1
h(x) = 4−x = 4 − x = − 1 x + 4 . We see the degree of h is 1, the leading term is − 5 x, the
5
5
5
5
5
1
4
leading coeﬃcient is − 5 and the constant term is 5 .
4. It may seem that we have some work ahead of us to get p in the form of Deﬁnition 4.2.
However, it is possible to glean the information requested about p without multiplying out
the entire expression (2x − 1)3 (x − 2)(3x + 2). The leading term of p will be the term which
has the highest power of x. The way to get this term is to multiply the terms with the highest
power of x from each factor together  in other words, the leading term of p(x) is the product of
the leading terms of the factors of p(x). Hence, the leading term of p is (2x)3 (x)(3x) = 24x5 .
This means the degree of p is 5 and the leading coeﬃcient is 24. As for the constant term,
we can perform a similar trick. The constant term is obtained by multiplying the constant
terms from each of the factors (−1)3 (−2)(2) = 4.
Our next example shows how polynomials of higher degree arise ‘naturally’4 in even the most
basic geometric applications.
Example 4.1.3. A box with no top is to be fashioned from a 10 inch × 12 inch piece of cardboard
by cutting out congruent squares from each corner of the cardboard and then folding the resulting
tabs. Let x denote the length of the side of the square which is removed from each corner.
x x x x 12 in
height
x x
x depth
width x 10 in 1. Find the volume V of the box as a function of x. Include an appropriate applied domain.
2. Use a graphing calculator to graph y = V (x) on the domain you found in part 1 and approximate the dimensions of the box with maximum volume to two decimal places. What is the
maximum volume?
Solution.
4 this is a dangerous word... 4.1 Graphs of Polynomials 219 1. From Geometry, we know Volume = width × height × depth. The key is to now ﬁnd each of
these quantities in terms of x. From the ﬁgure, we see the height of the box is x itself. The
cardboard piece is initially 10 inches wide. Removing squares with a side length of x inches
from each corner leaves 10 − 2x inches for the width.5 As for the depth, the cardboard is
initially 12 inches long, so after cutting out x inches from each side, we would have 12 − 2x
inches remaining. As a function6 of x, the volume is
V (x) = x(10 − 2x)(12 − 2x) = 4x3 − 44x2 + 120x.
To ﬁnd a suitable applied domain, we note that to make a box at all we need x > 0. Also the
shorter of the two dimensions of the cardboard is 10 inches, and since we are removing 2x
inches from this dimension, we also require 10 − 2x > 0 or x < 5. Hence, our applied domain
is 0 < x < 5.
2. Using a graphing calculator, we see the graph of y = V (x) has a relative maximum. For
0 < x < 5, this is also the absolute maximum. Using the ‘Maximum’ feature of the calculator,
we get x ≈ 1.81, y ≈ 96.77. The height, x ≈ 1.81 inches, the width, 10 − 2x ≈ 6.38 inches,
and the depth 12 − 2x ≈ 8.38 inches. The y coordinate is the maximum volume, which is
approximately 96.77 cubic inches (also written in3 ). In order to solve Example 4.1.3, we made good use of the graph of the polynomial y = V (x).
So we ought to turn our attention to graphs of polynomials in general. Below are the graphs of
y = x2 , y = x4 , and y = x6 , sidebyside. We have omitted the axes so we can see that as the
exponent increases, the ‘bottom’ becomes ‘ﬂatter’ and the ‘sides’ become ‘steeper.’ If you take the
the time to graph these functions by hand,7 you will see why.
5
There’s no harm in taking an extra step here and making sure this makes sense. If we chopped out a 1 inch
square from each side, then the width would be 8 inches, so chopping out x inches would leave 10 − 2x inches.
6
When we write V (x), it is in the context of function notation, not the volume V times the quantity x.
7
Make sure you choose some xvalues between −1 and 1. 220 Polynomial Functions y = x2 y = x4 y = x6 All of these functions are even, (Do you remember how to show this?) and it is exactly because
the exponent is even.8 One of the most important features of these functions which we can be
seen graphically is their end behavior. The end behavior of a function is a way to describe what
is happening to the function values as the x values approach the ‘ends’ of the xaxis:9 that is,
as they become small without bound10 (written x → −∞) and, on the ﬂip side, as they become
large without bound11 (written x → ∞). For example, given f (x) = x2 , as x → −∞, we imagine
substituting x = −100, x = −1000, etc., into f to get f (−100) = 10000, f (−1000) = 1000000, and
so on. Thus the function values are becoming larger and larger positive numbers (without bound).
To describe this behavior, we write: as x → −∞, f (x) → . If we study the behavior of f as
x → ∞, we see that in this case, too, f (x) → ∞. The same can be said for any function of the
form f (x) = xn where n is an even natural number. If we generalize just a bit to include vertical
scalings and reﬂections across the xaxis,12 we have
End Behavior of functions f (x) = axn , n even.
Suppose f (x) = axn where a = 0 is a real number and n is an even natural number. The end
behavior of the graph of y = f (x) matches one of the following: a>0 a<0 We now turn our attention to functions of the form f (x) = xn where n ≥ 3 is an odd natural
number.13 Below we have graphed y = x3 , y = x5 , and y = x7 . The ‘ﬂattening’ and ‘steepening’
that we saw with the even powers presents itself here as well, and, it should come as no surprise
that all of these functions are odd.14 The end behavior of these functions is all the same, with
8 Herein lies one of the possible origins of the term ‘even’ when applied to functions.
Of course, there are no ends to the xaxis.
10
We think of x as becoming a very large negative number far to the left of zero.
11
We think of x as moving far to the right of zero and becoming a very large positive number.
12
See Theorems 2.4 and 2.5 in Section 2.5.
13
We ignore the case when n = 1, since the graph of f (x) = x is a line and doesn’t ﬁt the general pattern of
higherdegree odd polynomials.
14
And are, perhaps, the inspiration for the moniker ‘odd function’.
9 4.1 Graphs of Polynomials 221 f (x) → −∞ as x → −∞ and f (x) → ∞ as x → ∞. y = x5 y = x3 y = x7 As with the even degreed functions we studied earlier, we can generalize their end behavior. End Behavior of functions f (x) = axn , n odd.
Suppose f (x) = axn where a = 0 is a real number and n ≥ 3 is an odd natural number. The
end behavior of the graph of y = f (x) matches one of the following: a>0 a<0 Despite having diﬀerent end behavior, all functions of the form f (x) = axn for natural numbers
n share two properties which help distinguish them from other animals in the algebra zoo: they are
continuous and smooth. While these concepts are formally deﬁned using Calculus,15 informally,
graphs of continuous functions have no ‘breaks’ or ‘holes’ in their graphs, and smooth functions have
no ‘sharp turns.’ It turns out that these traits are preserved when functions are added together, so
general polynomial functions inherit these qualities. Below we ﬁnd the graph of a function which is
neither smooth nor continuous, and to its right we have a graph of a polynomial, for comparison.
The function whose graph appears on the left fails to be continuous where it has a ‘break’ or ‘hole’
in the graph; everywhere else, the function is continuous. The function is continuous at the ‘corner’
and the ‘cusp’, but we consider these ‘sharp turns’, so these are places where the function fails
to be smooth. Apart from these four places, the function is smooth and continuous. Polynomial
functions are smooth and continuous everywhere, as exhibited in graph on the right. 15 In fact, if you take Calculus, you’ll ﬁnd that smooth functions are automatically continuous, so that saying
‘polynomials are continuous and smooth’ is redundant. 222 Polynomial Functions ‘hole’
‘corner’
‘cusp’
‘break’
Pathologies not found on graphs of polynomials
The graph of a polynomial The notion of smoothness is what tells us graphically that, for example, f (x) = x, whose graph
is the characteric ‘∨’ shape, cannot be a polynomial. The notion of continuity is what allowed us
to construct the sign diagram for quadratic inequalities as we did in Section 3.4. This last result is
formalized in the following theorem. Theorem 4.1. The Intermediate Value Theorem (Polynomial Zero Version): If f is
a polynomial where f (a) and f (b) have diﬀerent signs, then f has at least one zero between
x = a and x = b; that is, for at least one real number c such that a < c < b, we have f (c) = 0. The Intermediate Value Theorem is extremely profound; it gets to the heart of what it means to
be a real number, and is one of the most oft used and under appreciated theorems in Mathematics.
With that being said, most students see the result as common sense, since it says, geometrically,
that the graph of a polynomial function cannot be above the xaxis at one point and below the
xaxis at another point without crossing the xaxis somewhere in between. The following example
uses the Intermediate Value Theorem to establish a fact that that most students take for granted.
Many students, and sadly some instructors, will ﬁnd it silly.
Example 4.1.4. Use the Intermediate Value Theorem to establish that √ 2 is a real number. Solution. Consider the polynomial function f (x) = x2 − 2. Then f (1) = −1 and f (3) = 7.
Since f (1) and f (3) have diﬀerent signs, the Intermediate Value√
Theorem guarantees us a real
2 − 2 = 0 then c = ± 2. Since c is between 1 and 3,
number c between 1√ 3 with f (c) = 0. If c
and
c is positive, so c = 2.
Our primary use of the Intermediate Value Theorem is in the construction of sign diagrams,
as in Section 3.4, since it guarantees us that polynomial functions are always positive (+) or
always negative (−) on intervals which do not contain any of its zeros. The general algorithm for
polynomials is given below. 4.1 Graphs of Polynomials 223 Steps for Constructing a Sign Diagram for a Polynomial Function
Suppose f is a polynomial function.
1. Find the zeros of f and place them on the number line with the number 0 above them.
2. Choose a real number, called a test value, in each of the intervals determined in step 1.
3. Determine the sign of f (x) for each test value in step 2, and write that sign above the
corresponding interval. Example 4.1.5. Construct a sign diagram for f (x) = x3 (x − 3)2 (x + 2) x2 + 1 . Use it to give a
rough sketch of the graph of y = f (x).
Solution. First, we ﬁnd the zeros of f by solving x3 (x − 3)2 (x +2) x2 + 1 = 0. We get x = 0,
x = 3, and x = −2. (The equation x2 + 1 = 0 produces no real solutions.) These three points
divide the real number line into four intervals: (−∞, −2), (−2, 0), (0, 3) and (3, ∞). We select the
test values x = −3, x = −1, x = 1, and x = 4. We ﬁnd f (−3) is (+), f (−1) is (−) and f (1) is (+)
as is f (4). Wherever f is (+), its graph is above the xaxis; wherever f is (−), its graph is below
the xaxis. The xintercepts of the graph of f are (−2, 0), (0, 0) and (3, 0). Knowing f is smooth
and continuous allows us to sketch its graph.
y (+) 0 (−) 0 (+) 0 (+)
−3 −2
0
3
−1
1 x 4
A sketch of y = f (x) A couple of notes about the Example 4.1.5 are in order. First, note that we purposefully did
not label the y axis in the sketch of the graph of y = f (x). This is because the sign diagram
gives us the zeros and the relative position of the graph  it doesn’t give us any information as to
how high or low the graph strays from the xaxis. Furthermore, as we have mentioned earlier in
the text, without Calculus, the values of the relative maximum and minimum can only be found
approximately using a calculator. If we took the time to ﬁnd the leading term of f , we would ﬁnd
it to be x8 . Looking at the end behavior of f , we notice it matches the end behavior of y = x8 .
This is no accident, as we ﬁnd out in the next theorem.
Theorem 4.2. End Behavior for Polynomial Functions: The end behavior of a polynomial
f (x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 with an = 0 matches the end behavior of y = an xn . 224 Polynomial Functions To see why Theorem 4.2 is true, let’s ﬁrst look at a speciﬁc example. Consider f (x) = 4x3 − x +5.
If we wish to examine end behavior, we look to see the behavior of f as x → ±∞. Since we’re
concerned with x’s far down the xaxis, we are far away from x = 0 and so can rewrite f (x) for
these values of x as
1
5
f (x) = 4 x3 1 − 2 + 3
4x
4x
As x becomes unbounded (in either direction), the terms
0, as the table below indicates.
1
4x2 x
−1000 1
4x2 and 5
4x3 become closer and closer to 5
4x3 0.00000025 −0.00000000125 −100 0.000025 −0.00000125 −10 0.0025 −0.00125 10 0.0025 0.00125 100 0.000025 0.00000125 1000 0.00000025 0.00000000125 In other words, as x → ±∞, f (x) ≈ 4x3 (1 − 0 + 0) = 4x3 , which is the leading term of f . The
formal proof of Theorem 4.2 works in much the same way. Factoring out the leading term leaves
f (x) = an xn 1 + an−1
a2
a1
a0
+ ... +
+
+
an x
an xn−2 an xn−1 an xn As x → ±∞, any term with an x in the denominator becomes closer and closer to 0, and we have
f (x) ≈ an xn . Geometrically, Theorem 4.2 says that if we graph y = f (x), say, using a graphing
calculator, and continue to ‘zoom out,’ the graph of it and its leading term become indistinguishable.
Below are the graphs of y = 4x3 − x + 5 (the thicker line) and y = 4x3 (the thinner line) in two
diﬀerent windows. A view ‘close’ to the origin. A ‘zoomed out’ view. Let’s return to the function in Example 4.1.5, f (x) = x3 (x − 3)2 (x + 2) x2 + 1 , whose sign
diagram and graph are reproduced below for reference. Theorem 4.2 tells us that the end behavior 4.1 Graphs of Polynomials 225 is the same as that of its leading term, x8 . This tells us that the graph of y = f (x) starts and ends
above the xaxis. In other words, f (x) is (+) as x → ±∞, and as a result, we no longer need to
evaluate f at the test values x = −3 and x = 4. Is there a way to eliminate the need to evaluate f
at the other test values? What we would really need to know is how the function behaves near its
zeros – does it cross through the xaxis at these points, as it does at x = −2 and x = 0, or does it
simply touch and rebound like it does at x = 3. From the sign diagram, the graph of f will cross
the xaxis whenever the signs on either side of the zero switch (like they do at x = −2 and x = 0);
it will touch when the signs are the same on either side of the zero (as is the case with x = 3).
What we need to determine is the reason behind whether or not the sign change occurs.
y (+) 0 (−) 0 (+) 0 (+)
−3 −2
0
3
−1
1 x 4
A sketch of y = f (x) Fortunately, f was given to us in factored form: f (x) = x3 (x − 3)2 (x + 2). When we attempt
to determine the sign of f (−4), we are attempting to ﬁnd the sign of the number (−4)3 (−7)2 (−2),
which works out to be (−)(+)(−) which is (+). If we move to the other side of x = −2, and ﬁnd
the sign of f (−1), we are determining the sign of (−1)3 (−4)2 (+1), which is (−)(+)(+) which gives
us the (−). Notice that signs of the ﬁrst two factors in both expressions are the same in f (−4) and
f (−1). The only factor which switches sign is the third factor, (x + 2), precisely the factor which
gave us the zero x = −2. If we move to the other side of 0 and look closely at f (1), we get the sign
pattern (+1)3 (−2)2 (+3) or (+)(+)(+) and we note that, once again, going from f (−1) to f (1),
the only factor which changed sign was the ﬁrst factor, x3 , which corresponds to the zero x = 0.
Finally, to ﬁnd f (4), we substitute to get (+4)3 (+2)2 (+5) which is (+)(+)(+) or (+). The sign
didn’t change for the middle factor (x − 3)2 . Even though this is the factor which corresponds to
the zero x = 3, the fact that the quantity is squared kept the sign of the middle factor the same
on either side of 3. If we look back at the exponents on the factors (x + 2) and x3 , we note they
are both odd  so as we substitute values to the left and right of the corresponding zeros, the signs
of the corresponding factors change which results in the sign of the function value changing. This
is the key to the behavior of the function near the zeros. We need a deﬁnition and then a theorem. Definition 4.3. Suppose f is a polynomial function and m is a natural number. If (x − c)m is
a factor of f (x) but (x − c)m+1 is not, then we say x = c is a zero of multiplicity m. Hence, rewriting f (x) = x3 (x − 3)2 (x + 2) as f (x) = (x − 0)3 (x − 3)2 (x − (−2))1 , we see that
x = 0 is a zero of multiplicity 3, x = 3 is a zero of multiplicity 2, and x = −2 is a zero of multiplicity
1. 226 Polynomial Functions Theorem 4.3. The Role of Multiplicity: Suppose f is a polynomial function and x = c is
a zero of multiplicity m.
If m is even, the graph of y = f (x) touches and rebounds from the xaxis as (c, 0).
If m is odd, the graph of y = f (x) crosses through the xaxis as (c, 0). 4.1 Graphs of Polynomials 227 Our last example shows how end behavior and multiplicity allow us to sketch a decent graph
without appealing to a sign diagram.
Example 4.1.6. Sketch the graph of f (x) = −3(2x − 1)(x + 1)2 using end behavior and the
multiplicity of its zeros.
Solution. The end behavior of the graph of f will match that of its leading term. To ﬁnd
the leading term, we multiply by the leading terms of each factor to get (−3)(2x)(x)2 = −6x3 .
This tells us the graph will start above the xaxis, in Quadrant II, and ﬁnish below the xaxis, in
Quadrant IV. Next, we ﬁnd the zeros of f . Fortunately for us, f is factored.16 Setting each factor
1
equal to zero gives is x = 2 and x = −1 as zeros. To ﬁnd the multiplicity of x = 1 we note that
2
it corresponds to the factor (2x − 1). This isn’t strictly in the form required in Deﬁnition 4.3. If
1
we factor out the 2, however, we get (2x − 1) = 2 x − 2 , and we see the multiplicity of x = 1 is
2
1. Since 1 is an odd number, we know from Theorem 4.3 that the graph of f will cross through
1
the xaxis at 2 , 0 . Since the zero x = −1 corresponds to the factor (x + 1)2 = (x − (−1))2 , we
see its multiplicity is 2 which is an even number. As such, the graph of f will touch and rebound
from the xaxis at (−1, 0). Though we’re not asked to, we can ﬁnd the y intercept by ﬁnding
f (0) = −3(2(0) − 1)(0 + 1)2 = 3. Thus (0, 3) is an additional point on the graph. Putting this
together gives us the graph below.
y x 16 Obtaining the factored form of a polynomial is the main focus of the next few sections. 228 4.1.1 Polynomial Functions Exercises 1. For each polynomial given below, ﬁnd the degree, the leading term, the leading coeﬃcient,
the constant term and the end behavior.
√
(a) f (x) = 3x17 + 22.5x10 − πx7 + 1
(d) s(t) = −4.9t2 + v0 t + s0
3
(b) p(t) = −t2 (3 − 5t)(t2 + t + 4)
(c) Z (b) = 42b − (e) P (x) = (x − 1)(x − 2)(x − 3)(x − 4) b3 (f) q (r) = 1 − 16r4 2. For each polynomial given below, ﬁnd its real zeros and their corresponding multiplicities.
Use this information along with a sign chart to provide a rough sketch of the graph of the
polynomial.
(a) a(x) = x(x + 2)2 (d) Z (b) = b(42 − b2 ) (b) F (x) = x3 (x + 2)2 (e) Q(x) = (x + 5)2 (x − 3)4 (c) P (x) = (x − 1)(x − 2)(x − 3)(x − 4) (f) g (x) = x(x + 2)3 3. According to US Postal regulations, a rectangular shipping box must satisfy the inequality
“Length + Girth ≤ 130 inches” for Parcel Post and “Length + Girth ≤ 108 inches” for other
services.17 Let’s assume we have a closed rectangular box with a square face of side length
x as drawn below. The length is the longest side and is clearly labeled. The girth is the
distance around the box in the other two dimensions so in our case it is the sum of the four
sides of the square, 4x.
(a) Assuming that we’ll be mailing a box via Parcel Post where Length + Girth = 130
inches, express the length of the box in terms of x and then express the volume, V , of
the box in terms of x.
(b) Find the dimensions of the box of maximum volume that can be shipped via Parcel Post.
(c) Repeat parts 3a and 3b if the box is shipped using “other services”. x
length x 17 See here for details. 4.1 Graphs of Polynomials 229 4. Use transformations to sketch the graphs of the following polynomials.
(a) f (x) = (x + 2)3 + 1 (c) h(x) = −x5 − 3 (b) g (x) = (x + 2)4 + 1 (d) j (x) = 2 − 3(x − 1)4 5. Use the Intermediate Value Theorem to ﬁnd intervals of length 1 which contain the real zeros
of f (x) = x3 − 9x + 5.
6. The original function used to model the cost of producing x PortaBoys Game Systems given in
Example 3.1.5 was C (x) = 80x +150. While developing their newest game, Sasquatch Attack!,
the makers of the PortaBoy revised their cost function using a cubic polynomial. The new
cost of producing x PortaBoys is given by C (x) = .03x3 − 4.5x2 + 225x + 250. Market
research indicates that the demand function p(x) = −1.5x + 250 remains unchanged. Find
the production level x that maximizes the proﬁt made by producing and selling x PortaBoys.
7. Here is a chart of the number of hours of daylight they get on the 21st of each month in Fairbanks, Alaska based on the 2009 sunrise and sunset data found on the U.S. Naval Observatory
website. We let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so
on.
Month
Number 1 2 3 4 5 6 7 8 9 10 11 12 5.8 9.3 12.4 15.9 19.4 21.8 19.4 15.6 12.4 9.1 5.6 3.3 Hours of
Daylight Find cubic (third degree) and quartic (fourth degree) polynomials which model this data and
comment on the goodness of ﬁt for each. What can we say about using either model to make
predictions about the year 2020? (Hint: Think about the end behavior of polynomials.) Use
the models to see how many hours of daylight they got on your birthday and then check the
website to see how accurate the models are. Knowing that Sasquatch are largely nocturnal,
what days of the year according to your models are going to allow for at least 14 hours of
darkness for ﬁeld research on the elusive creatures?
8. An electric circuit is built with a variable resistor installed. For each of the following resistance values (measured in kiloohms, k Ω), the corresponding power to the load (measured in
milliwatts, mW ) is given in the table below. 18
Resistance: (k Ω) 2.199 3.275 4.676 6.805 9.975 Power: (mW )
18 1.012
1.063 1.496 1.610 1.613 1.505 1.314 The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem
and generating the accompanying data set. 230 Polynomial Functions
(a) Make a scatter diagram of the data using the Resistance as the independent variable
and Power as the dependent variable.
(b) Use your calculator to ﬁnd quadratic (2nd degree), cubic (3rd degree) and quartic (4th
degree) regression models for the data and judge the reasonableness of each.
(c) For each of the models found above, ﬁnd the predicted maximum power that can be
delivered to the load. What is the corresponding resistance value?
(d) Discuss with your classmates the limitations of these models  in particular, discuss the
end behavior of each. 9. Show that the end behavior of a linear function f (x) = mx + b is as it should be according
to the results we’ve established in the section for polynomials of odd degree. (That is, show
that the graph of a linear function is “up on one side and down on the other” just like the
graph of y = an xn for odd numbers n.)
10. There is one subtlety about the role of multiplicity that we need to discuss further; speciﬁcally
we need to see ‘how’ the graph crosses the xaxis at a zero of odd multiplicity. In the section,
we deliberately excluded the function f (x) = x from the discussion of the end behavior of
f (x) = xn for odd numbers n and we said at the time that it was due to the fact that f (x) = x
didn’t ﬁt the pattern we were trying to establish. You just showed in the previous exercise
that the end behavior of a linear function behaves like every other polynomial of odd degree,
so what doesn’t f (x) = x do that g (x) = x3 does? It’s the ‘ﬂattening’ for values of x near zero.
It is this local behavior that will distinguish between a zero of multiplicity 1 and one of higher
odd multiplicity. Look again closely at the graphs of a(x) = x(x + 2)2 and F (x) = x3 (x + 2)2
from Exercise 2. Discuss with your classmates how the graphs are fundamentally diﬀerent
at the origin. It might help to use a graphing calculator to zoom in on the origin to see
the diﬀerent crossing behavior. Also compare the behavior of a(x) = x(x + 2)2 to that of
g (x) = x(x + 2)3 near the point (−2, 0). What do you predict will happen at the zeros of
f (x) = (x − 1)(x − 2)2 (x − 3)3 (x − 4)4 (x − 5)5 ?
11. Here are a few other questions for you to discuss with your classmates.
(a) How many local extrema could a polynomial of degree n have? How few local extrema
can it have?
(b) Could a polynomial have two local maxima but no local minima?
(c) If a polynomial has two local maxima and two local minima, can it be of odd degree?
Can it be of even degree?
(d) Can a polynomial have local extrema without having any real zeros?
(e) Why must every polynomial of odd degree have at least one real zero?
(f) Can a polynomial have two distinct real zeros and no local extrema?
(g) Can an xintercept yield a local extrema? Can it yield an absolute extrema?
(h) If the y intercept yields an absolute minimum, what can we say about the degree of the
polynomial and the sign of the leading coeﬃcient? 4.1 Graphs of Polynomials 4.1.2 231 Answers √
1. (a) f (x) = 3x17 + 22.5x10 − πx7 +
Degree 17
√
Leading term 3x17
√
Leading coeﬃcient 3
Constant term 1
3
As x → −∞, f (x) → −∞
As x → ∞, f (x) → ∞ 1
3 (d) s(t) = −4.9t2 + v0 t + s0
Degree 2
Leading term −4.9t2
Leading coeﬃcient −4.9
Constant term s0
As t → −∞, s(t) → −∞
As t → ∞, s(t) → −∞ (b) p(t) = −t2 (3 − 5t)(t2 + t + 4)
Degree 5
Leading term 5t5
Leading coeﬃcient 5
Constant term 0
As t → −∞, p(t) → −∞
As t → ∞, p(t) → ∞ (e) P (x) = (x − 1)(x − 2)(x − 3)(x − 4)
Degree 4
Leading term x4
Leading coeﬃcient 1
Constant term 24
As x → −∞, P (x) → ∞
As x → ∞, P (x) → ∞ (c) Z (b) = 42b − b3
Degree 3
Leading term −b3
Leading coeﬃcient −1
Constant term 0
As b → −∞, Z (b) → ∞
As b → ∞, Z (b) → −∞ (f) q (r) = 1 − 16r4
Degree 4
Leading term −16r4
Leading coeﬃcient −16
Constant term 1
As r → −∞, q (r) → −∞
As r → ∞, q (r) → −∞ 2. (a) a(x) = x(x + 2)2
x = 0 multiplicity 1
x = −2 multiplicity 2 (b) F (x) = x3 (x + 2)2
x = 0 multiplicity 3
x = −2 multiplicity 2 y −2 −1 y x −2 −1 x 232 Polynomial Functions
(c) P (x) = (x − 1)(x − 2)(x − 3)(x − 4)
x = 1 multiplicity 1
x = 2 multiplicity 1
x = 3 multiplicity 1
x = 4 multiplicity 1 (e) Q(x) = (x + 5)2 (x − 3)4
x = −5 multiplicity 2
x = 3 multiplicity 4
y y −5 −4 −3 −2 −1
1 2 3 1 2 3 4 5 x x 4 (f) g (x) = x(x + 2)3
x = 0 multiplicity 1
x = −2 multiplicity 3 (d) Z (b) = b(42 − b2 )
√
b = − 42 multiplicity 1
b = √multiplicity 1
0
b = 42 multiplicity 1 y Z −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 b
−2 −1 x 3. (a) Our ultimate goal is to maximize the volume, so we’ll start with the maximum Length
+ Girth of 130. This means the length is 130 − 4x. The volume of a rectangular box is
always length × width × height so we get V (x) = x2 (130 − 4x) = −4x3 + 130x2 .
(b) Graphing y = V (x) on [0, 33] × [0, 21000] shows a maximum at (21.67, 20342.59) so the
dimensions of the box with maximum volume are 21.67in. × 21.67in. × 43.32in. for a
volume of 20342.59in.3 .
(c) If we start with Length + Girth = 108 then the length is 108 − 4x and the volume
is V (x) = −4x3 + 108x2 . Graphing y = V (x) on [0, 27] × [0, 11700] shows a maximum at (18.00, 11664.00) so the dimensions of the box with maximum volume are
18.00in. × 18.00in. × 36in. for a volume of 11664.00in.3 . (Calculus will conﬁrm that the
measurements which maximize the volume are exactly 18in. by 18in. by 36in., however,
as I’m sure you are aware by now, we treat all calculator results as approximations and
list them as such.) 4.1 Graphs of Polynomials 233
(c) h(x) = −x5 − 3 4. (a) f (x) = (x + 2)3 + 1
y y 12
11
10
9
8
7
6
5
4
3
2
1
−4 −3 −2 10
9
8
7
6
5
4
3
2
1 x −1 −1
−2
−3
−4
−5
−6
−7
−8
−9
−10 (b) g (x) = (x + 2)4 + 1 −2 −1 x y 21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
−3 1 (d) j (x) = 2 − 3(x − 1)4 y −4 −1 −1
−2
−3
−4
−5
−6
−7
−8
−9
−10 2
1
−1
−2
−3
−4
−5
−6
−7
−8
−9
−10
−11
−12
−13 1 2 x x 5. We have f (−4) = −23, f (−3) = 5, f (0) = 5, f (1) = −3, f (2) = −5 and f (3) = 5 so the
Intermediate Value Theorem tells us that f (x) = x3 − 9x + 5 has real zeros in the intervals
[−4, −3], [0, 1] and [2, 3].
6. Making and selling 71 PortaBoys yields a maximized proﬁt of $5910.67.
7. The cubic regression model is p3 (x) = 0.0226x3 − 0.9508x2 + 8.615x − 3.446. It has R2 =
0.93765 which isn’t bad. The graph of y = p3 (x) in the viewing window [−1, 13] × [0, 24]
along with the scatter plot is shown below on the left. Notice that p3 hits the xaxis at about
x = 12.45 making this a bad model for future predictions. To use the model to approximate
the number of hours of sunlight on your birthday, you’ll have to ﬁgure out what decimal value 234 Polynomial Functions
of x is close enough to your birthday and then plug it into the model. My (Jeﬀ’s) birthday
is July 31 which is 10 days after July 21 (x = 7). Assuming 30 days in a month, I think
x = 7.33 should work for my birthday and p3 (7.33) ≈ 17.5. The website says there will be
about 18.25 hours of daylight that day. To have 14 hours of darkness we need 10 hours of
daylight. We see that p3 (1.96) ≈ 10 and p3 (10.05) ≈ 10 so it seems reasonable to say that
we’ll have at least 14 hours of darkness from December 21, 2008 (x = 0) to February 21, 2009
(x = 2) and then again from October 21,2009 (x = 10) to December 21, 2009 (x = 12).
The quartic regression model is p4 (x) = 0.0144x4 − 0.3507x3 + 2.259x2 − 1.571x + 5.513. It has
R2 = 0.98594 which is good. The graph of y = p4 (x) in the viewing window [−1, 15] × [0, 35]
along with the scatter plot is shown below on the right. Notice that p4 (15) is above 24 making
this a bad model as well for future predictions. However, p4 (7.33) ≈ 18.71 making it much
better at predicting the hours of daylight on July 31 (my birthday). This model says we’ll
have at least 14 hours of darkness from December 21, 2008 (x = 0) to about March 1, 2009
(x = 2.30) and then again from October 10, 2009 (x = 9.667) to December 21, 2009 (x = 12). y = p3 (x) y = p4 (x) 8. (a) The scatter plot is shown below with each of the three regression models.
(b) The quadratic model is P2 (x) = −0.02x2 + 0.241x + 0.956 with R2 = 0.77708.
The cubic model is P3 (x) = 0.005x3 − 0.103x2 + 0.602x + 0.573 with R2 = 0.98153.
The quartic model is P4 (x) = −0.000969x4 + 0.0253x3 − 0.240x2 + 0.944x + 0.330 with
R2 = 0.99929.
(c) The maximums predicted by the three models are P2 (5.737) ≈ 1.648, P3 (4.232) ≈ 1.657
and P4 (3.784) ≈ 1.630, respectively. y = P2 (x) y = P3 (x) y = P4 (x) 4.2 The Factor Theorem and The Remainder Theorem 4.2 235 The Factor Theorem and The Remainder Theorem Suppose we wish to ﬁnd the zeros of f (x) = x3 + 4x2 − 5x − 14. Setting f (x) = 0 results in the
polynomial equation x3 + 4x2 − 5x − 14 = 0. Despite all of the factoring techniques we learned1
in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing
calculator, we get The graph suggests that x = 2 is a zero, and we can verify f (2) = 0. The other two zeros seem to
be less friendly, and, even though we could use the ‘Zero’ command to ﬁnd decimal approximations
for these, we seek a method to ﬁnd the remaining zeros exactly. Based on our experience, if x = 2
is a zero, it seems that there should be a factor of (x − 2) lurking around in the factorization of
f (x). In other words, it seems reasonable to expect that x3 + 4x2 − 5x − 14 = (x − 2) q (x), where
q (x) is some other polynomial. How could we ﬁnd such a q (x), if it even exists? The answer comes
from our old friend, polynomial division. Dividing x3 + 4x2 − 5x − 14 by x − 2 gives
x2 + 6x + 7
x− 2 x3 + 4x2 − 5x − 14
− x3 − 2x2
6x2 − 5x
− 6x2 − 12x)
7x − 14
− (7x − 14)
0
As you may recall, this means x3 + 4x2 − 5x − 14 = (x − 2) x2 + 6x + 7 , and so to ﬁnd the zeros
of f , we now can solve (x − 2) x2 + 6x + 7 = 0. We get x − 2 = 0 (which gives us our known
zero, x = 2) as well as x2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic
√
Formula to get x = −3 ± 2. The point of this section is to generalize the technique applied here.
1
2 and probably forgot
pun intended 236 Polynomial Functions First up is a friendly reminder of what we can expect when we divide polynomials.
Theorem 4.4. Polynomial Division: Suppose d(x) and p(x) are nonzero polynomials where
the degree of p is greater than or equal to the degree of d. There exist two unique polynomials,
q (x) and r(x), such that p(x) = d(x) q (x) + r(x), where either r(x) = 0 or the degree of r is
strictly less than the degree of d. As you may recall, all of the polynomials in Theorem 4.4 have special names. The polynomial
p is called the dividend; d is the divisor; q is the quotient; r is the remainder. If r(x) = 0 then
d is called a factor of p. The proof of Theorem 4.4 is usually relegated to a course in Abstract
Algebra,3 but we will use the result to establish two important facts which are the basis of the rest
of the chapter.
Theorem 4.5. The Remainder Theorem: Suppose p is a polynomial of degree at least 1
and c is a real number. When p(x) is divided by x − c the remainder is p(c). The proof of Theorem 4.5 is a direct consequence of Theorem 4.4. When a polynomial is divided
by x − c, the remainder is either 0 or has degree less than the degree of x − c. Since x − c is degree
1, this means the degree of the remainder must be 0, which means the remainder is a constant.
Hence, in either case, p(x) = (x − c) q (x) + r, where r, the remainder, is a real number, possibly
0. It follows that p(c) = (c − c) q (c) + r = 0 · q (c) + r = r, and so we get r = p(c), as required.
There is one last ‘low hanging fruit’4 to collect  it is an immediate consequence of The Remainder
Theorem.
Theorem 4.6. The Factor Theorem: Suppose p is a nonzero polynomial. The real number
c is a zero of p if and only if (x − c) is a factor of p(x). The proof of The Factor Theorem is a consequence of what we already know. If (x − c) is a
factor of p(x), this means p(x) = (x − c) q (x) for some polynomial q . Hence, p(c) = (c − c) q (c) = 0,
and so c is a zero of p. Conversely, if c is a zero of p, then p(c) = 0. In this case, The Remainder
Theorem tells us the remainder when p(x) is divided by (x − c), namely p(c), is 0, which means
(x − c) is a factor of p. What we have established is the fundamental connection between zeros of
polynomials and factors of polynomials.
Of the things The Factor Theorem tells us, the most pragmatic is that we had better ﬁnd a
more eﬃcient way to divide polynomials by quantities of the form x − c. Fortunately, people like
3
4 Yes, Virginia, there are algebra courses more abstract than this one.
Jeﬀ hates this expression and Carl included it just to annoy him. 4.2 The Factor Theorem and The Remainder Theorem 237 Ruﬃni and Horner have already blazed this trail. Let’s take a closer look at the long division we
performed at the beginning of the section and try to streamline it. First oﬀ, let’s change all of the
subtractions into additions by distributing through the −1s. x2 + 6 x + 7
x−2 x3 + 4x2 − 5x −14
−x3 + 2x2
6x2 − 5x
−6x2 + 12x
7x −14
−7x+14
0
Next, observe that the terms −x3 , −6x2 , and −7x are the exact opposite of the terms above
them. The algorithm we use ensures this is always the case, so we can omit them without losing
any information. Also note that the terms we ‘bring down’ (namely the −5x and −14) aren’t really
necessary to recopy, and so we omit them, too. x2 + 6 x + 7
x−2 x3 +4x2 − 5x −14
2x2
6x2
12x
7x
14
0
Now, let’s move things up a bit and, for reasons which will become clear in a moment, copy the 238 Polynomial Functions x3 into the last row.
x2 + 6 x + 7
x−2 x3 +4x2 − 5x −14
2x2 12x 14
x3 6x2 7x 0 Note that by arranging things in this manner, each term in the last row is obtained by adding the
two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of
the ﬁrst three terms in the last row by x and adding the results. If you take the time to work back
through the original division problem, you will ﬁnd that this is exactly the way we determined the
quotient polynomial. This means that we no longer need to write the quotient polynomial down,
nor the x in the divisor, to determine our answer. −2 x3 +4x2 − 5x −14
2x2 12x 14
x3 6x2 7x 0 We’ve streamlined things quite a bit so far, but we can still do more. Let’s take a moment to
remind ourselves where the 2x2 , 12x, and 14 came from in the second row. Each of these terms
was obtained by multiplying the terms in the quotient, x2 , 6x and 7, respectively, by the −2 in
x − 2, then by −1 when we changed the subtraction to addition. Multiplying by −2 then by −1
is the same as multiplying by 2, and so we replace the −2 in the divisor by 2. Furthermore, the
coeﬃcients of the quotient polynomial match the coeﬃcients of the ﬁrst three terms in the last row,
so we now take the plunge and write only the coeﬃcients of the terms to get
2 1 4 −5 −14
2 12 14 16 7 0 We have constructed is the synthetic division tableau for this polynomial division problem.
Let’s rework our division problem using this tableau to see how it greatly streamlines the division
process. To divide x3 + 4x2 − 5x − 14 by x − 2, we write 2 in the place of the divisor and the
coeﬃcients of x3 + 4x2 − 5x − 14 in for the dividend. Then ‘bring down’ the ﬁrst coeﬃcient of the
dividend. 4.2 The Factor Theorem and The Remainder Theorem
2 1 4 −5 −14 2 239 1 4 −5 −14
↓
1 Next, take the 2 from the divisor and multiply by the 1 that was ‘brought down’ to get 2. Write
this underneath the 4, then add to get 6.
2 1 4 −5 −14 2 1 4 −5 −14 ↓2 ↓2 1 16 Now take the 2 from the divisor times the 6 to get 12, and add it to the −5 to get 7.
2 1 4 −5 −14
↓2 2 1 4 −5 −14
↓2 16 12 16 12 7 Finally, take the 2 in the divisor times the 7 to get 14, and add it to the −14 to get 0.
2 1 4 −5 −14
↓2 12 16 7 14 2 1 4 −5
↓2 12 16 7 −14
14
0 The ﬁrst three numbers in the last row of our tableau are the coeﬃcients of the quotient
polynomial. Remember, we started with a third degree polynomial and divided by a ﬁrst degree
polynomial, so the quotient is a second degree polynomial. Hence the quotient is x2 + 6x + 7. The
number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials
by divisors of the form x − c. It is important to note that it works only for these kinds of divisors.5
Also take note that when a polynomial (of degree at least 1) is divided by x − c, the result will be
a polynomial of exactly one less degree. Finally, it is worth the time to trace each step in synthetic
division back to its corresponding step in long division. While the authors have done their best to
indicate where the algorithm comes from, there is no substitute for working through it yourself.
Example 4.2.1. Use synthetic division to perform the following polynomial divisions. Find the
quotient and the remainder polynomials, then write the dividend, quotient and remainder in the
form given in Theorem 4.4.
5 You’ll need to use good oldfashioned polynomial long division for divisors of degree larger than 1. 240 Polynomial Functions 1. 5x3 − 2x2 + 1 ÷ (x − 3) 3. 2. x3 + 8 ÷ (x + 2) 4 − 8x − 12x2
2x − 3 Solution.
1. When setting up the synthetic division tableau, we need to enter 0 for the coeﬃcient of x in
the dividend. Doing so gives
3 5 −2 0 1 ↓ 15 39 117 5 13 39 118 Since the dividend was a third degree polynomial, the quotient is a quadratic polynomial
with coeﬃcients 5, 13 and 39. Our quotient is q (x) = 5x2 + 13x + 39 and the remainder is
r(x) = 118. According to Theorem 4.4, we have 5x3 − 2x2 +1 = (x − 3) 5x2 + 13x + 39 + 118.
2. For this division, we rewrite x + 2 as x − (−2) and proceed as before
−2 1 00 8 ↓ −2 4 −8 1 −2 4 0 We get the quotient q (x) = x2 − 2x + 4 and the remainder r(x) = 0. Relating the dividend,
quotient and remainder gives x3 + 8 = (x + 2) x2 − 2x + 4 .
3. To divide 4 − 8x − 12x2 by 2x − 3, two things must be done. First, we write the dividend
in descending powers of x as −12x2 − 8x + 4. Second, since synthetic division works only
3
for factors of the form x − c, we factor 2x − 3 as 2 x − 2 . Our strategy is to ﬁrst divide
−12x2 − 8x + 4 by 2, to get −6x2 − 4x + 2. Next, we divide by x − 3 . The tableau becomes
2
3
2 −6 −4 2 ↓ −9 − 39
2 −6 −13 − 35
2 3
From this, we get −6x2 − 4x + 2 = x − 2 (−6x − 13) − 35 . Multiplying both sides by 2 and
2
distributing gives −12x2 − 8x + 4 = (2x − 3) (−6x − 13) − 35. At this stage, we have written
−12x2 − 8x + 4 in the form (2x − 3)q (x) + r(x), but how can we be sure the quotient polynomial is 4.2 The Factor Theorem and The Remainder Theorem 241 −6x − 13 and the remainder is −35? The answer is the word ‘unique’ in Theorem 4.4. The theorem
states that there is only one way to decompose −12x2 − 8x + 4 into a multiple of (2x − 3) plus a
constant term. Since we have found such a way, we can be sure it is the only way.
The next example pulls together all of the concepts discussed in this section.
Example 4.2.2. Let p(x) = 2x3 − 5x + 3.
1. Find p(−2) using The Remainder Theorem. Check your answer by substitution.
2. Use the fact that x = 1 is a zero of p to factor p(x) and ﬁnd all of the real zeros of p.
Solution.
1. The Remainder Theorem states p(−2) is the remainder when p(x) is divided by x − (−2).
We set up our synthetic division tableau below. We are careful to record the coeﬃcient of x2
as 0, and proceed as above.
−2 2 0 −5 ↓ −4 8 2 −4 3 3
−6
−3 According to the Remainder Theorem, p(−2) = −3. We can check this by direct substitution
into the formula for p(x): p(−2) = 2(−2)3 − 5(−2) + 3 = −16 + 10 + 3 = −3.
2. The Factor Theorem tells us that since x = 1 is a zero of p, x − 1 is a factor of p(x). To factor
p(x), we divide
1 2 0 −5
↓2 2 2 2 −3 3
−3
0 We get a remainder of 0 which veriﬁes that, indeed, p(1) = 0. Our quotient polynomial is a
second degree polynomial with coeﬃcients 2, 2, and −3. So q (x) = 2x2 + 2x − 3. Theorem
4.4 tells us p(x) = (x − 1) 2x2 + 2x − 3 . To ﬁnd the remaining real zeros of p, we need to
solve 2x2 + 2x − 3 = 0 for x. Since this doesn’t factor nicely, we use the quadratic formula to
√
ﬁnd that the remaining zeros are x = −1± 7 .
2
In Section 4.1, we discussed the notion of the multiplicity of a zero. Roughly speaking, a zero
with multiplicity 2 can be divided twice into a polynomial; multiplicity 3, three times and so on.
This is illustrated in the next example. 242 Polynomial Functions Example 4.2.3. Let p(x) = 4x4 − 4x3 − 11x2 + 12x − 3. Given that x = 1 is a zero of multiplicity
2
2, ﬁnd all of the real zeros of p.
Solution. We set up for synthetic division. Since we are told the multiplicity of 1 is two, we
2
continue our tableau and divide 1 into the quotient polynomial
2
1
2 12 −3 ↓
1
2 4 −4 −11
−1 −6 3 4 −2 −12 6 2 ↓ 2 4 0 −12 0 −6 0 0 1
From the ﬁrst division, we get 4x4 − 4x3 − 11x2 + 12x − 3 = x − 2 4x3 − 2x2 − 12x + 6 . The
3 − 2x2 − 12x + 6 = x − 1
2 − 12 . Combining these results, we
second division tells us 4x
4x
2
4 − 4x3 − 11x2 + 12x − 3 = x − 1 2 4x2 − 12 . To ﬁnd the remaining zeros of p, we set
have 4x
2
√
4x2 − 12 = 0 and get x = ± 3. A couple of things about the last example are worth mentioning. First, the extension of the
synthetic division tableau for repeated divisions will be a common site in the sections to come.
Typically, we will start with a higher order polynomial and peel oﬀ one zero at a time until we are
left with a quadratic, whose roots can always be found using the Quadratic Formula. Secondly, we
√
√
√
found x = ± 3 are zeros of p. The Factor Theorem guarantees x − 3 and x − − 3 are
both factors of p. We can certainly put the Factor Theorem to the test and continue the synthetic
division tableau from above to see what happens.
1
2 3 √
−3 −11 12 −3 2 −1 −6 3 4 −2 −12 6 ↓
√ −4 ↓
1
2 4 2 0 −6 0
√
↓
43
√
4
43
√
↓ −4 3 −12 4 4 0 0 12
0 0 This gives us 4x4 − 4x3 − 11x2 + 12x − 3 = x − 12
2 x− √ 3 √
x− − 3 (4), or, when written 4.2 The Factor Theorem and The Remainder Theorem 243 with the constant in front
p(x) = 4 x − 1
2 2 x− √ 3 √
x− − 3 We have shown that p is a product of its leading term times linear factors of the form (x − c) where
c are zeros of p. It may surprise and delight the reader that, in theory, all polynomials can be
reduced to this kind of factorization. We leave that discussion to Section 10.1, because the zeros
may not be real numbers. Our ﬁnal theorem in the section gives us an upper bound on the number
of real zeros.
Theorem 4.7. Suppose f is a polynomial of degree n, n ≥ 1. Then f has at most n real zeros,
counting multiplicities. Theorem 4.7 is a consequence of the Factor Theorem and polynomial multiplication. Every zero
c of f gives us a factor of the form (x − c) for f (x). Since f has degree n, there can be at most n
of these factors. The next section provides us some tools which not only help us determine where
the real zeros are to be found, but which real numbers they may be.
We close this section with a summary of several concepts previously presented. You should take
the time to look back through the text to see where each concept was ﬁrst introduced and where
each connection to the other concepts was made. Connections Between Zeros, Factors and Graphs of Polynomial Functions
Suppose p is a polynomial function of degree n ≥ 1. The following statements are equivalent:
The real number c is a zero of p
p(c) = 0
x = c is a solution to the polynomial equation p(x) = 0
(x − c) is a factor of p(x)
The point (c, 0) is an xintercept of the graph of y = p(x) 244 4.2.1 Polynomial Functions Exercises 1. (An Intermediate Algebra review exercise) Use polynomial long division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q (x) + r(x).
(a) (5x4 − 3x3 + 2x2 − 1) ÷ (x2 + 4) (c) (9x3 + 5) ÷ (2x − 3) (b) (−x5 + 7x3 − x) ÷ (x3 − x2 + 1) (d) (4x2 − x − 23) ÷ (x2 − 1) 2. Use synthetic division and the Remainder Theorem to test whether or not the given number
is a zero of the polynomial p(x) = 15x5 − 121x4 + 17x3 − 73x2 + 2x + 48.
2
3 (a) c = −1 (d) c = (b) c = 8 (e) c = 0 (c) c = 1
2 3
(f) c = − 5 3. For each polynomial given below, you are given one of its zeros. Use the techniques in this
section to ﬁnd the rest of the real zeros and factor the polynomial.
(a) x3 − 6x2 + 11x − 6, c = 1
(b) x3 − 24x2 + 192x − 512, c = 8
(c) 4x4 − 28x3 + 61x2 − 42x + 9, c =
(d) 3x3 + 4x2 − x − 2, c =
(e) x4 − x2 , c = 0
(f) x2 − 2x − 2, c = 1 − 1
2 2
3 √
3 3
(g) 125x5 − 275x4 − 2265x3 − 3213x2 − 1728x − 324, c = − 5 4. Create a polynomial p with the following attributes.
As x → −∞, p(x) → ∞.
The point (−2, 0) yields a local maximum.
The degree of p is 5.
The point (3, 0) is one of the xintercepts of the graph of p. √
3
29
5. Find a quadratic polynomial with integer coeﬃcients which has x = ±
as its real zeros.
5
5 4.2 The Factor Theorem and The Remainder Theorem 4.2.2 Answers 1. (a) 5x4 − 3x3 + 2x2 − 1 = (x2 + 4)(5x2 − 3x − 18) + (12x + 71)
(b) −x5 + 7x3 − x = (x3 − x2 + 1)(−x2 − x + 6) + (7x2 − 6)
9
(c) 9x3 + 5 = (2x − 3)( 2 x2 + 27
4x + 81
8) + 283
8 (d) 4x2 − x − 23 = (x2 − 1)(4) + (−x − 19)
2. (a) p(−1) − 180
(b) p(8) = 0
(c) p( 1 ) =
2 825
32 (d) p( 2 ) = 0
3
(e) p(0) = 48
(f) p(− 3 ) = 0
5
3. (a) x3 − 6x2 + 11x − 6 = (x − 1)(x − 2)(x − 3)
(b) x3 − 24x2 + 192x − 512 = (x − 8)3
(c) 4x4 − 28x3 + 61x2 − 42x + 9 = 4(x − 1 )2 (x − 3)2
2
2
(d) 3x3 + 4x2 − x − 2 = 3(x − 3 )(x + 1)2 (e) x4 − x2 = x2 (x − 1)(x + 1)
√
√
(f) x2 − 2x − 2 = (x − (1 − 3))(x − (1 + 3))
3
(g) 125x5 − 275x4 − 2265x3 − 3213x2 − 1728x − 324 = 125(x + 5 )3 (x − 6)(x + 2) 4. Something like p(x) = −(x + 2)2 (x − 3)(x + 3)(x − 4) will work.
5. q (x) = 5x2 − 6x − 4 245 246 Polynomial Functions Chapter 5 Rational Functions
5.1 Introduction to Rational Functions If we add, subtract or multiply polynomial functions according to the function arithmetic rules
deﬁned in Section 2.3, we will produce another polynomial function. If, on the other hand, we
divide two polynomial functions, the result may not be a polynomial. In this chapter we study
rational functions  functions which are ratios of polynomials.
Definition 5.1. A rational function is a function which is the ratio of polynomial functions.
Said diﬀerently, r is a rational function if it is of the form
r(x) = p(x)
,
q (x) where p and q are polynomial functionsa
a According to this deﬁnition, all polynomial functions are also rational functions. (Take q (x) = 1). As we recall from Section 2.2, we have domain issues anytime the denominator of a fraction is
zero. In the example below, we review this concept as well as some of the arithmetic of rational
expressions.
Example 5.1.1. Find the domain of the following rational functions. Write them in the form
for polynomial functions p and q and simplify.
1. f (x) = 2x − 1
x+1 2. g (x) = 2 − 3
x+1 3. h(x) = 2x2 − 1 3x − 2
−2
x2 − 1
x −1 4. r(x) = 2x2 − 1 3x − 2
÷2
x2 − 1
x −1 p(x)
q (x) 248 Rational Functions Solution. 1. To ﬁnd the domain of f , we proceed as we did in Section 2.2: we ﬁnd the zeros of the
denominator and exclude them from the domain. Setting x + 1 = 0 results in x = −1. Hence,
our domain is (−∞, −1) ∪ (−1, ∞). The expression f (x) is already in the form requested and
when we check for common factors among the numerator and denominator we ﬁnd none, so
we are done. 2. Proceeding as before, we determine the domain of g by solving x + 1 = 0. As before, we ﬁnd
the domain of g is (−∞, −1) ∪ (−1, ∞). To write g (x) in the form requested, we need to get
a common denominator g (x) = 2 − 3
x+1 = 2
3
−
1 x+1 = (2)(x + 1)
3
−
(1)(x + 1) x + 1 = (2x + 2) − 3
x+1 = 2x − 1
x+1 This formula is also completely simpliﬁed. 3. The denominators in the formula for h(x) are both x2 − 1 whose zeros are x = ±1. As a
result, the domain of h is (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). We now proceed to simplify h(x).
Since we have the same denominator in both terms, we subtract the numerators. We then
factor the resulting numerator and denominator, and cancel out the common factor. 5.1 Introduction to Rational Functions 249 2x2 − 1 3x − 2
−2
x2 − 1
x −1 h(x) = 2x2 − 1 − (3x − 2)
x2 − 1 =
= 2x2 − 1 − 3x + 2
x2 − 1 = 2x2 − 3x + 1
x2 − 1 = (2x − 1)(x − 1)
(x + 1)(x − 1) = $
(2x − 1)$$$
(x − 1)
$
(x − 1)
(x + 1)$$$ = 2x − 1
x+1 4. To ﬁnd the domain of r, it may help to temporarily rewrite r(x) as
2x2 − 1
2
r (x) = x − 1
3x − 2
x2 − 1
We need to set all of the denominators equal to zero which means we need to solve not only
2
−2
x2 − 1 = 0, but also 3x−1 = 0. We ﬁnd x = ±1 for the former and x = 3 for the latter. Our
x2
2
domain is (−∞, −1) ∪ −1, 3 ∪ 2 , 1 ∪ (1, ∞). We simplify r(x) by rewriting the division as
3
multiplication by the reciprocal and then simplifying
r(x) =
= 2x2 − 1 3x − 2
÷2
x2 − 1
x −1
2x2 − 1 x2 − 1
·
x2 − 1 3x − 2 = 2x2 − 1 x2 − 1
(x2 − 1) (3x − 2) = $
2x2 − 1 $$− $
x2 $ 1
$
x2 $$
$$− 1 (3x − 2) = 2x2 − 1
3x − 2 250 Rational Functions A few remarks about Example 5.1.1 are in order. Note that the expressions for f (x), g (x) and
h(x) work out to be the same. However, only two of these functions are actually equal. Recall that
functions are ultimately sets of ordered pairs,1 and so for two functions to be equal, they need,
among other things, to have the same domain. Since f (x) = g (x) and f and g have the same
domain, they are equal functions. Even though the formula h(x) is the same as f (x), the domain
of h is diﬀerent than the domain of f , and thus they are diﬀerent functions.
We now turn our attention to the graphs of rational functions. Consider the function f (x) =
from Example 5.1.1. Using a graphing calculator, we obtain 2x−1
x+1 Two behaviors of the graph are worthy of further discussion. First, note that the graph appears
to ‘break’ at x = −1. We know from our last example that x = −1 is not in the domain of f
which means f (−1) is undeﬁned. When we make a table of values to study the behavior of f near
x = −1 we see that we can get ‘near’ x = −1 from two directions. We can choose values a little
less than −1, for example x = −1.1, x = −1.01, x = −1.001, and so on. These values are said to
‘approach −1 from the left.’ Similarly, the values x = −0.9, x = −0.99, x = −0.999, etc., are said
to ‘approach −1 from the right.’ If we make two tables, we ﬁnd that the numerical results conﬁrm
what we see graphically.
x f (x) (x, f (x)) x f (x) (x, f (x)) −1.1 32 (−1.1, 32) −0.9 −28 (−0.9, −28) −1.01 302 (−1.01, 302) −0.99 −298 (−0.99, −298) −1.001 3002 (−1.001, 3002) −0.999 −2998 (−0.999, −2998) 30002 (−1.001, 30002) −0.9999 −1.0001 −29998 (−0.9999, −29998) As the x values approach −1 from the left, the function values become larger and larger positive
numbers.2 We express this symbolically by stating as x → −1− , f (x) → ∞. Similarly, using
analogous notation, we conclude from the table that as x → −1+ , f (x) → −∞. For this type of
unbounded behavior, we say the graph of y = f (x) has a vertical asymptote of x = −1. Roughly
speaking, this means that near x = −1, the graph looks very much like the vertical line x = −1.
1
2 You should review Sections 1.2 and 2.1 if this statement caught you oﬀ guard.
We would need Calculus to conﬁrm this analytically. 5.1 Introduction to Rational Functions 251 Another feature worthy of note about the graph of y = f (x) is it seems to ‘level oﬀ’ on the left
and right hand sides of the screen. This is a statement about the end behavior of the function. As
we discussed in Section 4.1, the end behavior of a function is its behavior as x as x attains larger3
and larger negative values without bound, x → −∞, and as x becomes large without bound,
x → ∞. Making tables of values, we ﬁnd x f (x) (x, f (x)) x f (x) (x, f (x)) −10 ≈ 2.3333 ≈ (−10, 2.3333) 10 ≈ 1.7273 ≈ (10, 1.7273) −100 ≈ 2.0303 ≈ (−100, 2.0303) 100 ≈ 1.9703 ≈ (100, 1.9703) −1000 ≈ 2.0030 ≈ (−1000, 2.0030) 1000 ≈ 1.9970 ≈ (1000, 1.9970) ≈ 2.0003 ≈ (−10000, 2.0003) 10000 −10000 ≈ 1.9997 ≈ (10000, 1.9997) From the tables, we see as x → −∞, f (x) → 2+ and as x → ∞, f (x) → 2− . Here the ‘+’ means
‘from above’ and the ‘−’ means ‘from below’. In this case, we say the graph of y = f (x) has a
horizontal asymptote of y = 2. This means that the end behavior of f resembles the horizontal
line y = 2, which explains the ‘leveling oﬀ’ behavior we see in the calculator’s graph. We formalize
the concepts of vertical and horizontal asymptotes in the following deﬁnitions. Definition 5.2. The line x = c is called a vertical asymptote of the graph of a function
y = f (x) if as x → c− or as x → c+ , either f (x) → ∞ or f (x) → −∞. Definition 5.3. The line y = c is called a horizontal asymptote of the graph of a function
y = f (x) if as x → −∞ or as x → ∞, either f (x) → c− or f (x) → c+ . In our discussion following Example 5.1.1, we determined that, despite the fact that the formula
for h(x) reduced to the same formula as f (x), the functions f and h are diﬀerent, since x = 1 is in
x2 −
−2
the domain of f , but x = 1 is not in the domain of h. If we graph h(x) = 2x2 −11 − 3x−1 using a
x2
graphing calculator, we are surprised to ﬁnd that the graph looks identical to the graph of y = f (x).
There is a vertical asymptote at x = −1, but near x = 1, everything seem ﬁne. Tables of values
provide numerical evidence which supports the graphical observation.
3 Here, the word ‘larger’ means larger in absolute value. 252 Rational Functions
x h(x) (x, h(x)) 0.9 ≈ 0.4210 ≈ (0.9, 0.4210) 0.99 ≈ 0.4925 0.999 ≈ 0.4992 0.9999 x h(x) (x, h(x)) 1.1 ≈ 0.5714 ≈ (1.1, 0.5714) ≈ (0.99, 0.4925) 1.01 ≈ 0.5075 ≈ (1.01, 0.5075) ≈ (0.999, 0.4992) 1.001 ≈ 0.5007 ≈ (1.001, 0.5007) ≈ 0.4999 ≈ (0.9999, 0.4999) 1.0001 ≈ 0.5001 ≈ (1.0001, 0.5001) We see that as x → 1− , h(x) → 0.5− and as x → 1+ , h(x) → 0.5+ . In other words, the points
on the graph of y = h(x) are approaching (1, 0.5), but since x = 1 is not in the domain of h, it
would be inaccurate to ﬁll in a point at (1, 0.5). As we’ve done in past sections when something
like this occurs,4 we put an open circle (also called a ‘hole’ in this case5 ) at (1, 0.5). Below is a
detailed graph of y = h(x), with the vertical and horizontal asymptotes as dashed lines.
y
8
7
6
5
4
3 1 −4 −3 −2 1 2 3 4 x −1
−2
−3
−4
−5
−6 Neither x = −1 nor x = 1 are in the domain of h, yet we see the behavior of the graph of
y = h(x) is drastically diﬀerent near these points. The reason for this lies in the second to last
x−1)(x−1)
step when we simpliﬁed the formula for h(x) in Example 5.1.1. We had h(x) = (2x+1)(x−1) . The
(
reason x = −1 is not in the domain of h is because the factor (x + 1) appears in the denominator
of h(x); similarly, x = 1 is not in the domain of h because of the factor (x − 1) in the denominator
of h(x). The major diﬀerence between these two factors is that (x − 1) cancels with a factor in
4 For instance, graphing piecewise deﬁned functions in Section 2.4.
Stay tuned. In Calculus, we will see how these ‘holes’ can be ‘plugged’ when embarking on a more advanced
study of contintuity.
5 5.1 Introduction to Rational Functions 253 the numerator whereas (x + 1) does not. Loosely speaking, the trouble caused by (x − 1) in the
denominator is canceled away while the factor (x + 1) remains to cause mischief. This is why the
graph of y = h(x) has a vertical asymptote at x = −1 but only a hole at x = 1. These observations
are generalized and summarized in the theorem below, whose proof is found in Calculus.
Theorem 5.1. Location of Vertical Asymptotes and Holes:a Suppose r is a rational
(x)
function which can be written as r(x) = p(x) where p and q have no common zeros.b Let c be
q
a real number which is not in the domain of r.
(c)
If q (c) = 0, then the graph of y = r(x) has a hole at c, p(c) .
q If q (c) = 0, then the the line x = c is a vertical asymptote of the graph of y = r(x).
a
b Or, ‘How to tell your asymptote from a hole in the graph.’
In other words, r(x) is in lowest terms. In English, Theorem 5.1 says if x = c is not in the domain of r but, when we simplify r(x), it
no longer makes the denominator 0, then we have a hole at x = c. Otherwise, we have a vertical
asymptote.
Example 5.1.2. Find the vertical asymptotes of, and/or holes in, the graphs of the following
rational functions. Verify your answers using a graphing calculator.
1. f (x) = 2x
2−3
x 3. h(x) = x2 − x − 6
x2 + 9 2. g (x) = x2 − x − 6
x2 − 9 4. r(x) = x2 − x − 6
x2 + 4 x + 4 Solution.
1. To use Theorem 5.1, we ﬁrst ﬁnd all of the √ numbers which aren’t in the domain of f . To
real
do so, we solve x2 − 3 = 0 and get x = ± 3. Since the expression f (xis in lowest √
)
terms,
there is no cancellation possible, and we conclude that the lines x = − 3 and x = 3 are
vertical asymptotes to the graph of y = f (x). The calculator veriﬁes this claim.
−
x
x+2
2. Solving x2 − 9 = 0 gives x = ±3. In lowest terms g (x) = x x−−−6 = (x−3)(x+2) = x+3 . Since
29
(x 3)(x+3)
x = −3 continues to make trouble in the denominator, we know the line x = −3 is a vertical
asymptote of the graph of y = g (x). Since x = 3 no longer produces a 0 in the denominator,
we have a hole at x = 3. To ﬁnd the y coordinate of the hole, we substitute x = 3 into x+2
x+3
5
and ﬁnd the hole is at 3, 6 . When we graph y = g (x) using a calculator, we clearly see the
vertical asymptote at x = −3, but everything seems calm near x = 3.
2 254 Rational Functions The graph of y = f (x) The graph of y = g (x) 3. The domain of h is all real numbers, since x2 + 9 = 0 has no real solutions. Accordingly, the
graph of y = h(x) is devoid of both vertical asymptotes and holes.
4. Setting x2 + 4x + 4 = 0 gives us x = −2 as the only real number of concern. Simplifying,
x2 −
we see r(x) = x2 +4x−6 = (x−3)(x+2) = x−3 . Since x = −2 continues to produce a 0 in the
x+2
x+4
(x+2)2
denominator of the reduced function, we know x = −2 is a vertical asymptote to the graph,
which the calculator conﬁrms. The graph of y = h(x) The graph of y = r(x) Our next example gives us a physical interpretation of a vertical asymptote. This type of model
arises from a family of equations cheerily named ‘doomsday’ equations.6 The unfortunate name
will make sense shortly.
Example 5.1.3. A mathematical model for the population P , in thousands, of a certain species of
bacteria, t days after it is introduced to an environment is given by P (t) = (5100)2 , 0 ≤ t < 5.
−t
1. Find and interpret P (0).
2. When will the population reach 100,000?
3. Determine the behavior of P as t → 5− . Interpret this result graphically and within the
context of the problem.
6 This is a class of Calculus equations in which a population grows very rapidly. 5.1 Introduction to Rational Functions 255 Solution. 1. Substituting t = 0 gives P (0) =
into the environment. 100
(5−0)2 = 4, which means 4000 bacteria are initially introduced 2. To ﬁnd when the population reaches 100,000, we ﬁrst need to remember that P (t) is measured
in thousands. In other words, 100,000 bacteria corresponds to P (t) = 100. Substituting
for P (t) gives the equation (5100)2 = 100. Clearing denominators and dividing by 100 gives
−t
(5 − t)2 = 1, which, after extracting square roots, produces t = 4 or t = 6. Of these two
solutions, only t = 4 in in our domain, so this is the solution we keep. Hence, it takes 4 days
for the population of bacteria to reach 100,000. 3. To determine the behavior of P as t → 5− , we can make a table t P (t) 4.9 10000 4.99 1000000 4.999 100000000 4.9999 10000000000 In other words, as t → 5− , P (t) → ∞. Graphically, the line t = 5 is a vertical asymptote of
the graph of y = P (t). Physically, this means the population of bacteria is increasing without
bound as we near 5 days, which cannot physically happen. For this reason, t = 5 is called
the ‘doomsday’ for this population. There is no way any environment can support inﬁnitely
many bacteria, so shortly before t = 5 the environment would collapse. Now that we have thoroughly investigated vertical asymptotes, we now turn our attention to
horizontal asymptotes. The next theorem tells us when to expect horizontal asymptotes. 256 Rational Functions Theorem 5.2. Location of Horizontal Asymptotes: Suppose r is a rational function and
(x)
r(x) = p(x) , where p and q are polynomial functions with leading coeﬃcients a and b, respecq
tively.
If the degree of p(x) is the same as the degree of q (x), then y =
asymptote of the graph of y = r(x). a
b is thea horizontal If the degree of p(x) is less than the degree of q (x), then y = 0 is the horizontal asymptote
of the graph of y = r(x).
If the degree of p(x) is greater than the degree of q (x), then the graph of y = r(x) has no
horizontal asymptotes. a The use of the deﬁnite article will be justiﬁed momentarily. Like Theorem 5.1, Theorem 5.2 is proved using Calculus. Nevertheless, we can understand
x−
the idea behind it using our example f (x) = 2x+11 . If we interpret f (x) as a division problem,
(2x − 1) ÷ (x + 1), we ﬁnd the quotient is 2 with a remainder of −3. Using what we know about
polynomial division, speciﬁcally Theorem 4.4, we get 2x − 1 = 2(x + 1) − 3. Dividing both sides
x−
3
by (x + 1) gives 2x+11 = 2 − x+1 . (You may remember this as the formula for g (x) in Example
3
5.1.1.) As x becomes unbounded in either direction, the quantity x+1 gets closer and closer to zero
so that the values of f (x) become closer and closer to 2. In symbols, as x → ±∞, f (x) → 2, and
we have the result.7 Notice that the graph gets close to the same y value as x → −∞ or x → ∞.
This means that the graph can have only one horizontal asymptote if it is going to have one at all.
Thus we were justiﬁed in using ‘the’ in the previous theorem. (By the way, using long division to
determine the asymptote will serve us well in the next section so you might want to review that
topic.)
Alternatively, we can use what we know about end behavior of polynomials to help us understand
this theorem. From Theorem 4.2, we know the end behavior of a polynomial is determined by its
leading term. Applying this to the numerator and denominator of f (x), we get that as x → ±∞,
x−
x
f (x) = 2x+11 ≈ 2x = 2. This last approach is useful in Calculus, and, indeed, is made rigorous there.
(Keep this in mind for the remainder of this paragraph.) Applying this reasoning to the general
(x)
case, suppose r(x) = p(x) where a is the leading coeﬃcient of p(x) and b is the leading coeﬃcient
q
n ax
of q (x). As x → ±∞, r(x) ≈ bxm , where n and m are the degrees of p(x) and q (x), respectively.
If the degree of p(x) and the degree of q (x) are the same, then n = m so that r(x) ≈ a , which
b
means y = a is the horizontal asymptote in this case. If the degree of p(x) is less than the degree
b
a
of q (x), then n < m, so m − n is a positive number, and hence, r(x) ≈ bxm−n → 0 as x → ±∞. If
the degree of p(x) is greater than the degree of q (x), then n > m, and hence n − m is a positive Note that as x → −∞, f (x) → 2+ , whereas as x → ∞, f (x) → 2− . We write f (x) → 2 if we are unconcerned
from which direction the function values f (x) approach the number 2.
7 5.1 Introduction to Rational Functions 257 n−m number and r(x) ≈ ax b , which becomes unbounded as x → ±∞. As we said before, if a rational
function has a horizontal asymptote, then it will have only one. (This is not true for other types
of functions we shall see in later chapters.)
Example 5.1.4. Determine the horizontal asymptotes, if any, of the graphs of the following functions. Verify your answers using a graphing calculator.
1. f (x) = 5x
+1 x2 2. g (x) = x2 − 4
x+1 3. h(x) = 6x3 − 3x + 1
5 − 2x3 Solution.
1. The numerator of f (x) is 5x, which is degree 1. The denominator of f (x) is x2 + 1, which
is degree 2. Applying Theorem 5.2, y = 0 is the horizontal asymptote. Sure enough, as
x → ±∞, the graph of y = f (x) gets closer and closer to the xaxis.
2. The numerator of g (x), x2 − 4, is degree 2, but the degree of the denominator, x + 1, is degree
1. By Theorem 5.2, there is no horizontal asymptote. From the graph, we see the graph of
y = g (x) doesn’t appear to level oﬀ to a constant value, so there is no horizontal asymptote.8
3. The degrees of the numerator and denominator of h(x) are both three, so Theorem 5.2 tells
6
us y = −2 = −3 is the horizontal asymptote. The calculator conﬁrms this. The graph of y = f (x) The graph of y = g (x) The graph of y = h(x) Our last example of the section gives us a realworld application of a horizontal asymptote.
Though the population below is more accurately modeled with the functions in Chapter 7, we
approximate it9 using a rational function.
Example 5.1.5. The number of students, N , at local college who have had the ﬂu t months after
450
the semester begins can be modeled by the formula N (t) = 500 − 1+3t for t ≥ 0.
1. Find and interpret N (0).
8
The graph does, however, seem to resemble a nonconstant line as x → ±∞. We will discuss this phenomenon
in the next section.
9
Using techniques you’ll see in Calculus. 258 Rational Functions 2. How long will it take until 300 students will have had the ﬂu?
3. Determine the behavior of N as t → ∞. Interpret this result graphically and within the
context of the problem.
Solution.
450
1. N (0) = 500 − 1+3(0) = 50. This means that at the beginning of the semester, 50 students
have had the ﬂu.
450
450
2. We set N (t) = 300 to get 500 − 1+3t = 300 and solve. Isolating the fraction gives 1+3t = 200.
5
Clearing denominators gives 450 = 200(1 + 3t). Finally, we get t = 12 . This means it will
5
take 12 months, or about 13 days, for 300 students to have had the ﬂu. 3. To determine the behavior of N as t → ∞, we can use a table.
t N (t) 10 ≈ 485.48 100 ≈ 498.50 1000 ≈ 499.85 10000 ≈ 499.98 The table suggests that as t → ∞, N (t) → 500. (More speciﬁcally, 500− .) This means as
time goes by, only a total of 500 students will have ever had the ﬂu. 5.1 Introduction to Rational Functions 5.1.1 259 Exercises 1. For each rational function f given below:
Find the domain of f .
Identify any vertical asymptotes of the graph of y = f (x) and describe the behavior of
the graph near them using proper notation.
Identify any holes in the graph.
Find the horizontal asymptote, if it exists, and describe the end behavior of f using
proper notation. x
3x − 6
3 + 7x
(b) f (x) =
5 − 2x
x
(c) f (x) = 2
x + x − 12
x
(d) f (x) = 2
x +1
(a) f (x) = x+7
(x + 3)2
x3 + 1
(f) f (x) = 2
x −1
4x
(g) f (x) = 2
x +4
4x
(h) f (x) = 2
x −4 x2 − x − 12
x2 + x − 6
3x2 − 5x − 2
(j) f (x) =
x2 − 9
3 + 2 x2 + x
x
(k) f (x) = 2
x −x−2 (e) f (x) = (i) f (x) = 2. In Exercise 11 in Section 2.2, the population of Sasquatch in Portage County was modeled
by the function P (t) = t150t , where t = 0 represents the year 1803. Find the horizontal
+15
asymptote of the graph of y = P (t) and explain what it means.
3. In Exercise 8 in Section 4.1, we ﬁt a few polynomial models to the following electric circuit
data. (The circuit was built with a variable resistor. For each of the following resistance values
(measured in kiloohms, k Ω), the corresponding power to the load (measured in milliwatts,
mW ) is given in the table below.)10
Resistance: (k Ω) 1.012 2.199 3.275 4.676 6.805 9.975 Power: (mW ) 1.063 1.496 1.610 1.613 1.505 1.314 Using some fundamental laws of circuit analysis mixed with a healthy dose of algebra, we can
25
derive the actual formula relating power to resistance. For this circuit, it is P (x) = (x+3x9)2 ,
.
where x is the resistance value, x ≥ 0.
(a) Graph the data along with the function y = P (x) on your calculator.
(b) Approximate the maximum power that can be delivered to the load. What is the corresponding resistance value?
(c) Find and interpret the end behavior of P (x) as x → ∞.
10 The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem
and generating the accompanying data set. 260 Rational Functions 4. In his now famous 1919 dissertation The Learning Curve Equation, Louis Leon Thurstone
presents a rational function which models the number of words a person can type in four
minutes as a function of the number of pages of practice one has completed. (This paper,
which is now in the public domain and can be found here, is from a bygone era when students
at business schools took typing classes on manual typewriters.) Using his original notation
L X +P
and original language, we have Y = (X(+P )+) where L is the predicted practice limit in terms
R
of speed units, X is pages written, Y is writing speed in terms of words in four minutes, P is
equivalent previous practice in terms of pages and R is the rate of learning. In Figure 5 of the
X +19)
paper, he graphs a scatter plot and the curve Y = 216(+148 . Discuss this equation with your
X
classmates. How would you update the notation? Explain what the horizontal asymptote of
the graph means. You should take some time to look at the original paper. Skip over the
computations you don’t understand yet and try to get a sense of the time and place in which
the study was conducted. 5.1 Introduction to Rational Functions 5.1.2 Answers
x
3x − 6
Domain: (−∞, 2) ∪ (2, ∞)
Vertical asymptote: x = 2
As x → 2− , f (x) → −∞
As x → 2+ , f (x) → ∞
No holes in the graph
Horizontal asymptote: y =
1−
As x → −∞, f (x) → 3
+
As x → ∞, f (x) → 1
3 1
3 3 + 7x
5 − 2x
Domain: (−∞, 5 ) ∪ ( 5 , ∞)
2
2
Vertical asymptote: x = 5
2
−
As x → 5 , f (x) → ∞
2
+
As x → 5 , f (x) → −∞
2
No holes in the graph
7
Horizontal asymptote: y = − 2
+
As x → −∞, f (x) → − 7
2
−
As x → ∞, f (x) → − 7
2 (b) f (x) = x
x
=
+ x − 12
(x + 4)(x − 3)
Domain: (−∞, −4) ∪ (−4, 3) ∪ (3, ∞)
Vertical asymptotes: x = −4, x = 3
As x → −4− , f (x) → −∞
As x → −4+ , f (x) → ∞
As x → 3− , f (x) → −∞
As x → 3+ , f (x) → ∞
No holes in the graph
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (c) f (x) = x2 x
+1
Domain: (−∞, ∞)
No vertical asymptotes
No holes in the graph
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (d) f (x) = 1. (a) f (x) = 11 261 x2 x+7
(x + 3)2
Domain: (−∞, −3) ∪ (−3, ∞)
Vertical asymptote: x = −3
As x → −3− , f (x) → ∞
As x → −3+ , f (x) → ∞
No holes in the graph
Horizontal asymptote: y = 0
11 As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (e) f (x) = x3 + 1
x2 − x + 1
=
x2 − 1
x−1
Domain: (−∞, −1) ∪ (−1, 1) ∪ (1, ∞)
Vertical asymptote: x = 1
As x → 1− , f (x) → −∞
As x → 1+ , f (x) → ∞
Hole at (−1, − 3 )
2
No horizontal asymptote
As x → −∞, f (x) → −∞
As x → ∞, f (x) → ∞ (f) f (x) = 4x
x2 + 4
Domain: (−∞, ∞)
No vertical asymptotes
No holes in the graph
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (g) f (x) = This is hard to see on the calculator, but trust me, the graph is below the xaxis to the left of x = −7. 262 Rational Functions
4x
4x
=
x2 − 4
(x + 2)(x − 2)
Domain: (−∞, −2) ∪ (−2, 2) ∪ (2, ∞)
Vertical asymptotes: x = −2, x = 2
As x → −2− , f (x) → −∞
As x → −2+ , f (x) → ∞
As x → 2− , f (x) → −∞
As x → 2+ , f (x) → ∞
No holes in the graph
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (h) f (x) = x2 − x − 12
x−4
=
x2 + x − 6
x−2
Domain: (−∞, −3) ∪ (−3, 2) ∪ (2, ∞)
Vertical asymptote: x = 2
As x → 2− , f (x) → ∞
As x → 2+ , f (x) → −∞
7
Hole at −3, 5
Horizontal asymptote: y = 1
As x → −∞, f (x) → 1+
As x → ∞, f (x) → 1− (i) f (x) = (3x + 1)(x − 2)
3x2 − 5x − 2
=
2−9
x
(x + 3)(x − 3)
Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
Vertical asymptotes: x = −3, x = 3
As x → −3− , f (x) → ∞
As x → −3+ , f (x) → −∞
As x → 3− , f (x) → −∞
As x → 3+ , f (x) → ∞
No holes in the graph
Horizontal asymptote: y = 3
As x → −∞, f (x) → 3+
As x → ∞, f (x) → 3− (j) f (x) = x3 + 2 x2 + x
x(x + 1)
=
2−x−2
x
x−2
Domain: (−∞, −1) ∪ (−1, 2) ∪ (2, ∞)
Vertical asymptote: x = 2
As x → 2− , f (x) → −∞
As x → 2+ , f (x) → −∞
Hole at (−1, 0)
No horizontal asymptote
As x → −∞, f (x) → −∞
As x → ∞, f (x) → ∞ (k) f (x) = 2. The horizontal asymptote of the graph of P (t) = t150t is y = 150 and it means that the model
+15
predicts the population of Sasquatch in Portage County will never exceed 150. 3. (a)
(b) The maximum power is approximately 1.603 mW which corresponds to 3.9 k Ω.
(c) As x → ∞, P (x) → 0+ which means as the resistance increases without bound, the
power diminishes to zero. 5.2 Graphs of Rational Functions 5.2 263 Graphs of Rational Functions In this section, we take a closer look at graphing rational functions. In Section 5.1, we learned that
the graphs of rational functions may include vertical asymptotes, holes in the graph, and horizontal
asymptotes. Theorems 5.1 and 5.2 tell us exactly when and where these behaviors will occur, and
if we combine these results with what we already know about graphing functions, we will quickly
be able to generate reasonable graphs of rational functions.
One of the standard tools we will use is the sign diagram which was ﬁrst introduced in Section
3.4, and then revisited in Section 4.1. In those sections, we operated under the belief that a function
couldn’t change its sign without its graph crossing through the xaxis. The major theorem we
used to justify this belief was the Intermediate Value Theorem, Theorem 4.1. It turns out the
Intermediate Value Theorem applies to all continuous functions,1 not just polynomials. Although
rational functions are continuous on their domains,2 Theorem 5.1 tells us vertical asymptotes and
holes occur at the values excluded from their domains. In other words, rational functions aren’t
continuous at these excluded values which leaves open the possibility that the function could change
sign without crossing through the xaxis. Consider the graph of y = h(x) from Example 5.1.1,
1
recorded below for convenience. We have added its xintercept at 2 , 0 for the discussion that
follows. Suppose we wish to construct a sign diagram for h(x). Recall that the intervals where
h(x) > 0, or (+), correspond to the xvalues where the graph of y = h(x) is above the xaxis; the
intervals on which h(x) < 0, or (−) correspond to where the graph is below the xaxis.
y 8
7
6
5
4
3 (+) (−) 0 (+) (+)
1 −4 −3 −2
−1 −1
11
2 2 3 4 1
2 1 x −2
−3
−4
−5
−6 As we examine the graph of y = h(x), reading from left to right, we note that from (−∞, −1),
1
2 Recall that, for our purposes, this means the graphs are devoid of any breaks, jumps or holes
Another result from Calculus. 264 Rational Functions the graph is above the xaxis, so h(x) is (+) there. At x = −1, we have a vertical asymptote, at
which point the graph ‘jumps’ across the xaxis. On the interval −1, 1 , the graph is below the
2
xaxis, so h(x) is (−) there. The graph crosses through the xaxis at 1 , 0 and remains above the
2
xaxis until x = 1, where we have a ‘hole’ in the graph. Since h(1) is undeﬁned, there is no sign
1
here. So we have h(x) as (+) on the interval 2 , 1 . Continuing, we see that on (1, ∞), the graph
of y = h(x) is above the xaxis, and so we mark (+) there. To construct a sign diagram from this
information, we not only need to denote the zero of h, but also the places not in the domain of
1
h. As is our custom, we write ‘0’ above 2 on the sign diagram to remind us that it is a zero of h.
We need a diﬀerent notation for −1 and 1, and we have chosen to use ‘’  a nonstandard symbol
called the interrobang. We use this symbol to convey a sense of surprise, caution, and wonderment
 an appropriate attitude to take when approaching these points. The moral of the story is that
when constructing sign diagrams for rational functions, we include the zeros as well as the values
excluded from the domain. Steps for Constructing a Sign Diagram for a Rational Function
Suppose r is a rational function.
1. Place any values excluded from the domain of r on the number line with an ‘’ above
them.
2. Find the zeros of r and place them on the number line with the number 0 above them.
3. Choose a test value in each of the intervals determined in steps 1 and 2.
4. Determine the sign of r(x) for each test value in step 3, and write that sign above the
corresponding interval. We now present our procedure for graphing rational functions and apply it to a few exhaustive
examples. Please note that we decrease the amount of detail given in the explanations as we move
through the examples. The reader should be able to ﬁll in any details in those steps which we have
abbreviated. 5.2 Graphs of Rational Functions 265 Steps for Graphing Rational Functions
Suppose r is a rational function.
1. Find the domain of r.
2. Reduce r(x) to lowest terms, if applicable.
3. Find the x and y intercepts of the graph of y = r(x), if they exist.
4. Determine the location of any vertical asymptotes or holes in the graph, if they exist.
Analyze the behavior of r on either side of the vertical asymptotes, if applicable.
5. Analyze the end behavior of r. Use long division, as needed.
6. Use a sign diagram and plot additional points, as needed, to sketch the graph of y = r(x). 3x
.
−4
Solution. We follow the six step procedure outlined above. Example 5.2.1. Sketch a detailed graph of f (x) = x2 1. As usual, we set the denominator equal to zero to get x2 − 4 = 0. We ﬁnd x = ±2, so our
domain is (−∞, −2) ∪ (−2, 2) ∪ (2, ∞).
2. To reduce f (x) to lowest terms, we factor the numerator and denominator which yields
3x
f (x) = (x−2)(x+2) . There are no common factors which means f (x) is already in lowest
terms.
3x
3. To ﬁnd the xintercepts of the graph of y = f (x), we set y = f (x) = 0. Solving (x−2)(x+2) = 0
results in x = 0. Since x = 0 is in our domain, (0, 0) is the xintercept. To ﬁnd the y intercept,
we set x = 0 and ﬁnd y = f (0) = 0, so that (0, 0) is our y intercept as well.3 4. The two numbers excluded from the domain of f are x = −2 and x = 2. Since f (x) didn’t
reduce at all, both of these values of x still cause trouble in the denominator, and so, by
Theorem 5.1, x = −2 and x = 2 are vertical asymptotes of the graph. We can actually go
a step farther at this point and determine exactly how the graph approaches the asymptote
near each of these values. Though not absolutely necessary,4 it is good practice for those
heading oﬀ to Calculus. For the discussion that follows, it is best to use the factored form of
3x
f (x) = (x−2)(x+2) .
3
As we mentioned at least once earlier, since functions can have at most one y intercept, once we ﬁnd (0, 0) is on
the graph, we know it is the y intercept.
4
The sign diagram in step 6 will also determine the behavior near the vertical asymptotes. 266 Rational Functions
The behavior of y = f (x) as x → −2: Suppose x → −2− . If we were to build a table of
values, we’d use xvalues a little less than −2, say −2.1, −2.01 and −2.001. While there
is no harm in actually building a table like we did in Section 5.1, we want to develop a
‘number sense’ here. Let’s think about each factor in the formula of f (x) as we imagine
substituting a number like x = −2.000001 into f (x). The quantity 3x would be very
close to −6, the quantity (x − 2) would be very close to −4, and the factor (x + 2) would
be very close to 0. More speciﬁcally, (x + 2) would be a little less than 0, in this case,
−0.000001. We will call such a number a ‘very small (−)’, ‘very small’ meaning close to
zero in absolute value. So, mentally, as x → −2− , we estimate f (x) = 3x
−6
3
≈
=
(x − 2)(x + 2)
(−4) (very small (−))
2 (very small (−)) Now, the closer x gets to −2, the smaller (x + 2) will become, and so even though we
are multiplying our ‘very small (−)’ by 2, the denominator will continue to get smaller
and smaller, and remain negative. The result is a fraction whose numerator is positive,
but whose denominator is very small and negative. Mentally,
f (x) ≈ 3
3
≈
≈ very big (−)
2 (very small (−))
very small (−) The term ‘very big (−)’ means a number with a large absolute value which is negative.5
What all of this means is that as x → −2− , f (x) → −∞. Now suppose we wanted to
determine the behavior of f (x) as x → −2+ . If we imagine substituting something a
little larger than −2 in for x, say −1.999999, we mentally estimate
f (x) ≈ −6
3
3
=
≈
≈ very big (+)
(−4) (very small (+))
2 (very small (+))
very small (+) We conclude that as x → −2+ , f (x) → ∞.
The behavior of y = f (x) as x → 2: Consider x → 2− . We imagine substituting
x = 1.999999. Approximating f (x) as we did above, we get f (x) ≈ 6
3
3
=
≈
≈ very big (−)
(very small (−)) (4)
2 (very small (−))
very small (−) We conclude that as x → 2− , f (x) → −∞. Similarly, as x → 2+ , we imagine substituting
3
x = 2.000001, we get f (x) ≈ very small (+) ≈ very big (+). So as x → 2+ , f (x) → ∞.
Graphically, we have that near x = −2 and x = 2 the graph of y = f (x) looks like6
5 The actual retail value of f (−2.000001) is approximately −1,500,000.
We have deliberately left oﬀ the labels on the y axis because we know only the behavior near x = ±2, not the
actual function values.
6 5.2 Graphs of Rational Functions 267
y −3 −1 1 3 x 5. Next, we determine the end behavior of the graph of y = f (x). Since the degree of the
numerator is 1, and the degree of the denominator is 2, Theorem 5.2 tells us that y = 0
is the horizontal asymptote. As with the vertical asymptotes, we can glean more detailed
3x
information using ‘number sense’. For the discussion below, we use the formula f (x) = x2 −4 .
The behavior of y = f (x) as x → −∞: If we were to make a table of values to discuss
the behavior of f as x → −∞, we would substitute very ‘large’ negative numbers in for
x, say, for example, x = −1 billion. The numerator 3x would then be −3 billion, whereas
the denominator x2 − 4 would be (−1 billion)2 − 4, which is pretty much the same as
1(billion)2 . Hence, f (−1 billion) ≈ −3 billion
3
≈−
≈ very small (−)
2
1(billion)
billion Notice that if we substituted in x = −1 trillion, essentially the same kind of cancellation
would happen, and we would be left with an even ‘smaller’ negative number. This not
only conﬁrms the fact that as x → −∞, f (x) → 0, it tells us that f (x) → 0− . In other
words, the graph of y = f (x) is a little bit below the xaxis as we move to the far left.
The behavior of y = f (x) as x → ∞: On the ﬂip side, we can imagine substituting very
large positive numbers in for x and looking at the behavior of f (x). For example, let
x = 1 billion. Proceeding as before, we get f (1 billion) ≈ 3 billion
3
≈
≈ very small (+)
2
1(billion)
billion The larger the number we put in, the smaller the positive number we would get out. In
other words, as x → ∞, f (x) → 0+ , so the graph of y = f (x) is a little bit above the
xaxis as we look toward the far right.
Graphically, we have7
7 As with the vertical asymptotes in the previous step, we know only the behavior of the graph as x → ±∞. For
that reason, we provide no xaxis labels. 268 Rational Functions
y
1 x
−1 6. Lastly, we construct a sign diagram for f (x). The xvalues excluded from the domain of f
are x = ±2, and the only zero of f is x = 0. Displaying these appropriately on the number
line gives us four test intervals, and we choose the test values8 we x = −3, x = −1, x = 1,
and x = 3. We ﬁnd f (−3) is (−), f (−1) is (+), f (1) is (−), and f (3) is (+). Combining this
with our previous work, we get the graph of y = f (x) below.
y
3
2 (−) (+) 0 (−) (+) 1 −2
0
2
−3 −1
1
3 −5 −4 −3 −1 1 3 4 5 x −1
−2
−3 A couple of notes are in order. First, the graph of y = f (x) certainly seems to possess symmetry
with respect to the origin. In fact, we can check f (−x) = −f (x) to see that f is an odd function.
In some textbooks, checking for symmetry is part of the standard procedure for graphing rational
functions; but since it happens comparatively rarely9 we’ll just point it out when we see it. Also
note that while y = 0 is the horizontal asymptote, the graph of f nevertheless crosses the xaxis
at (0, 0). The myth that graphs of rational functions can’t cross their horizontal asymptotes is
completely false, as we shall see again in our next example.
Example 5.2.2. Sketch a detailed graph of g (x) = 2x2 − 3x − 5
.
x2 − x − 6 Solution.
1. Setting x2 − x − 6 = 0 gives x = −2 and x = 3. Our domain is (−∞, −2) ∪ (−2, 3) ∪ (3, ∞).
2. Factoring g (x) gives g (x) = (2x−5)(x+1)
(x−3)(x+2) . There is no cancellation, so g (x) is in lowest terms. 8
In this particular case, we can eschew test values, since our analysis of the behavior of f near the vertical
asymptotes and our end behavior analysis have given us the signs on each of the test intervals. In general, however,
this won’t always be the case, so for demonstration purposes, we continue with our usual construction.
9
And Jeﬀ doesn’t think much of it to begin with... 5.2 Graphs of Rational Functions 269 3. To ﬁnd the xintercept we set y = g (x) = 0. Using the factored form of g (x) above, we ﬁnd
the zeros to be the solutions of (2x − 5)(x + 1) = 0. We obtain x = 5 and x = −1. Since
2
both of these numbers are in the domain of g , we have two xintercepts, 5 , 0 and (−1, 0).
2
To ﬁnd the y intercept, we set x = 0 and ﬁnd y = g (0) = 5 , so our y intercept is 0, 5 .
6
6
4. Since g (x) was given to us in lowest terms, we have, once again by Theorem 5.1 vertical
x−5)(x+1)
asymptotes x = −2 and x = 3. Keeping in mind g (x) = (2x−3)(x+2) , we proceed to our
(
analysis near each of these values.
The behavior of y = g (x) as x → −2: As x → −2− , we imagine substituting a number
a little bit less than −2. We have g (x) ≈ (−9)(−1)
9
≈
≈ very big (+)
(−5)(very small (−))
very small (+) so as x → −2− , g (x) → ∞. On the ﬂip side, as x → −2+ , we get
g (x) ≈ 9
≈ very big (−)
very small (−) so g (x) → −∞.
The behavior of y = g (x) as x → 3: As x → 3− , we imagine plugging in a number just
shy of 3. We have g (x) ≈ (1)(4)
4
≈
≈ very big (−)
( very small (−))(5)
very small (−) Hence, as x → 3− , g (x) → −∞. As x → 3+ , we get
g (x) ≈ 4
≈ very big (+)
very small (+) so g (x) → ∞.
Graphically, we have (again, without labels on the y axis)
y −3 −1 1 2 4 x 270 Rational Functions 5. Since the degrees of the numerator and denominator of g (x) are the same, we know from
Theorem 5.2 that we can ﬁnd the horizontal asymptote of the graph of g by taking the
2
ratio of the leading terms coeﬃcients, y = 1 = 2. However, if we take the time to do a
more detailed analysis, we will be able to reveal some ‘hidden’ behavior which would be lost
otherwise.10 As in the discussion following Theorem 5.2, we use the result of the long division
2
2x2 − 3x − 5 ÷ x2 − x − 6 to rewrite g (x) = 2x2 −3x−5 as g (x) = 2 − x2x−7 6 . We focus our
x −x−6
−x−
attention on the term x2x−7 6 .
− x−
The behavior of y = g (x) as x → −∞: If imagine substituting x = −1 billion into
x−7
, we estimate x2x−7 6 ≈ −1 billion ≈ very small (−).11 Hence,
x2 −x−6
−x−
1billion2 g (x) = 2 − x2 x−7
≈ 2 − very small (−) = 2 + very small (+)
−x−6 In other words, as x → −∞, the graph of y = g (x) is a little bit above the line y = 2.
The behavior of y = g (x) as x → ∞. To consider x2x−7 6 as x → ∞, we imagine
−x−
substituting x = 1 billion and, going through the usual mental routine, ﬁnd x−7
≈ very small (+)
x2 − x − 6
Hence, g (x) ≈ 2 − very small (+), in other words, the graph of y = g (x) is just below
the line y = 2 as x → ∞.
On y = g (x), we have (again, without labels on the xaxis)
y 1 x
−1 6. Finally we construct our sign diagram. We place an ‘’ above x = −2 and x = 3, and a ‘0’
above x = 5 and x = −1. Choosing test values in the test intervals gives us f (x) is (+) on
2
5
5
the intervals (−∞, −2), −1, 2 , and (3, ∞), and (−) on the intervals (−2, −1) and 2 , 3 .
As we piece together all of the information, we note that the graph must cross the horizontal
asymptote at some point after x = 3 in order for it to approach y = 2 from underneath. This
is the subtlety that we would have missed had we skipped the long division and subsequent
end behavior analysis. We can, in fact, ﬁnd exactly when the graph crosses y = 2. As a result
10 That is, if you use a calculator to graph. Once again, Calculus is the ultimate graphing power tool.
In the denominator, we would have (1billion)2 − 1billion − 6. It’s easy to see why the 6 is insigniﬁcant, but to
ignore the 1 billion seems criminal. However, compared to (1 billion)2 , it’s on the insigniﬁcant side; it’s 1018 versus
109 . We are once again using the fact that for polynomials, end behavior is determined by the leading term, so in
the denominator, the x2 term wins out over the x term.
11 5.2 Graphs of Rational Functions 271 of the long division, we have g (x) = 2 − x2x−7 6 . For g (x) = 2, we would need x2x−7 6 = 0.
−x−
−x−
This gives x − 7 = 0, or x = 7. Note that x − 7 is the remainder when 2x2 − 3x − 5 is divided
by x2 − x − 6, and so it makes sense that for g (x) to equal the quotient 2, the remainder from
the division must be 0. Sure enough, we ﬁnd g (7) = 2. Moreover, it stands to reason that g
must attain a relative minimum at some point past x = 7. Calculus veriﬁes that at x = 13,
we have such a minimum at exactly (13, 1.96). The reader is challenged to ﬁnd calculator
windows which show the graph crossing its horizontal asymptote on one window, and the
relative minimum in the other.
y
8
7
6
5
4 (+) (−) 0 (+) 0 (−) (+)
−2 −1 5
2 3 3 1
−9 −8 −7 −6 −5 −4 −3 −1
−1 1 2 4 5 6 7 8 9 x −2
−3
−4 Our next example gives us not only a hole in the graph, but also some slightly diﬀerent end
behavior.
Example 5.2.3. Sketch a detailed graph of h(x) = 2x3 + 5 x2 + 4 x + 1
.
x2 + 3 x + 2 Solution.
1. For domain, you know the drill. Solving x2 + 3x + 2 = 0 gives x = −2 and x = −1. Our
answer is (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞).
2. To reduce h(x), we need to factor the numerator and denominator. We get
1 h(x) = 2x3 5 x2 1)2 2
!
+
+ 4x + 1
(2x + 1)(x +
(2x + 1)(x + 1)¡
(2x + 1)(x + 1)
=
=
$
$=
2 + 3x + 2
x
(x + 2)(x + 1)
(x + 2)$$ 1)
(x +
x+2 We will use this reduced formula for h(x) as long as we’re not substituting x = −1. To make
this exclusion speciﬁc, we write h(x) = (2x+1)(x+1) , x = −1.
x+2
3. To ﬁnd the xintercepts, as usual, we set h(x) = 0 and solve. Solving (2x+1)(x+1) = 0 yields
x+2
x = − 1 and x = −1. The latter isn’t in the domain of h, so we exclude it. Our only x2
1
intercept is − 2 , 0 . To ﬁnd the y intercept, we set x = 0. Since 0 = −1, we can use the
reduced formula for h(x) and we get h(0) = 1 for a y intercept of 0, 1 .
2
2 272 Rational Functions 4. From Theorem 5.1, we know that since x = −2 still poses a threat in the denominator of
the reduced function, we have a vertical asymptote there. As for x = −1, we note the factor
(x + 1) was canceled from the denominator when we reduced h(x), and so it no longer causes
trouble there. This means we get a hole when x = −1. To ﬁnd the y coordinate of the hole,
we substitute x = −1 into (2x+1)(x+1) , per Theorem 5.1 and get 0. Hence, we have a hole on
x+2
the xaxis at (−1, 0). It should make you uncomfortable plugging x = −1 into the reduced
formula for h(x), especially since we’ve made such a big deal concerning the stipulation about
not letting x = −1 for that formula. What we are really doing is taking a Calculus shortcut
to the more detailed kind of analysis near x = −1 which we will show below. Speaking of
which, for the discussion that follows, we will use the formula h(x) = (2x+1)(x+1) , x = −1.
x+2
The behavior of y = h(x) as x → −2: As x → −2− , we imagine substituting a number
(−3)(−
3
a little bit less than −2. We have h(x) ≈ (very small1)−)) ≈ (very small (−)) ≈ very big (−)
(
and so as x → −2− , h(x) → −∞. On the other side of −2, as x → −2+ , we ﬁnd that
3
h(x) ≈ very small (+) ≈ very big (+), so h(x) → ∞.
The behavior of y = h(x) as x → −1. As x → −1− , we imagine plugging in a number
a bit less than x = −1. We have h(x) ≈ (−1)(very 1small (−)) = very small (+) Hence, as
x → −1− , h(x) → 0+ . This means, as x → −1− , the graph is a bit above the point
(−1, 0). As x → −1+ , we get h(x) ≈ (−1)(very 1small (+)) = very small (−). This gives us
that as x → −1+ , h(x) → 0− , so the graph is a little bit lower than (−1, 0) here. Graphically, we have
y −3 x 5. For end behavior, we note that the degree of the numerator of h(x), 2x3 + 5x2 + 4x + 1 is 3,
and the degree of the denominator, x2 + 3x + 2, is 2. Theorem 5.2 is of no help here, since the
degree of the numerator is greater than the degree of the denominator. That won’t stop us,
however, in our analysis. Since for end behavior we are considering values of x as x → ±∞,
we are far enough away from x = −1 to use the reduced formula, h(x) = (2x+1)(x+1) , x = −1.
x+2
To perform long division, we multiply out the numerator and get h(x) = 2x2 +3x+1
,
x+2 x = −1, 5.2 Graphs of Rational Functions 273 3
and, as a result, we rewrite h(x) = 2x − 1 + x+2 , x = −1. As in the previous example, we
3
focus our attention on the term generated from the remainder, x+2 . 3
The behavior of y = h(x) as x → −∞: Substituting x = −1 billion into x+2 , we get the
3
3
estimate −1 billion ≈ very small (−). Hence, h(x) = 2x−1+ x+2 ≈ 2x−1+very small (−).
This means the graph of y = h(x) is a little bit below the line y = 2x − 1 as x → −∞.
3
The behavior of y = h(x) as x → ∞: If x → ∞, then x+2 ≈ very small (+). This means
h(x) ≈ 2x − 1 + very small (+), or that the graph of y = h(x) is a little bit above the
line y = 2x − 1 as x → ∞. This is end behavior unlike any we’ve ever seen. Instead of approaching a horizontal line,
the graph is approaching a slanted line. For this reason, y = 2x − 1 is called a slant
asymptote12 of the graph of y = h(x). A slant asymptote will always arise when the degree
of the numerator is exactly one more than the degree of the denominator, and there’s no way
to determine exactly what it is without going through the long division. Graphically we have y
4
3
2
1 x
−1
−2
−3
−4 6. To make our sign diagram, we place an ‘’ above x = −2 and x = −1 and a ‘0’ above x = − 1 .
2
On our four test intervals, we ﬁnd h(x) is (+) on (−2, −1) and − 1 , ∞ and h(x) is (−) on
2
1
(−∞, −2) and −1, − 2 . Putting all of our work together yields the graph below.
12 Also called an ‘oblique’ asymptote in some texts. 274 Rational Functions
y
9
8
7
6
5
4
3
2
1 (−) (+)
−2 −1 (−) 0 (+) −4 −3 1
−2 −1 −1 1 2 3 4 x −2
−3
−4
−5
−6
−7
−8
−9
−10
−11
−12
−13
−14 We could ask whether the graph of y = h(x) crosses its slant asymptote. From the formula
3
3
h(x) = 2x − 1 + x+2 , x = −1, we see that if h(x) = 2x − 1, we would have x+2 = 0. Since this will
13
never happen, we conclude the graph never crosses its slant asymptote.
We end this section with an example that shows it’s not all pathological weirdness when it
comes to rational functions and technology still has a role to play in studying their graphs at this
level.
Example 5.2.4. Sketch the graph of r(x) = x4 + 1
.
x2 + 1 Solution.
1. The denominator x2 + 1 is never zero so the domain is (−∞, ∞).
2. With no real zeros in the denominator, x2 + 1 is an irreducible quadratic. Our only hope of
reducing r(x) is if x2 + 1 is a factor of x4 + 1. Performing long division gives us
x4 + 1
2
= x2 − 1 + 2
2+1
x
x +1
The remainder is not zero so r(x) is already reduced.
3. To ﬁnd the xintercept, we’d set r(x) = 0. Since there are no real solutions to
have no xintercepts. Since r(0) = 1, so we get (0, 1) for the y intercept.
4. This step doesn’t apply to r, since its domain is all real numbers.
13 But rest assured, some graphs do! x4 +1
x2 +1 = 0, we 5.2 Graphs of Rational Functions 275 5. For end behavior, once again, since the degree of the numerator is greater than that of the
denominator, Theorem 5.2 doesn’t apply. We know from our attempt to reduce r(x) that we
can rewrite r(x) = x2 − 1 + x22 , and so we focus our attention on the term corresponding
+1
to the remainder, x22 It should be clear that as x → ±∞, x22 ≈ very small (+), which
+1
+1
means r(x) ≈ x2 − 1 + very small (+). So the graph y = r(x) is a little bit above the graph
of the parabola y = x2 − 1 as x → ±∞. Graphically,
y
5
4
3
2
1 x 6. There isn’t much work to do for a sign diagram for r(x), since its domain is all real numbers
and it has no zeros. Our sole test interval is (−∞, ∞), and since we know r(0) = 1, we
conclude r(x) is (+) for all real numbers. At this point, we don’t have much to go on for
a graph. Below is a comparison of what we have determined analytically versus what the
calculator shows us. We have no way to detect the relative extrema analytically14 apart from
brute force plotting of points, which is done more eﬃciently by the calculator.
6 y 5
4
3
2
1 −3 14 −1 1 Without appealing to Calculus, of course. 2 3
x 276 5.2.1 Rational Functions Exercises 1. Find the slant asymptote of the graph of the rational function.
x3 − 3x + 1
x2 + 1
2 + 5x − 3
2x
(b) f (x) =
3x + 2
(a) f (x) = −5x4 − 3x3 + x2 − 10
x3 − 3x2 + 3x − 1
−x3 + 4x
(d) f (x) =
x2 − 9
(c) f (x) = 2. Use the sixstep procedure to graph each rational function given. Be sure to draw any
asymptotes as dashed lines.
(a) f (x) =
(b) f (x) =
(c) f (x) =
(d) f (x) =
(e) f (x) =
(f) f (x) =
(g) f (x) = 4
x+2
5x
6 − 2x
1
x2
1
2 + x − 12
x
2x − 1
−2x2 − 5x + 3
x
2 + x − 12
x
4x
x2 + 4 (h) f (x) = 4x
x2 − 4 (i) f (x) = x2 − x − 12
x2 + x − 6 (j) f (x) = 3x2 − 5x − 2
x2 − 9 (k) f (x) = x3 + 2 x2 + x
x2 − x − 2 (l) f (x) = −x3 + 4x
x2 − 9 (m) 15 f (x) = x2 − 2x + 1
x3 + x2 − 2x 3. Example 5.2.4 showed us that the sixstep procedure cannot tell us everything of importance
about the graph of a rational function. Without Calculus, we need to use our graphing
calculators to reveal the hidden mysteries of rational function behavior. Working with your
classmates, use a graphing calculator to examine the graphs of the following rational functions.
Compare and contrast their features. Which features can the sixstep process reveal and which
features cannot be detected by it?
(a) f (x) =
(b) f (x) = 15 x2 1
+1 x2 x
+1 x2
x2 + 1
x3
(d) f (x) = 2
x +1
(c) f (x) = Once you’ve done the sixstep procedure, use your calculator to graph this function on the viewing window
[0, 12] × [0, 0.25]. What do you see? 5.2 Graphs of Rational Functions 277 4. Graph the following rational functions by applying transformations to the graph of y =
1
x−2
3
(b) g (x) = 1 −
x
(a) f (x) = 1
.
x −2x + 1
(Hint: Divide)
x
3x − 7
(d) j (x) =
(Hint: Long division)
x−2
(c) h(x) = ax + b
. What restrictions must
cx + d
1
be placed on a, b, c and d so that the graph is indeed a transformation of y = ?
x
Discuss with your classmates how you would graph f (x) = 3 5. In Example 4.1.1 in Section 4.1 we showed that p(x) = 4x+x is not a polynomial even though
x
its formula reduced to 4 + x2 for x = 0. However, it is a rational function similar to those
studied in the section. With the help of your classmates, graph p(x).
x4 − 8x3 + 24x2 − 72x + 135
. With the help of your classmates, ﬁnd the x and
x3 − 9x2 + 15x − 7
y  intercepts of the graph of g . Find the intervals on which the function is increasing, the
intervals on which it is decreasing and the local extrema. Find all of the asymptotes of the
graph of g and any holes in the graph, if they exist. Be sure to show all of your work including
any polynomial or synthetic division. Sketch the graph of g , using more than one picture if
necessary to show all of the important features of the graph. 6. Let g (x) = 278 5.2.2 Rational Functions Answers
(c) y = −5x − 18 1. (a) y = x
2
(b) y = 3 x + 11
9 4
x+2
Domain: (−∞, −2) ∪ (−2, ∞)
No xintercepts
y intercept: (0, 2)
Vertical asymptote: x = −2
As x → −2− , f (x) → −∞
As x → −2+ , f (x) → ∞
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (d) y = −x 2. (a) f (x) = 5x
6 − 2x
Domain: (−∞, 3) ∪ (3, ∞)
xintercept: (0, 0)
y intercept: (0, 0)
Vertical asymptote: x = 3
As x → 3− , f (x) → ∞
As x → 3+ , f (x) → −∞
5
Horizontal asymptote: y = − 2
+
As x → −∞, f (x) → − 5
2
−
As x → ∞, f (x) → − 5
2 y
5
4
3
2
1
2 3 4 5 x 5 6 7 8 9 x −2
−3
−4
−5 (b) f (x) = 1
x2
Domain: (−∞, 0) ∪ (0, ∞)
No xintercepts
No y intercepts
Vertical asymptote: x = 0
As x → 0− , f (x) → ∞
As x → 0+ , f (x) → ∞
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0+
As x → ∞, f (x) → 0+ 1 −7 −6 −5 −4 −3 −2 −1
−1 y
3
2
1
−3 −2 −1
−1 1 2 3 4 −2
−3
−4
−5
−6
−7 y (c) f (x) = 5
4
3
2
1 −4 −3 −2 −1 1 2 3 4 x 5.2 Graphs of Rational Functions
1
x2 + x − 12
Domain: (−∞, −4) ∪ (−4, 3) ∪ (3, ∞)
No xintercepts
1
y intercept: (0, − 12 )
Vertical asymptotes: x = −4 and x = 3
As x → −4− , f (x) → ∞
As x → −4+ , f (x) → −∞
As x → 3− , f (x) → −∞
As x → 3+ , f (x) → ∞
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0+
As x → ∞, f (x) → 0+ 279 (d) f (x) = 2x − 1
−2x2 − 5x + 3
1
Domain: (−∞, −3) ∪ (−3, 2 ) ∪ ( 1 , ∞)
2
No xintercepts
y intercept: (0, − 1 )
3
−1
f (x) =
, x= 1
2
x+3
2
1
Hole in the graph at ( 2 , − 7 )
Vertical asymptote: x = −3
As x → −3− , f (x) → ∞
As x → −3+ , f (x) → −∞
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0+
As x → ∞, f (x) → 0+ y 1 −6 −5 −4 −3 −2 −1 1 2 3 x 4 −1 y (e) f (x) = x
x2 + x − 12
Domain: (−∞, −4) ∪ (−4, 3) ∪ (3, ∞)
xintercept: (0, 0)
y intercept: (0, 0)
Vertical asymptotes: x = −4 and x = 3
As x → −4− , f (x) → −∞
As x → −4+ , f (x) → ∞
As x → 3− , f (x) → −∞
As x → 3+ , f (x) → ∞
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ 1 x
−7 −6 −5 −4 −3 −2 −1 1 2 −1 (f) f (x) = y 1 −6 −5 −4 −3 −2 −1 −1 1 2 3 4 5 x 280 Rational Functions
4x
x2 + 4
Domain: (−∞, ∞)
xintercept: (0, 0)
y intercept: (0, 0)
No vertical asymptotes
No holes in the graph
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (g) f (x) = 4x
4x
=
2−4
x
(x + 2)(x − 2)
Domain: (−∞, −2) ∪ (−2, 2) ∪ (2, ∞)
xintercept: (0, 0)
y intercept: (0, 0)
Vertical asymptotes: x = −2, x = 2
As x → −2− , f (x) → −∞
As x → −2+ , f (x) → ∞
As x → 2− , f (x) → −∞
As x → 2+ , f (x) → ∞
No holes in the graph
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ y 1 −7 −6 −5 −4 −3 −2 −1 2 3 4 5 6 7 x −1 y (h) f (x) = x2 − x − 12
x−4
=
2+x−6
x
x−2
Domain: (−∞, −3) ∪ (−3, 2) ∪ (2, ∞)
xintercept: (4, 0)
y intercept: (0, 2)
Vertical asymptote: x = 2
As x → 2− , f (x) → ∞
As x → 2+ , f (x) → −∞
7
Hole at −3, 5
Horizontal asymptote: y = 1
As x → −∞, f (x) → 1+
As x → ∞, f (x) → 1− 1 5
4
3
2
1 −5 −4 −3 −2 −1 1 2 3 4 5 x 1 2 3 4 5 x −1
−2
−3
−4
−5 y (i) f (x) = 5
4
3
2
1 −5 −4 −3 −2 −1
−1
−2
−3
−4
−5 5.2 Graphs of Rational Functions
(3x + 1)(x − 2)
3x2 − 5x − 2
=
2−9
x
(x + 3)(x − 3)
Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
1
xintercepts: − 3 , 0 , (2, 0)
2
y intercept: 0, 9
Vertical asymptotes: x = −3, x = 3
As x → −3− , f (x) → ∞
As x → −3+ , f (x) → −∞
As x → 3− , f (x) → −∞
As x → 3+ , f (x) → ∞
No holes in the graph
Horizontal asymptote: y = 3
As x → −∞, f (x) → 3+
As x → ∞, f (x) → 3− 281 (j) f (x) = x3 + 2x2 + x
x(x + 1)
=
2−x−2
x
x−2
Domain: (−∞, −1) ∪ (−1, 2) ∪ (2, ∞)
xintercept: (0, 0)
y intercept: (0, 0)
Vertical asymptote: x = 2
As x → 2− , f (x) → −∞
As x → 2+ , f (x) → −∞
Hole at (−1, 0)
Slant asymptote: y = x + 3
As x → −∞, f (x) → −∞
As x → ∞, f (x) → ∞ y
9
8
7
6
5
4
3
2
1
−9−8−7−6−5−4−3−2−11
− 123456789 −2
−3
−4
−5
−6
−7
−8
−9 y (k) f (x) = 18
16
14
12
10
8
6
4
2
−9 8 7 6 5 4 3 2 1
−−−−−−−−
−2 123456789 −4
−6
−8
−10 −x3 + 4x
x2 − 9
Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
xintercepts: (−2, 0), (0, 0), (2, 0)
y intercept: (0, 0)
Vertical asymptotes: x = −3, x = 3
As x → −3− , f (x) → ∞
As x → −3+ , f (x) → −∞
As x → 3− , f (x) → ∞
As x → 3+ , f (x) → −∞
Slant asymptote: y = −x
As x → −∞, f (x) → ∞
As x → ∞, f (x) → −∞ (l) f (x) = y
7
6
5
4
3
2
1
−6 −5 −4 −3 −2 −1
−1
−2
−3
−4
−5
−6
−7 1 2 3 4 5 6 x x x 282 Rational Functions
x2 − 2x + 1
x3 + x2 − 2x
Domain: (−∞, −2) ∪ (−2, 0) ∪ (0, 1) ∪ (1, ∞)
x−1
f (x) =
, x=1
x(x + 2)
No xintercepts
No y intercepts
Vertical asymptotes: x = −2 and x = 0
As x → −2− , f (x) → −∞
As x → −2+ , f (x) → ∞
As x → 0− , f (x) → ∞
As x → 0+ , f (x) → −∞
Hole in the graph at (1, 0)
Horizontal asymptote: y = 0
As x → −∞, f (x) → 0−
As x → ∞, f (x) → 0+ (m) f (x) = y
5
4
3
2
1 −5 −4 −3 −2 −1 1 2 3 4 −1
−2
−3
−4
−5 y 1
x−2
1
Shift the graph of y =
x
to the right 2 units. 4. (a) f (x) = 3
2
1 −1 1 2 3 4 5 x −1
−2
−3 3
x
1
Vertically stretch the graph of y =
x
by a factor of 3.
3
Reﬂect the graph of y =
x
about the xaxis.
3
Shift the graph of y = −
x
up 1 unit. (b) g (x) = 1 − y
7
6
5
4
3
2
1
−6 −5 −4 −3 −2 −1
−1
−2
−3
−4
−5 1 2 3 4 5 6 x 5 x 5.2 Graphs of Rational Functions
−2x + 1
1
= −2 +
x
x
1
Shift the graph of y =
x
down 2 units. 283
y (c) h(x) = 1 −3 −2 −1 1 2 3 x −1
−2
−3
−4
−5 y 3x − 7
1
=3−
x−2
x−2
1
Shift the graph of y =
x
to the right 2 units. (d) j (x) = Reﬂect the graph of y =
about the xaxis.
Shift the graph of y = −
up 3 units. 7
6
5 1
x−2 4
3 1
x−2 2
1 −3 −2 −1 1
−1 2 3 4 5 x 284 5.3 Rational Functions Rational Inequalities and Applications In this section, we use sign diagrams to solve rational inequalities including some that arise from
realworld applications. Our ﬁrst example showcases the critical diﬀerence in procedure between
solving a rational equation and a rational inequality.
Example 5.3.1.
1. Solve x3 − 2x + 1
1
= x − 1.
x−1
2 2. Solve x3 − 2x + 1
1
≥ x − 1.
x−1
2 3. Use your calculator to graphically check your answers to 1 and 2.
Solution.
1. To solve the equation, we clear denominators
x3 − 2x + 1
x−1
x3 − 2x + 1
x−1 = 1
x−1
2 · 2(x − 1) = 1
x − 1 · 2(x − 1)
2 2x3 − 4x + 2 = x2 − 3x + 2 expand 2x3 − x2 − x = 0
x(2x + 1)(x − 1) = 0 factor x = − 1 , 0, 1
2
Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we
see that x = 1 does not satisfy the original equation and must be discarded. Our solutions
are x = − 1 and x = 0.
2
2. To solve the inequality, it may be tempting to begin as we did with the equation − namely
by multiplying both sides by the quantity (x − 1). The problem is that, depending on x,
(x − 1) may be positive (which doesn’t aﬀect the inequality) or (x − 1) could be negative
(which would reverse the inequality). Instead of working by cases, we collect all of the terms
on one side of the inequality with 0 on the other and make a sign diagram using the technique
given on page 264 in Section 5.2. 5.3 Rational Inequalities and Applications 285 x3 − 2x + 1
x−1 ≥ 1
x−1
2 x3 − 2x + 1 1
− x+1 ≥ 0
x−1
2
2 x3 − 2x + 1 − x(x − 1) + 1(2(x − 1))
2(x − 1) ≥0 get a common denominator 2x3 − x2 − x
2x − 2 ≥0 expand Viewing the left hand side as a rational function r(x) we make a sign diagram. The only
value excluded from the domain of r is x = 1 which is the solution to 2x − 2 = 0. The zeros
of r are the solutions to 2x3 − x2 − x = 0, which we have already found to be x = 0, x = − 1
2
and x = 1, the latter was discounted as a zero because it is not in the domain. Choosing test
values in each test interval, we construct the sign diagram below.
(+) 0 (−) 0 (+) (+)
−1
2 0 1 1
We are interested in where r(x) ≥ 0. We ﬁnd r(x) > 0, or (+), on the intervals −∞, − 2 ,
(0, 1) and (1, ∞). We add to these intervals the zeros of r, − 1 and 0, to get our ﬁnal solution:
2
−∞, − 1 ∪ [0, 1) ∪ (1, ∞).
2
3 1
2
3. Geometrically, if we set f (x) = x −−x+1 and g (x) = 2 x − 1, the solutions to f (x) = g (x) are
x1
the xcoordinates of the points where the graphs of y = f (x) and y = g (x) intersect. The
solution to f (x) ≥ g (x) represents not only where the graphs meet, but the intervals over
which the graph of y = f (x) is above (>) the graph of g (x). We obtain the graphs below. The ‘Intersect’ command conﬁrms that the graphs cross when x = − 1 and x = 0. It is clear
2
from the calculator that the graph of y = f (x) is above the graph of y = g (x) on −∞, − 1
2
1
as well as on (0, ∞). According to the calculator, our solution is then −∞, − 2 ∪ [0, ∞) 286 Rational Functions
which almost matches the answer we found analytically. We have to remember that f is
not deﬁned at x = 1, and, even though it isn’t shown on the calculator, there is a hole1 in
the graph of y = f (x) when x = 1 which is why x = 1 needs to be excluded from our ﬁnal
answer. Our next example deals with the average cost function of PortaBoy Game systems from
Example 3.1.5 in Section 3.1.
Example 5.3.2. Given a cost function C (x), which returns the total cost of producing x products,
(
the average cost function, AC (x) = C xx) , computes the cost per item. Recall that the cost C , in
dollars, to produce x PortaBoy game systems for a local retailer is C (x) = 80x + 150, x ≥ 0.
1. Find an expression for the average cost function AC (x). Determine an appropriate applied
domain for AC .
2. Find and interpret AC (10).
3. Solve AC (x) < 100 and interpret.
4. Determine the behavior of AC (x) as x → ∞ and interpret.
Solution.
(
1. From AC (x) = C xx) , we obtain AC (x) = 80x+150 . The domain of C is x ≥ 0, but since x = 0
x
causes problems for AC (x), we get our domain to be x > 0, or (0, ∞). 2. We ﬁnd AC (10) =
per system. 80(10)+150
10 = 95, so the average cost to produce 10 game systems is $95 3. Solving AC (x) < 100 means we solve 80x+150
x 80x + 150
x
80x + 150
− 100
x
80x + 150 − 100x
x
150 − 20x
x < 100. We proceed as in the previous example. < 100
<0
<0 common denominator <0 If we take the left hand side to be a rational function r(x), we need to keep in mind the the
applied domain of the problem is x > 0. This means we consider only the positive half of the
number line for our sign diagram. On (0, ∞), r is deﬁned everywhere so we need only look
for zeros of r. Setting r(x) = 0 gives 150 − 20x = 0, so that x = 15 = 7.5. The test intervals
2
on our domain are (0, 7.5) and (7.5, ∞). We ﬁnd r(x) < 0 on (7.5, ∞).
1 There is no asymptote at x = 1 since the graph is well behaved near x = 1. According to Theorem 5.1, there
must be a hole there. 5.3 Rational Inequalities and Applications 287 (+) 0 (−) 0 7.5 In the context of the problem, x represents the number of PortaBoy games systems produced
and AC (x) is the average cost to produce each system. Solving AC (x) < 100 means we are
trying to ﬁnd how many systems we need to produce so that the average cost is less than $100
per system. Our solution, (7.5, ∞) tells us that we need to produce more than 7.5 systems to
achieve this. Since it doesn’t make sense to produce half a system, our ﬁnal answer is [8, ∞).
4. We can apply Theorem 5.2 to AC (x) and we ﬁnd y = 80 is a horizontal asymptote to the
graph of y = AC (x). To more precisely determine the behavior of AC (x) as x → ∞, we
ﬁrst use long division2 and rewrite AC (x) = 80 + 150 . As x → ∞, 150 → 0+ , which means
x
x
AC (x) ≈ 80 + very small (+). Thus the average cost per system is getting closer to $80
per system. If we set AC (x) = 80, we get 150 = 0, which is impossible, so we conclude
x
that AC (x) > 80 for all x > 0. This means the average cost per system is always greater
than $80 per system, but the average cost is approaching this amount as more and more
systems are produced. Looking back at Example 3.1.5, we realize $80 is the variable cost
per system − the cost per system above and beyond the ﬁxed initial cost of $150. Another
way to interpret our answer is that ‘inﬁnitely’ many systems would need to be produced to
eﬀectively counterbalance the ﬁxed cost.
Our next example is another classic ‘box with no top’ problem.
Example 5.3.3. A box with a square base and no top is to be constructed so that it has a volume
of 1000 cubic centimeters. Let x denote the width of the box, in centimeters. Refer to the ﬁgure
below. height depth
width, x 1. Express the height h in centimeters as a function of the width x and state the applied domain.
2. Solve h(x) ≥ x and interpret.
3. Find and interpret the behavior of h(x) as x → 0+ and as x → ∞.
4. Express the surface area S of the box as a function of x and state the applied domain.
5. Use a calculator to approximate (to two decimal places) the dimensions of the box which
minimize the surface area.
2 In this case, long division amounts to termbyterm division. 288 Rational Functions Solution.
1. We are told the volume of the box is 1000 cubic centimeters and that x represents the width,
in centimeters. From geometry, we know Volume = width × height × depth. Since the base
of the box is to be a square, the width and the depth are both x centimeters. Using h for the
height, we have 1000 = x2 h, so that h = 1000 . Using function notation,3 h(x) = 1000 As for
x2
x2
the applied domain, in order for there to be a box at all, x > 0, and since every such choice
of x will return a positive number for the height h we have no other restrictions and conclude
our domain is (0, ∞).
2. To solve h(x) ≥ x, we proceed as before and collect all nonzero terms on one side of the
inequality and use a sign diagram.
h(x) ≥ x
1000
≥x
x2
1000
−x ≥ 0
x2
1000 − x3
x2 ≥ 0 common denominator We consider the left hand side of the inequality as our rational function r(x). We see r is
undeﬁned at x = 0, but, as in the previous example, the applied domain of the problem is
x > 0, so we are considering only the behavior of r on (0, ∞). The sole zero of r comes when
1000 − x3 = 0, which is x = 10. Choosing test values in the intervals (0, 10) and (10, ∞) gives
the following diagram.
(+) 0 (−) 0 10 We see r(x) > 0 on (0, 10), and since r(x) = 0 at x = 10, our solution is (0, 10]. In the context
of the problem, h represents the height of the box while x represents the width (and depth)
of the box. Solving h(x) ≥ x is tantamount to ﬁnding the values of x which result in a box
where the height is at least as big as the width (and, in this case, depth.) Our answer tells
us the width of the box can be at most 10 centimeters for this to happen.
3. As x → 0+ , h(x) = 1000 → ∞. This means the smaller the width x (and, in this case, depth),
x2
the larger the height h has to be in order to maintain a volume of 1000 cubic centimeters. As
x → ∞, we ﬁnd h(x) → 0+ , which means to maintain a volume of 1000 cubic centimeters,
the width and depth must get bigger the smaller the height becomes.
3 That is, h(x) means ‘h of x’, not ‘h times x’ here. 5.3 Rational Inequalities and Applications 289 4. Since the box has no top, the surface area can be found by adding the area of each of the
sides to the area of the base. The base is a square of dimensions x by x, and each side has
dimensions x by h. We get the surface area, S = x2 + 4xh. To get S as a function of x, we
000
substitute h = 1000 to obtain S = x2 + 4x 1x2 . Hence, as a function of x, S (x) = x2 + 4000 .
x
x2
The domain of S is the same as h, namely (0, ∞), for the same reasons as above.
5. A ﬁrst attempt at the graph of y = S (x) on the calculator may lead to frustration. Chances
are good that the ﬁrst window chosen to view the graph will suggest y = S (x) has the xaxis
as a horizontal asymptote. From the formula S (x) = x2 + 4000 , however, we get S (x) ≈ x2 as
x
x → ∞, so S (x) → ∞. Readjusting the window, we ﬁnd S does possess a relative minimum
at x ≈ 12.60. As far as we can tell,4 this is the only relative extremum, and so it is the
absolute minimum as well. This means the width and depth of the box should each measure
approximately 12.60 centimeters. To determine the height, we ﬁnd h(12.60) ≈ 6.30, so the
height of the box should be approximately 6.30 centimeters. In many instances in the sciences, rational functions are encountered as a result of fundamental
natural laws which are typically a result of assuming certain basic relationships between variables.
These basic relationships are summarized in the deﬁnition below.
Definition 5.4. Suppose x, y , and z are variable quantities. We say
y varies directly with (or is directly proportional to) x if there is a constant k such
that y = kx.
y varies inversely with (or is inversely proportional to) x if there is a constant k
k
such that y = x .
z varies jointly with (or is jointly proportional to) x and y if there is a constant k
such that z = kxy . The constant k in the above deﬁnitions is called the constant of proportionality. 4 without Calculus, that is... 290 Rational Functions Example 5.3.4. Translate the following into mathematical equations using Deﬁnition 5.4.
1. Hooke’s Law: The force F exerted on a spring is directly proportional the extension x of the
spring.
2. Boyle’s Law: At a constant temperature, the pressure P of an ideal gas is inversely proportional to its volume V .
3. The volume V of a right circular cone varies jointly with the height h of the cone and the
square of the radius r of the base.
4. Ohm’s Law: The current I through a conductor between two points is directly proportional to
the voltage V between the two points and inversely proportional to the resistance R between
the two points.
5. Newton’s Law of Universal Gravitation: Suppose two objects, one of mass m and one of mass
M , are positioned so that the distance between their centers of mass is r. The gravitational
force F exerted on the two objects varies directly with the product of the two masses and
inversely with the square of the distance between their centers of mass.
Solution.
1. Applying the deﬁnition of direct variation, we get F = kx for some constant k .
2. Since P and V are inversely proportional, we write P = k
V . 3. There is a bit of ambiguity here. It’s clear the volume and height of the cone is represented by
the quantities V and h, respectively, but does r represent the radius of the base or the square
of the radius of the base? It is the former. Usually, if an algebraic operation is speciﬁed
(like squaring), it is meant to be expressed in the formula. We apply Deﬁnition 5.4 to get
V = khr2 .
4. Even though the problem doesn’t use the phrase ‘varies jointly’, the fact that the current I
is given as relating to two diﬀerent quantities implies this. Since I varies directly with V but
inversely with R, we write I = kV .
R
5. We write the product of the masses mM and the square of the distance as r2 . We have F
varies directly with mM and inversely with r2 , so that F = kmM .
r2 5.3 Rational Inequalities and Applications 5.3.1 291 Exercises 1. Solve each rational equation. Be sure to check for extraneous solutions.
x
=3
5x + 4
3x − 1
(b) 2
=1
x +1
1
1
x2 − 3
(c)
+
=2
x+3 x−3
x −9 2x + 17
=x+5
x+1
x2 − 2x + 1
(e) 3
=1
x + x2 − 2x
−x3 + 4x
(f)
= 4x
x2 − 9 (a) (d) 2. Solve each rational inequality. Express your answer using interval notation.
(a)
(b)
(c)
(d)
(e)
(n) 1
≥0
x+2
x−3
≤0
x+2
x
>0
2−1
x
4x
≤0
x2 + 4
4x
≥0
2−4
x (f) x2 − x − 12
>0
x2 + x − 6 (j) 3x − 1
≤1
x2 + 1 (g) 3x2 − 5x − 2
<0
x2 − 9 (k) 2x + 17
>x+5
x+1 (h) x3 + 2 x2 + x
≥0
x2 − x − 2 (l) −x3 + 4x
≥ 4x
x2 − 9 (i) x2 + 5 x + 6
>0
x2 − 1 (m) x4 − 4x3 + x2 − 2x − 15
≥x
x3 − 4x2 (o) x2 1
<0
+1 5x3 − 12x2 + 9x + 10
≥ 3x − 1
x2 − 1 3. Another Classic Problem: A can is made in the shape of a right circular cylinder and is to
hold one pint. (For dry goods, one pint is equal to 33.6 cubic inches.)5
(a) Find an expression for the volume V of the can based on the height h and the base
radius r.
(b) Find an expression for the surface area S of the can based on the height h and the base
radius r. (Hint: The top and bottom of the can are circles of radius r and the side of
the can is really just a rectangle that has been bent into a cylinder.)
(c) Using the fact that V = 33.6, write S as a function of r and state its applied domain.
(d) Use your graphing calculator to ﬁnd the dimensions of the can which has minimal surface
area.
4. In Exercise 11 in Section 2.2, the population of Sasquatch in Portage County was modeled
by the function P (t) = t150t , where t = 0 represents the year 1803. When were there fewer
+15
than 100 Sasquatch in the county?
5 According to www.dictionary.com, there are diﬀerent values given for this conversion. We will stick with 33.6in3
for this problem. 292 Rational Functions 5. The cost C in dollars to remove p% of the invasive species of Ippizuti ﬁsh from Sasquatch
1770
Pond is given by C (p) = 100−p where 0 ≤ p < 100.
p
(a) Find and interpret C (25) and C (95).
(b) What does the vertical asymptote at x = 100 mean within the context of the problem?
(c) What percentage of the Ippizuti ﬁsh can you remove for $40000?
6. Translate the following into mathematical equations.
(a) At a constant pressure, the temperature T of an ideal gas is directly proportional to its
volume V . (This is Charles’s Law)
(b) The frequency of a wave f is inversely proportional to the wavelength of the wave λ.
(c) The density d of a material is directly proportional to the mass of the object m and
inversely proportional to its volume V .
(d) The square of the orbital period of a planet P is directly proportional to the cube of the
semimajor axis of its orbit a. (This is Kepler’s Third Law of Planetary Motion )
(e) The drag of an object traveling through a ﬂuid D varies jointly with the density of the
ﬂuid ρ and the square of the velocity of the object ν .
(f) Suppose two electric point charges, one with charge q and one with charge Q, are positioned r units apart. The electrostatic force F exerted on the charges varies directly with
the product of the two charges and inversely with the square of the distance between
the charges. (This is Coulomb’s Law)
1
T
2L µ
where T is the tension, µ is the linear mass6 of the string and L is the length of the vibrating
part of the string. Express this relationship using the language of variation. 7. According to this webpage, the frequency f of a vibrating string is given by f = 8. According to the Centers for Disease Control and Prevention www.cdc.gov, a person’s Body
Mass Index B is directly proportional to his weight W in pounds and inversely proportional
to the square of his height h in inches.
(a) Express this relationship as a mathematical equation.
(b) If a person who was 5 feet, 10 inches tall weighed 235 pounds had a Body Mass Index
of 33.7, what is the value of the constant of proportionality?
(c) Rewrite the mathematical equation found in part 8a to include the value of the constant
found in part 8b and then ﬁnd your Body Mass Index.
9. We know that the circumference of a circle varies directly with its radius with 2π as the
constant of proportionality. (That is, we know C = 2πr.) With the help of your classmates,
compile a list of other basic geometric relationships which can be seen as variations.
6 Also known as the linear density. It is simply a measure of mass per unit length. 5.3 Rational Inequalities and Applications 5.3.2 293 Answers 6
1. (a) x = − 7
(b) x = 1, x = 2 2. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h) (c) x = −1
(d) x = −6, x = 2 (−2, ∞)
(−2, 3]
(−1, 0) ∪ (1, ∞)
(−∞, 0]
(−2, 0] ∪ (2, ∞)
(−∞, −3) ∪ (−3, 2) ∪ (4, ∞)
1
−3, − 3 ∪ (2, 3)
(−1, 0] ∪ (2, ∞) (i) (−∞, −3) ∪ (−2, −1) ∪ (1, ∞)
(j) (−∞, 1] ∪ [2, ∞)
(k) (−∞, −6) ∪ (−1, 2)
√
√
(l) (−∞, −3) ∪ −2 2, 0 ∪ 2 2, 3
(m) No solution
(n) [−3, 0) ∪ (0, 4) ∪ [5, ∞)
1
(o) −1, − 2 ∪ (1, ∞) 3. (a) V = πr2 h.
(b) S = 2πr2 (e) No solution
√
(f) x = 0, x = ±2 2 (c) S (r) = 2πr2 + 67.2
r, Domain r > 0 (d) r ≈ 1.749 in. and h ≈ 3.498 in. + 2πrh 4. P (30) = 100 so before 1903 there were fewer than 100 Sasquatch in Portage County.
5. (a) C (25) = 590 means it costs $590 to remove 25% of the ﬁsh and and C (95) = 33630
means it would cost $33630 to remove 95% of the ﬁsh from the pond.
(b) The vertical asymptote at x = 100 means that as we try to remove 100% of the ﬁsh from
the pond, the cost increases without bound; i.e., it’s impossible to remove all of the ﬁsh.
(c) For $40000 you could remove about 95.76% of the ﬁsh.
6. (a) T = kV
k
(b) 7 f = .
λ km
V
(d) P 2 = ka3
(c) d = (e) 8 (f) 9 D = kρν 2 .
kqQ
F= 2
r √
1
T
1
2T
as f = √ we see that the frequency f varies directly with the
7. Rewriting f =
2L µ
Lµ
square root of the tension and varies inversely with the length and the square root of the
linear mass.
8. (a) B = 7 kW
h2 (b) 10 k = 702.68 (c) B = 703W
h2 The character λ is the lower case Greek letter ‘lambda.’
Note: The characters ρ and ν are the lower case Greek letters ‘rho’ and ‘nu,’ respectively.
9
Note the similarity to this formula and Newton’s Law of Universal Gravitation as discussed in Example 5.
10
The CDC uses 703.
8 294 Rational Functions Chapter 6 Further Topics in Functions
6.1 Function Composition Before we embark upon any further adventures with functions, we need to take some time to gather
our thoughts and gain some perspective. Chapter 2 ﬁrst introduced us to functions in Section 2.1.
At that time, functions were speciﬁc kinds of relations  sets of points in the plane which passed the
Vertical Line Test, Theorem 2.1. In Section 2.2, we developed the idea that functions are processes
 rules which match inputs to outputs  and this gave rise to the concepts of domain and range.
We spoke about how functions could be combined in Section 2.3 using the four basic arithmetic
operations, took a more detailed look at their graphs in Section 2.4 and studied how their graphs
behaved under certain classes of transformations in Section 2.5. In Chapter 3, we took a closer
look linear functions (Section 3.1), and quadratic functions (Section 3.3). Linear and quadratic
functions were special cases of polynomial functions, which we studied in generality in Chapter 4.
(In Chapter 10 we will learn the Real Factorization Theorem, Theorem 10.5, which says that all
polynomial functions with real coeﬃcients can be thought of as products of linear and quadratic
functions.) Our next step was to enlarge our ﬁeld1 of study to rational functions in Chapter 5.
Being quotients of polynomials, we can ultimately view this family of functions as being built up
of linear and quadratic functions as well. So in some sense, Chapters 3, 4, and 5 can be thought
of as an exhaustive study of linear and quadratic functions and their arithmetic combinations as
√
described in Section 2.3. We now wish to study other algebraic functions, such as f (x) = x and
g (x) = x2/3 , and the purpose of the ﬁrst two sections of this chapter is to see how these kinds of
functions arise from polynomial and rational functions. To that end, we ﬁrst study a new way to
combine functions as deﬁned below.
Definition 6.1. Suppose f and g are two functions. The composite of g with f , denoted
g ◦ f , is deﬁned by the formula (g ◦ f )(x) = g (f (x)), provided x is an element of the domain of
f and f (x) is an element of the domain of g . 1 This is a really bad math pun. 296 Further Topics in Functions The quantity g ◦ f is also read ‘g composed with f ’ or, more simply ‘g of f .’ At its most basic
level, Deﬁnition 6.1 tells us to obtain the formula for (g ◦ f ) (x), we replace every occurrence of x
in the formula for g (x) with the formula we have for f (x). If we take a step back and look at this
from a procedural, ‘inputs and outputs’ perspective, Deﬁntion 6.1 tells us the output from g ◦ f is
found by taking the output from f , f (x), and then making that the input to g . The result, g (f (x)),
is the output from g ◦ f . From this perspective, we see g ◦ f as a two step process taking an input
x and ﬁrst applying the procedure f then applying the procedure g . Abstractly, we have
g f x f (x)
g (f (x)) g◦f In the expression g (f (x)), the function f is often called the ‘inside’ function while g is often
called the ‘outside’ function. There are two ways to go about evaluating composite functions ‘inside out’ and ‘outside in’  depending on which function we replace with its formula ﬁrst. Both
ways are demonstrated in the following example.
Example 6.1.1. Let f (x) = x2 − 4x, g (x) = 2 − √ x + 3, and h(x) = indicated composite functions. State the domain of each.
1. (g ◦ f )(x) 2x
. Find and simplify the
x+1 5. (h ◦ h)(x) 2. (f ◦ g )(x)
6. (h ◦ (g ◦ f ))(x)
3. (g ◦ h)(x)
4. (h ◦ g )(x) 7. ((h ◦ g ) ◦ f )(x) Solution.
1. By deﬁnition, (g ◦ f )(x) = g (f (x)). We now illustrate the two ways to evaluate this.
inside out : We insert the expression f (x) into g ﬁrst to get (g ◦ f )(x) = g (f (x)) = g x2 − 4x = 2 −
Hence, (g ◦ f )(x) = 2 − √ x2 − 4x + 3. (x2 − 4x) + 3 = 2 − x2 − 4x + 3 6.1 Function Composition 297 outside in : We use the formula for g ﬁrst to get (x2 − 4x) + 3 = 2 −
√
We get the same answer as before, (g ◦ f )(x) = 2 − x2 − 4x + 3.
(g ◦ f )(x) = g (f (x)) = 2 − f (x) + 3 = 2 − x2 − 4x + 3 To ﬁnd the domain of g ◦ f , we need to ﬁnd the elements in the domain of f whose outputs
f (x) are in the domain of g . We accomplish this by following the rule set forth in Section
2.2, that is, we ﬁnd the domain before we simplify. To that end, we examine (g ◦ f )(x) =
2 − (x2 − 4x) + 3. To keep the square root happy, we solve the inequality x2 − 4x + 3 ≥ 0
by creating a sign diagram. If we let r(x) = x2 − 4x + 3, we ﬁnd the zeros of r to be x = 1
and x = 3. We obtain
(+) 0 (−) 0 (+)
1 3 Our solution to x2 − 4x + 3 ≥ 0, and hence the domain of g ◦ f , is (−∞, 1] ∪ [3, ∞).
2. To ﬁnd (f ◦ g )(x), we ﬁnd f (g (x)).
inside out : We insert the expression g (x) into f ﬁrst to get (f ◦ g )(x) = = f (g (x))
√
f 2− x+3
√
√
2
2− x+3 −4 2− x+3
√
√
√
2
4−4 x+3+
x+3 −8+4 x+3 = 4+x+3−8 = x−1 =
= outside in : We use the formula for f (x) ﬁrst to get (f ◦ g )(x) = f (g (x)) =
= (g (x))2 − 4 (g (x))
√
√
2
2− x+3 −4 2− x+3 = x−1 same algebra as before Thus we get (f ◦ g )(x) = x − 1. To ﬁnd the domain of (f ◦ g ), we look to the step before 298 Further Topics in Functions
√
√
2
we did any simpliﬁcation and ﬁnd (f ◦ g )(x) = 2 − x + 3 − 4 2 − x + 3 . To keep the
square root happy, we set x + 3 ≥ 0 and ﬁnd our domain to be [−3, ∞). 3. To ﬁnd (g ◦ h)(x), we compute g (h(x)).
inside out : We insert the expression h(x) into g ﬁrst to get (g ◦ h)(x) = g (h(x)) = g = 2− = 2− 2x
3(x + 1)
+
x+1
x+1 = 2− 5x + 3
x+1 2x
x+1
2x
x+1 +3
get common denominators outside in : We use the formula for g (x) ﬁrst to get (g ◦ h)(x) = g (h(x)) = 2− h(x) + 3 = 2− 2x
x+1 = 2− 5x + 3
x+1 Hence, (g ◦ h)(x) = 2 −
simplify: (g ◦ h)(x) = 2 − 5x+3
x+1 . +3
get common denominators as before To ﬁnd the domain, we look to the step before we began to 2x
x+1 + 3. To avoid division by zero, we need x = −1. To keep the radical happy, we need to solve x2x + 3 ≥ 0. Getting common denominators as before,
+1
x+3
x+3
this reduces to 5x+1 ≥ 0. Deﬁning r(x) = 5x+1 , we have that r is undeﬁned at x = −1 and
3
r(x) = 0 at x = − 5 . We get
(+) (−) 0 (+)
−1
3
Our domain is (−∞, −1) ∪ − 5 , ∞ . −3
5 6.1 Function Composition 299 4. We ﬁnd (h ◦ g )(x) by ﬁnding h(g (x)).
inside out : We insert the expression g (x) into h ﬁrst to get (h ◦ g )(x) =
=
=
= h(g (x))
√
h 2− x+3
√
2 2− x+3
√
2− x+3 +1
√
4−2 x+3
√
3− x+3 outside in : We use the formula for h(x) ﬁrst to get (h ◦ g )(x) =
=
=
= h(g (x))
2 (g (x))
(g (x)) + 1
√
2 2− x+3
√
2− x+3 +1
√
4−2 x+3
√
3− x+3 √
4−2 x+3
√
. To ﬁnd the
3− x+3
√
2(2− x+3)
(h ◦ g )(x) = 2−√x+3 +1 . To
(
) Hence, (h ◦ g )(x) = domain of h ◦ g , we look to the step before any keep the square root happy, we require x + 3 ≥ 0
√
√
or x ≥ −3. Setting the denominator equal to zero gives 2 − x + 3 + 1 = 0 or x + 3 = 3.
Squaring both sides gives us x + 3 = 9, or x = 6. Since x = 6 checks in the original equation,
√
2 − x + 3 + 1 = 0, we know x = 6 is the only zero of the denominator. Hence, the domain
of h ◦ g is [−3, 6) ∪ (6, ∞).
simpliﬁcation: 5. To ﬁnd (h ◦ h)(x), we substitute the function h into itself, h(h(x)).
inside out : We insert the expression h(x) into h to get (h ◦ h)(x) = h(h(x)) = h 2x
x+1 300 Further Topics in Functions
2x
x+1
2x
+1
x+1
4x
x + 1 · (x + 1)
2x
(x + 1)
+1
x+1
4x
· (x + 1)
x+1
2x
· (x + 1) + 1 · (x + 1)
x+1
4x
$
· (x$$
+ 1)
X
$
$ 1$
(x +
$$ 1)
2x
$
· (x$$ + x + 1
+ 1)
X$
$$ 1
(x +
$$ 1)
4x
3x + 1
2 = = = = = outside in : This approach yields (h ◦ h)(x) = h(h(x)) = 2(h(x))
h(x) + 1
2x
x+1
2x
+1
x+1
4x
3x + 1
2 = = same algebra as before
2 To ﬁnd the domain of h ◦ h, we analyze (h ◦ h)(x) = 2x
x+1 2x
x+1 x + 1 happy, we need x = −1. Setting the denominator
1
domain is (−∞, −1) ∪ −1, − 3 ∪ − 1 , ∞ .
3 . To keep the denominator +1 2x
x+1 1
+ 1 = 0 gives x = − 3 . Our 6. The expression (h ◦ (g ◦ f ))(x) indicates that we ﬁrst ﬁnd the composite, g ◦ f√
and compose
the function h with the result. We know from number 1 that (g ◦ f )(x) = 2 − x2 − 4x + 3.
We now proceed as usual. 6.1 Function Composition 301 inside out : We insert the expression (g ◦ f )(x) into h ﬁrst to get (h ◦ (g ◦ f ))(x) =
=
=
= h((g ◦ f )(x))
√
h 2 − x2 − 4x + 3
√
2 2 − x2 − 4x + 3
√
2 − x2 − 4x + 3 + 1
√
4 − 2 x2 − 4x + 3
√
3 − x2 − 4x + 3 outside in : We use the formula for h(x) ﬁrst to get (h ◦ (g ◦ f ))(x) =
= =
=
So we get (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x))
2 ((g ◦ f )(x))
((g ◦ f )(x)) + 1
√
2 2 − x2 − 4x + 3
√
2 − x2 − 4x + 3 + 1
√
4 − 2 x2 − 4x + 3
√
3 − x2 − 4x + 3 √
4−2 x2 −4x+3
√
.
3− x2 −4x+3 To ﬁnd the domain, we look at the step before
√
2(2− x2 −4x+3)
we began to simplify, (h ◦ (g ◦ f ))(x) = 2−√x2 −4x+3 +1 . For the square root, we need
(
)
x2 − 4x + 3 ≥ 0, which we determined in number 1 to be (−∞, 1] ∪ [3, ∞). Next, we set
√
√
the denominator to zero and solve: 2 − x2 − 4x + 3 + 1 = 0. We get x2 − 4x + 3 = 3,
and, after squaring both sides, we have √ − 4x + 3 = 9. To solve x2 − 4x − 6 = 0, we use
x2
the quadratic formula and get x = 2 ± 10. The reader is encouraged to check that both
√
of these numbers satisfy the original equation, 2 − x2 − 4x + 3 + 1 = 0. Hence we must
exclude these numbers from the domain of h ◦ (g ◦ f ). Our ﬁnal domain for h ◦ (f ◦ g ) is
√
√
√
√
(−∞, 2 − 10) ∪ (2 − 10, 1] ∪ 3, 2 + 10 ∪ 2 + 10, ∞ .
7. The expression ((h ◦ g ) ◦ f )(x) indicates that we ﬁrst ﬁnd the composite h ◦ g and then compose
√
x+3
that with f . From number 4, we gave (h ◦ g )(x) = 43−2√x+3 . We now proceed as before.
−
inside out : We insert the expression f (x) into h ◦ g ﬁrst to get ((h ◦ g ) ◦ f )(x) = (h ◦ g )(f (x)) 302 Further Topics in Functions
=
=
= (h ◦ g ) x2 − 4x
4 − 2 (x2 − 4x) + 3
(x2 − 4x) + 3
√
4 − 2 x2 − 4x + 3
√
3 − x2 − 4x + 3
3− outside in : We use the formula for (h ◦ g )(x) ﬁrst to get ((h ◦ g ) ◦ f )(x) =
=
=
= (h ◦ g )(f (x))
4−2 (f (x)) + 3 3− f (x)) + 3 4−2 (x2 − 4x) + 3 (x2 − 4x) + 3
√
4 − 2 x2 − 4x + 3
√
3 − x2 − 4x + 3
3− We note that the formula for ((h ◦ g ) ◦ f )(x) before simpliﬁcation is identical to that of
(h ◦ (g ◦ f ))(x) before √ simpliﬁed it. Hence, the two functions have the same domain,
we
√
√
√
h ◦ (f ◦ g ) is (−∞, 2 − 10) ∪ (2 − 10, 1] ∪ 3, 2 + 10 ∪ 2 + 10, ∞ . It should be clear from Example 6.1.1 that, in general, when you compose two functions, such
as f and g above, the order matters.2 We found that the functions f ◦ g and g ◦ f were diﬀerent as
were g ◦ h and h ◦ g . Thinking of functions as processes, this isn’t all that surprising. If we think of
one process as putting on our socks, and the other as putting on our shoes, the order in which we
do these two tasks does matter.3 Also note the importance of ﬁnding the domain of the composite
function before simplifying. For instance, the domain of f ◦ g is much diﬀerent than its simpliﬁed
formula would indicate. Composing a function with itself, as in the case of h ◦ h, may seem odd.
Looking at this from a procedural perspective, however, this merely indicates performing a task
h and then doing it again  like setting the washing machine to do a ‘double rinse’. Composing a
function with itself is called ‘iterating’ the function, and we could easily spend an entire course on
just that. The last two problems in Example 6.1.1 serve to demonstrate the associative property
of functions. That is, when composing three (or more) functions, as long as we keep the order the
same, it doesn’t matter which two functions we compose ﬁrst. This property as well as another
important property are listed in the theorem below.
2 This shows us function composition isn’t commutative. An example of an operation we perform on two functions
which is commutative is function addition, which we deﬁned in Section 2.3. In other words, the functions f + g and
g + f are always equal. Which of the remaining operations on functions we have discussed are commutative?
3
A more mathematical example in which the order of two processes matters can be found in Section 2.5. In fact,
all of the transformations in that section can be viewed in terms of composing functions with linear functions. 6.1 Function Composition 303 Theorem 6.1. Properties of Function Composition: Suppose f , g , and h are functions.
h ◦ (g ◦ f ) = (h ◦ g ) ◦ f , provided the composite functions are deﬁned.
If I is deﬁned as I (x) = x for all real numbers x, then I ◦ f = f ◦ I = f . By repeated applications of Deﬁnition 6.1, we ﬁnd (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h(g (f (x))).
Similarly, ((h ◦ g ) ◦ f )(x) = (h ◦ g )(f (x)) = h(g (f (x))). This establishes that the formulas for the
two functions are the same. We leave it to the reader to think about why the domains of these
two functions are identical, too. These two facts establish the equality h ◦ (g ◦ f ) = (h ◦ g ) ◦ f .
A consequence of the associativity of function composition is that there is no need for parentheses
when we write h ◦ g ◦ f . The second property can also be veriﬁed using Deﬁnition 6.1. Recall that
the function I (x) = x is called the identity function and was introduced in Exercise 12 in Section
3.1. If we compose the function I with a function f , then we have (I ◦ f )(x) = I (f (x)) = f (x),
and a similar computation shows (f ◦ I )(x) = f (x). This establishes that we have an identity
for function composition much in the same way the real number 1 is an identity for real number
multiplication. That is, just as for any real number x, 1 · x = x · 1 = x , we have for any function
f , I ◦ f = f ◦ I = f . We shall see the concept of an identity take on great signiﬁcance in the next
section. Out in the wild, function composition is often used to relate two quantities which may not
be directly related, but have a variable in common, as illustrated in our next example.
Example 6.1.2. The surface area S of a sphere is a function of its radius r and is given by the
formula S (r) = 4πr2 . Suppose the sphere is being inﬂated so that the radius of the sphere is
increasing according to the formula r(t) = 3t2 , where t is measured in seconds, t ≥ 0, and r is
measured in inches. Find and interpret (S ◦ r)(t).
Solution. If we look at the functions S (r) and r(t) individually, we see the former gives the
surface area of a sphere of a given radius while the latter gives the radius at a given time. So,
given a speciﬁc time, t, we could ﬁnd the radius at that time, r(t) and feed that into S (r) to ﬁnd
the surface area at that time. From this we see that the surface area S is ultimately a function of
2
time t and we ﬁnd (S ◦ r)(t) = S (r(t)) = 4π (r(t))2 = 4π 3t2 = 36πt4 . This formula allows us to
compute the surface area directly given the time without going through the ‘middle man’ r.
A useful skill in Calculus is to be able to take a complicated function and break it down into a
composition of easier functions which our last example illustrates.
Example 6.1.3. Write each of the following functions as a composition of two or more (nonidentity)
functions. Check your answer by performing the function composition.
1. F (x) = 3x − 1
2. G(x) = x2 2
+1 304 Further Topics in Functions √
x+1
3. H (x) = √
x−1
Solution. There are many approaches to this kind of problem, and we showcase a diﬀerent
methodology in each of the solutions below.
1. Our goal is to express the function F as F = g ◦ f for functions g and f . From Deﬁnition
6.1, we know F (x) = g (f (x)), and we can think of f (x) as being the ‘inside’ function and g
as being the ‘outside’ function. Looking at F (x) = 3x − 1 from an ‘inside versus outside’
perspective, we can think of 3x − 1 being inside the absolute value symbols. Taking this
cue, we deﬁne f (x) = 3x − 1. At this point, we have F (x) = f (x). What is the outside
function? The function which takes the absolute value of its input, g (x) = x. Sure enough,
(g ◦ f )(x) = g (f (x)) = f (x) = 3x − 1 = F (x), so we are done.
2. We attack deconstructing G from an operational approach. Given an input x, the ﬁrst step
is to square x, then add 1, then divide the result into 2. We will assign each of these steps a
function so as to write G as a composite of three functions: f , g and h. Our ﬁrst function,
f , is the function that squares its input, f (x) = x2 . The next function is the function that
adds 1 to its input, g (x) = x + 1. Our last function takes its input and divides it into 2,
2
h(x) = x . The claim is that G = h ◦ g ◦ f . We ﬁnd (h ◦ g ◦ f )(x) = h(g (f (x))) = h(g x2 ) =
h x2 + 1 = x22 = G(x).
+1
√ 3. If we look H (x) = √x+1 with an eye towards building a complicated function from simpler
x− 1
√
functions, we see the expression x is a simple piece of the larger function. If we deﬁne
√
f (x)+1
f (x) = x, we have H (x) = f (x)−1 . If we want to decompose H = g ◦ f , then we can glean
the formula from g (x) by looking at what is being done to f (x). We ﬁnd g (x) = x+1 . We
x−1
check (g ◦ f )(x) = g (f (x)) = f (x)+1
f (x)−1 = √
√x+1
x−1 = H (x), as required. 6.1 Function Composition 6.1.1 305 Exercises √
1
1. Let f (x) = 3x − 6, g (x) = x, h(x) = x and k (x) = . Find and simplify the indicated
x
composite functions. State the domain of each.
(a) (f ◦ g )(x) (h) (k ◦ f )(x) (b) (g ◦ f )(x) (i) (h ◦ k )(x) (c) (f ◦ h)(x) (j) (k ◦ h)(x) (d) (h ◦ f )(x) (k) (f ◦ g ◦ h)(x) (e) (g ◦ h)(x) (l) (h ◦ g ◦ k )(x) (f) (h ◦ g )(x) (m) (k ◦ h ◦ f )(x) (g) (f ◦ k )(x) (n) (h ◦ k ◦ g ◦ f )(x) 2. Let f (x) = 2x + 1, g (x) = x2 − x − 6 and h(x) =
composite functions. Find the domain of each. x+6
. Find and simplify the indicated
x−6 (a) (g ◦ f ) (x) (c) (h ◦ g ) (x) (b) (h ◦ f ) (x) (d) (h ◦ h) (x) 3. Let f (x) = √ x − 3, g (x) = 4x + 3 and h(x) = composite functions. Find the domain of each.
(a)
(b)
(c)
(d)
(e) (f ◦ g ) (x)
(g ◦ f ) (x)
(f ◦ h) (x)
(h ◦ f ) (x)
(g ◦ h) (x) x−2
. Find and simplify the indicated
x+3 (f) (h ◦ g ) (x)
(g) (f ◦ f ) (x)
(h) (g ◦ g ) (x)
(i) (h ◦ h) (x) √
4. Let f (x) = 9 − x2 and g (x) = x2 − 9. Find and simplify the indicated composite functions.
State the domain of each.
(a) (f ◦ f )(x) (c) (g ◦ f )(x) (b) (g ◦ g )(x) (d) (f ◦ g )(x) 306 Further Topics in Functions 5. Let f be the function deﬁned by f = {(−3, 4), (−2, 2), (−1, 0), (0, 1), (1, 3), (2, 4), (3, −1)} and
let g be the function deﬁned g = {(−3, −2), (−2, 0), (−1, −4), (0, 0), (1, −3), (2, 1), (3, 2)}.
Find the each of the following values if it exists.
(a)
(b)
(c)
(d)
(e)
(f)
(g) (h) (g ◦ f )(−2) (f ◦ g )(3)
f (g (−1))
(f ◦ f )(0)
(f ◦ g )(−3)
(g ◦ f )(3)
g (f (−3))
(g ◦ g )(−2) (i) g (f (g (0)))
(j) f (f (f (−1)))
(k) f (f (f (f (f (1)))))
n times
(l) (g ◦ g ◦ · · · ◦ g )(0) 6. Let g (x) = −x, h(x) = x + 2, j (x) = 3x and k (x) √ x − 4. In what order must these functions
=
√
be composed with f (x) = x to create F (x) = 3 −x + 2 − 4?
7. What linear functions could be used to transform f (x) = x3 into F (x) = − 1 (2x − 7)3 + 1?
2
What is the proper order of composition?
8. Write the following as a composition of two or more nonidentity functions.
(a) h(x) =
(b) r(x) = √ (c) F (x) = x2 − 1 2x − 1 2
5x + 1 9. Write the function F (x) = (d) R(x) = 3 2x3 + 1
x3 − 1 x3 + 6
as a composition of three or more nonidentity functions.
x3 − 9 10. The volume V of a cube is a function of its side length x. Let’s assume that x = t + 1 is
also a function of time t, where x is measured in inches and t is measured in minutes. Find
a formula for V as a function of t.
11. Suppose a local vendor charges $2 per hot dog and that the number of hot dogs sold per hour
x is given by x(t) = −4t2 + 20t + 92, where t is the number of hours since 10 AM, 0 ≤ t ≤ 4.
(a) Find an expression for the revenue per hour R as a function of x.
(b) Find and simplify (R ◦ x) (t). What does this represent?
(c) What is the revenue per hour at noon?
12. Discuss with your classmates how ‘realworld’ processes such as ﬁlling out federal income tax
forms or computing your ﬁnal course grade could be viewed as a use of function composition.
Find a process for which composition with itself (iteration) makes sense. 6.1 Function Composition 6.1.2 307 Answers
1
x
Domain: (0, ∞) 1. (a) (f ◦ g )(x) = 3x − 6
Domain: (−∞, ∞) (i) (h ◦ k )(x) = (b) (g ◦ f )(x) = 3x − 6
Domain: (−∞, ∞)
√
(c) (f ◦ h)(x) = 3 x − 6
Domain: [0, ∞)
√
(d) (h ◦ f )(x) = 3x − 6
Domain: [2, ∞)
√
(e) (g ◦ h)(x) = x
Domain: [0, ∞) 1
(j) (k ◦ h)(x) = √
x
Domain: (0, ∞)
√
(k) (f ◦ g ◦ h)(x) = 3 x − 6
Domain: [0, ∞) (f) (h ◦ g )(x) = x
Domain: (−∞, ∞)
3
(g) (f ◦ k )(x) = − 6
x
Domain: (−∞, 0) ∪ (0, ∞)
1
(h) (k ◦ f )(x) =
3x − 6
Domain: (−∞, 2) ∪ (2, ∞)
2. (a) (g ◦ f ) (x) = 4x2 + 2x − 6
Domain: (−∞, ∞)
2x + 7
2x − 5
Domain: −∞, 5 ∪
2 1
x
Domain: (−∞, 0) ∪ (0, ∞) (l) (h ◦ g ◦ k )(x) = (b) (h ◦ f ) (x) = 5
2, √
3. (a) (f ◦ g ) (x) = 2 x
Domain: [0, ∞)
√
(b) (g ◦ f ) (x) = 4 x − 3 + 3
Domain: [3, ∞)
−2x − 11
(c) (f ◦ h) (x) =
x+3
11
Domain: − 2 , −3
√
x−3−2
(d) (h ◦ f ) (x) = √
x−3+3
Domain: [3, ∞)
7x + 1
(e) (g ◦ h) (x) =
x+3
Domain: (−∞, −3) ∪ (−3, ∞) (m) (k ◦ h ◦ f )(x) = √ 1
3x − 6 Domain: (2, ∞)
1
3x − 6
Domain: (−∞, 2) ∪ (2, ∞) (n) (h ◦ k ◦ g ◦ f )(x) = x2 − x
x2 − x − 12
Domain: (−∞, −3) ∪ (−3, 4) ∪ (4, ∞)
7x − 30
(d) (h ◦ h) (x) = −
5x − 42
Domain: (−∞, 6) ∪ 6, 42 ∪ 452 , ∞
5
(c) (h ◦ g ) (x) = 4x + 1
4x + 6
3
Domain: −∞, − 3 ∪ − 2 , ∞
2
√
(g) (f ◦ f ) (x) =
x−3−3
(f) (h ◦ g ) (x) = Domain: [12, ∞)
(h) (g ◦ g ) (x) = 16x + 15
Domain: (−∞, ∞)
(i) (h ◦ h) (x) = −x − 8
4x + 7 Domain:
(−∞, −3) ∪ −3, − 7 ∪ − 7 , ∞
4
4 308 Further Topics in Functions (b) (g ◦ g )(x) = x4 − 18x2 + 72
Domain: (−∞, ∞) (c) (g ◦ f )(x) = −x2
Domain: [−3, 3]
√
(d) (f ◦ g )(x) =√ −x4 √ 18x2√ 72
+
−√
Domain: [− 12, − 6] ∪ [ 6, 12]4 5. (a) (f ◦ g )(3) = f (g (3)) = f (2) = 4 (h) (g ◦ f )(−2) = g (f (−2)) = g (2) = 1 4. (a) (f ◦ f )(x) = x
Domain: [−3, 3] (b) f (g (−1)) = f (−4) which is undeﬁned (i) g (f (g (0))) = g (f (0)) = g (1) = −3 (c) (f ◦ f )(0) = f (f (0)) = f (1) = 3 (j) f (f (f (−1))) = f (f (0)) = f (1) = 3 (d) (f ◦ g )(−3) = f (g (−3)) = f (−2) = 2
(f) g (f (−3)) = g (4) which is undeﬁned (k) f (f (f (f (f (1))))) = f (f (f (f (3)))) =
f (f (f (−1))) = f (f (0)) = f (1) = 3
n times (g) (g ◦ g )(−2) = g (g (−2)) = g (0) = 0 (l) (g ◦ g ◦ · · · ◦ g )(0) = 0 (e) (g ◦ f )(3) = g (f (3)) = g (−1) = −4 √
6. F (x) = 3 −x + 2 − 4 = k (j (f (h(g (x)))))
1
7. One possible solution is F (x) = − 2 (2x − 7)3 + 1 = k (j (f (h(g (x))))) where g (x) = 2x, h(x) =
1
x − 7, j (x) = − 2 x and k (x) = x + 1. You could also have F (x) = H (f (G(x))) where
1
G(x) = 2x − 7 and H (x) = − 2 x + 1. 8. (a) h(x) = (g ◦ f ) (x) where f (x) = 2x − 1
√
and g (x) = x. (c) F (x) = (g ◦ f ) (x) where f (x) = x2 − 1
and g (x) = x3 . (b) r(x) = (g ◦ f ) (x) where f (x) = 5x + 1
2
and g (x) = .
x (d) R(x) = (g ◦ f ) (x) where f (x) = x3 and
2x + 1
g (x) =
.
x−1 9. F (x) = √
x3 + 6
x+6
= (h(g (f (x))) where f (x) = x3 , g (x) =
and h(x) = x.
x3 − 9
x−9 10. V (x) = x3 so V (x(t)) = (t + 1)3
11. (a) R(x) = 2x
(b) (R ◦ x) (t) = −8t2 + 40t + 184, 0 ≤ t ≤ 4. This gives the revenue per hour as a function
of time.
(c) Noon corresponds to t = 2, so (R ◦ x) (2) = 232. The hourly revenue at noon is $232
per hour. 4 The quantity −x4 + 18x2 − 72 is a ‘quadratic in disguise’ which factors nicely. 6.2 Inverse Functions 6.2 309 Inverse Functions Thinking of a function as a process like we did in Section 2.2, in this section we seek another
function which might reverse that process. As in real life, we will ﬁnd that some processes (like
putting on socks and shoes) are reversible while some (like cooking a steak) are not. We start by
discussing a very basic function which is reversible, f (x) = 3x + 4. Thinking of f as a process, we
start with an input x and apply two steps, as we saw in Section 2.2
1. multiply by 3
2. add 4
To reverse this process, we seek a function g which will undo each of these steps and take the
output from f , 3x + 4, and return the input x. If we think of the realworld reversible twostep
process of ﬁrst putting on socks then putting on shoes, to reverse the process, we ﬁrst take oﬀ the
shoes, and then we take oﬀ the socks. In much the same way, the function g should undo the second
step of f ﬁrst. That is, the function g should
1. subtract 4
2. divide by 3
Following this procedure, we get g (x) = x−4 . Let’s check to see if the function g does the
3
job. If x = 5, then f (5) = 3(5) + 4 = 15 + 4 = 19. Taking the output 19 from f , we substitute
−
it into g to get g (19) = 193 4 = 15 = 5, which is our original input to f . To check that g does
3
the job for all x in the domain of f , we take the generic output from f , f (x) = 3x + 4, and
x
substitute that into g . That is, g (f (x)) = g (3x + 4) = (3x+4)−4 = 33 = x, which is our original
3
input to f . If we carefully examine the arithmetic as we simplify g (f (x)), we actually see g ﬁrst
‘undoing’ the addition of 4, and then ‘undoing’ the multiplication by 3. Not only does g undo
f , but f also undoes g . That is, if we take the output from g , g (x) = x−4 , and put that into
3
f , we get f (g (x)) = f x−4 = 3 x−4 + 4 = (x − 4) + 4 = x. Using the language of function
3
3
composition developed in Section 6.1, the statements g (f (x)) = x and f (g (x)) = x can be written
as (g ◦ f )(x) = x and (f ◦ g )(x) = x, respectively. Abstractly, we can visualize the relationship
between f and g in the diagram below.
f x = g (f (x)) y = f (x) g 310 Further Topics in Functions The main idea to get from the diagram is that g takes the outputs from f and returns them to
their respective inputs, and conversely, f takes outputs from g and returns them to their respective
inputs. We now have enough background to state the central deﬁnition of the section.
Definition 6.2. Suppose f and g are two functions such that
1. (g ◦ f )(x) = x for all x in the domain of f and
2. (f ◦ g )(x) = x for all x in the domain of g .
Then f and g are said to be inverses of each other. The functions f and g are said to be
invertible. Our ﬁrst result of the section formalizes the concepts that inverse functions exchange inputs
and outputs and is a consequence of Deﬁnition 6.2 and the Fundamental Graphing Principle for
Functions.
Theorem 6.2. Properties of Inverse Functions: Suppose f and g are inverse functions.
The rangea of f is the domain of g and the domain of f is the range of g
f (a) = b if and only if g (b) = a
(a, b) is on the graph of f if and only if (b, a) is on the graph of g a Recall this is the set of all outputs of a function. The third property in Theorem 6.2 tells us that the graphs of inverse functions are reﬂections
about the line y = x. For a proof of this, we refer the reader to Example 1.1.6 in Section 1.1. A
plot of the inverse functions f (x) = 3x + 4 and g (x) = x−4 conﬁrms this to be the case.
3
y
2 y = f (x) y=x 1 x
−2 −1 1 2 −1 y = g (x)
−2 If we abstract one step further, we can express the sentiment in Deﬁnition 6.2 by saying that
f and g are inverses if and only if g ◦ f = I1 and f ◦ g = I2 where I1 is the identity function 6.2 Inverse Functions 311 restricted1 to the domain of f and I2 is the identity function restricted to the domain of g . In other
words, I1 (x) = x for all x in the domain of f and I2 (x) = x for all x in the domain of g . Using
this description of inverses along with the properties of function composition listed in Theorem 6.1,
we can show that function inverses are unique.2 Suppose g and h are both inverses of a function
f . By Theorem 6.2, the domain of g is equal to the domain of h, since both are the range of f .
This means the identity function I2 applies both to the domain of h and the domain of g . Thus
h = h ◦ I2 = h ◦ (f ◦ g ) = (h ◦ f ) ◦ g = I1 ◦ g = g , as required.3 We summarize the discussion of the
last two paragraphs in the following theorem.4 Theorem 6.3. Uniqueness of Inverse Functions and Their Graphs : Suppose f is an
invertible function.
There is exactly one inverse function for f , denoted f −1 (read f inverse)
The graph of y = f −1 (x) is the reﬂection of the graph of y = f (x) across the line y = x. The notation f −1 is an unfortunate choice since you’ve been programmed since Elementary
1
Algebra to think of this as f . This is most deﬁnitely not the case since, for instance, f (x) = 3x + 4
1
has as its inverse f −1 (x) = x−4 , which is certainly diﬀerent than f (x) = 3x1 . Why does this
3
+4
confusing notation persist? As we mentioned in Section 6.1, the identity function I is to function
composition what the real number 1 is to real number multiplication. The choice of notation f −1
alludes to the property that f −1 ◦ f = I1 and f ◦ f −1 = I2 , in much the same way as 3−1 · 3 = 1
and 3 · 3−1 = 1.
Let’s turn our attention to the function f (x) = x2 . Is f invertible? A likely candidate for
√
the inverse is the function g (x) = x. Checking the composition yields (g ◦ f )(x) = g (f (x)) =
√
x2 = x, which is not equal to x for all x in the domain (−∞, ∞). For example, when x = −2,
√
f (−2) = (−2)2 = 4, but g (4) = 4 = 2, which means g failed to return the input −2 from its
output 4. What g did, however, is match the output 4 to a diﬀerent input, namely 2, which
satisﬁes f (2) = 4. This issue is presented schematically in the picture below.
1 The identity function I , which was introduced in Section 3.1 and mentioned in Theorem 6.1, has a domain of all
real numbers. In general, the domains of f and g are not all real numbers, which necessitates the restrictions listed
here.
2
In other words, invertible functions have exactly one inverse.
3
It is an excellent exercise to explain each step in this string of equalities.
4
In the interests of full disclosure, the authors would like to admit that much of the discussion in the previous
paragraphs could have easily been avoided had we appealed to the description of a function as a set of ordered pairs.
We make no apology for our discussion from a function composition standpoint, however, since it exposes the reader
to more abstract ways of thinking of functions and inverses. We will revisit this concept again in Chapter ??. 312 Further Topics in Functions
f x = −2
4 x=2 g
We see from the diagram that since both f (−2) and f (2) are 4, it is impossible to construct
a function which takes 4 back to both x = 2 and x = −2. (By deﬁnition, a function matches
a real number with exactly one other real number.) From a graphical standpoint, we know that
if y = f −1 (x) exists, its graph can be obtained by reﬂecting y = x2 about the line y = x, in
accordance with Theorem 6.3. Doing so produces
y
7 y
(4, 2) 6
5 2 4 (−2, 4) 1 (2, 4)
3 1 2
1 −2 −1 2 3 4 5 6 7 x −1
−2
1 y = f ( x) = x 2 2 x reﬂect across y = x −− − − − − − −
− − − − − − −→
switch x and y coordinates (4, −2) y = f −1 ( x ) ? We see that the line x = 4 intersects the graph of the supposed inverse twice  meaning the
graph fails the Vertical Line Test, Theorem 2.1, and as such, does not represent y as a function of
x. The vertical line x = 4 on the graph on the right corresponds to the horizontal line y = 4 on the
graph of y = f (x). The fact that the horizontal line y = 4 intersects the graph of f twice means
two diﬀerent inputs, namely x = −2 and x = 2, are matched with the same output, 4, which is
the cause of all of the trouble. In general, for a function to have an inverse, diﬀerent inputs must
go to diﬀerent outputs, or else we will run into the same problem we did with f (x) = x2 . We give
this property a name.
Definition 6.3. A function f is said to be onetoone if f matches diﬀerent inputs to diﬀerent
outputs. Equivalently, f is onetoone if and only if whenever f (c) = f (d), then c = d. Graphically, we detect onetoone functions using the test below. 6.2 Inverse Functions 313 Theorem 6.4. The Horizontal Line Test: A function f is onetoone if and only if no
horizontal line intersects the graph of f more than once. We say that the graph of a function passes the Horizontal Line Test if no horizontal line
intersects the graph more than once; otherwise, we say the graph of the function fails the Horizontal
Line Test. We have argued that if f is invertible, then f must be onetoone, otherwise the graph
given by reﬂecting the graph of y = f (x) about the line y = x will fail the Vertical Line Test. It
turns out that being onetoone is also enough to guarantee invertibility. To see this, we think of f
as the set of ordered pairs which constitute its graph. If switching the x and y coordinates of the
points results in a function, then f is invertible and we have found f −1 . This is precisely what the
Horizontal Line Test does for us: it checks to see whether or not a set of points describes x as a
function of y . We summarize these results below. Theorem 6.5. Equivalent Conditions for Invertibility: Suppose f is a function. The
following statements are equivalent.
f is invertible.
f is onetoone.
The graph of f passes the Horizontal Line Test. We put this result to work in the next example.
Example 6.2.1. Determine if the following functions are onetoone in two ways: (a) analytically
using Deﬁnition 6.3 and (b) graphically using the Horizontal Line Test.
1 − 2x
5
2x
2. g (x) =
1−x
1. f (x) = 3. h(x) = x2 − 2x + 4
4. F = {(−1, 1), (0, 2), (2, 1)} Solution.
1. (a) To determine if f is onetoone analytically, we assume f (c) = f (d) and attempt to
deduce that c = d. 314 Further Topics in Functions f (c) = f (d)
1 − 2c
5 = 1 − 2d
5 1 − 2c = 1 − 2d
−2c = −2d
c=d
Hence, f is onetoone.
(b) To check if f is onetoone graphically, we look to see if the graph of y = f (x) passes the
Horizontal Line Test. We have that f is a nonconstant linear function, which means its
graph is a nonhorizontal line. Thus the graph of f passes the Horizontal Line Test as
seen below.
2. (a) We begin with the assumption that g (c) = g (d) and try to show c = d.
g (c) = g (d)
2c
1−c = 2d
1−d 2c(1 − d) = 2d(1 − c)
2c − 2cd = 2d − 2dc
2c = 2d
c=d
We have shown that g is onetoone.
(b) We can graph g using the six step procedure outlined in Section 5.2. We get the sole
intercept at (0, 0), a vertical asymptote x = 1 and a horizontal asymptote (which the
graph never crosses) y = −2. We see from that the graph of g passes the Horizontal
Line Test.
y y
4
3
2
1 2
1 x
−2 −1 1
−1
−2 y = f (x) 2 −2 −1 −1 x
1 −2
−3
−4
−5
−6 y = g (x) 2 6.2 Inverse Functions 315 3. (a) We begin with h(c) = h(d). As we work our way through the problem, we encounter a
nonlinear equation. We move the nonzero terms to the left, leave a 0 on the right and
factor accordingly.
h(c) = h(d)
c2 − 2c + 4 = d2 − 2d + 4
c2 − 2c = d2 − 2d
c2 − d2 − 2c + 2d = 0
(c + d)(c − d) − 2(c − d) = 0
(c − d)((c + d) − 2) = 0 factor by grouping c − d = 0 or c + d − 2 = 0
c = d or c = 2 − d
We get c = d as one possibility, but we also get the possibility that c = 2 − d. This
suggests that f may not be onetoone. Taking d = 0, we get c = 0 or c = 2. With
f (0) = 4 and f (2) = 4, we have produced two diﬀerent inputs with the same output
meaning f is not onetoone.
(b) We note that h is a quadratic function and we graph y = h(x) using the techniques
presented in Section 3.3. The vertex is (1, 3) and the parabola opens upwards. We see
immediately from the graph that h is not onetoone, since there are several horizontal
lines which cross the graph more than once.
4. (a) The function F is given to us as a set of ordered pairs. The condition F (c) = F (d)
means the outputs from the function (the y coordinates of the ordered pairs) are the
same. We see that the points (−1, 1) and (2, 1) are both elements of F with F (−1) = 1
and F (2) = 1. Since −1 = 2, we have established that F is not onetoone.
(b) Graphically, we see the horizontal line y = 1 crosses the graph more than once. Hence,
the graph of F fails the Horizontal Line Test.
y
6 y 5
4 2 3 1 x
2
−2 −1 1 1 y = F (x)
1 2 −1 y = h(x) x 2 316 Further Topics in Functions We have shown that the functions f and g in Example 6.2.1 are onetoone. This means they
are invertible, so it is natural to wonder what f −1 (x) and g −1 (x) would be. For f (x) = 1−2x ,
5
we can think our way through the inverse since there is only one occurrence of x. We can track
stepbystep what is done to x and reverse those steps as we did at the beginning of the chapter.
The function g (x) = 12xx is a bit trickier since x occurs in two places. When one evaluates g (x) for
−
a speciﬁc value of x, which is ﬁrst, the 2x or the 1 − x? We can imagine functions more complicated
than these so we need to develop a general methodology to attack this problem. Theorem 6.2 tells
us equation y = f −1 (x) is equivalent to f (y ) = x and this is the basis of our algorithm. Steps for ﬁnding the Inverse of a Onetoone Function
1. Write y = f (x)
2. Interchange x and y
3. Solve x = f (y ) for y to obtain y = f −1 (x) Note that we could have simply written ‘Solve x = f (y ) for y ’ and be done with it. The act of
interchanging the x and y is there to remind us that we are ﬁnding the inverse function by switching
the inputs and outputs. Example 6.2.2. Find the inverse of the following onetoone functions. Check your answers analytically using function composition and graphically. 1. f (x) = 1 − 2x
5 2. g (x) = 2x
1−x Solution. 1. As we mentioned earlier, it is possible to think our way through the inverse of f by recording
the steps we apply to x and the order in which we apply them and then reversing those steps
in the reverse order. We encourage the reader to do this. We, on the other hand, will practice
the algorithm. We write y = f (x) and proceed to switch x and y 6.2 Inverse Functions 317 y = f (x)
1 − 2x
y=
5
1 − 2y
x=
5 switch x and y 5x = 1 − 2y
5x − 1 = −2y
5x − 1
=y
−2
y = −5x +
2 1
2 1
We have f −1 (x) = − 5 x+ 2 . To check this answer analytically, we ﬁrst check that f −1 ◦ f (x) =
2
x for all x in the domain of f , which is all real numbers. f −1 ◦ f (x) = f −1 (f (x))
1
2 = − 5 f (x) +
2
= −5
2 1 − 2x
5 = − 1 (1 − 2x) +
2
1
= −2 + x + + 1
2 1
2 1
2 =x
We now check that f ◦ f −1 (x) = x for all x in the range of f which is also all real numbers.
(Recall that the domain of f −1 ) is the range of f .)
f ◦ f −1 (x) = f (f −1 (x))
=
=
=
=
= 1 − 2f −1 (x)
5
5
1 − 2 −2x +
5
1 + 5x − 1
5
5x
5
x 1
2 318 Further Topics in Functions
To check our answer graphically, we graph y = f (x) and y = f −1 (x) on the same set of axes.5
They appear to be reﬂections across the line y = x.
y 2 y=x
1 −4 −3 −2 −1 1 2 3 4 x −1 y = f (x)
−2 y = f −1 (x) 2. To ﬁnd g −1 (x), we start with y = g (x). We note that the domain of g is (−∞, 1) ∪ (1, ∞).
y = g (x)
2x
1−x
2y
x=
1−y
x(1 − y ) = 2y
y= switch x and y x − xy = 2y
x = xy + 2y
x = y (x + 2)
y= factor x
x+2 x
We obtain g −1 (x) = x+2 . To check this analytically, we ﬁrst check g −1 ◦ g (x) = x for all x
in the domain of g , that is, for all x = 1. g −1 ◦ g (x) g −1 (g (x)) =
5 = g −1 2x
1−x Note that if you perform your check on a calculator for more sophisticated functions, you’ll need to take advantage
of the ‘ZoomSquare’ feature to get the correct geometric perspective. 6.2 Inverse Functions 319
2x
1−x
2x
+2
1−x = = = 2x
(1 − x)
1−x
·
2x
(1 − x)
+2
1−x
2x
2x + 2(1 − x) = 2x
2x + 2 − 2x = 2x
2 = clear denominators x Next, we check g g −1 (x) = x for all x in the range of g . From the graph of g in Example
6.2.1, we have that the range of g is (−∞, −2) ∪ (−2, ∞). This matches the domain we get
x
from the formula g −1 (x) = x+2 , as it should. 320 Further Topics in Functions g ◦ g −1 (x) = g g −1 (x)
x
x+2 =g = x
x+2
x
1−
x+2 = x
x+2
x
1−
x+2 = 2x
(x + 2) − x = 2x
2 2 2 · (x + 2)
(x + 2) clear denominators =x Graphing y = g (x) and y = g −1 (x) on the same set of axes is busy, but we can see the symmetric relationship if we thicken the curve for y = g −1 (x). Note that the vertical asymptote
x = 1 of the graph of g corresponds to the horizontal asymptote y = 1 of the graph of g −1 ,
as it should since x and y are switched. Similarly, the horizontal asymptote y = −2 of the
graph of g corresponds to the vertical asymptote x = −2 of the graph of g −1 . 6.2 Inverse Functions 321
y
6
5
4 y=x 3
2
1 −6 −5 −4 −3 −2 −1 1 2 3 4 5 x 6 −1
−2
−3
−4
−5
−6 y = g (x) and y = g −1 (x) We now return to f (x) = x2 . We know that f is not onetoone, and thus, is not invertible.
However, if we restrict the domain of f , we can produce a new function g which is onetoone. If
we deﬁne g (x) = x2 , x ≥ 0, then we have
y y 7
6 6 5 5 4 4 3 3 2 2 1 −2 −1 7 1 1 2 y = f (x) = x2 −2 −1 x
restrict domain to x ≥ 0 −− − − − − − −
− − − − − − −→ 1 2 x y = g (x) = x2 , x ≥ 0 The graph of g passes the Horizontal Line Test. To ﬁnd an inverse of g , we proceed as usual 322 Further Topics in Functions y = g (x)
y = x2 , x ≥ 0
x = y2, y ≥ 0
√
y=±x
√
y=
x switch x and y since y ≥ 0 √
We get g −1 (x) = x. At ﬁrst it looks like we’ll run into the same trouble as before, but
when we check the composition, the domain restriction on g saves the day. We get g −1 ◦ g (x) =
√
g −1 (g (x)) = g −1 x2 = x2 = x = x, since x ≥ 0. Checking g ◦ g −1 (x) = g g −1 (x) =
√
√2
g ( x) = ( x) = x. Graphing6 g and g −1 on the same set of axes shows that they are reﬂections
about the line y = x.
y
y = g (x) 8 y=x 7
6
5 y = g −1 (x) 4
3
2
1 1 2 3 4 5 6 7 8 x Our next example continues the theme of domain restriction.
Example 6.2.3. Graph the following functions to show they are onetoone and ﬁnd their inverses.
Check your answers analytically using function composition and graphically.
1. j (x) = x2 − 2x + 4, x ≤ 1. 2. k (x) = √ x+2−1 Solution.
1. The function j is a restriction of the function h from Example 6.2.1. Since the domain of j
is restricted to x ≤ 1, we are selecting only the ‘left half’ of the parabola. We see that the
graph of j passes the Horizontal Line Test and thus j is invertible.
6 We graphed y = √
x in Section 2.5. 6.2 Inverse Functions 323
y
6
5
4
3
2
1 1 x 2 −1 y = j (x) We now use our algorithm to ﬁnd j −1 (x). y = j (x)
y = x2 − 2x + 4 , x ≤ 1
x = y 2 − 2y + 4 , y ≤ 1
0 = y 2 − 2y + 4 − x
2 ± (−2)2 − 4(1)(4 − x)
y=
2(1)
√
2 ± 4x − 12
y=
2
2 ± 4(x − 3)
y=
2
√
2±2 x−3
y=
2
√
2 1± x−3
y=
2
√
y = 1± x−3
√
y = 1− x−3 switch x and y quadratic formula, c = 4 − x since y ≤ 1. √
We have j −1 (x) = 1 − x − 3. When we simplify j −1 ◦ j (x), we need to remember that
the domain of j is x ≤ 1. 324 Further Topics in Functions j −1 ◦ j (x) = j −1 (j (x))
= j −1 x2 − 2x + 4 , x ≤ 1
(x2 − 2x + 4) − 3 = 1−
= 1− √ x2 − 2x + 1
(x − 1)2 = 1− = 1 −  x − 1
= 1 − (−(x − 1)) since x ≤ 1 =x
Checking j ◦ j −1 , we get
j ◦ j −1 (x) = j j −1 (x)
√
= j 1− x−3
√
√
2
= 1− x−3 −2 1− x−3 +4
√
√
√
2
= 1−2 x−3+
x−3 −2+2 x−3+4
= 3+x−3
=x
We can use what we know from Section 2.5 to graph y = j −1 (x) on the same axes as y = j (x)
to get
y
6
5 y = j (x) y=x 4
3
2
1 1 2 3 4 5 6 x −1 y = j −1 (x) 2. We graph y = k (x) = √ x + 2 − 1 using what we learned in Section 2.5 and see k is onetoone. 6.2 Inverse Functions 325
y
2
1 −2 −1 1 2 x −1
−2 y = k (x) We now try to ﬁnd k −1 . y = k (x)
√
x+2−1
y=
√
x=
y+2−1
√
x+1 =
y+2
√
2
(x + 1)2 =
y+2 switch x and y x2 + 2 x + 1 = y + 2
y = x2 + 2 x − 1
We have k −1 (x) = x2 + 2x − 1. Based on our experience, we know something isn’t quite right.
We determined k −1 is a quadratic function, and we have seen several times in this section
that these are not onetoone unless their domains are suitably restricted. Theorem 6.2 tells
us that the domain of k −1 is the range of k . From the graph of k , we see that the range
is [−1, ∞), which means we restrict the domain of k −1 to x ≥ −1. We now check that this
works in our compositions.
k −1 ◦ k (x) = k −1 (k (x))
√
= k −1
x + 2 − 1 , x ≥ −2
√
√
2
=
x+2−1 +2 x+2−1 −1
√
√
√
2
=
x+2 −2 x+2+1+2 x+2−2−1
= x+2−2
=x
and 326 Further Topics in Functions k ◦ k −1 (x) = k x2 + 2x − 1 (x2 + 2x − 1) + 2 − 1 =
= x ≥ −1 √ x2 + 2 x + 1 − 1
(x + 1)2 − 1 = = x + 1 − 1
= x+1−1 since x ≥ −1 =x
Graphically, everything checks out as well, provided that we remember the domain restriction
on k −1 means we take the right half of the parabola.
y
2
y = k(x) 1 −2 −1 1
−1 2 x y = k−1 (x) −2 Our last example of the section gives an application of inverse functions.
Example 6.2.4. Recall from Section 3.1 that the pricedemand equation for the PortaBoy game
system is p(x) = −1.5x + 250 for 0 ≤ x ≤ 166, where x represents the number of systems sold
weekly and p is the price per system in dollars.
1. Explain why p is onetoone and ﬁnd a formula for p−1 (x). State the restricted domain.
2. Find and interpret p−1 (220).
3. Recall from Section 3.3 that the weekly proﬁt P , in dollars, as a result of selling x systems is
given by P (x) = −1.5x2 + 170x − 150. Find and interpret P ◦ p−1 (x).
4. Use your answer to part 3 to determine the price per PortaBoy which would yield the maximum proﬁt. Compare with Example 3.3.3.
Solution. 6.2 Inverse Functions 327 1. We leave to the reader to show the graph of p(x) = −1.5x + 250, 0 ≤ x ≤ 166, is a line
segment from (0, 250) to (166, 1), and as such passes the Horizontal Line Test. Hence, p is
−
onetoone. We ﬁnd the expression for p−1 (x) as usual and get p−1 (x) = 5003 2x . The domain
of p−1 should match the range of p, which is [1, 250], and as such, we restrict the domain of
p−1 to 1 ≤ x ≤ 250.
2(220)
2. We ﬁnd p−1 (220) = 500−3
= 20. Since the function p took as inputs the weekly sales and
furnished the price per system as the output, p−1 takes the price per system and returns the
weekly sales as its output. Hence, p−1 (220) = 20 means 20 systems will be sold in a week if
the price is set at $220 per system.
2 −
−
−
3. We compute P ◦ p−1 (x) = P p−1 (x) = P 5003 2x = −1.5 5003 2x + 170 5003 2x − 150.
7 we obtain P ◦ p−1 (x) = − 2 x2 +220x − 40450 .
After a hefty amount of Elementary Algebra,
3
3
To understand what this means, recall that the original proﬁt function P gave us the weekly
proﬁt as a function of the weekly sales. The function p−1 gives us the weekly sales as a
function of the price. Hence, P ◦ p−1 takes as its input a price. The function p−1 returns the
weekly sales, which in turn is fed into P to return the weekly proﬁt. Hence, P ◦ p−1 (x)
gives us the weekly proﬁt (in dollars) as a function of the price per system, x, using the weekly
sales p−1 (x) as the ‘middle man’. 4. We know from Section 3.3 that the graph of y = P ◦ p−1 (x) is a parabola opening downwards. The maximum proﬁt is realized at the vertex. Since we are concerned only with the
price per system, we need only ﬁnd the xcoordinate of the vertex. Identifying a = − 2 and
3
b = 220, we get, by the Vertex Formula, Equation 3.4, x = − 2ba = 165. Hence, weekly proﬁt
is maximized if we set the price at $165 per system. Comparing this with our answer from
Example 3.3.3, there is a slight discrepancy to the tune of $0.50. We leave it to the reader to
balance the books appropriately. 7 It is good review to actually do this! 328 6.2.1 Further Topics in Functions Exercises 1. Show that the following functions are onetoone and ﬁnd the inverse. Check your answers
algebraically and graphically. Verify the range of f is the domain of f −1 and viceversa.
(a) f (x) = 6x − 2
(b) f (x) = 5x − 3
4 + 3x
(c) f (x) = 1 −
√5
(d) f (x) = − x − 5 + 2
√
(e) f (x) = 3x − 1 + 5
√
(f) f (x) = 5 3x − 1
(g) f (x) = x2 − 10x, x ≥ 5
(h) f (x) = 3(x + 4)2 − 5, x ≤ −4
(i) f (x) = x2 − 6x + 5, x ≤ 3 (j) f (x) = 4x2 + 4x + 1, x < −1
3
(k) f (x) =
4−x
x
(l) f (x) =
1 − 3x
2x − 1
(m) f (x) =
3x + 4
4x + 2
(n) f (x) =
3x − 6
−3x − 2
(o) f (x) =
x+3 2. Show that the Fahrenheit to Celsius conversion function found in Exercise 3 in Section 3.1 is
invertible and that its inverse is the Celsius to Fahrenheit conversion function.
3. Analytically show that the function f (x) = x3 + 3x + 1 is onetoone. Since ﬁnding a formula
for its inverse is beyond the scope of this textbook, use Theorem 6.2 to help you compute
f −1 (1), f −1 (5), and f −1 (−3).
4. With the help of your classmates, ﬁnd a formula for the inverse of the following.
(a) f (x) = ax + b, a = 0
√
(b) f (x) = a x − h + k, a = 0, x ≥ h (c) f (x) =
(d) f (x) = ax+b
cx+d , a = 0, b = 0, c = 0, d = 0
ax2 + bx + c where a = 0, x ≥ − 2ba . 2x
5. Let f (x) = x2 −1 . Using the techniques in Section 5.2, graph y = f (x). Verify f is onetoone on the interval (−1, 1). Use the procedure outlined on Page 316 and your graphing
calculator to ﬁnd the formula for f −1 (x). Note that since f (0) = 0, it should be the case that
f −1 (0) = 0. What goes wrong when you attempt to substitute x = 0 into f −1 (x)? Discuss
with your classmates how this problem arose and possible remedies. 6. Suppose f is an invertible function. Prove that if graphs of y = f (x) and y = f −1 (x) intersect
at all, they do so on the line y = x.
7. With the help of your classmates, explain why a function which is either strictly increasing
or strictly decreasing on its entire domain would have to be onetoone, hence invertible.
8. Let f and g be invertible functions. With the help of your classmates show that (f ◦ g ) is
onetoone, hence invertible, and that (f ◦ g )−1 (x) = (g −1 ◦ f −1 )(x).
9. What graphical feature must a function f possess for it to be its own inverse? 6.2 Inverse Functions 6.2.2 329 Answers
x+2
6
x+3
(b) f −1 (x) =
5
−1 (x) = − 5 x +
(c) f
3 (i) f −1 (x) = 3 − 1. (a) f −1 (x) = (d) f −1 (x) = (x − 1
3
2)2 + 5, x ≤ 2 1
(f) f −1 (x) = 1 x5 + 3
3
√
(g) f −1 (x) = 5 + x + 25
x+5
3 x+4 √ 1
(e) f −1 (x) = 3 (x − 5)2 + 1 , x ≥ 5
3 (h) f −1 (x) = − √ −4 (j) f −1 (x) = − x+1 , x > 1
2
4x − 3
(k) f −1 (x) =
x
x
−1 (x) =
(l) f
3x + 1
4x + 1
(m) f −1 (x) =
2 − 3x
6x + 2
(n) f −1 (x) =
3x − 4
−3x − 2
(o) f −1 (x) =
x+3 3. Given that f (0) = 1, we have f −1 (1) = 0. Similarly f −1 (5) = 1 and f −1 (−3) = −1 330
20 Function Composition 21 Inverse Functions Further Topics in Functions Chapter 7 Exponential and Logarithmic
Functions 7.1 Introduction to Exponential and Logarithmic Functions Of all of the functions we study in this text, exponential and logarithmic functions are possibly
the ones which impact everyday life the most.1 This section will introduce us to these functions
while the rest of the chapter will more thoroughly explore their properties. Up to this point, we
have dealt with functions which involve terms like x2 or x2/3 , in other words, terms of the form xp
where the base of the term, x, varies but the exponent of each term, p, remains constant. In this
chapter, we study functions of the form f (x) = bx where the base b is a constant and the exponent
x is the variable. We start our exploration of these functions with f (x) = 2x . (Apparently this is a
tradition. Every College Algebra book we have ever read starts with f (x) = 2x .) We make a table
of values, plot the points and connect them in a pleasing fashion. 1 Take a class in Diﬀerential Equations and you’ll see why. 332 Exponential and Logarithmic Functions x
−3 f (x) (x, f (x))
2−3 = 1
8 −3, 1
8 y
8 2−2 = 1
4 −2, 1
4 7 −1 2−1 −1, 1
2 5 0 20 = 1 (0, 1) 1 21 = 2 (1, 2) 2 22 =4 (2, 4) 3 23 = 8 (2, 8) −2 6 4
3
2
1
−3 −2 −1 1 2 3 x y = f (x) = 2x A few remarks about the graph of f (x) = 2x which we have constructed are in order. As
x → −∞ and attains values like x = −100 or x = −1000, the function f (x) = 2x takes on values
1
1
like f (−100) = 2−100 = 2100 or f (−1000) = 2−1000 = 21000 . In other words, as x → −∞,
2x ≈ 1
≈ very small (+)
very big (+) So as x → −∞, 2x → 0+ . This is represented graphically using the xaxis (the line y = 0) as a
horizontal asymptote. On the ﬂip side, as x → ∞, we ﬁnd f (100) = 2100 , f (1000) = 21000 , and so
on, thus 2x → ∞. As a result, our graph suggests the range of f is (0, ∞). The graph of f passes
the Horizontal Line Test which means f is onetoone and hence invertible. We also note that when
we ‘connected the dots in a pleasing fashion’, we have made the implicit assumption that f (x) = 2x
is continuous2 and has a domain of all real numbers. In particular, we have suggested that things
√
√
like 2 3 exist as real numbers. We should take a moment to discuss what something like 2 3 might
mean, and refer the interested reader to a solid course in Calculus for a more rigorous explanation.
√
The number 3 = 1.73205 . . . is an irrational number3 and as√
such, its decimal representation
neither repeats nor terminates. We can, however, approximate 3 by terminating decimals, and
√
√
it stands to reason4 we can use these to approximate√ 3 . For example, if we approximate 3
2
√
173
100 173
by 1.73, we can approximate 2 3 ≈ 21.73 = 2 100 =
2 . It is not, by any means, a pleasant
number, but it is at least a number that we understand in terms of powers and roots. It also stands
2 Recall that this means there are no holes or other kinds of breaks in the graph.
You can actually prove this by considering the polynomial p(x) = x2 − 3 and showing it has no rational zeros by
applying Theorem ??.
4
This is where Calculus and continuity come into play.
3 7.1 Introduction to Exponential and Logarithmic Functions 333 √
to reason that better √
and better approximations of 3 yield better and better approximations of
√
2 3 , so the value of 2 3 should be the result of this sequence of approximations.5
Suppose we wish to study the family of functions f (x) = bx . Which bases b make sense to
study? We ﬁnd that we run into diﬃculty if b < 0. For example, if b = −2, then the function
√
f (x) = (−2)x has trouble, for instance, at x = 1 since (−2)1/2 = −2 is not a real number. In
2
general, if x is any rational number with an even denominator, then (−2)x is not deﬁned, so we
must restrict our attention to bases b ≥ 0. What about b = 0? The function f (x) = 0x is undeﬁned
for x ≤ 0 because we cannot divide by 0 and 00 is an indeterminant form. For x > 0, 0x = 0 so
the function f (x) = 0x is the same as the function f (x) = 0, x > 0. We know everything we can
possibly know about this function, so we exclude it from our investigations. The only other base
we exclude is b = 1, since the function f (x) = 1x = 1 is, once again, a function we have already
studied. We are now ready for our deﬁnition of exponential functions.
Definition 7.1. A function of the form f (x) = bx where b is a ﬁxed real number, b > 0, b = 1
is called a base b exponential function. We leave it to the reader to verify6 that if b > 1, then the exponential function f (x) = bx
will share the same basic shape and characteristics as f (x) = 2x . What if 0 < b < 1? Consider
1x
g (x) = 2 . We could certainly build a table of values and connect the points, or we could take a
x
x
step back and note that g (x) = 1 = 2−1 = 2−x = f (−x), where f (x) = 2x . Thinking back
2
to Section 2.5, the graph of f (−x) is obtained from the graph of f (x) by reﬂecting it across the
y axis. As such, we have
y y
8
7 7 6 6 5 5 4 4 3 3 2 2 1
−3 −2 −1 8 1
1 2 y = f (x) = 2x 3 x reﬂect across y axis −− − − − −→
−−−−−− multiply each xcoordinate by −1 −3 −2 −1 1 2 y = g (x) = 2−x = 3x
` 1 ´x
2 We see that the domain and range of g match that of f , namely (−∞, ∞) and (0, ∞), respectively. Like f , g is also onetoone. Whereas f is always increasing, g is always decreasing. As
a result, as x → −∞, g (x) → ∞, and on the ﬂip side, as x → ∞, g (x) → 0+ . It shouldn’t be
too surprising that for all choices of the base 0 < b < 1, the graph of y = bx behaves similarly to
the graph of g . We summarize these observations, and more, in the following theorem whose proof
ultimately requires Calculus.
5
6 Want more information? Look up “convergent sequences” on the Internet.
Meaning, graph some more examples on your own. 334 Exponential and Logarithmic Functions Theorem 7.1. Properties of Exponential Functions: Suppose f (x) = bx .
The domain of f is (−∞, ∞) and the range of f is (0, ∞).
(0, 1) is on the graph of f and y = 0 is a horizontal asymptote to the graph of f .
f is onetoone, continuous and smootha
If 0 < b < 1: If b > 1: – f is always increasing – f is always decreasing – As x → −∞, f (x) → 0+ – As x → −∞, f (x) → ∞ – As x → ∞, f (x) → ∞ – As x → ∞, f (x) → 0+ – The graph of f resembles: – The graph of f resembles: y = bx , b > 1 a y = bx , 0 < b < 1 Recall that this means the graph of f has no sharp turns or corners. Of all of the bases for exponential functions, two occur the most often in scientiﬁc circles.
The ﬁrst, base 10, is often called the common base. The second base is an irrational number,
e ≈ 2.718, called the natural base. We will more formally discuss the origins of this number in
Section 7.5. For now, it is enough to know that since e > 1, f (x) = ex is an increasing exponential
function. The following examples oﬀer a glimpse as to the kind of realworld phenomena these
functions can model.
Example 7.1.1. The value of a car can be modeled by V (x) = 25
car in years and V (x) is the value in thousands of dollars. 4x
5, where x ≥ 0 is age of the 1. Find and interpret V (0).
2. Sketch the graph of y = V (x) using transformations.
3. Find and interpret the horizontal asymptote of the graph you found in 2.
Solution. 7.1 Introduction to Exponential and Logarithmic Functions 335 0 1. To ﬁnd V (0), we replace x with 0 to obtain V (0) = 25 4 = 25. Since x represents the age
5
of the car in years, x = 0 corresponds to the car being brand new. Since V (x) is measured
in thousands of dollars, V (0) = 25 corresponds to a value of $25, 000. Putting it all together,
we interpret V (0) = 25 to mean the purchase price of the car was $25, 000.
x x 4
2. To graph y = 25 5 , we start with the basic exponential function f (x) = 4 . Since the
5
4
base b = 5 is between 0 and 1, the graph of y = f (x) is decreasing. We plot the y intercept
(0, 1) and two other points, −1, 5 and 1, 4 , and label the horizontal asymptote y = 0.
4
5
4x
To obtain V (x) = 25 5 , x ≥ 0, we multiply the output from f by 25, in other words,
V (x) = 25f (x). In accordance with Theorem 2.5, this results in a vertical stretch by a factor
of 25. We multiply all of the y values in the graph by 25 (including the y value of the
horizontal asymptote) and obtain the points −1, 125 , (0, 25) and (1, 20). The horizontal
4
asymptote remains y = 0. Finally, we restrict the domain to [0, ∞) to ﬁt with the applied
domain given to us. We have the result below.
y
y
30 2 (0, 25)
(0, 1) 20
15
10 −3 −2 −1 1 2 3 vertical scale by a factor of 25 H.A. y = 0 y = f (x) = 5 x ` 4 ´x
5 −− − − − − − − − − −
− − − − − − − − − −→
multiply each y coordinate by 25 1 2 3456
H.A. y = 0 x y = V (x) = 25f (x), x ≥ 0 3. We see from the graph of V that its horizontal asymptote is y = 0. (We leave it to reader to
verify this analytically by thinking about what happens as we take larger and larger powers
4
of 5 .) This means as the car gets older, its value diminishes to 0.
The function in the previous example is often called a ‘decay curve’. Increasing exponential
functions are used to model ‘growth curves’ and we shall see several diﬀerent examples of those
in Section 7.5. For now, we present another common decay curve which will serve as the basis
for further study of exponential functions. Although it may look more complicated than the previous example, it is actually just a basic exponential function which has been modiﬁed by a few
transformations from Section 2.5.
Example 7.1.2. According to Netwon’s Law of Cooling7 the temperature of coﬀee T (in degrees
Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t .
1. Find and interpret T (0).
2. Sketch the graph of y = T (t) using transformations.
3. Find and interpret the horizontal asymptote of the graph.
7 We will discuss this in greater detail in Section 7.5. 336 Exponential and Logarithmic Functions Solution.
1. To ﬁnd T (0), we replace every occurrence of the independent variable t with 0 to obtain
T (0) = 70 + 90e−0.1(0) = 160. This means that the coﬀee was served at 160◦ F.
2. To graph y = T (t) using transformations, we start with the basic function, f (t) = et . As we
have already remarked, e ≈ 2.718 > 1 so the graph of f is an increasing exponential with
y intercept (0, 1) and horizontal asymptote y = 0. The points −1, e−1 ≈ (−1, 0.37) and
(1, e) ≈ (1, 2.72) are also on the graph. Since the formula T (t) looks rather complicated, we
rewrite T (t) in the form presented in Theorem 2.7 and use that result to track the changes to
our three points and the horizontal asymptote. We have T (t) = 90e−0.1t + 70 = 90f (−0.1t) +
70. Multiplication of the input to f , t, by −0.1 results in a horizontal expansion by a factor
of 10 as well as a reﬂection about the y axis. We divide each of the x values of our points by
−0.1 (which amounts to multiplying them by −10) to obtain 10, e−1 , (0, 1), and (−10, e).
Since none of these changes aﬀected the y values, the horizontal asymptote remains y = 0.
Next, we see that the output from f is being multiplied by 90. This results in a vertical
stretch by a factor of 90. We multiply the y coordinates by 90 to obtain 10, 90e−1 , (0, 90),
and (−10, 90e). We also multiply the y value of the horizontal asymptote y = 0 by 90, and it
remains y = 0. Finally, we add 70 to all of the y coordinates, which shifts the graph upwards to
obtain 10, 90e−1 + 70 ≈ (10, 103.11), (0, 160), and (−10, 90e + 70) ≈ (−10, 314.64). Adding
70 to the horizontal asymptote shifts it upwards as well to y = 70. We connect these three
points using the same shape in the same direction as in the graph of f and, last but not least,
we restrict the domain to match the applied domain [0, ∞). The result is below.
y
180 y
7 160 6 140 5 120 4 100 3 80 2 60 H.A. y = 70 40 (0, 1) 20
−3 −2 −1
12
H.A. y = 0 y = f (t) = et 3 t
2 −− − − − −→
−−−−−− 4 6 8 10 12 14 16 18 20 t y = T (t) 3. From the graph, we see that the horizontal asymptote is y = 70. It is worth a moment or two
of our time to see how this happens analytically and to review some of the ‘number sense’
developed in Chapter 5. As t → ∞, We get T (t) = 70 + 90e−0.1t ≈ 70 + 90every big (−) . Since
1
e > 1, every big (−) = every 1 (+) ≈ very big (+) ≈ very small (+). The larger t becomes, the smaller
big
e−0.1t becomes, so the term 90e−0.1t ≈ very small (+). Hence, T (t) ≈ 70 + very small (+)
which means the graph is approaching the horizontal line y = 70 from above. This means
that as time goes by, the temperature of the coﬀee is cooling to 70◦ F, presumably room
temperature. 7.1 Introduction to Exponential and Logarithmic Functions 337 As we have already remarked, the graphs of f (x) = bx all pass the Horizontal Line Test.
Thus the exponential functions are invertible. We now turn our attention to these inverses, the
logarithmic functions, which are called ‘logs’ for short. Definition 7.2. The inverse of the exponential function f (x) = bx is called the base b logarithm function, and is denoted f −1 (x) = logb (x) The expression logb (x) is read ‘log base b of
x.’ We have special notations for the common base, b = 10, and the natural base, b = e. Definition 7.3. The common logarithm of a real number x is log10 (x) and is usually written
log(x). The natural logarithm of a real number x is loge (x) and is usually written ln(x). Since logs are deﬁned as the inverses of exponential functions, we can use Theorems 6.2 and 6.3
to tell us about logarithmic functions. For example, we know that the domain of a log function is the
range of an exponential function, namely (0, ∞), and that the range of a log function is the domain
of an exponential function, namely (−∞, ∞). Since we know the basic shapes of y = f (x) = bx for
the diﬀerent cases of b, we can obtain the graph of y = f −1 (x) = logb (x) by reﬂecting the graph of
f across the line y = x as shown below. The y intercept (0, 1) on the graph of f corresponds to
an xintercept of (1, 0) on the graph of f −1 . The horizontal asymptotes y = 0 on the graphs of the
exponential functions become vertical asymptotes x = 0 on the log graphs. y = bx , b > 1
y = logb (x), b > 1 y = bx , 0 < b < 1
y = logb (x), 0 < b < 1 On a procedural level, logs undo the exponentials. Consider the function f (x) = 2x . When
we evaluate f (3) = 23 = 8, the input 3 becomes the exponent on the base 2 to produce the real
number 8. The function f −1 (x) = log2 (x) then takes the number 8 as its input and returns the
exponent 3 as its output. In symbols, log2 (8) = 3. More generally, log2 (x) is the exponent you
put on 2 to get x. Thus, log2 (16) = 4, because 24 = 16. The following theorem summarizes the
basic properties of logarithmic functions, all of which come from the fact that they are inverses of
exponential functions. 338 Exponential and Logarithmic Functions Theorem 7.2. Properties of Logarithmic Functions: Suppose f (x) = logb (x).
The domain of f is (0, ∞) and the range of f is (−∞, ∞).
(1, 0) is on the graph of f and x = 0 is a vertical asymptote of the graph of f .
f is onetoone, continuous and smooth
ba = c if and only if logb (c) = a. That is, logb (c) is the exponent you put on b to obtain c.
logb (bx ) = x for all x and blogb (x) = x for all x > 0
If b > 1: If 0 < b < 1: – f is always increasing – f is always decreasing 0+ , – As x → 0+ , f (x) → ∞ – As x → f (x) → −∞ – As x → ∞, f (x) → ∞ – As x → ∞, f (x) → −∞ – The graph of f resembles: – The graph of f resembles: y = logb (x), b > 1 y = logb (x), 0 < b < 1 As we have mentioned, Theorem 7.2 is a consequence of Theorems 6.2 and 6.3. However, it is
worth the reader’s time to understand Theorem 7.2 from an exponential perspective. For instance,
we know that the domain of g (x) = log2 (x) is (0, ∞). Why? Because the range of f (x) = 2x is
(0, ∞). In a way, this says everything, but at the same time, it doesn’t. For example, if we try to
ﬁnd log2 (−1), we are trying to ﬁnd the exponent we put on 2 to give us −1. In other words, we are
looking for x that satisﬁes 2x = −1. There is no such real number, since all powers of 2 are positive.
While what we have said is exactly the same thing as saying ‘the domain of g (x) = log2 (x) is (0, ∞)
because the range of f (x) = 2x is (0, ∞)’, we feel it is in a student’s best interest to understand
the statements in Theorem 7.2 at this level instead of just merely memorizing the facts.
Example 7.1.3. Simplify the following. 7.1 Introduction to Exponential and Logarithmic Functions
1. log3 (81)
2. log2 4. ln √
3 339 e2 5. log(0.001) 1
8 6. 2log2 (8)
7. 117− log117 (6) 3. log√5 (25)
Solution. 1. The number log3 (81) is the exponent we put on 3 to get 81. As such, we want to write 81 as
a power of 3. We ﬁnd 81 = 34 , so that log3 (81) = 4.
2. To ﬁnd log2 1
, we need rewrite
8 1
8 as a power of 2. We ﬁnd 1
8 = 1
23 = 2−3 , so log2 1
8 = −3. √
3. To determine log√5 (25), we need to express 25 as a power of 5. We know 25 = 52 , and
√2
√ 22
√4
5=
5 , so we have 25 =
5
5 . We get log√5 (25) = 4.
=
√
√
32
32
4. First, recall that the notation ln
e means loge
e , so we are looking for the exponent
√
√
√
3
3
32
to put on e to obtain e2 . Rewriting e2 = e2/3 , we ﬁnd ln
e = ln e2/3 = 2 .
3
5. Rewriting log(0.001) as log10 (0.001), we see that we need to write 0.001 as a power of 10. We
1
1
have 0.001 = 1000 = 103 = 103 . Hence, log(0.001) = log 10−3 = −3.
6. We can use Theorem 7.2 directly to simplify 2log2 (8) = 8. We can also understand this problem
by ﬁrst ﬁnding log2 (8). By deﬁnition, log2 (8) is the exponent we put on 2 to get 8. Since
8 = 23 , we have log2 (8) = 3. We now substitute to ﬁnd 2log2 (8) = 23 = 8.
7. We note that we cannot apply Theorem 7.2 directly to 117− log117 (6) . (Why not?) We use
1
a property of exponents to rewrite 117− log117 (6) as
. At this point, we can apply
log117 (6)
117
1
Theorem 7.2 to get 117log117 (6) = 6 and thus 117− log117 (6) =
= 1 . It is worth
6
log117 (6)
117
a moment of your time to think your way through why 117log117 (6) = 6. By deﬁnition,
log117 (6) is the exponent we put on 117 to get 6. What are we doing with this exponent?
We are putting it on 117. By deﬁnition we get 6. In other words, the exponential function
f (x) = 117x undoes the logarithmic function g (x) = log117 (x).
Up until this point, restrictions on the domains of functions came from avoiding division by
zero and keeping negative numbers from beneath even radicals. With the introduction of logs, we
now have another restriction. Since the domain of f (x) = logb (x) is (0, ∞), the argument8 of the
log must be strictly positive.
8 See page 87 if you’ve forgotten what this term means. 340 Exponential and Logarithmic Functions Example 7.1.4. Find the domain of the following functions. Check your answers graphically using
the calculator.
1. f (x) = 2 log(3 − x) − 1
2. g (x) = ln x
x−1 Solution.
1. We set 3−x > 0 to obtain x < 3, or (−∞, 3). The graph from the calculator below veriﬁes this.
Note that we could have graphed f using transformations. Taking a cue from Theorem 2.7, we
rewrite f (x) = 2 log10 (−x + 3) − 1 and ﬁnd the main function involved is y = h(x) = log10 (x).
1
We select three points to track, 10 , −1 , (1, 0) and (10, 1), along with the vertical asymptote
x = 0. Since f (x) = 2h(−x + 3) − 1, Theorem 2.7 tells us that to obtain the destinations of
these points, we ﬁrst subtract 3 from the xcoordinates (shifting the graph left 3 units), then
divide (multiply) by the xcoordinates by −1 (causing a reﬂection across the y axis). These
transformations apply to the vertical asymptote x = 0 as well. Subtracting 3 gives us x = −3
as our asymptote, then multplying by −1 gives us the vertical asymptote x = 3. Next, we
multiply the y coordinates by 2 which results in a vertical stretch by a factor of 2, then we
ﬁnish by subtracting 1 from the y coordinates which shifts the graph down 1 unit. We leave
it to the reader to perform the indicated arithmetic on the points themselves and to verify
the graph produced by the calculator below.
x
2. To ﬁnd the domain of g , we set x−1 > 0 and use a sign diagram to solve this inequality. We
x
deﬁne r(x) = x−1 ﬁnd its domain to be r is (−∞, 1) ∪ (1, ∞). Setting r(x) = 0 gives x = 0. (+) 0 (−) (+)
0 1 x
We ﬁnd x−1 > 0 on (−∞, 0) ∪ (1, ∞) to get the domain of g . The graph of y = g (x) conﬁrms
this. We can tell from the graph of g that it is not the result of Section 2.5 transformations
being applied to the graph y = ln(x), so barring a more detailed analysis using Calculus, the
calculator graph is the best we can do. One thing worthy of note, however, is the end behavior
of g . The graph suggests that as x → ±∞, g (x) → 0. We can verify this analytically. Using
x
results from Chapter 5 and continuity, we know that as x → ±∞, x−1 ≈ 1. Hence, it makes sense that g (x) = ln x
x−1 ≈ ln(1) = 0. 7.1 Introduction to Exponential and Logarithmic Functions y = f (x) = 2 log(3 − x) − 1 y = g (x) = ln 341 x
x−1 While logarithms have some interesting applications of their own which you’ll explore in the
exercises, their primary use to us will be to undo exponential functions. (This is, after all, how
they were deﬁned.) Our last example solidiﬁes this and reviews all of the material in the section.
Example 7.1.5. Let f (x) = 2x−1 − 3.
1. Graph f using transformations and state the domain and range of f .
2. Explain why f is invertible and ﬁnd a formula for f −1 (x).
3. Graph f −1 using transformations and state the domain and range of f −1 .
4. Verify f −1 ◦ f (x) = x for all x in the domain of f and f ◦ f −1 (x) = x for all x in the
domain of f −1 .
5. Graph f and f −1 on the same set of axes and check the symmetry about the line y = x.
Solution.
1
1. If we identify g (x) = 2x , we see f (x) = g (x − 1) − 3. We pick the points −1, 2 , (0, 1)
and (1, 2) on the graph of g along with the horizontal asymptote y = 0 to track through
the transformations. By Theorem 2.7 we ﬁrst add 1 to the xcoordinates of the points on
1
the graph of g (shifting g to the right 1 unit) to get 0, 2 , (1, 1) and (2, 2). The horizontal
asymptote remains y = 0. Next, we subtract 3 from the y coordinates, shifting the graph
5
down 3 units. We get the points 0, − 2 , (1, −2) and (2, −1) with the horizontal asymptote
now at y = −3. Connecting the dots in the order and manner as they were on the graph of
g , we get the graph below. We see that the domain of f is the same as g , namely (−∞, ∞),
but that the range of f is (−3, ∞). 342 Exponential and Logarithmic Functions y y
8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 −3 −2 −1
−1 1 2 3 4 x −3 −2 −1
−1 1 2 3 4 x −2 −2
−3 y = h(x) = 2x −− − − − −→
−−−−−− y = f (x) = 2x−1 − 3 2. The graph of f passes the Horizontal Line Test so f is onetoone, hence invertible. To ﬁnd
a formula for f −1 (x), we normally set y = f (x), interchange the x and y , then proceed to
solve for y . Doing so in this situation leads us to the equation x = 2y−1 − 3. We have yet
to discuss how to solve this kind of equation, so we will attempt to ﬁnd the formula for f −1
from a procedural perspective. If we break f (x) = 2x−1 − 3 into a series of steps, we ﬁnd f
takes an input x and applies the steps
(a) subtract 1
(b) put as an exponent on 2
(c) subtract 3
Clearly, to undo subtracting 1, we will add 1, and similarly we undo subtracting 3 by adding
3. How do we undo the second step? The answer is we use the logarithm. By deﬁnition,
log2 (x) undoes exponentiation by 2. Hence, f −1 should
(a) add 3
(b) take the logarithm base 2
(c) add 1
In symbols, f −1 (x) = log2 (x + 3) + 1.
3. To graph f −1 (x) = log2 (x + 3) + 1 using transformations, we start with j (x) = log2 (x). We
1
track the points 2 , −1 , (1, 0) and (2, 1) on the graph of j along with the vertical asymptote
x = 0 through the transformations using Theorem 2.7. Since f −1 (x) = j (x + 3) + 1, we ﬁrst
5
subtract 3 from each of the x values (including the vertical asymptote) to obtain − 2 , −1 ,
(−2, 0) and (−1, 1) with a vertical asymptote x = −3. Next, we add 1 to the y values on the
5
graph and get − 2 , 0 , (−2, 1) and (−1, 2). If you are experiencing d´j` vu, there is a good
ea
reason for it but we leave it to the reader to determine the source of this uncanny familiarity.
We obtain the graph below. The domain of f −1 is (−3, ∞), which matches the range of f ,
and the range of f −1 is (−∞, ∞), which matches the domain of f . 7.1 Introduction to Exponential and Logarithmic Functions 343 y y
4 4 3 3 2 2 1 1 −3 −2 −1
−1 1 2 3 4 5 6 7 8 x −2 −1
−1 −2
−3 1 2 3 4 5 6 7 8 x −2
−3 y = j (x) = log2 (x) −− − − − −→
−−−−−− y = f −1 (x) = log2 (x + 3) + 1 4. We now verify that f (x) = 2x−1 − 3 and f −1 (x) = log2 (x + 3) + 1 satisfy the composition
requirement for inverses. For all real numbers x, f −1 ◦ f (x) = f −1 (f (x))
= f −1 2x−1 − 3
= log2 2x−1 − 3 + 3 + 1 = log2 2x−1 + 1
= (x − 1) + 1 Since log2 (2u ) = u for all real numbers u =x
For all real numbers x > −3, we have9 f ◦ f −1 (x) = f f −1 (x)
= f (log2 (x + 3) + 1)
= 2(log2 (x+3)+1)−1 − 3
= 2log2 (x+3) − 3
= (x + 3) − 3 Since 2log2 (u) = u for all real numbers u > 0 =x
9 Pay attention  can you spot in which step below we need x > −3? 344 Exponential and Logarithmic Functions 5. Last, but certainly not least, we graph y = f (x) and y = f −1 (x) on the same set of axes and
see the symmetry about the line y = x.
y
8
7
6
5
4
3
2
1
−3 −2 −1
−1 1 2 3 4 5 6 7 8 −2 y = f (x) = 2x−1 − 3
y = f −1 (x) = log2 (x + 3) + 1 x 7.1 Introduction to Exponential and Logarithmic Functions 7.1.1 345 Exercises 1. Evaluate the expression.
(a)
(b)
(c)
(d)
(e)
(f)
(g) log3 (27)
log6 (216)
log2 (32)
1
log6 36
log8 (4)
log36 (216)
log 1 (625)
5 (h) log 1 (216)
6 (i) log36 (36) 1
log 1000000
log(0.01)
ln e3
log4 (8)
log6 (1)
√
log13
13
√
4
36
(p) log36 (s) log36 36216 (j)
(k)
(l)
(m)
(n)
(o) (t) ln e5
√
9
1011
(u) log
√
3
(v) log
105
(w) ln 1
√
e (x) log5 3log3 (5) (q) 7log7 (3)
(r) 36log36 (216) (y) log eln(100) 2. Find the domain of the function.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h) f (x) = ln(x2 + 1)
f (x) = log7 (4x + 8)
f (x) = ln(4x − 20)
f (x) = log x2 + 9x + 18
x+2
f (x) = log
x2 − 1
x2 + 9x + 18
f (x) = log
4x − 20
f (x) = ln(7 − x) + ln(x − 4)
f (x) = ln(4x − 20) + ln x2 + 9x + 18 (i) f (x) = log x2 + x + 1
(j) f (x) = 4 log4 (x) (k) f (x) = log9 (x + 3 − 4)
√
(l) f (x) = ln( x − 4 − 3)
1
(m) f (x) =
3 − log5 (x)
√
−1 − x
(n) f (x) =
log 1 (x)
2 (o) f (x) = ln(−2x3 − x2 + 13x − 6) 3. For each function given below, ﬁnd its inverse from the ‘procedural perspective’ discussed in
Example 7.1.5 and graph the function and its inverse on the same set of axes.
(a) f (x) = 3x+2 − 4
(b) f (x) = log4 (x − 1) (c) f (x) = −2−x + 1
(d) f (x) = 5 log(x) − 2 4. Show that logb 1 = 0 and logb b = 1 for every b > 0, b = 1.
5. (Crazy bonus question) Without using your calculator, determine which is larger: eπ or π e .
6. (The Logarithmic Scales) There are three widely used measurement scales which involve
common logarithms: the Richter scale, the decibel scale and the pH scale. The computations
involved in all three scales are nearly identical so pay close attention to the subtle diﬀerences. 346 Exponential and Logarithmic Functions
(a) Earthquakes are complicated events and it is not our intent to provide a complete discussion of the science involved in them. Instead, we refer the interested reader to a
solid course in Geology10 or the U.S. Geological Survey’s Earthquake Hazards Program
found here and present only a simpliﬁed version of the Richter scale. The Richter scale
measures the magnitude of an earthquake by comparing the amplitude of the seismic
waves of the given earthquake to those of a “magnitude 0 event”, which was chosen to
be a seismograph reading of 0.001 millimeters recorded on a seismometer 100 kilometers
from the earthquake’s epicenter. Speciﬁcally, the magnitude of an earthquake is given
by
x
M (x) = log
0.001
where x is the seismograph reading in millimeters of the earthquake recorded 100 kilometers from the epicenter.
i. Show that M (0.001) = 0.
ii. Compute M (80, 000).
iii. Show that an earthquake which registered 6.7 on the Richter scale had a seismograph
reading ten times larger than one which measured 5.7.
iv. Find two news stories about recent earthquakes which give their magnitudes on the
Richter scale. How many times larger was the seismograph reading of the earthquake
with larger magnitude?
(b) While the decibel scale can be used in many disciplines,11 we shall restrict our attention
to its use in acoustics, speciﬁcally its use in measuring the intensity level of sound.12
The Sound Intensity Level L (measured in decibels) of a sound intensity I (measured in
watts per square meter) is given by
L(I ) = 10 log I
10−12 . W
Like the Richter scale, this scale compares I to baseline: 10−12 m2 is the threshold of
human hearing. i. Compute L(10−6 ).
ii. Damage to your hearing can start with short term exposure to sound levels around
115 decibels. What intensity I is needed to produce this level?
iii. Compute L(1). How does this compare with the threshold of pain which is around
140 decibels?
10 Rocksolid, perhaps?
See this webpage for more information.
12
As of the writing of this exercise, the Wikipedia page given here states that it may not meet the “general notability
guideline” nor does it cite any references or sources. I ﬁnd this odd because it is this very usage of the decibel scale
which shows up in every College Algebra book I have read. Perhaps those other books have been wrong all along
and we’re just blindly following tradition.
11 7.1 Introduction to Exponential and Logarithmic Functions 347 (c) The pH of a solution is a measure of its acidity or alkalinity. Speciﬁcally, pH = − log[H+ ]
where [H+ ] is the hydrogen ion concentration in moles per liter. A solution with a pH
less than 7 is an acid, one with a pH greater than 7 is a base (alkaline) and a pH of 7 is
regarded as neutral.
i. The hydrogen ion concentration of pure water is [H+ ] = 10−7 . Find its pH.
ii. Find the pH of a solution with [H+ ] = 6.3 × 10−13 .
iii. The pH of gastric acid (the acid in your stomach) is about 0.7. What is the corresponding hydrogen ion concentration? 348 7.1.2 Exponential and Logarithmic Functions Answers 1. (a)
(b)
(c)
(d)
(e)
(f)
(g) 1
log 1000000 = −6
log(0.01) = −2
ln e3 = 3
3
log4 (8) = 2
log6 (1) = 0
√
13 = 1
log13
2
√
4
(p) log36
36 = 1
4 log3 (27) = 3
log6 (216) = 3
log2 (32) = 5
1
log6 36 = −2
log8 (4) = 2
3
log36 (216) = 3
2
log 1 (625) = −4 (s) log36 36216 = 216 (j)
(k)
(l)
(m)
(n)
(o) 5 (h) log 1 (216) = −3 (t) ln(e5 ) = 5
√
9
(u) log
1011 = 11
9
√
3
5
(v) log
105 = 3 6 = −1
2 (x) log5 3log3 5 = 1 (q) 7log7 (3) = 3
(r) 36log36 (216) = 216 (i) log36 (36) = 1 1
√
e (w) ln (y) log eln(100) = 2 (f) (−6, −3) ∪ (5, ∞) (k) (−∞, −7) ∪ (1, ∞) (b) (−2, ∞) (g) (4, 7) (l) (13, ∞) (c) (5, ∞) (h) (5, ∞) (m) (0, 125) ∪ (125, ∞) (d) (−∞, −6) ∪ (−3, ∞) (i) (−∞, ∞) (n) No domain (e) (−2, −1) ∪ (1, ∞) (j) [1, ∞) (o) (−∞, −3) ∪ 2. (a) (−∞, ∞) 3. (a) f (x) = 3x+2 − 4
f −1 (x) = log3 (x + 4) − 2 (b) f (x) = log4 (x − 1)
f −1 (x) = 4x + 1 y y 6 6 5 5 4 4 3 3 2 2 1
−4 −3 −2 −1
−1 1
1 2 3 4 5 6 x −2 −2 −1
−1 1 2 3 4 5 6 −2 −3
−4 y = f (x) = 3x+2 − 4
y = f −1 (x) = log3 (x + 4) − 2 y = f (x) = log4 (x − 1)
y = f −1 (x) = 4x + 1 x 1
2, 2 7.1 Introduction to Exponential and Logarithmic Functions
(c) f (x) = −2−x + 1
f −1 (x) = − log2 (1 − x) 349 (d) f (x) = 5 log(x) − 2
x+2
f −1 (x) = 10 5 y y
5 2 4
3 1 2
−2 −1 1 2 x −1
−2 1
−4 −3 −2 −1
−1 1 2 3 4 5 x −2
−3 y = f (x) = −2−x + 1
y = f −1 (x) = − log2 (1 − x) −4 y = f (x) = 5 log(x) − 2
y = f −1 (x) = 10 6. (a) x+2
5 0.001
= log(1) = 0.
0.001
80, 000
ii. M (80, 000) = log
= log(80, 000, 000) ≈ 7.9.
0.001
i. M (0.001) = log (b) i. L(10−6 ) = 60 decibels.
ii. I = 10−.5 ≈ 0.316 watts per square meter.
iii. Since L(1) = 120 decibels and L(100) = 140 decibels, a sound with intensity level
140 decibels has an intensity 100 times greater than a sound with intensity level 120
decibels. (c) i. The pH of pure water is 7.
ii. If [H+ ] = 6.3 × 10−13 then the solution has a pH of 12.2.
iii. [H+ ] = 10−0.7 ≈ .1995 moles per liter. 350 7.2 Exponential and Logarithmic Functions Properties of Logarithms In Section 7.1, we introduced the logarithmic functions as inverses of exponential functions and
discussed a few of their functional properties from that perspective. In this section, we explore
the algebraic properties of logarithms. Historically, these have played a huge rule in the scientiﬁc
development of our society since, among other things, they were used to develop analog computing
devices called slide rules which enabled scientists and engineers to perform accurate calculations
leading to such things as space travel and the moon landing. As we shall see shortly, logs inherit
analogs of all of the properties of exponents you learned in Elementary and Intermediate Algebra.
We ﬁrst extract two properties from Theorem 7.2 to remind us of the deﬁnition of a logarithm as
the inverse of an exponential function. Theorem 7.3. (Inverse Properties of Exponential and Log Functions) Let b > 0, b = 1.
ba = c if and only if logb (c) = a
logb (bx ) = x for all x and blogb (x) = x for all x > 0 Next, we spell out in more detail what it means for exponential and logarithmic functions to
be onetoone. Theorem 7.4. (Onetoone Properties of Exponential and Log Functions) Let f (x) =
bx and g (x) = logb (x) where b > 0, b = 1. Then f and g are onetoone. In other words:
bu = bw if and only if u = w for all real numbers u and w.
logb (u) = logb (w) if and only if u = w for all real numbers u > 0, w > 0. We now state the algebraic properties of exponential functions which will serve as a basis for
the properties of logarithms. While these properties may look identical to the ones you learned
in Elementary and Intermediate Algebra, they apply to real number exponents, not just rational
exponents. Note that in the theorem that follows, we are interested in the properties of exponential
functions, so the base b is restricted to b > 0, b = 1. An added beneﬁt of this restriction is that it
3/2
eliminates the pathologies discussed in Section ?? when, for example, we simpliﬁed x2/3
and
obtained x instead of what we had expected from the arithmetic in the exponents, x1 = x. 7.2 Properties of Logarithms 351 Theorem 7.5. (Algebraic Properties of Exponential Functions) Let f (x) = bx be an
exponential function (b > 0, b = 1) and let u and w be real numbers.
Product Rule: f (u + w) = f (u)f (w). In other words, bu+w = bu bw
Quotient Rule: f (u − w) = f (u)
bu
. In other words, bu−w = w
f (w)
b Power Rule: (f (u))w = f (uw). In other words, (bu )w = buw While the properties listed in Theorem 7.5 are certainly believable based on similar properties
of integer and rational exponents, the full proofs require Calculus. To each of these properties of
exponential functions corresponds an analogous property of logarithmic functions. We list these
below in our next theorem.
Theorem 7.6. (Algebraic Properties of Logarithm Functions) Let g (x) = logb (x) be a
logarithmic function (b > 0, b = 1) and let u > 0 and w > 0 be real numbers.
Product Rule: g (uw) = g (u) + g (w). In other words, logb (uw) = logb (u) + logb (w)
Quotient Rule: g u
w = g (u) − g (w). In other words, logb u
w = logb (u) − logb (w) Power Rule: g (uw ) = wg (u). In other words, logb (uw ) = w logb (u) There are a couple of diﬀerent ways to understand why Theorem 7.6 is true. Consider the
product rule: logb (uw) = logb (u) + logb (w). Let a = logb (uw), c = logb (u), and d = logb (w). Then,
by deﬁnition, ba = uw, bc = u and bd = w. Hence, ba = uw = bc bd = bc+d , so that ba = bc+d . By
the onetoone property of bx , we have a = c + d. In other words, logb (uw) = logb (u) + logb (w).
The remaining properties are proved similarly. From a purely functional approach, we can see
the properties in Theorem 7.6 as an example of how inverse functions interchange the roles of
inputs in outputs. For instance, the Product Rule for exponential functions given in Theorem 7.5,
f (u + w) = f (u)f (w), says that adding inputs results in multiplying outputs. Hence, whatever f −1
is, it must take the products of outputs from f and return them to the sum of their respective inputs.
Since the outputs from f are the inputs to f −1 and viceversa, we have that that f −1 must take
products of its inputs to the sum of their respective outputs. This is precisely what the Product Rule
for Logarithmic functions states in Theorem 7.6: g (uw) = g (u) + g (w). The reader is encouraged to
view the remaining properties listed in Theorem 7.6 similarly. The following examples help build
familiarity with these properties. In our ﬁrst example, we are asked to ‘expand’ the logarithms.
This means that we read the properties in Theorem 7.6 from left to right and rewrite products
inside the log as sums outside the log, quotients inside the log as diﬀerences outside the log, and 352 Exponential and Logarithmic Functions powers inside the log as factors outside the log. While it is the opposite process, which we will
practice later, that is most useful in Algebra, the utility of expanding logarithms becomes apparent
in Calculus.
Example 7.2.1. Expand the following using the properties of logarithms and simplify. Assume
when necessary that all quantities represent positive real numbers.
1. log2 8
x 4. log 3 100x2
yz 5 2. log0.1 10x2
3. ln 3
ex 2 5. log117 x2 − 4 Solution.
1. To expand log2 8
x , we use the Quotient Rule identifying u = 8 and w = x and simplify. log2 8
x = log2 (8) − log2 (x) Quotient Rule
= 3 − log2 (x) Since 23 = 8 = − log2 (x) + 3
2. In the expression log0.1 10x2 , we have a power (the x2 ) and a product. In order to use the
Product Rule, the entire quantity inside the logarithm must be raised to the same exponent.
Since the exponent 2 applies only to the x, we ﬁrst apply the Product Rule with u = 10 and
w = x2 . Once we get the x2 by itself inside the log, we may apply the Power Rule with u = x
and w = 2 and simplify. log0.1 10x2 = log0.1 (10) + log0.1 x2 Product Rule = log0.1 (10) + 2 log0.1 (x) Power Rule = −1 + 2 log0.1 (x)
= 2 log0.1 (x) − 1 Since (0.1)−1 = 10 7.2 Properties of Logarithms 353
2 3
3. We have a power, quotient and product occurring in ln ex . Since the exponent 2 applies
3
to the entire quantity inside the logarithm, we begin with the Power Rule with u = ex and
w = 2. Next, we see the Quotient Rule is applicable, with u = 3 and w = ex, so we replace
3
3
ln ex with the quantity ln(3) − ln(ex). Since ln ex is being multiplied by 2, the entire
quantity ln(3) − ln(ex) is multiplied by 2. Finally, we apply the Product Rule with u = e and
w = x, and replace ln(ex) with the quantity ln(e) + ln(x), and simplify, keeping in mind that
the natural log is log base e. 3
ex 2 3
ex Power Rule = 2 [ln(3) − ln(ex)] ln Quotient Rule = 2 ln = 2 ln(3) − 2 ln(ex)
= 2 ln(3) − 2 [ln(e) + ln(x)] Product Rule = 2 ln(3) − 2 ln(e) − 2 ln(x)
= 2 ln(3) − 2 − 2 ln(x) Since e1 = e = −2 ln(x) + 2 ln(3) − 2 4. In Theorem 7.6, there is no mention of how to deal with radicals. However, thinking back to
Deﬁnition ??, we can rewrite the cube root as a 1 exponent. We begin by using the Power
3 354 Exponential and Logarithmic Functions
Rule1 , and we keep in mind that the common log is log base 10. log 3 100x2
yz 5 = log 100x2
yz 5 1/3 100x2
yz 5 = 1
3 = 1
3 log 100x2 − log y z 5 = 1
3 1
log 100x2 − 3 log y z 5 = 1
3 = 1
3 1
1
log(100) + 3 log x2 − 3 log(y ) − 1 log z 5
3 = 1
3 2
log(100) + 3 log(x) − 1 log(y ) − 5 log(z )
3
3 = 2
3 1
+ 2 log(x) − 3 log(y ) − 5 log(z )
3
3 = 2
3 log(x) − 1 log(y ) − 5 log(z ) +
3
3 log log(100) + log x2 Power Rule − Quotient Rule 1
3 log(y ) + log z 5 Product Rule Power Rule
Since 102 = 100 2
3 5. At ﬁrst it seems as if we have no means of simplifying log117 x2 − 4 , since none of the
properties of logs addresses the issue of expanding a diﬀerence inside the logarithm. However,
we may factor x2 − 4 = (x + 2)(x − 2) thereby introducing a product which gives us license
to use the Product Rule. log117 x2 − 4 = log117 [(x + 2)(x − 2)] Factor = log117 (x + 2) + log117 (x − 2) Product Rule A couple of remarks about Example 7.2.1 are in order. First, while not explicitly stated in
the above example, a general rule of thumb to determine which log property to apply ﬁrst to a
complicated problem is ‘reverse order of operations.’ For example, if we were to substitute a number
for x into the expression log0.1 10x2 , we would ﬁrst square the x, then multiply by 10. The last
step is the multiplication, which tells us the ﬁrst log property to apply is the Product Rule. In a
1 At this point in the text, the reader is encouraged to carefully read through each step and think of which quantity
is playing the role of u and which is playing the role of w as we apply each property. 7.2 Properties of Logarithms 355 multistep problem, this rule can give the required guidance on which log property to apply at each
step. The reader is encouraged to look through the solutions to Example 7.2.1 to see this rule in
action. Second, while we were instructed to assume when necessary that all quantities represented
positive real numbers, the authors would be committing a sin of omission if we failed to point out
that, for instance, the functions f (x) = log117 x2 − 4 and g (x) = log117 (x + 2) + log117 (x − 2)
have diﬀerent domains, and, hence, are diﬀerent functions. We leave it to the reader to verify the
domain of f is (−∞, −2) ∪ (2, ∞) whereas the domain of g is (2, ∞). In general, when using log
properties to expand a logarithm, we may very well be restricting the domain as we do so. One
last comment before we move to reassembling logs from their various bits and pieces. The authors
are well aware of the propensity for some students to become overexcited and invent their own
properties of logs like log117 x2 − 4 = log117 x2 − log117 (4), which simply isn’t true, in general.
The unwritten2 property of logarithms is that if it isn’t written in a textbook, it probably isn’t
true.
Example 7.2.2. Use the properties of logarithms to write the following as a single logarithm.
1. log3 (x − 1) − log3 (x + 1) 3. 4 log2 (x) + 3 2. log(x) + 2 log(y ) − log(z ) 4. − ln(x) − 1
2 Solution. Whereas in Example 7.2.1 we read the properties in Theorem 7.6 from left to right
to expand logarithms, in this example we read them from right to left.
1. The diﬀerence of logarithms requires the Quotient Rule: log3 (x − 1) − log3 (x +1) = log3 x− 1
x+1 . 2. In the expression, log(x) + 2 log(y ) − log(z ), we have both a sum and diﬀerence of logarithms.
However, before we use the product rule to combine log(x) + 2 log(y ), we note that we need
to somehow deal with the coeﬃcient 2 on log(y ). This can be handled using the Power Rule.
We can then apply the Product and Quotient Rules as we move from left to right. Putting it
all together, we have
log(x) + 2 log(y ) − log(z ) = log(x) + log y 2 − log(z ) Power Rule = log xy 2 − log(z ) Product Rule xy 2
z Quotient Rule = log 3. We can certainly get started rewriting 4 log2 (x) + 3 by applying the Power Rule to 4 log2 (x)
to obtain log2 x4 , but in order to use the Product Rule to handle the addition, we need to
rewrite 3 as a logarithm base 2. From Theorem 7.3, we know 3 = log2 23 , so we get
2 The authors relish the irony involved in writing what follows. 356 Exponential and Logarithmic Functions 4 log2 (x) + 3 = log2 x4 + 3 Power Rule = log2 x4 + log2 23 Since 3 = log2 23 = log2 x4 + log2 (8)
= log2 8x4 Product Rule 1
4. To get started with − ln(x) − 2 , we rewrite − ln(x) as (−1) ln(x). We can then use the Power
1
Rule to obtain (−1) ln(x) = ln x−1 . In order to use the Quotient Rule, we need to write 2
1
1/2 = ln (√e). We have
as a natural logarithm. Theorem 7.3 gives us 2 = ln e − ln(x) − 1
2 = (−1) ln(x) −
= ln x−1 − 1
2 1
2 = ln x−1 − ln e1/2 Power Rule
Since 1
2 = ln e1/2 √
= ln x−1 − ln ( e)
= ln x−1
√
e = ln 1
√
xe Quotient Rule As we would expect, the rule of thumb for reassembling logarithms is the opposite of what
it was for dismantling them. That is, if we are interested in rewriting an expression as a single
logarithm, we apply log properties following the usual order of operations: deal with multiples of
logs ﬁrst with the Power Rule, then deal with addition and subtraction using the Product and
Quotient Rules, respectively. Additionally, we ﬁnd that using log properties in this fashion can
increase the domain of the expression. For example, we leave it to the reader to verify the domain
x−1
of f (x) = log3 (x−1)−log3 (x+1) is (1, ∞) but the domain of g (x) = log3 x+1 is (−∞, −1)∪(1, ∞).
We will need to keep this in mind when we solve equations involving logarithms in Section 7.4  it
is precisely for this reason we will have to check for extraneous solutions. 7.2 Properties of Logarithms 357 The two logarithm buttons commonly found on calculators are the ‘LOG’ and ‘LN’ buttons
which correspond to the common and natural logs, respectively. Suppose we wanted an approximation to log2 (7). The answer should be a little less than 3, (Can you explain why?) but how do
we coerce the calculator into telling us a more accurate answer? We need the following theorem.
Theorem 7.7. (Change of Base) Let a, b > 0, a, b = 1.
ax = bx logb (a) for all real numbers x.
loga (x) = logb (x)
for all real numbers x > 0.
logb (a) The proofs of the Change of Base formulas are a result of the other properties studied in this
section. If we start with bx logb (a) and use the Power Rule in the exponent to rewrite x logb (a) as
logb (ax ) and then apply one of the Inverse Properties in Theorem 7.3, we get
x) bx logb (a) = blogb (a = ax , as required. To verify the logarithmic form of the property, we also use the Power Rule and an
Inverse Property. We note that
loga (x) · logb (a) = logb aloga (x) = logb (x),
and we get the result by dividing through by logb (a). Of course, the authors can’t help but point
out the inverse relationship between these two change of base formulas. To change the base of
an exponential expression, we multiply the input by the factor logb (a). To change the base of a
logarithmic expression, we divide the output by the factor logb (a). While, in the grand scheme
of things, both change of base formulas are really saying the same thing, the logarithmic form is
the one usually encountered in Algebra while the exponential form isn’t usually introduced until
Calculus.3 What Theorem 7.7 really tells us is that all exponential and logarithmic functions are
just scalings of one another. Not only does this explain why their graphs have similar shapes, but
it also tells us that we could do all of mathematics with a single base  be it 10, e, 42, or 117. Your
Calculus teacher will have more to say about this when the time comes.
Example 7.2.3. Use an appropriate change of base formula to convert the following expressions to
ones with the indicated base. Verify your answers using a calculator, as appropriate. 3
The authors feel so strongly about showing students that every property of logarithms comes from and corresponds
to a property of exponents that we have broken tradition with the vast majority of other authors in this ﬁeld. This
isn’t the ﬁrst time this happened, and it certainly won’t be the last. 358 Exponential and Logarithmic Functions 1. 32 to base 10 3. log4 (5) to base e 2. 2x to base e 4. ln(x) to base 10 Solution.
1. We apply the Change of Base formula with a = 3 and b = 10 to obtain 32 = 102 log(3) . Typing
the latter in the calculator produces an answer of 9 as required.
2. Here, a = 2 and b = e so we have 2x = ex ln(2) . To verify this on our calculator, we can graph
f (x) = 2x and g (x) = ex ln(2) . Their graphs are indistinguishable which provides evidence
that they are the same function. y = f (x) = 2x and y = g (x) = ex ln(2)
3. Applying the change of base with a = 4 and b = e leads us to write log4 (5) = ln(5)
ln(4) . Evaluating ln(5)
ln(4) this in the calculator gives
≈ 1.16. How do we check this really is the value of log4 (5)?
By deﬁnition, log4 (5) is the exponent we put on 4 to get 5. The calculator conﬁrms this.4
)
4. We write ln(x) = loge (x) = log(x) . We graph both f (x) = ln(x) and g (x) =
log(e
both graphs appear to be identical. log(x)
log(e) y = f (x) = ln(x) and y = g (x) = 4 Which means if it is lying to us about the ﬁrst answer it gave us, at least it is being consistent. and ﬁnd log(x)
log(e) 7.2 Properties of Logarithms 7.2.1 359 Exercises 1. Expand the following using the properties of logarithms and simplify. Assume when necessary
that all quantities represent positive real numbers.
(a) ln(x3 y 2 )
128
(b) log2
x2 + 4
z3
(c) log5
25
(d) log(1.23 × 1037 )
√
z
(e) ln
xy (h) log 1 (9x(y 3 − 8))
3 (i) log 1000x3 y 5
(j) log3
4 (k) ln
(l) log6 (f) log5 x2 − 25
(g) log√2 4x3 (m) ln x2
81y 4
xy
ez 216
x3 y
√
3
x
√
10 yz 4 2. Use the properties of logarithms to write the following as a single logarithm.
(a)
(b)
(c)
(d)
(e)
(f)
(g) 4 ln(x) + 2 ln(y )
3 − log(x)
log2 (x) + log2 (y ) − log2 (z )
log3 (x) − 2 log3 (y )
1
2 log3 (x) − 2 log3 (y ) − log3 (z )
2 ln(x) − 3 ln(y ) − 4 ln(z )
1
log(x) − 3 log(z ) + 1 log(y )
2 1
1
(h) − 3 ln(x) − 1 ln(y ) + 3 ln(z )
3 (i) log2 (x) + log 1 (x − 1)
2 (j) log2 (x) + log4 (x − 1)
(k) log5 (x) − 3
(l) log7 (x) + log7 (x − 3) − 2
(m) ln(x) + 1
2 3. Use an appropriate change of base formula to convert the following expressions to ones with
the indicated base.
(a) 7x−1 to base e
(b) log3 (x + 2) to base 10 2x
to base e
3
(d) log(x2 + 1) to base e
(c) 4. Use the appropriate change of base formula to approximate the following logarithms.
(a) log3 (12)
(b) log5 (80)
(c) log6 (72) 1
10
(e) log 3 (1000) (d) log4 5 (f) log 2 (50)
3 360 Exponential and Logarithmic Functions 5. Compare and contrast the graphs of y = ln(x2 ) and y = 2 ln(x).
6. Prove the Quotient Rule and Power Rule for Logarithms.
7. Give numerical examples to show that, in general,
(a) logb (x + y ) = logb (x) + logb (y )
(b) logb (x − y ) = logb (x) − logb (y )
logb (x)
x
=
(c) logb
y
logb (y )
8. The HendersonHasselbalch Equation: Suppose HA represents a weak acid. Then we have a
reversible chemical reaction
HA
H + + A− .
The acid disassociation constant, Ka , is given by
Kα = [H + ][A− ]
[A− ]
= [H + ]
,
[HA]
[HA] where the square brackets denote the concentrations just as they did in Exercise 6c in Section
7.1. The symbol pKa is deﬁned similarly to pH in that pKa = − log(Ka ). Using the deﬁnition
of pH from Exercise 6c and the properties of logarithms, derive the HendersonHasselbalch
Equation which states
[A− ]
pH = pKa + log
[HA]
9. Research the history of logarithms including the origin of the word ‘logarithm’ itself. Why is
the abbreviation of natural log ‘ln’ and not ‘nl’ ?
10. There is a scene in the movie ‘Apollo 13’ in which several people at Mission Control use slide
rules to verify a computation. Was that scene accurate? Look for other pop culture references
to logarithms and slide rules. 7.2 Properties of Logarithms 7.2.2 361 Answers 1. (a)
(b)
(c)
(d)
(e)
(f)
(g) 3 ln(x) + 2 ln(y )
7 − log2 (x2 + 4)
3 log5 (z ) − 6
log(1.23) + 37
1
2 ln(z ) − ln(x) − ln(y )
log5 (x − 5) + log5 (x + 5)
3 log√2 (x) + 4 2. (a) ln(x4 y 2 )
1000
(b) log
x
xy
(c) log2
z
x
(d) log3
y2
√
x
(e) log3
y2z
(f) ln x2
y3z4 1
(h) −2+log 1 (x)+log 1 (y − 2)+log 3 (y 2 +2y +4)
3
3 (i) 3 + 3 log(x) + 5 log(y )
(j) 2 log3 (x) − 4 − 4 log3 (y )
(k) 1
4 ln(x) + 1 ln(y ) −
4 1
4 (l) 12 − 12 log6 (x) − 4 log6 (y )
(m) 1
3 1
ln(x) − ln(10) − 1 ln(y ) − 2 ln(z )
2 √
xy
(g) log √
3
z
z
(h) ln 3
xy
x
x−1
√
log2 x x − 1
x
log5
125
x(x − 3)
log7
49
√
ln (x e) (i) log2
(j)
(k)
(l)
(m) x 3. (a) 7x−1 = e(x−1) ln(7)
log(x + 2)
(b) log3 (x + 2) =
log(3) (d) log(x2 + 1) = 4. (a) log3 (12) ≈ 2.26186 (d) log4 (b) log5 (80) ≈ 2.72271
(c) log6 (72) ≈ 2.38685 − 1 ln(z )
4 (c) 2
3 2 = ex ln( 3 )
ln(x2 + 1)
ln(10) 1
≈ −1.66096
10
(e) log 3 (1000) ≈ −13.52273
5 (f) log 2 (50) ≈ −9.64824
3 362 7.3 Exponential and Logarithmic Functions Exponential Equations and Inequalities In this section we will develop techniques for solving equations involving exponential functions.
Suppose, for instance, we wanted to solve the equation 2x = 128. After a moment’s calculation, we
ﬁnd 128 = 27 , so we have 2x = 27 . The onetoone property of exponential functions, detailed in
Theorem 7.4, tells us that 2x = 27 if and only if x = 7. This means that not only is x = 7 a solution
to 2x = 27 , it is the only solution. Now suppose we change the problem ever so slightly to 2x = 129.
We could use one of the inverse properties of exponentials and logarithms listed in Theorem 7.3 to
write 129 = 2log2 (129) . We’d then have 2x = 2log2 (129) , which means our solution is x = log2 (129).
This makes sense because, after all, the deﬁnition of log2 (129) is ‘the exponent we put on 2 to get
129.’ Indeed we could have obtained this solution directly by rewriting the equation 2x = 129 in
its logarithmic form log2 (129) = x. Either way, in order to get a reasonable decimal approximation
to this number, we’d use the change of base formula, Theorem 7.7, to give us something more
calculator friendly,1 say log2 (129) = ln(129) . Another way to arrive at this answer is as follows
ln(2)
2x = 129
ln (2x ) = ln(129) Take the natural log of both sides. x ln(2) = ln(129) Power Rule x= ln(129)
ln(2) ‘Taking the natural log’ of both sides is akin to squaring both sides: since f (x) = ln(x) is
a function, as long as two quantities are equal, their natural logs are equal.2 Also note that we
treat ln(2) as any other nonzero real number and divide it through3 to isolate the variable x. We
summarize below the two common ways to solve exponential equations, motivated by our examples.
Steps for Solving an Equation involving Exponential Functions
1. Isolate the exponential function.
2. (a) If convenient, express both sides with a common base and equate the exponents.
(b) Otherwise, take the natural log of both sides of the equation and use the Power Rule. Example 7.3.1. Solve the following equations. Check your answer graphically using a calculator.
1 You can use natural logs or common logs. We choose natural logs. (In Calculus, you’ll learn these are the most
‘mathy’ of the logarithms.)
2
This is also the ‘if’ part of the statement logb (u) = logb (w) if and only if u = w in Theorem 7.4.
3
Please resist the temptation to divide both sides by ‘ln’ instead of ln(2). Just like it wouldn’t make sense to
√
√
divide both sides by the square root symbol ‘ ’ when solving x 2 = 5, it makes no sense to divide by ‘ln’. 7.3 Exponential Equations and Inequalities
1. 23x = 161−x 3. 9 · 3x = 72x 2. 2000 = 1000 · 3−0.1t 4. 75 = 100
1+3e−2t 363
5. 25x = 5x + 6
6. ex −e−x
2 =5 Solution.
1−x 1. Since 16 is a power of 2, we can rewrite 23x = 161−x as 23x = 24
. Using properties of
exponents, we get 23x = 24(1−x) . Using the onetoone property of exponential functions, we
4
get 3x = 4(1 − x) which gives x = 7 . To check graphically, we set f (x) = 23x and g (x) = 161−x
and see that they intersect at x = 4 ≈ 0.5714.
7
2. We begin solving 2000 = 1000 · 3−0.1t by dividing both sides by 1000 to isolate the exponential
which yields 3−0.1t = 2. Since it is inconvenient to write 2 as a power of 3, we use the natural
log to get ln 3−0.1t = ln(2). Using the Power Rule, we get −0.1t ln(3) = ln(2), so we
divide both sides by −0.1 ln(3) to get t = − 0.ln(2) = − 10 ln(2) . On the calculator, we graph
1 ln(3)
ln(3)
f (x) = 2000 and g (x) = 1000 · 3−0.1x and ﬁnd that they intersect at x = − 10 ln(2) ≈ −6.3093.
ln(3) y = f (x) = 23x and y = f (x) = 2000 and y = g (x) = 161−x y = g (x) = 1000 · 3−0.1x 3. We ﬁrst note that we can rewrite the equation 9 · 3x = 72x as 32 · 3x = 72x to obtain 3x+2 = 72x .
Since it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use
the natural log: ln 3x+2 = ln 72x . The power rule gives (x + 2) ln(3) = 2x ln(7). Even
though this equation appears very complicated, keep in mind that ln(3) and ln(7) are just
constants. The equation (x + 2) ln(3) = 2x ln(7) is actually a linear equation and as such we
gather all of the terms with x on one side, and the constants on the other. We then divide
both sides by the coeﬃcient of x, which we obtain by factoring. 364 Exponential and Logarithmic Functions (x + 2) ln(3) = 2x ln(7)
x ln(3) + 2 ln(3) = 2x ln(7)
2 ln(3) = 2x ln(7) − x ln(3)
2 ln(3) = x(2 ln(7) − ln(3)) Factor.
x= 2 ln(3)
2 ln(7)−ln(3) Graphing f (x) = 9 · 3x and g (x) = 72x on the calculator, we see that these two graphs intersect
2 ln(3)
at x = 2 ln(7)−ln(3) ≈ 0.7866.
100
4. Our objective in solving 75 = 1+3e−2t is to ﬁrst isolate the exponential. To that end, we
clear denominators and get 75 1 + 3e−2t = 100. From this we get 75 + 225e−2t = 100,
which leads to 225e−2t = 25, and ﬁnally, e−2t = 1 . Taking the natural log of both sides
9
gives ln e−2t = ln 1 . Since natural log is log base e, ln e−2t = −2t. We can also use
9
the Power Rule to write ln 1 = − ln(9). Putting these two steps together, we simplify
9
1
ln e−2t = ln 9 to −2t = − ln(9). We arrive at our solution, t = ln(9) which simpliﬁes to
2
t = ln(3). (Can you explain why?) The calculator conﬁrms the graphs of f (x) = 75 and
100
g (x) = 1+3e−2x intersect at x = ln(3) ≈ 1.099. y = f (x) = 9 · 3x and
y = g (x) = 72x y = f (x) = 75 and
y = g (x) = 100
1+3e−2x
x 5. We start solving 25x = 5x + 6 by rewriting 25 = 52 so that we have 52 = 5x + 6, or
52x = 5x + 6. Even though we have a common base, having two terms on the right hand side
of the equation foils our plan of equating exponents or taking logs. If we stare at this long
enough, we notice that we have three terms with the exponent on one term exactly twice that
of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting u = 5x ,
we have u2 = (5x )2 = 52x so the equation 52x = 5x + 6 becomes u2 = u + 6. Solving this as
u2 − u − 6 = 0 gives u = −2 or u = 3. Since u = 5x , we have 5x = −2 or 5x = 3. Since
5x = −2 has no real solution, (Why not?) we focus on 5x = 3. Since it isn’t convenient to
express 3 as a power of 5, we take natural logs and get ln (5x ) = ln(3) so that x ln(5) = ln(3) 7.3 Exponential Equations and Inequalities
ln(3)
ln(5) . When
ln(3)
ln(5) ≈ 0.6826. or x =
x= 365 we graph f (x) = 25x and g (x) = 5x + 6, we see that they intersect at x −x 6. At ﬁrst, it’s unclear how to proceed with e −e = 5, besides clearing the denominator to
2
obtain ex − e−x = 10. Of course, if we rewrite e−x = e1 , we see we have another denominator
x
lurking in the problem: ex − e1 = 10. Clearing this denominator gives us e2x − 1 = 10ex ,
x
and once again, we have an equation with three terms where the exponent on one term is
exactly twice that of another  a ‘quadratic in disguise.’ If we let u = ex , then u2 = e2x so the
equation e2x − 1 = 10ex can be viewed as u2 − 1 = 10u. Solving u2 − 10u − 1 = 0, we obtain
√
√
√
by the quadratic formula u = 5 ± √ From this, we have ex = 5 ± 26. Since 5 − 26 < 0,
26.
√
we get no real solution to ex = 5 − 26, but for ex = 5 + 26, we take natural logs to obtain
√
x
−x
x = ln 5 + 26 . If we graph f (x) = e −e and g (x) = 5, we see that the graphs intersect
2
√
at x = ln 5 + 26 ≈ 2.312 y = f (x) = 25x and y = f (x) = y = g (x) = 5x + 6 ex −e−x
2 and y = g (x) = 5 The authors would be remiss not to mention that Example 7.3.1 still holds great educational
value. Much can be learned about logarithms and exponentials by verifying the solutions obtained
in Example 7.3.1 analytically. For example, to verify our solution to 2000 = 1000 · 3−0.1t , we
substitute t = − 10 ln(2) and obtain
ln(3)
? 2000 = 1000 · 3
? ”
“
10 ln(2)
−0.1 − ln(3)
ln(2) 2000 = 1000 · 3 ln(3)
? 2000 = 1000 · 3log3 (2)
? 2000 = 1000 · 2 Change of Base
Inverse Property 2000 = 2000
The other solutions can be veriﬁed by using a combination of log and inverse properties. Some
fall out quite quickly, while others are more involved. We leave them to the reader. 366 Exponential and Logarithmic Functions Since exponential functions are continuous on their domains, the Intermediate Value Theorem
4.1 applies. As with the algebraic functions in Section ??, this allows us to solve inequalities using
sign diagrams as demonstrated below.
Example 7.3.2. Solve the following inequalities. Check your answer graphically using a calculator.
1. 2x 2 −3x − 16 ≥ 0 2. ex
≤3
ex − 4 3. xe2x < 4x
Solution.
2 1. Since we already have 0 on one side of the inequality, we set r(x) = 2x −3x − 16. The domain
of r is all real numbers, so in order to construct our sign diagram, we seed to ﬁnd the zeros of
2
2
2
r. Setting r(x) = 0 gives 2x −3x − 16 = 0 or 2x −3x = 16. Since 16 = 24 we have 2x −3x = 24 ,
so by the onetoone property of exponential functions, x2 − 3x = 4. Solving x2 − 3x − 4 = 0
gives x = 4 and x = −1. From the sign diagram, we see r(x) ≥ 0 on (−∞, −1] ∪ [4, ∞), which
2
corresponds to where the graph of y = r(x) = 2x −3x − 16, is on or above the xaxis. (+) 0 (−) 0 (+)
−1 4
y = r(x) = 2x 2 −3x − 16 x e
2. The ﬁrst step we need to take to solve ex −4 ≤ 3 is to get 0 on one side of the inequality. To
that end, we subtract 3 from both sides and get a common denominator ex
ex − 4 ≤3 ex
−3 ≤ 0
ex − 4
ex
3 (ex − 4)
−
ex − 4
ex − 4
12 − 2ex
ex − 4 ≤ 0 Common denomintors. ≤0 7.3 Exponential Equations and Inequalities 367 x −2
We set r(x) = 12x −e and we note that r is undeﬁned when its denominator ex − 4 = 0, or
e4
x = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4), ∞). To
when e
ﬁnd the zeros of r, we solve r(x) = 0 and obtain 12 − 2ex = 0. Solving for ex , we ﬁnd ex = 6,
or x = ln(6). When we build our sign diagram, ﬁnding test values may be a little tricky since
we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4
which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the
sign of r (ln(3)) may be very unsettling, remember that eln(3) = 3, so 12 − 2(3)
12 − 2eln(3)
=
= −6
ln(3) − 4
3−4
e
We determine the signs of r (ln(5)) and r (ln(7)) similarly.5 From the sign diagram, we
ﬁnd our answer to be (−∞, ln(4)) ∪ [ln(6), ∞). Using the calculator, we see the graph of
ex
f (x) = ex −4 is below the graph of g (x) = 3 on (−∞, ln(4)) ∪ (ln(6), ∞), and they intersect
at x = ln(6) ≈ 1.792.
r (ln(3)) = (−) (+) 0 (−)
ln(4) ln(6) y = f (x) = ex
ex −4 y = g (x) = 3
3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x − 4x < 0.
We set r(x) = xe2x − 4x and since there are no denominators, evenindexed radicals, or logs,
the domain of r is all real numbers. Setting r(x) = 0 produces xe2x − 4x = 0. With x both
in and out of the exponent, this could cause some diﬃculty. However, before panic sets in,
we factor out the x to obtain x e2x − 4 = 0 which gives x = 0 or e2x − 4 = 0. To solve the
latter, we isolate the exponential and take logs to get 2x = ln(4), or x = ln(4) = ln(2). (Can
2
you explain the last equality using properties of logs?) As in the previous example, we need
to be careful about choosing test values. Since ln(1) = 0, we choose ln 1 , ln 3 and ln(3).
2
2
1
Evaluating,6 we have r(ln 1 ) = ln 1 e2 ln( 2 ) − 4 ln 1 . Applying the Power Rule to the log
2 4 2 2 This is because the base of ln(x) is e > 1. If the base b were in the interval 0 < b < 1, then logb (x) would
decreasing.
5
We could, of course, use the calculator, but what fun would that be?
6
A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. 368 Exponential and Logarithmic Functions
2 1
1
1
in the exponent, we obtain ln 1 eln( 2 ) − 4 ln 1 = ln 1 eln( 4 ) − 4 ln 2 . Using the inverse
2
2
2
1
1
1
properties of logs, this reduces to 4 ln 1 − 4 ln 2 = − 15 ln 1 . Since 1 < 1, ln 2 < 0
2
4
2
2
1
and we get r(ln 2 ) is (+). Continuing in this manner, we ﬁnd r(x) < 0 on (0, ln(2)). The
calculator conﬁrms that the graph of f (x) = xe2x is below the graph of g (x) = 4x on this
intervals.7 (+) 0 (−) 0 (+)
0 ln(2) y = f (x) = xe2x and
y = g (x) = 4x Example 7.3.3. Recall from Example 7.1.2 that the temperature of coﬀee T (in degrees Fahrenheit)
t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t . When will the coﬀee be warmer
than 100◦ F?
Solution. We need to ﬁnd when T (t) > 100, or in other words, we need to solve the inequality
70 + 90e−0.1t > 100. Getting 0 on one side of the inequality, we have 90e−0.1t − 30 > 0, and
we set r(t) = 90e−0.1t − 30. The domain of r is artiﬁcially restricted due to the context of the
problem to [0, ∞), so we proceed to ﬁnd the zeros of r. Solving 90e−0.1t − 30 = 0 results in
1
e−0.1t = 3 so that t = −10 ln 1 which, after a quick application of the Power Rule leaves us with
3
t = 10 ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1 < 3,
0 = ln(1) < ln(3) so that 10 ln(3) > 0. So we choose t = 0 as a test value in [0, 10 ln(3)). Since
3 < 4, 10 ln(3) < 10 ln(4), so the latter is our choice of a test value for the interval (10 ln(3), ∞).
Our sign diagram is below, and next to it is our graph of t = T (t) from Example 7.1.2 with the
horizontal line y = 100.
7 Note: ln(2) ≈ 0.693. 7.3 Exponential Equations and Inequalities 369
y
180
160
140 (+)
0 120 0 (−) y = 100 80 10 ln(3) 60 H.A. y = 70 40
20
2 4 6 8 10 12 14 16 18 20 t y = T (t) In order to interpret what this means in the context of the real world, we need a reasonable
approximation of the number 10 ln(3) ≈ 10.986. This means it takes approximately 11 minutes for
the coﬀee to cool to 100◦ F. Until then, the coﬀee is warmer than that.8
We close this section by ﬁnding the inverse of a function which is a composition of a rational
function with an exponential function.
5ex
is onetoone. Find a formula for f −1 (x) and check
ex + 1
your answer graphically using your calculator.
Solution. We start by writing y = f (x), and interchange the roles of x and y . To solve for y ,
we ﬁrst clear denominators and then isolate the exponential function. Example 7.3.4. The function f (x) = y= 5ex
ex + 1 x= 5ey
ey + 1 Switch x and y x (ey + 1) = 5ey
xey + x = 5ey
x = 5ey − xey
x = ey (5 − x)
ey = x
5−x ln (ey ) = ln
y = ln
8 x
5−x
x
5−x Critics may point out that since we needed to use the calculator to interpret our answer anyway, why not use it
earlier to simplify the computations? It is a fair question which we answer unfairly: it’s our book. 370 Exponential and Logarithmic Functions We claim f −1 (x) = ln x
5−x . To verify this analytically, we would need to verify the com f −1 positions
◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in
the domain of f −1 . We leave this to the reader. To verify our solution graphically, we graph
x
x
y = f (x) = e5e and y = g (x) = ln 5−x on the same set of axes and observe the symmetry about
x +1
the line y = x. Note the domain of f is the range of g and viceversa. y = f (x) = 5ex
ex +1 and y = g (x) = ln x
5 −x 7.3 Exponential Equations and Inequalities 7.3.1 371 Exercises 1. Solve the following equations analytically.
(a) 3(x−1) = 27
(b) 3(x−1) = 29
(c) 3(x−1) = 2x (h) (l) 2(x 7e2x = (m) (d)
=
x= 1
(e) 8
128
(f) 37x = 814−2x 0.06 12t
12
−5730k = 1
e
2 (j) 1 +
(k) 3 − x) e2x =1 = 2ex (n) 70 + 90e−0.1t = 75 (i) 73+7x = 34−2x 1 (x+5)
2 3(x−1) 1 2x
9
28e−6x (g) 9 · 37x = =3 (o)
(p) 150
= 75
1+29e−0.8t
x
4
25 5 = 10 2. Solve the following inequalities analytically.
(a) ex > 53
(b) 1000 1 +
(c) 3
2(x −x) 4x
≥ 10
5
150
≤ 130
1+29e−0.8t
−0.1t ≤ 75
70 + 90e (d) 25
0.06 12t
12 ≥ 3000 (e) <1 (f) 3. Use your calculator to help you solve the following equations and inequalities.
(a) ex < x3 − x
(b) 2x = x2 √ (c) e x = x + 1
(d) e−x − xe−x ≥ 0 (e) 3(x−1) < 2x
(f) ex = ln(x) + 5 4. Since f (x) = ln(x) is a strictly increasing function, if 0 < a < b then ln(a) < ln(b). Use this
fact to solve the inequality e(3x−1) > 6 without a sign diagram.
5. Compute the inverse of f (x) = ex − e−x
. State the domain and range of both f and f −1 .
2 6. In Example 7.3.4, we found that the inverse of f (x) = 5ex
was f −1 (x) = ln
ex + 1 x
5−x but we left a few loose ends for you to tie up.
(a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for
all x in the domain of f −1 .
(b) Find the range of f by ﬁnding the domain of f −1 .
5x
and h(x) = ex . Show that f = g ◦ h and that (g ◦ h)−1 = h−1 ◦ g −1 .
(c) Let g (x) =
x+1 9 7. With the help of your classmates, solve the inequality ex > xn for a variety of natural
numbers n. What might you conjecture about the “speed” at which f (x) = ex grows versus
any polynomial?
9 We know this is true in general by Exercise 8 in Section 6.2, but it’s nice to see a speciﬁc example of the property. 372 7.3.2 Exponential and Logarithmic Functions Answers 1. (a) x = 4
ln(29) + ln(3)
(b) x =
ln(3)
ln(3)
(c) x =
ln(3) − ln(2)
ln(3) + 5 ln 1
2
(d) x =
ln(3) − ln 1
2
7
(e) x = − 3
16
(f) x = 15
2
(g) x = − 11
1
(h) x = − 1 ln 1 = 4 ln(2)
8
4
4 ln(3) − 3 ln(7)
(i) x =
7 ln(7) + 2 ln(3)
2. (a) (ln(53), ∞)
ln(3)
(b)
,∞
12 ln(1.005)
(c) (−∞, −1) ∪ (0, 1)
(d) −∞, ln
ln 2
5
4
5 3. (a) (2.3217, 4.3717)
(b) x ≈ −0.76666, x = 2, x = 4
(c) x = 0 (j) t = ln(3)
12 ln(1.005) ln 1
2
−5730
(l) x = −1, 0, 1 (k) k = (m) x = ln(2)
(n) t = 10 ln(18)
1
ln 29
−0.8
ln 2
5
(p) x =
ln 4
5 (o) t = 2
ln 377
−0.8 (e) −∞, (f) 1
ln 18
,∞
−0.1 = −∞, − 5 ln
4 2
377 = [10 ln(18), ∞) (d) (−∞, 1]
(e) (−∞, 2.7095)
(f) x ≈ 0.01866, x ≈ 1.7115 4. x > 1 (ln(6) + 1)
3
5. f −1 = ln x + √ x2 + 1 . Both f and f −1 have domain (−∞, ∞) and range (−∞, ∞). 7.4 Logarithmic Equations and Inequalities 7.4 373 Logarithmic Equations and Inequalities In Section 7.3 we solved equations and inequalities involving exponential functions using one of
two basic strategies. We now turn our attention to equations and inequalities involving logarithmic
functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose
we wish to solve log2 (x) = log2 (5). Theorem 7.4 tells us that the only solution to this equation
is x = 5. Now suppose we wish to solve log2 (x) = 3. If we want to use Theorem 7.4, we need to
rewrite 3 as a logarithm base 2. We can use Theorem 7.3 to do just that: 3 = log2 23 = log2 (8).
Our equation then becomes log2 (x) = log2 (8) so that x = 8. However, we could have arrived at the
same answer, in fewer steps, by using Theorem 7.3 to rewrite the equation log2 (x) = 3 as 23 = x,
or x = 8. We summarize the two common ways to solve log equations below. Steps for Solving an Equation involving Logarithmic Fuctions
1. Isolate the logarithmic function.
2. (a) If convenient, express both sides as logs with the same base and equate the arguments
of the log functions.
(b) Otherwise, rewrite the log equation as an exponential equation. Example 7.4.1. Solve the following equations. Check your solutions graphically using a calculator.
1. log117 (1 − 3x) = log117 x2 − 3 4. log7 (1 − 2x) = 1 − log7 (3 − x) 2. 2 − ln(x − 3) = 1 5. log2 (x + 3) = log2 (6 − x) + 3 3. log6 (x + 4) + log6 (3 − x) = 1 6. 1 + 2 log4 (x + 1) = 2 log2 (x) Solution.
1. Since we have the same base on both sides of the equation log117 (1 − 3x) = log117 x2 − 3 ,
we equate what’s inside the logs to get 1 − 3x = x2 − 3. Solving x2 + 3x − 4 = 0 gives
x = −4 and x = 1. To check these answers using the calculator, we make use of the change
ln(x2 −3)
of base formula and graph f (x) = ln(1−3x) and g (x) = ln(117) and we see they intersect only
ln(117)
at x = −4. To see what happened to the solution x = 1, we substitute it into our original
equation to obtain log117 (−2) = log117 (−2). While these expressions look identical, neither
is a real number,1 which means x = 1 is not in the domain of the original equation, and is
not a solution.
1 They do, however, represent the same family of complex numbers. We stop ourselves at this point and refer the
reader to a good course in Complex Variables. 374 Exponential and Logarithmic Functions 2. Our ﬁrst objective in solving 2 − ln(x − 3) = 1 is to isolate the logarithm. We get ln(x − 3) = 1,
which, as an exponential equation, is e1 = x − 3. We get our solution x = e + 3. On the
calculator, we see the graph of f (x) = 2 − ln(x − 3) intersects the graph of g (x) = 1 at
x = e + 3 ≈ 5.718. y = f (x) = log117 (1 − 3x) and y = f (x) = 2 − ln(x − 3) and y = g (x) = log117 x2 − 3 y = g (x) = 1 3. We can start solving log6 (x + 4) + log6 (3 − x) = 1 by using the Product Rule for logarithms to
rewrite the equation as log6 [(x + 4)(3 − x)] = 1. Rewriting this as an exponential equation,
we get 61 = (x + 4)(3 − x). This reduces to x2 + x − 6 = 0, which gives x = −3 and x = 2.
Graphing y = f (x) = ln(x+4) + ln(3−x) and y = g (x) = 1, we see they intersect twice, at
ln(6)
ln(6)
x = −3 and x = 2. y = f (x) = log6 (x + 4) + log6 (3 − x) and y = g (x) = 1
4. Taking a cue from the previous problem, we begin solving log7 (1 − 2x) = 1 − log7 (3 − x) by
ﬁrst collecting the logarithms on the same side, log7 (1 − 2x) + log7 (3 − x) = 1, and then using
the Product Rule to get log7 [(1 − 2x)(3 − x)] = 1. Rewriting this as an exponential equation
gives 71 = (1 − 2x)(3 − x) which gives the quadratic equation 2x2 − 7x − 4 = 0. Solving, we ﬁnd
1
x = − 2 and x = 4. Graphing, we ﬁnd y = f (x) = ln(1−2x) and y = g (x) = 1 − ln(3−x) intersect
ln(7)
ln(7)
1
only at x = − 2 . Checking x = 4 in the original equation produces log7 (−7) = 1 − log7 (−1),
which is a clear domain violation.
5. Starting with log2 (x + 3) = log2 (6 − x) + 3, we gather the logarithms to one side and get
x+3
log2 (x + 3) − log2 (6 − x) = 3, and then use the Quotient Rule to obtain log2 6−x = 3. 7.4 Logarithmic Equations and Inequalities
Rewriting this as an exponential equation gives 23 = 375
x+3
6−x . This reduces to the linear equation 8(6 − x) = x + 3, which gives us x = 5. When we graph f (x) =
we ﬁnd they intersect at x = 5. ln(x+3)
ln(2) and g (x) = ln(6−x)
ln(2) y = f (x) = log7 (1 − 2x) and y = f (x) = log2 (x + 3) and y = g (x) = 1 − log7 (3 − x) + 3, y = g (x) = log2 (6 − x) + 3 6. Starting with 1 + 2 log4 (x + 1) = 2 log2 (x), we gather the logs to one side to get the equation
1 = 2 log2 (x) − 2 log4 (x + 1). Before we can combine the logarithms, however, we need a
common base. Since 4 is a power of 2, we use change of base to convert log4 (x+1) = log2 (x+1) =
log (4)
1
2 log2 (x + 1). Hence, our original equation becomes 1 = 2 log2 (x) − 2 1
2 2 log2 (x + 1) or 1 = 2 log2 (x) − log2 (x + 1). Using the Power and Quotient Rules, we obtain 1 = log2 x2
x+1 . x2 Rewriting this in exponential form, we get x+1 = 2 or x2 − 2x − 2 = 0. Using the quadratic
√
ln(x
formula, we get x = 1 ± 3. Graphing f (x) = 1 + 2 ln(x+1) and g (x) = 2ln(2)) , we see the
ln(4)
√
√
graphs intersect only at x = 1 + 3 ≈ 2.732. The solution√ = 1 − 3 < 0, which means if
x
substituted into the original equation, the term 2 log2 1 − 3 is undeﬁned. y = f (x) = 1 + 2 log4 (x + 1) and y = g (x) = 2 log2 (x) If nothing else, Example 7.4.1 demonstrates the importance of checking for extraneous solutions2
when solving equations involving logarithms. Even though we checked our answers graphically,
2 Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation. 376 Exponential and Logarithmic Functions extraneous solutions are easy to spot  any supposed solution which causes a negative number
inside a logarithm needs to be discarded. As with the equations in Example 7.3.1, much can be
learned from checking all of the answers in Example 7.4.1 analytically. We leave this to the reader
and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions
are continuous on their domains, we can use sign diagrams.
Example 7.4.2. Solve the following inequalities. Check your answer graphically using a calculator.
1. 1
≤1
ln(x) + 1 2. (log2 (x))2 < 2 log2 (x) + 3 3. x log(x + 1) ≥ x Solution.
1. We start solving 1
ln(x)+1 1
ln(x)+1 − 1
−
reduces to ln(ln(x)
x)+1 ≤ 1 by getting 0 on one side of the inequality:
1
ln(x)+1
ln(x)
ln(x)+1 and Getting a common denominator yields
ln(x)
ln(x)+1 − ln(x)+1
ln(x)+1 ≤ 0 which ≤ 0.
≤ 0, or
≥ 0. We deﬁne r(x) =
set about ﬁnding the domain and the zeros
of r. Due to the appearance of the term ln(x), we require x > 0. In order to keep the
denominator away from zero, we solve ln(x) + 1 = 0 so ln(x) = −1, so x = e−1 = 1 . Hence,
e
ln(x)
the domain of r is 0, 1 ∪ 1 , ∞ . To ﬁnd the zeros of r, we set r(x) = ln(x)+1 = 0 so that
e
e
ln(x) = 0, and we ﬁnd x = e0 = 1. In order to determine test values for r without resorting
to the calculator, we need to ﬁnd numbers between 0, 1 , and 1 which have a base of e. Since
e
1
e ≈ 2.718 > 1, 0 < e1 < 1 < √e < 1 < e. To determine the sign of r e1 , we use the fact that
2
2
e
ln e1 = ln e−2 = −2, and ﬁnd r e1 = −−2 = 2, which is (+). The rest of the test values
2
2
2+1
are determined similarly. From our sign diagram, we ﬁnd the solution to be 0, 1 ∪ [1, ∞).
e
1
Graphing f (x) = ln(x)+1 and g (x) = 1, we see the the graph of f is below the graph of g on
the solution intervals, and that the graphs intersect at x = 1. (+) (−) 0 (+)
0 1
e 1 y = f (x) = 1
ln(x)+1 and y = g (x) = 1 2. Moving all of the nonzero terms of (log2 (x))2 < 2 log2 (x) + 3 to one side of the inequality,
we have (log2 (x))2 − 2 log2 (x) − 3 < 0. Deﬁning r(x) = (log2 (x))2 − 2 log2 (x) − 3, we get
the domain of r is (0, ∞), due to the presence of the logarithm. To ﬁnd the zeros of r, we 7.4 Logarithmic Equations and Inequalities 377 set r(x) = (log2 (x))2 − 2 log2 (x) − 3 = 0 which results in a ‘quadratic in disguise.’ We set
u = log2 (x) so our equation becomes u2 − 2u − 3 = 0 which gives us u = −1 and u = 3. Since
u = log2 (x), we get log2 (x) = −1, which gives us x = 2−1 = 1 , and log2 (x) = 3, which yields
2
1
x = 23 = 8. We use test values which are powers of 2: 0 < 1 < 2 < 1 < 8 < 16, and from our
4
sign diagram, we see r(x) < 0 on
is below the graph of y = g (x) = 1
2, 8
2 ln(x)
ln(2) . Geometrically, we see the graph of f (x) = ln(x)
ln(2) 2 + 3 on the solution interval. (+) 0 (−) 0 (+)
1
2 0 8
y = f (x) = (log2 (x))2 and y = g (x) = 2 log2 (x) + 3 3. We begin to solve x log(x +1) ≥ x by subtracting x from both sides to get x log(x +1) − x ≥ 0.
We deﬁne r(x) = x log(x +1) − x and due to the presence of the logarithm, we require x +1 > 0,
or x > −1. To ﬁnd the zeros of r, we set r(x) = x log(x + 1) − x = 0. Factoring, we get
x (log(x + 1) − 1) = 0, which gives x = 0 or log(x +1) − 1 = 0. The latter gives log(x +1) = 1,
or x + 1 = 101 , which admits x = 9. We select test values x so that x + 1 is a power of 10,
√
and we obtain −1 < −0.9 < 0 < 10 − 1 < 9 < 99. Our sign diagram gives the solution to
be (−1, 0] ∪ [9, ∞). The calculator indicates the graph of y = f (x) = x log(x + 1) is above
y = g (x) = x on the solution intervals, and the graphs intersect at x = 0 and x = 9. (+) 0 (−) 0 (+)
−1 0 9
y = f (x) = x log(x + 1) and y = g (x) = x 378 Exponential and Logarithmic Functions Near x = 0 Near x = 9 Our next example revisits the concept of pH as ﬁrst introduced in the exercises in Section 7.1.
Example 7.4.3. In order to successfully breed Ippizuti ﬁsh the pH of a freshwater tank must be
at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion
concentration.
Solution. Recall from Exercise 6c in Section 7.1 that pH = − log[H+ ] where [H+ ] is the
hydrogen ion concentration in moles per liter. We require 7.8 ≤ − log[H+ ] ≤ 8.5 or −7.8 ≥
log[H+ ] ≥ −8.5. To solve this compound inequality we solve −7.8 ≥ log[H+ ] and log[H+ ] ≥ −8.5
and take the intersection of the solution sets.3 The former inequality yields 0 < [H+ ] ≤ 10−7.8
and the latter yields [H+ ] ≥ 10−8.5 . Taking the intersection gives us our ﬁnal answer 10−8.5 ≤
[H+ ] ≤ 10−7.8 . (Your Chemistry professor may want the answer written as 3.16 × 10−9 ≤ [H+ ] ≤
1.58 × 10−8 .) After carefully adjusting the viewing window on the graphing calculator we see
that the graph of f (x) = − log(x) lies between the lines y = 7.8 and y = 8.5 on the interval
[3.16 × 10−9 , 1.58 × 10−8 ]. The graphs of y = f (x) = − log(x), y = 7.8 and y = 8.5 We close this section by ﬁnding an inverse of a onetoone function which involves logarithms.
3 Refer to page 202 for a discussion of what this means. 7.4 Logarithmic Equations and Inequalities 379 log(x)
is onetoone. Find a formula for f −1 (x) and
1 − log(x)
check your answer graphically using your calculator. Example 7.4.4. The function f (x) = Solution. We ﬁrst write y = f (x) then interchange the x and y and solve for y . y = f (x)
log(x)
y=
1 − log(x)
log(y )
x=
1 − log(y ) Interchange x and y . x (1 − log(y )) = log(y )
x − x log(y ) = log(y )
x = x log(y ) + log(y )
x = (x + 1) log(y )
x
= log(y )
x+1
x
y = 10 x+1 Rewrite as an exponential equation. 380 Exponential and Logarithmic Functions
x We have f −1 (x) = 10 x+1 . Graphing f and f −1 on the same viewing window yields y = f (x) = x
log(x)
and y = g (x) = 10 x+1
1 − log(x) 7.4 Logarithmic Equations and Inequalities 7.4.1 381 Exercises 1. Solve the following equations analytically.
x
= 150
10−12
(j) log3 (x) = log 1 (x) + 8 (a) log 1 x = −3 (i) 10 log 2 (b) ln(x2 ) = (ln(x))2 3 (c) log3 (x − 4) + log3 (x + 4) = 2
(d) log5 (2x + 1) + log5 (x + 2) = 1
(e) log2 (x3 ) = log2 (x)
(f) log169 (3x + 7) − log169 (5x − 9) =
x
(g) log
= 4.7
10−3
(h) − log(x) = 5.4 1
2 1
3x − 2
=
(k) log125
2x + 3
3
(l) ln(x + 1) − ln(x) = 3
(m) ln(ln(x)) = 3
(n) 2 log7 (x) = log7 (2) + log7 (x + 12)
(o) log(x) − log(2) = log(x + 8) − log(x + 2) 2. Solve the following inequalities analytically.
(a) x ln(x) − x > 0
x
(b) 5.6 ≤ log
≤ 7.1
10−3
x
≥ 90
(c) 10 log
10−12 (d) 2.3 < − log(x) < 5.4
1 − ln(x)
<0
(e)
x2
(f) ln(x2 ) ≤ (ln(x))2 3. Use your calculator to help you solve the following equations and inequalities.
(a) ln(x) = e−x
(b) ln(x2 + 1) ≥ 5 √
(c) ln(x) = 4 x
(d) ln(−2x3 − x2 + 13x − 6) < 0 4. Since f (x) = ex is a strictly increasing function, if a < b then ea < eb . Use this fact to solve
the inequality ln(2x + 1) < 3 without a sign diagram. Also, compare this exercise to question
4 in Section 7.3.
5. Solve ln(3 − y ) − ln(y ) = 2x + ln(5) for y .
6. In Example 7.4.4 we found the inverse of f (x) = x
log(x)
to be f −1 (x) = 10 x+1 .
1 − log(x) (a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for
all x in the domain of f −1 .
(b) Find the range of f by ﬁnding the domain of f −1 .
x
(c) Let g (x) =
and h(x) = log(x). Show that f = g ◦ h and (g ◦ h)−1 = h−1 ◦ g −1 .
1−x
(We know this is true in general by Exercise 8 in Section 6.2, but it’s nice to see a speciﬁc
example of the property.) 382 7. Let f (x) = Exponential and Logarithmic Functions
1
ln
2 1+x
. Compute f −1 (x) and ﬁnd its domain and range.
1−x 8. Explain the equation in Exercise 1g and the inequality in Exercise 2b above in terms of the
Richter scale for earthquake magnitude. (See Exercise 6a in Section 7.1.)
9. Explain the equation in Exercise 1i and the inequality in Exercise 2c above in terms of sound
intensity level as measured in decibels. (See Exercise 6b in Section 7.1.)
10. Explain the equation in Exercise 1h and the inequality in Exercise 2d above in terms of the
pH of a solution. (See Exercise 6c in Section 7.1.)
√
11. With the help of your classmates, solve the inequality n x > ln(x) for a variety of natural
numbers n. What might you conjecture about the “speed” at which f (x) = ln(x) grows
versus any principal nth root function? 7.4 Logarithmic Equations and Inequalities 7.4.2 383 Answers 1. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h) x=8
x = 1, x = e2
x=5
1
x= 2
x=1
x=2
x = 101.7
x = 10−5.4 (i) x = 103
(j) x = 81
(k) x = − 17
7
(l) x = 1
e3 −1
3 (m) x = ee
(n) x = 6
(o) x = 4 2. (a) (e, ∞)
(b) 102.6 , 104.1
(c) 10−3 , ∞ (d) 10−5.4 , 10−2.3
(e) (e, ∞)
(f) (0, 1] ∪ [e2 , ∞) 3. (a) x ≈ 1.3098
(b) (−∞, −12.1414) ∪ (12.1414, ∞) (c) x ≈ 4.177, x ≈ 5503.665
(d) (−3.0281, −3) ∪ (0.5, 0.5991) ∪ (1.9299, 2) 1
e3 − 1
4. − < x <
2
2
5. y = 3
5e2x +1 e2x − 1
ex − e−x
=x
. (The reason for this rewriting will be explained much later
e2x + 1
e + e−x
in the text.) The domain of f −1 is (−∞, ∞) and its range is the same as the domain of f ,
namely (−1, 1). 7. f −1 (x) = 384 Exponential and Logarithmic Functions 7.5 Applications of Exponential and Logarithmic Functions As we mentioned in Section 7.1, exponential and logarithmic functions are used to model a wide
variety of behaviors in the real world. In the examples that follow, note that while the applications
are drawn from many diﬀerent disciplines, the mathematics remains essentially the same. Due to
the applied nature of the problems we will examine in this section, the calculator is often used to
express our answers as decimal approximations. 7.5.1 Applications of Exponential Functions Perhaps the most wellknown application of exponential functions comes from the ﬁnancial world.
Suppose you have $100 to invest at your local bank and they are oﬀering a whopping 5 % annual
percentage interest rate. This means that after one year, the bank will pay you 5% of that $100,
or $100(0.05) = $5 in interest, so you now have $105.1 This is in accordance with the formula
for simple interest which you have undoubtedly run across at some point in your mathematical
upbringing.
Equation 7.1. Simple Interest The amount of interest I accrued at an annual rate r on an
investmenta P after t years is
I = P rt
The amount A in the account after t years is given by
A = P + I = P + P rt = P (1 + rt)
a Called the principal Suppose, however, that six months into the year, you hear of a better deal at a rival bank.2
Naturally, you withdraw your money and try to invest it at the higher rate there. Since six months
is one half of a year, that initial $100 yields $100(0.05) 1 = $2.50 in interest. You take your
2
$102.50 oﬀ to the competitor and ﬁnd out that those restrictions which may apply actually do
apply to you, and you return to your bank which happily accepts your $102.50 for the remaining
six months of the year. To your surprise and delight, at the end of the year your statement reads
$105.06, not $105 as you had expected.3 Where did those extra six cents come from? For the ﬁrst
six months of the year, interest was earned on the original principal of $100, but for the second
six months, interest was earned on $102.50, that is, you earned interest on your interest. This is
the basic concept behind compound interest. In the previous discussion, we would say that the
interest was compounded twice, or semiannually.4 If more money can be earned by earning interest
1 How generous of them!
Some restrictions may apply.
3
Actually, the ﬁnal balance should be $105.0625.
4
Using this convention, simple interest after one year is the same as compounding the interest only once.
2 7.5 Applications of Exponential and Logarithmic Functions 385 on interest already earned, a natural question to ask is what happens if the interest is compounded
more often, say 4 times a year, which is every three months, or ‘quarterly.’ In this case, the
money is in the account for three months, or 1 of a year, at a time. After the ﬁrst quarter, we
4
have A = P (1 + rt) = $100 1 + 0.05 · 1 = $101.25. We now invest the $101.25 for the next three
4
months and ﬁnd that at the end of the second quarter, we have A = $101.25 1 + 0.05 · 1 ≈ $102.51.
4
Continuing in this manner, the balance at the end of the third quarter is $103.79, and, at last, we
obtain $105.08. The extra two cents hardly seems worth it, but we see that we do in fact get more
money the more often we compound. In order to develop a formula for this phenomenon, we need
to do some abstract calculations. Suppose we wish to invest our principal P at an annual rate r and
1 th
compound the interest n times per year. This means the money sits in the account n of a year
between compoundings. Let Ak denote the amount in the account after the k th compounding. Then
1
r
A1 = P 1 + r n which simpliﬁes to A1 = P 1 + n . After the second compounding, we use A1
r
r
r
r2
as our new principal and get A2 = A1 1 + n = P 1 + n
1 + n = P 1 + n . Continuing in
3 4 k r
r
r
this fashion, we get A3 = P 1 + n , A4 = P 1 + n , and so on, so that Ak = P 1 + n . Since
we compound the interest n times per year, after t years, we have nt compoundings. We have just
derived the general formula for compound interest below. Equation 7.2. Compounded Interest: If an initial principal P is invested at an annual rate
r and the interest is compounded n times per year, the amount A in the account after t years
is
r nt
A=P 1+
n 4t 05
If we take P = 100, r = 0.05, and n = 4, Equation 7.2 becomes A = 100 1 + 0.4
which
4t . This equation deﬁnes the amount A as an exponential function of
reduces to A = 100(1.0125)
1
time t, A(t). To check this against our previous calculations, we ﬁnd A 1 = 100(1.0125)4( 4 ) =
4
101.25, A 1 ≈ $102.51, A 3 ≈ $103.79, and A(1) ≈ $105.08.
2
4 Example 7.5.1. Suppose $2000 is invested in an account which oﬀers 7.125% compounded monthly.
1. Express the amount A in the account as a function of the term of the investment t in years.
2. How much is in the account after 5 years?
3. How long will it take for the initial investment to double?
4. Find and interpret the average rate of change5 of the amount in the account from the end of
the fourth year to the end of the ﬁfth year, and from the end of the thirtyfourth year to the
end of the thirtyﬁfth year.
Solution.
5 See Deﬁnition 3.3 in Section 3.1. 386 Exponential and Logarithmic Functions 1. Substituting P = 2000, r = 0.07125, and n = 12 (monthly) into Equation 7.2 yields A =
12t
2000 1 + 0.07125
. Using function notation, we get A(t) = 2000(1.0059375)12t .
12
2. Since t represents the length of the investment, we substitute t = 5 into A(t) to ﬁnd A(5) =
2000(1.0059375)12(5) ≈ 2852.92. After 5 years, we have approximately $2852.92.
3. Our initial investment is $2000, so to ﬁnd the time it takes this to double, we need to ﬁnd t
when A(t) = 4000. We get 2000(1.0059375)12t = 4000, or (1.0059375)12t = 2. Taking natural
ln(2)
logs as in Section 7.3, we get t = 12 ln(1.0059375) ≈ 9.75. Hence, it takes approximately 9 years
9 months for the investment to double.
4. To ﬁnd the average rate of change of A from the end of the fourth year to the end of the
ﬁfth year, we compute A(5)−A(4) ≈ 195.63. Similarly, the average rate of change of A from
5−4
the end of the thirtyfourth year to the end of the thirtyﬁfth year is A(35)−A(34) ≈ 1648.21.
35−34
This means that the value of the investment is increasing at a rate of approximately $195.63
per year between the end of the fourth and ﬁfth years, while that rate jumps to $1648.21 per
year between the end of the thirtyfourth and thirtyﬁfth years. So, not only is it true that
the longer you wait, the more money you have, but also the longer you wait, the faster the
money increases.6
We have observed that the more times you compound the interest per year, the more money
you will earn in a year. Let’s push this notion to the limit.7 Consider an investment of $1 invested
at 100% interest for 1 year compounded n times a year. Equation 7.2 tells us that the amount of
1n
money in the account after 1 year is A = 1 + n . Below is a table of values relating n and A.
n A 1 2 2 2.25 4 ≈ 2.4414 12 ≈ 2.6130 360 ≈ 2.7145 1000 ≈ 2.7169 10000 ≈ 2.7181 100000 ≈ 2.7182 As promised, the more compoundings per year, the more money there is in the account, but we
also observe that the increase in money is greatly diminishing. We are witnessing a mathematical
6
In fact, the rate of increase of the amount in the account is exponential as well. This is the quality that really
deﬁnes exponential functions and we refer the reader to a course in Calculus.
7
Once you’ve had a semester of Calculus, you’ll be able to fully appreciate this very lame pun. 7.5 Applications of Exponential and Logarithmic Functions 387 ‘tug of war’. While we are compounding more times per year, and hence getting interest on our
interest more often, the amount of time between compoundings is getting smaller and smaller, so
there is less time to build up additional interest. With Calculus, we can show8 that as n → ∞,
1n
A = 1 + n → e, where e is the natural base ﬁrst presented in Section 7.1. Taking the number
of compoundings per year to inﬁnity results in what is called continuously compounded interest. Theorem 7.8. If you invest $1 at 100% interest compounded continuously, then you will have
$e at the end of one year. Using this deﬁnition of e and a little Calculus, we can take Equation 7.2 and produce a formula
for continuously compounded interest. Equation 7.3. Continuously Compounded Interest: If an initial principal P is invested
at an annual rate r and the interest is compounded continuously, the amount A in the account
after t years is
A = P ert If we take the scenario of Example 7.5.1 and compare monthly compounding to continuous compounding over 35 years, we ﬁnd that monthly compounding yields A(35) = 2000(1.0059375)12(35)
which is about $24,035.28, whereas continuously compounding gives A(35) = 2000e0.07125(35) which
is about $24,213.18  a diﬀerence of less than 1%.
Equations 7.2 and 7.3 both use exponential functions to describe the growth of an investment.
Curiously enough, the same principles which govern compound interest are also used to model short
term growth of populations. In Biology, The Law of Uninhibited Growth states as its premise
that the instantaneous rate at which a population increases at any time is directly proportional to
the population at that time.9 In other words, the more organisms there are at a given moment,
the faster they reproduce. Formulating the law as stated results in a diﬀerential equation, which
requires Calculus to solve. Its solution is stated below. 8 Or deﬁne, depending on your point of view.
The average rate of change of a function over an interval was ﬁrst introduced in Section 3.1. Instantaneous rates
of change are the business of Calculus, as is mentioned on Page 167.
9 388 Exponential and Logarithmic Functions Equation 7.4. Uninhibited Growth: If a population increases according to The Law of
Uninhibited Growth, the number of organisms N at time t is given by the formula
N (t) = N0 ekt ,
where N (0) = N0 (read ‘N nought’) is the initial number of organisms and k > 0 is the constant
of proportionality which satisﬁes the equation
(instantaneous rate of change of N (t) at time t) = k N (t) It is worth taking some time to compare Equations 7.3 and 7.4. In Equation 7.3, we use P
to denote the initial investment; in Equation 7.4, we use N0 to denote the initial population. In
Equation 7.3, r denotes the annual interest rate, and so it shouldn’t be too surprising that the k
in Equation 7.4 corresponds to a growth rate as well. While Equations 7.3 and 7.4 look entirely
diﬀerent, they both represent the same mathematical concept. Example 7.5.2. In order to perform arthrosclerosis research, epithelial cells are harvested from
discarded umbilical tissue and grown in the laboratory. A technician observes that a culture of
twelve thousand cells grows to ﬁve million cells in one week. Assuming that the cells follow The
Law of Uninhibited Growth, ﬁnd a formula for the number of cells, N , in thousands, after t days.
Solution. We begin with N (t) = N0 ekt . Since N is to give the number of cells in thousands,
we have N0 = 12, so N (t) = 12ekt . In order to complete the formula, we need to determine the
growth rate k . We know that after one week, the number of cells has grown to ﬁve million. Since t
measures days and the units of N are in thousands, this translates mathematically to N (7) = 5000.
1250
t
1
We get the equation 12e7k = 5000 which gives k = 7 ln 1250 . Hence, N (t) = 12e 7 ln( 3 ) . Of
3
course, in practice, we would approximate k to some desired accuracy, say k ≈ 0.8618, which we
can interpret as an 86.18% daily growth rate for the cells.
Whereas Equations 7.3 and 7.4 model the growth of quantities, we can use equations like them to
describe the decline of quantities. One example we’ve seen already is Example 7.1.1 in Section 7.1.
There, the value of a car declined from its purchase price of $25,000 to nothing at all. Another real
world phenomenon which follows suit is radioactive decay. There are elements which are unstable
and emit energy spontaneously. In doing so, the amount of the element itself diminishes. The
assumption behind this model is that the rate of decay of an element at a particular time is directly
proportional to the amount of the element present at that time. In other words, the more of the
element there is, the faster the element decays. This is precisely the same kind of hypothesis which
drives The Law of Uninhibited Growth, and as such, the equation governing radioactive decay is
hauntingly similar to Equation 7.4 with the exception that the rate constant k is negative. 7.5 Applications of Exponential and Logarithmic Functions 389 Equation 7.5. Radioactive Decay The amount of a radioactive element A at time t is given
by the formula
A(t) = A0 ekt ,
where A(0) = A0 is the initial amount of the element and k < 0 is the constant of proportionality
which satisﬁes the equation
(instantaneous rate of change of A(t) at time t) = k A(t) Example 7.5.3. Iodine131 is a commonly used radioactive isotope used to help detect how well
the thyroid is functioning. Suppose the decay of Iodine131 follows the model given in Equation 7.5,
and that the halflife10 of Iodine131 is approximately 8 days. If 5 grams of Iodine131 is present
initially, ﬁnd a function which gives the amount of Iodine131, A, in grams, t days later.
Solution. Since we start with 5 grams initially, Equation 7.5 gives A(t) = 5ekt . Since the
halflife is 8 days, it takes 8 days for half of the Iodine131 to decay, leaving half of it behind.
1
Hence, A(8) = 2.5 which means 5e8k = 2.5. Solving, we get k = 8 ln 1 = − ln(2) ≈ −0.08664,
2
8
which we can interpret as a loss of material at a rate of 8.664% daily. Hence, A(t) = 5e−
5e−0.08664t . t ln(2)
8 ≈ We now turn our attention to some more mathematically sophisticated models. One such model
is Newton’s Law of Cooling, which we ﬁrst encountered in Example 7.1.2 of Section 7.1. In that
example we had a cup of coﬀee cooling from 160◦ F to room temperature 70◦ F according to the
formula T (t) = 70 + 90e−0.1t , where t was measured in minutes. In this situation, we know the
physical limit of the temperature of the coﬀee is room temperature,11 and the diﬀerential equation
which gives rise to our formula for T (t) takes this into account. Whereas the radioactive decay
model had a rate of decay at time t directly proportional to the amount of the element which
remained at time t, Newton’s Law of Cooling states that the rate of cooling of the coﬀee at a given
time t is directly proportional to how much of a temperature gap exists between the coﬀee at time
t and room temperature, not the temperature of the coﬀee itself. In other words, the coﬀee cools
faster when it is ﬁrst served, and as its temperature nears room temperature, the coﬀee cools ever
more slowly. Of course, if we take an item from the refrigerator and let it sit out in the kitchen,
the object’s temperature will rise to room temperature, and since the physics behind warming and
cooling is the same, we combine both cases in the equation below.
10 The time it takes for half of the substance to decay.
The Second Law of Thermodynamics states that heat can spontaneously ﬂow from a hotter object to a colder
one, but not the other way around. Thus, the coﬀee could not continue to release heat into the air so as to cool below
room temperature.
11 390 Exponential and Logarithmic Functions Equation 7.6. Newton’s Law of Cooling (Warming): The temperature T of an object at
time t is given by the formula
T (t) = Ta + (T0 − Ta ) e−kt ,
where T (0) = T0 is the initial temperature of the object, Ta is the ambient temperaturea and
k > 0 is the constant of proportionality which satisﬁes the equation
(instantaneous rate of change of T (t) at time t) = k (T (t) − Ta )
a That is, the temperature of the surroundings. If we reexamine the situation in Example 7.1.2 with T0 = 160, Ta = 70, and k = 0.1, we get,
according to Equation 7.6, T (t) = 70 + (160 − 70)e−0.1t which reduces to the original formula given.
The rate constant k = 0.1 indicates the coﬀee is cooling at a rate equal to 10% of the diﬀerence
between the temperature of the coﬀee and its surroundings. Note in Equation 7.6 that the constant
k is positive for both the cooling and warming scenarios. What determines if the function T (t)
is increasing or decreasing is if T0 (the initial temperature of the object) is greater than Ta (the
ambient temperature) or viceversa, as we see in our next example.
Example 7.5.4. A 40◦ F roast is cooked in a 350◦ F oven. After 2 hours, the temperature of the
roast is 125◦ F.
1. Assuming the temperature of the roast follows Newton’s Law of Warming, ﬁnd a formula for
the temperature of the roast T as a function of its time in the oven, t, in hours.
2. The roast is done when the internal temperature reaches 165◦ F. When will the roast be done?
Solution.
1. The initial temperature of the roast is 40◦ F, so T0 = 40. The environment in which we
are placing the roast is the 350◦ F oven, so Ta = 350. Newton’s Law of Warming tells us
T (t) = 350 + (40 − 350)e−kt , or T (t) = 350 − 310e−kt . To determine k , we use the fact that
after 2 hours, the roast is 125◦ F, which means T (2) = 125. This gives rise to the equation
350 − 310e−2k = 125 which yields k = − 1 ln 45 ≈ 0.1602. The temperature function is
2
62
t 45 T (t) = 350 − 310e 2 ln( 62 ) ≈ 350 − 310e−0.1602t .
2. To determine when the roast is done, we set T (t) = 165. This gives 350 − 310e−0.1602t = 165
1
whose solution is t = − 0.1602 ln 37 ≈ 3.22. It takes roughly 3 hours and 15 minutes to cook
62
the roast completely. 7.5 Applications of Exponential and Logarithmic Functions 391 If we had taken the time to graph y = T (t) in Example 7.5.4, we would have found the horizontal
asymptote to be y = 350, which corresponds to the temperature of the oven. We can also arrive
at this conclusion by applying a bit of ‘number sense’. As t → ∞, −0.1602t ≈ very big (−) so
that e−0.1602t ≈ very small (+). The larger the value of t, the smaller e−0.1602t becomes so that
T (t) ≈ 350 − very small (+), which indicates the graph of y = T (t) is approaching its horizontal
asymptote y = 350 from below. Physically, this means the roast will eventually warm up to 350◦ F.12
The function T is sometimes called a limited growth model, since the function T remains bounded
as t → ∞. If we apply the principles behind Newton’s Law of Cooling to a biological example, it
says the growth rate of a population is directly proportional to how much room the population has
to grow. In other words, the more room for expansion, the faster the growth rate. The logistic
growth model combines The Law of Uninhibited Growth with limited growth and states that the
rate of growth of a population varies jointly with the population itself as well as the room the
population has to grow.
Equation 7.7. Logistic Growth: If a population behaves according to the assumptions of
logistic growth, the number of organisms N at time t is given by the equation
N (t) = L
,
1 + Ce−kLt where N (0) = N0 is the initial population, L is the limiting populationa , C is a measure of how
much room there is to grow given by
C= L
− 1.
N0 and k > 0 is the constant of proportionality which satisﬁes the equation
(instantaneous rate of change of N (t) at time t) = k N (t) (L − N (t))
a That is, as t → ∞, N (t) → L The logistic function is used not only to model the growth of organisms, but is also often used
to model the spread of disease and rumors.13
Example 7.5.5. The number of people N , in hundreds, at a local community college who have
heard the rumor ‘Carl is afraid of Virginia Woolf’ can be modeled using the logistic equation
84
,
1 + 2799e−t
where t ≥ 0 is the number of days after April 1, 2009.
N (t) = 12
13 at which point it would be more toast than roast.
Which can be just as damaging as diseases. 392 Exponential and Logarithmic Functions 1. Find and interpret N (0).
2. Find and interpret the end behavior of N (t).
3. How long until 4200 people have heard the rumor?
4. Check your answers to 2 and 3 using your calculator.
Solution.
84
84
3
1. We ﬁnd N (0) = 1+2799e0 = 2800 = 100 . Since N (t) measures the number of people who have
heard the rumor in hundreds, N (0) corresponds to 3 people. Since t = 0 corresponds to April
1, 2009, we may conclude that on that day, 3 people have heard the rumor.14 2. We could simply note that N (t) is written in the form of Equation 7.7, and identify L = 84.
However, to see why the answer is 84, we proceed analytically. Since the domain of N is
restricted to t ≥ 0, the only end behavior of signiﬁcance is t → ∞. As we’ve seen before,15
84
as t → ∞, have 1997e−t → 0+ and so N (t) ≈
≈ 84. Hence, as t → ∞,
1+very small (+)
N (t) → 84. This means that as time goes by, the number of people who will have heard the
rumor approaches 8400.
3. To ﬁnd how long it takes until 4200 people have heard the rumor, we set N (t) = 42. Solving
84
= 42 gives t = ln(2799) ≈ 7.937. It takes around 8 days until 4200 people have
1+2799e−t
heard the rumor.
4. We graph y = N (x) using the calculator and see that the line y = 84 is the horizontal
asymptote of the graph, conﬁrming our answer to part 2, and the graph intersects the line
y = 42 at x = ln(2799) ≈ 7.937, which conﬁrms our answer to part 3. y = f (x) = 84
1+2799e−x y = 84 and y = f (x) = 84
1+2799e−x and y = 42 14
Or, more likely, three people started the rumor. I’d wager Jeﬀ, Jamie, and Jason started it. So much for telling
your best friends something in conﬁdence!
15
See, for example, Example 7.1.2. 7.5 Applications of Exponential and Logarithmic Functions 393 If we take the time to analyze the graph of y = N (x) above, we can see graphically how logistic
growth combines features of uninhibited and limited growth. The curve seems to rise steeply, then
at some point, begins to level oﬀ. The point at which this happens is called an inﬂection point
or is sometimes called the ‘point of diminishing returns’. At this point, even though the function is
still increasing, the rate at which it does so begins to decline. It turns out the point of diminishing
returns always occurs at half the limiting population. (In our case, when y = 42.) While these
concepts are more precisely quantiﬁed using Calculus, below are two views of the graph of y = N (x),
one on the interval [0, 8], the other on [8, 15]. The former looks strikingly like uninhibited growth;
the latter like limited growth. y = f (x) = 84
1+2799e−x for y = f (x) = 0≤x≤8 7.5.2 84
1+2799e−x for 8 ≤ x ≤ 16 Applications of Logarithms Just as many physical phenomena can be modeled by exponential functions, the same is true of
logarithmic functions. In Exercises 6a, 6b and 6c of Section 7.1, we showed that logarithms are
useful in measuring the intensities of earthquakes (the Richter scale), sound (decibels) and acids and
bases (pH). We now present yet a diﬀerent use of the a basic logarithm function, password strength.
Example 7.5.6. The information entropy H , in bits, of a randomly generated password consisting
of L characters is given by H = L log2 (N ), where N is the number of possible symbols for each
character in the password. In general, the higher the entropy, the stronger the password.
1. If a 7 character casesensitive16 password is comprised of letters and numbers only, ﬁnd the
associated information entropy.
2. How many possible symbol options per character is required to produce a 7 character password
with an information entropy of 50 bits?
Solution.
1. There are 26 letters in the alphabet, 52 if upper and lower case letters are counted as diﬀerent.
There are 10 digits (0 through 9) for a total of N = 62 symbols. Since the password is to be
ln(62)
7 characters long, L = 7. Thus, H = 7 log2 (62) = 7 ln(2) ≈ 41.68.
16 That is, upper and lower case letters are treated as diﬀerent characters. 394 Exponential and Logarithmic Functions 2. We have L = 7 and H = 50 and we need to ﬁnd N . Solving the equation 50 = 7 log2 (N )
gives N = 250/7 ≈ 141.323, so we would need 142 diﬀerent symbols to choose from.17
Chemical systems known as buﬀer solutions have the ability to adjust to small changes in
acidity to maintain a range of pH values. Buﬀer solutions have a wide variety of applications from
maintaining a healthy ﬁsh tank to regulating the pH levels in blood. Our next example shows how
the pH in a buﬀer solution is a little more complicated than the pH we ﬁrst encountered in Exercise
6c in Section 7.1.
Example 7.5.7. Blood is a buﬀer solution. When carbon dioxide is absorbed into the bloodstream
it produces carbonic acid and lowers the pH. The body compensates by producing bicarbonate, a
weak base to partially neutralize the acid. The equation18 which models blood pH in this situation
00
is pH = 6.1 + log 8x , where x is the partial pressure of carbon dioxide in arterial blood, measured
in torr. Find the partial pressure of carbon dioxide in arterial blood if the pH is 7.4.
00
00
Solution. We set pH = 7.4 and get 7.4 = 6.1 + log 8x , or log 8x = 1.3. Solving, we ﬁnd
800
x = 101.3 ≈ 40.09. Hence, the partial pressure of carbon dioxide in the blood is about 40 torr. 17
Since there are only 94 distinct ASCII keyboard characters, to achieve this strength, the number of characters in
the password should be increased.
18
Derived from the HendersonHasselbalch Equation. See Exercise 8 in Section 7.2. Hasselbalch himself was
studying carbon dioxide dissolving in blood  a process called metabolic acidosis. 7.5 Applications of Exponential and Logarithmic Functions 7.5.3 395 Exercises 1. On May, 31, 2009, the Annual Percentage Rate listed at my bank for regular savings accounts
was 0.25% compounded monthly. Use Equation 7.2 to answer the following.
(a) If P = 2000 what is A(8)?
(b) Solve the equation A(t) = 4000 for t.
(c) What principal P should be invested so that the account balance is $2000 is three years?
2. My bank also oﬀers a 36month Certiﬁcate of Deposit (CD) with an APR of 2.25%.
(a) If P = 2000 what is A(8)?
(b) Solve the equation A(t) = 4000 for t.
(c) What principal P should be invested so that the account balance is $2000 is three years?
(d) The Annual Percentage Yield is the simple interest rate that returns the same amount of
interest after one year as the compound interest does. With the help of your classmates,
compute the APY for this investment.
3. Use Equation 7.2 to show that the time it takes for an investment to double in value does
not depend on the principal P , but rather, depends only on the APR and the number of
compoundings per year. Let n = 12 and with the help of your classmates compute the
doubling time for a variety of rates r. Then look up the Rule of 72 and compare your answers
to what that rule says. If you’re really interested19 in ﬁnancial mathematics, you could also
compare and contrast the Rule of 72 with the Rule of 70 and the Rule of 69.
4. Use Equation 7.5 to show that k = − ln(2)
where h is the halflife of the radioactive isotope.
h 5. The halflife of the radioactive isotope Carbon14 is about 5730 years.
(a) Use Equation 7.5 to express the amount of Carbon14 left from an initial N milligrams
as a function of time t in years.
(b) What percentage of the original amount of Carbon14 is left after 20,000 years?
(c) If an old wooden tool is found in a cave and the amount of Carbon14 present in it is
estimated to be only 42% of the original amount, approximately how old is the tool?
(d) Radiocarbon dating is not as easy as these exercises might lead you to believe. With
the help of your classmates, research radiocarbon dating and discuss why our model is
somewhat oversimpliﬁed.
19 Awesome pun! 396 Exponential and Logarithmic Functions 6. Carbon14 cannot be used to date inorganic material such as rocks, but there are many other
methods of radiometric dating which estimate the age of rocks. One of them, RubidiumStrontium dating, uses Rubidium87 which decays to Strontium87 with a halflife of 50
billion years. Use Equation 7.5 to express the amount of Rubidium87 left from an initial 2.3
micrograms as a function of time t in billions of years. Research this and other radiometric
techniques and discuss the margins of error for various methods with your classmates.
x 7. In Example 7.1.1 in Section 7.1, the exponential function V (x) = 25 4 was used to model
5
the value of a car over time. Use the properties of logs and/or exponents to rewrite the model
in the form V (t) = 25ekt .
8. A pork roast was taken out of a hardwood smoker when its internal temperature had reached
180◦ F and it was allowed to rest in a 75◦ F house for 20 minutes after which its internal
temperature had dropped to 170◦ F.20 Assuming that the temperature of the roast follows
Newton’s Law of Cooling (Equation 7.6),
(a) Express the temperature T as a function of time t.
(b) Find the time at which the roast would have dropped to 140◦ F had it not been carved
and eaten. 20 This roast was enjoyed by Jeﬀ and his family on June 10, 2009. This is real data, folks! 7.5 Applications of Exponential and Logarithmic Functions 7.5.4 Answers 1. (a) A(8) = 2000 1 + 0.0025 12·8
12 ≈ $2040.40 ln(2)
≈ 277.29 years
12 ln 1 + 0.0025
12
2000
(c) P =
36 ≈ $1985.06
1 + 0.0025
12 (b) t = 2. (a) A(8) = 2000 1 + 0.0225 12·8
12 ≈ $2394.03 ln(2)
≈ 30.83 years
12 ln 1 + 0.0225
12
2000
(c) P =
36 ≈ $1869.57
1 + 0.0225
12 (b) t = (d) 1 + 0.0225 12
≈ 1.0227
12
”
“
ln(2)
− 5730 t 5. (a) A(t) = N e so the APY is 2.27% ≈ N e−0.00012097t (b) A(20000) ≈ 0.088978 · N so about 8.9% remains
ln(.42)
(c) t ≈
≈ 7171 years old
−0.00012097
6. A(t) = 2.3e−0.0138629t
4 7. V (t) = 25eln( 5 )t ≈ 25e−0.22314355t
8. (a) T (t) = 75 + 105e−0.005005t
(b) The roast would have cooled to 140◦ F in about 95 minutes. 397 398 Exponential and Logarithmic Functions Chapter 8 Systems of Equations
8.1 Systems of Linear Equations: Gaussian Elimination Up until now, when we concerned ourselves with solving diﬀerent types of equations there was only
one equation to solve at a time. Given an equation f (x) = g (x), we could check our solutions
geometrically by ﬁnding where the graphs of y = f (x) and y = g (x) intersect. The xcoordinates
of these intersection points correspond to the solutions to the equation f (x) = g (x), and the y coordinates were largely ignored. If we modify the problem and ask for the intersection points of
the graphs of y = f (x) and y = g (x), where both the solution to x and y are of interest, we have
what is known as a system of equations, usually written as y = f (x) y = g (x)
The ‘curly bracket’ notation means we are to ﬁnd all pairs of points (x, y ) which satisfy both
equations. We begin our study of systems of equations by reviewing some basic notions from
Intermediate Algebra.
Definition 8.1. A linear equation in two variables is an equation of the form a1 x + a2 y = c
where a1 , a2 and c are real numbers and at least one of a1 and a2 is nonzero. For reasons which will become clear later in the section, we are using subscripts in Deﬁnition 8.1
to indicate diﬀerent, but ﬁxed, real numbers and those subscripts have no mathematical meaning
beyond that. For example, 3x − y = 0.1 is a linear equation in two variables with a1 = 3, a2 = − 1
2
2
and c = 0.1. We can also consider x = 5 to be a linear equation in two variables by identifying
a1 = 1, a2 = 0, and c = 5.1 If a1 and a2 are both 0, then depending on c, we get either an
1
Critics may argue that x = 5 is clearly an equation in one variable. It can also be considered an equation in 117
variables with the coeﬃcients of 116 variables set to 0. As with many conventions in Mathematics, the context will
clarify the situation. 400 Systems of Equations equation which is always true, called an identity, or an equation which is never true, called a
contradiction. (If c = 0, then we get 0 = 0, which is always true. If c = 0, then we’d have
0 = 0, which is never true.) Even though identities and contradictions have a large role to play
in the upcoming sections, we do not consider them linear equations. The key to identifying linear
equations is to note that the variables involved are to the ﬁrst power and that the coeﬃcients of the
variables are numbers. Some examples of equations which are nonlinear are x2 + y = 1, xy = 5 and
e2x + ln(y ) = 1. We leave it to the reader to explain why these do not satisfy Deﬁnition 8.1. From
what we know from Sections 1.2 and 3.1, the graphs of linear equations are lines. If we couple two
or more linear equations together, in eﬀect to ﬁnd the points of intersection of two or more lines,
we obtain a system of linear equations in two variables. Our ﬁrst example reviews some of
the basic techniques ﬁrst learned in Intermediate Algebra.
Example 8.1.1. Solve the following systems of equations. Check your answer algebraically and
graphically. x 4y 2x − y = 1 6x + 3y = 9
7
−
=5
3
5
1.
5.
3. 4x + 2y = 12 2x y
y=3 +
=1
9 3x + 4y = −2
2. −3x − y =
5 3 2 2x − 4y = 6
4. 3x − 6y = 9 6. x−y = 0 x+y =
2 −2x + y = −2 Solution.
1. Our ﬁrst system is nearly solved for us. The second equation tells us that y = 3. To ﬁnd the
corresponding value of x, we substitute this value for y into the the ﬁrst equation to obtain
2x − 3 = 1, so that x = 2. Our solution to the system is (2, 3). To check this algebraically,
we substitute x = 2 and y = 3 into each equation and see that they are satisﬁed. We see
2(2) − 3 = 1, and 3 = 3, as required. To check our answer graphically, we graph the lines
2x − y = 1 and y = 3 and verify that they intersect at (2, 3).
2. To solve the second system, we use the addition method to eliminate the variable x. We
take the two equations as given and ‘add equals to equals’ to obtain
3x + 4y = −2
+ (−3x − y = 5) 3y = 3 This gives us y = 1. We now substitute y = 1 into either of the two equations, say −3x − y = 5,
to get −3x − 1 = 5 so that x = −2. Our solution is (−2, 1). Substituting x = −2 and y = 1 8.1 Systems of Linear Equations: Gaussian Elimination 401 into the ﬁrst equation gives 3(−2) + 4(1) = −2, which is true, and, likewise, when we check
(−2, 1) in the second equation, we get −3(−2) − 1 = 5, which is also true. Geometrically, the
lines 3x + 4y = −2 and −3x − y = 5 intersect at (−2, 1).
y
y
4
2
(2, 3) 1
(−2, 1) 2 x −4 −3 −2 −1 1 −1
−1 1 2 3 4 x −2 3x + 4y = −2
−3 x − y = 5 2x − y = 1
y=3 3. The equations in the third system are more approachable if we clear denominators. We
multiply both sides of the ﬁrst equation by 15 and both sides of the second equation by 18
to obtain the kinder, gentler system 5x − 12y = 21 4x + 6 y = 9
Adding these two equations directly fails to eliminate either of the variables, but we note
that if we multiply the ﬁrst equation by 4 and the second by −5, we will be in a position to
eliminate the x term
20x − 48y = 84 + (−20x − 30y = −45)
−78y = 39 1
From this we get y = − 2 . We can temporarily avoid too much unpleasantness by choosing to
1
substitute y = − 2 into one of the equivalent equations we found by clearing denominators,
say into 5x − 12y = 21. We get 5x + 6 = 21 which gives x = 3. Our answer is 3, − 1 .
2
At this point, we have no choice − in order to check an answer algebraically, we must see
if the answer satisﬁes both of the original equations, so we substitute x = 3 and y = − 1
2
y
x
1
into both x − 45 = 7 and 29 + y = 2 . We leave it to the reader to verify that the solution
3
5
3
is correct. Graphing both of the lines involved with considerable care yields an intersection
point of 3, − 1 .
2 402 Systems of Equations 4. An eerie calm settles over us as we cautiously approach our fourth system. Do its friendly
integer coeﬃcients belie something more sinister? We note that if we multiply both sides of
the ﬁrst equation by 3 and the both sides of the second equation by −2, we are ready to
eliminate the x 6x − 12y = 18 + (−6x + 12y = −18)
0= 0 We eliminated not only the x, but the y as well and we are left with the identity 0 = 0. This
means that these two diﬀerent linear equations are, in fact, equivalent. In other words, if an
ordered pair (x, y ) satisﬁes the equation 2x − 4y = 6, it automatically satisﬁes the equation
3x − 6y = 9. One way to describe the solution set to this system is to use the roster method2
and write {(x, y ) : 2x − 4y = 6}. While this is correct (and corresponds exactly to what’s
happening graphically, as we shall see shortly), we take this opportunity to introduce the
notion of a parametric solution. Our ﬁrst step is to solve 2x − 4y = 6 for one of the
1
1
variables, say y = 2 x − 3 . For each value of x, the formula y = 2 x − 3 determines the
2
2
corresponding y value of a solution. Since we have no restriction on x, it is called a free
1
variable. We let x = t, a socalled ‘parameter’, and get y = 2 t − 3 . Our set of solutions can
2
3
1
3 For speciﬁc values of t, we can generate
then be described as t, 2 t − 2 : −∞ < t < ∞ .
solutions. For example, t = 0 gives us the solution 0, − 3 ; t = 117 gives us (117, 57),
2
and while we can readily check each of these particular solutions satisfy both equations, the
question is how do we check our general answer algebraically? Same as always. We claim that
for any real number t, the pair t, 1 t − 3 satisﬁes both equations. Substituting x = t and
2
2
3
1
y = 1 t − 2 into 2x − 4y = 6 gives 2t − 4 2 t − 3 = 6. Simplifying, we get 2t − 2t +6 = 6, which
2
2
is always true. Similarly, when we make these substitutions in the equation 3x − 6y = 9, we
get 3t − 6 1 t − 3 = 9 which reduces to 3t − 3t + 9 = 9, so it checks out, too. Geometrically,
2
2
2x − 4y = 6 and 3x − 6y = 9 are the same line, which means that they intersect at every
point on their graphs. The reader is encouraged to think about how our parametric solution
says exactly that. 2 See Section 1.2 for a review of this.
Note that we could have just as easily chosen to solve 2x − 4y = 6 for x to obtain x = 2y + 3. Letting y be the
parameter t, we have that for any value of t, x = 2t + 3, which gives {(2t + 3, t) : −∞ < t < ∞}. There is no one
correct way to parameterize the solution set, which is why it is always best to check your answer.
3 8.1 Systems of Linear Equations: Gaussian Elimination 403
y y
2
1
−1
−1
−2 1
1 2
45
”
“
1
3, − 2 6 7 x
1 2 3 4 x −1 −3
−4 x
− 4y
3
5
2x
+y
9
3 =7
5
=1
2 2x − 4y = 6
3x − 6y = 9
(Same line.) 5. Multiplying both sides of the ﬁrst equation by 2 and the both sides of the second equation
by −3, we set the stage to eliminate x
12x + 6y = 18 + (−12x − 6y = −36)
−18 0= As in the previous example, both x and y dropped out of the equation, but we are left with
an irrevocable contradiction, 0 = −18. This tells us that it is impossible to ﬁnd a pair (x, y )
which satisﬁes both equations; in other words, the system has no solution. Graphically, we
see that the lines 6x + 3y = 9 and 4x + 2y = 12 are distinct and parallel, and as such do not
intersect.
6. We can begin to solve our last system by adding the ﬁrst two equations
x−y = 0 + (x + y = 2)
2x = 2 which gives x = 1. Substituting this into the ﬁrst equation gives 1 − y = 0 so that y = 1.
We seem to have determined a solution to our system, (1, 1). While this checks in the
ﬁrst two equations, when we substitute x = 1 and y = 1 into the third equation, we get
−2(1)+(1) = −2 which simpliﬁes to the contradiction −1 = −2. Graphing the lines x − y = 0,
x + y = 2, and −2x + y = −2, we see that the ﬁrst two lines do, in fact, intersect at (1, 1),
however, all three lines never intersect at the same point simultaneously, which is what is
required if a solution to the system is to be found. 404 Systems of Equations
y y
6
5
4
3
2
1
−1
−2
−3 1 1 2 6x + 3y = 9
4x + 2y = 12 x x
−1 y−x=0
y+x=2
−2x + y = −2 A few remarks about Example 8.1.1 are in order. It is clear that some systems of equations
have solutions, and some do not. Those which have solutions are called consistent, those with no
solution are called inconsistent. We also distinguish the two diﬀerent types of behavior among
consistent systems. Those which admit free variables are called dependent; those with no free
variables are called independent.4 Using this new vocabulary, we classify numbers 1, 2 and 3 in
Example 8.1.1 as consistent independent systems, number 4 is consistent dependent, and numbers
5 and 6 are inconsistent.5 The system in 6 above is called overdetermined, since we have more
equations than variables.6 Not surprisingly, a system with more variables than equations is called
underdetermined. While the system in number 6 above is overdetermined and inconsistent,
there exist overdetermined consistent systems (both dependent and independent) and we leave it
to the reader to think about what is happening algebraically and geometrically in these cases.
Likewise, there are both consistent and inconsistent underdetermined systems,7 but a consistent
underdetermined system of linear equations is necessarily dependent.8
In order to move this section beyond a review of Intermediate Algebra, we now deﬁne what is
meant by a linear equation in n variables.
4
In the case of systems of linear equations, regardless of the number of equations or variables, consistent independent systems have exactly one solution. The reader is encouraged to think about why this is the case for linear
equations in two variables. Hint: think geometrically.
5
The adjectives ‘dependent’ and ‘independent’ apply only to consistent systems  they describe the type of solutions.
6
If we think if each variable being an unknown quantity, then ostensibly, to recover two unknown quantities,
we need two pieces of information  i.e., two equations. Having more than two equations suggests we have more
information than necessary to determine the values of the unknowns. While this is not necessarily the case, it does
explain the choice of terminology ‘overdetermined’.
7
We need more than two variables to give an example of the latter.
8
Again, experience with systems with more variables helps to see this here, as does a solid course in Linear Algebra. 8.1 Systems of Linear Equations: Gaussian Elimination 405 Definition 8.2. A linear equation in n variables, x1 , x2 , . . . , xn is an equation of the form
a1 x1 + a2 x2 + . . . + an xn = c where a1 , a2 , . . . an and c are real numbers and at least one of
a1 , a2 , . . . , an is nonzero. Coupling more than one linear equation in n variables results in a
system of linear equations in n variables. Instead of using more familiar variables like x, y , and even z and/or w in Deﬁnition 8.2, we
use subscripts to distinguish the diﬀerent variables. We have no idea how many variables may be
involved, so we use numbers to distinguish them instead of letters. (There is an endless supply of
distinct numbers.) As an example, the linear equation 3x1 − x2 = 4 represents the same relationship
between the variables x1 and x2 as the equation 3x − y = 4 does between the variables x and y .
In addition, just as we cannot combine the terms in the expression 3x − y , we cannot combine the
terms in the expression 3x1 − x2 .
When solving a system, it becomes increasingly important to keep track of what operations are
performed to which equations and to develop a strategy based on the kind of manipulations we’ve
already employed. To this end, we ﬁrst remind ourselves of the maneuvers which can be applied to
a system of linear equations that result in an equivalent system.9
Theorem 8.1. Given a system of equations, the following moves will result in an equivalent
system of equations.
Interchange the position of any two equations.
Replace an equation with a nonzero multiple of itself.a
Replace an equation with itself plus a nonzero multiple of another equation.
a That is, an equation which results from multiplying both sides of the equation by the same nonzero number. We have seen plenty of instances of the second and third moves in Theorem 8.1 when we solved
the systems Example 8.1.1. The ﬁrst move, while it obviously admits an equivalent system, seems
silly. Our perception will change as we consider more equations and more variables in this, and
later sections.
Consider the system of equations x − 1y + 1z = 1 3
2 y − 1z =
4
2 z = −1
9 That is, a system with the same solution set. 406 Systems of Equations 1
Clearly z = −1, and we substitute this into the second equation y − 2 (−1) = 4 to obtain y = 7 .
2
1
7
Finally, we substitute y = 2 and z = −1 into the ﬁrst equation to get x − 3 7 + 1 (−1) = 1,
2
2
8
so that x = 3 . The reader can verify that these values of x, y and z satisfy all three original
equations. It is tempting for us to write the solution to this system by extending the usual (x, y )
notation to (x, y, z ) and list our solution as 8 , 7 , −1 . The question quickly becomes what does
32
7
an ‘ordered triple’ like 8 , 2 , −1 represent? Just as ordered pairs are used to locate points on the
3
twodimensional plane, ordered triples can be used to locate points in space.10 Moreover, just as
equations involving the variables x and y describe graphs of onedimensional lines and curves in the
twodimensional plane, equations involving variables x, y , and z describe objects called surfaces
in threedimensional space. Each of the equations in the above system can be visualized as a plane
situated in threespace. Geometrically, the system is trying to ﬁnd the intersection, or common
point, of all three planes. If you imagine three sheets of notebook paper each representing a portion
of these planes, you will start to see the complexities involved in how three such planes can intersect.
Below is a sketch of the three planes. It turns out that any two of these planes intersect in a line,11
so our intersection point is where all three of these lines meet. Since the geometry for equations involving more than two variables is complicated, we will focus
our eﬀorts on the algebra. Returning to the system x − 1y + 1z = 1 3
2 4
y − 1z =
2 z = −1
10 You were asked to think about this in Exercise 13 in Section 1.1.
In fact, these lines are described by the parametric solutions to the systems formed by taking any two of these
equations by themselves.
11 8.1 Systems of Linear Equations: Gaussian Elimination 407 we note the reason it was so easy to solve is that the third equation is solved for z , the second
equation involves only y and z , and since the coeﬃcient of y is 1, it makes it easy to solve for y
using our known value for z . Lastly, the coeﬃcient of x in the ﬁrst equation is 1 making it easy to
substitute the known values of y and z and then solve for x. We formalize this pattern below for
the most general systems of linear equations. Again, we use subscripted variables to describe the
general case. The variable with the smallest subscript in a given equation is typically called the
leading variable of that equation.
Definition 8.3. A system of linear equations with variables x1 , x2 , . . . xn is said to be in
triangular form provided all of the following conditions hold:
1. The subscripts of the variables in each equation are always increasing from left to right.
2. The leading variable in each equation has coeﬃcient 1.
3. The subscript on the leading variable in a given equation is greater than the subscript on
the leading variable in the equation above it.
4. Any equation without variablesa cannot be placed above an equation with variables. a necessarily an identity or contradiction In our previous system, if make the obvious choices x = x1 , y = x2 , and z = x3 , we see that the
system is in triangular form.12 An example of a more complicated system in triangular form is x1 − 4x3 + x4 − x6 = 6 x2 + 2 x3 = 1 x4 + 3 x5 − x6 = 8 x5 + 9x6 = 10
Our goal henceforth will be to transform a given system of linear equations into triangular form
using the moves in Theorem 8.1.
Example 8.1.2. Use Theorem 8.1 to put the following systems into triangular form and then solve
the system if possible. Classify each system as consistent independent, consistent dependent, or
inconsistent. 12 If letters are used instead of subscripted variables, Deﬁnition 8.3 can be suitably modiﬁed using alphabetical
order of the variables instead of numerical order on the subscripts of the variables. 408 1. Systems of Equations 3x − y + z = 3 2x − 4y + 3z = 16 x−y+z = 5 2x + 3y − z = 1 2.
10x − z = 2 4x − 9y + 2z = 5 3x1 + x2 + x4 = 6 3.
2x1 + x2 − x3 = 4 x − 3x − 2x = 0
2
3
4 Solution.
1. For deﬁnitiveness, we label the topmost equation in the system E 1, the equation beneath that
E 2, and so forth. We now attempt to put the system in triangular form using an algorithm
known as Gaussian Elimination. What this means is that, starting with x, we transform
the system so that conditions 2 and 3 in Deﬁnition 8.3 are satisﬁed. Then we move on to
the next variable, in this case y , and repeat. Since the variables in all of the equations have
a consistent ordering from left to right, our ﬁrst move is to get an x in E 1’s spot with a
coeﬃcient of 1. While there are many ways to do this, the easiest is to apply the ﬁrst move
listed in Theorem 8.1 and interchange E 1 and E 3. (E 1) 3x − y + z = 3 (E 2) 2x − 4y + 3z = 16 (E 3)
x−y+z = 5 (E 1) x−y+z = 5 Switch E 1 and E 3
−− − − − −
− − − − −→
(E 2) 2x − 4y + 3z = 16 (E 3)
3x − y + z = 3 To satisfy Deﬁnition 8.3, we need to eliminate the x’s from E 2 and E 3. We accomplish this
by replacing each of them with a sum of themselves and a multiple of E 1. To eliminate the
x from E 2, we need to multiply E 1 by −2 then add; to eliminate the x from E 3, we need to
multiply E 1 by −3 then add. Applying the third move listed in Theorem 8.1 twice, we get (E 1) x−y+z = 5 (E 2) 2x − 4y + 3z = 16 (E 3)
3x − y + z = 3 (E 1) x − y + z = 5 Replace E 2 with −2E 1 + E 2
−− − − − − − − −→
−−−−−−−−−
(E 2) −2y + z =
6 Replace E 3 with −3E 1 + E 3 (E 3)
2y − 2z = −12 Now we enforce the conditions stated in Deﬁnition 8.3 for the variable y . To that end we
need to get the coeﬃcient of y in E 2 equal to 1. We apply the second move listed in Theorem
8.1 and replace E 2 with itself times − 1 .
2 (E 1) x − y + z = 5 (E 2) −2y + z =
6 (E 3)
2y − 2z = −12 (E 1) x − y + z = 5 Replace E 2 with − 1 E 2
2
−−−−−−−→
−−−−−−−
(E 2)
y − 1 z = −3
2 (E 3)
2y − 2z = −12 8.1 Systems of Linear Equations: Gaussian Elimination 409 To eliminate the y in E 3, we add −2E 2 to it. (E 1) x − y + z = (E 2) (E 3) y − 1z =
2 5
−3 Replace E 3 with −2E 2 + E 3 −− − − − − − − −→
−−−−−−−−− 2y − 2z = −12 (E 1) x − y + z = (E 2) (E 3) 5 1
y − 2 z = −3 −z = −6 Finally, we apply the second move from Theorem 8.1 one last time and multiply E 3 by −1
to satisfy the conditions of Deﬁnition 8.3 for the variable z . (E 1) x − y + z = 5 (E 2)
y − 1 z = −3
2 (E 3)
−z = −6 (E 1) x − y + z = 5 Replace E 3 with −1E 3
1
−− − − − − −→
−−−−−−−
(E 2)
y − 2 z = −3 (E 3)
z=
6 Now we proceed to substitute. Plugging in z = 6 into E 2 gives y − 3 = −3 so that y = 0.
With y = 0 and z = 6, E 1 becomes x − 0 + 6 = 5, or x = −1. Our solution is (−1, 0, 6).
We leave it to the reader to check that substituting the respective values for x, y , and z into
the original system results in three identities. Since we have found a solution, the system is
consistent; since there are no free variables, it is independent.
2. Proceeding as we did in 1, our ﬁrst step is to get an equation with x in the E 1 position with
1
1 as its coeﬃcient. Since there is no easy ﬁx, we multiply E 1 by 2 . (E 1) 2x + 3y − z = 1 (E 2)
10x − z = 2 (E 3) 4x − 9y + 2z = 5 (E 1) x + 3 y − 1 z = 2
2 1
Replace E 1 with 2 E 1
−−−−−−−
−−−−−−→
(E 2)
10x − z = (E 3) 4x − 9y + 2z = 1
2 2
5 Now it’s time to take care of the x’s in E 2 and E 3. (E 1) 3
x + 2y − 1z =
2 1
2 10x − z = 2 (E 3) 4x − 9y + 2z = 5 (E 2) Replace E 2 with −10E 1 + E 2 −−−−−−−−−→
−−−−−−−−−
Replace E 3 with −4E 1 + E 3 (E 1) x + 3 y − 1 z = 2
2 1
2 −15y + 4z = 3 (E 2) (E 3) −15y + 4z = −3 Our next step is to get the coeﬃcient of y in E 2 equal to 1. To that end, we have 410 Systems of Equations (E 1) x + 3 y − 1 z = 2
2 1
Replace E 2 with − 15 E 2
4
−− − − − − − −
− − − − − − −→
(E 2)
y − 15 z = (E 3)
−15y + 4z = 1 (E 1) x + 3 y − 1 z = 2
2
2 (E 2)
−15y + 4z = −3 (E 3)
−15y + 4z =
3 1
2
1
5 3 Finally, we rid E 3 of y . (E 1) x + 3 y − 1 z = 2
2 4
(E 2)
y − 15 z = (E 3)
−15y + 4z = 1
2
1
5 3 (E 1) x − y + z = 5 Replace E 3 with 15E 2 + E 3
−− − − − − − − −
− − − − − − − −→
(E 2)
y − 1 z = −3
2 (E 3)
0=
6 The last equation, 0 = 6, is a contradiction so the system has no solution. According to
Theorem 8.1, since this system has no solutions, neither does the original, thus we have an
inconsistent system.
3. For our last system, we begin by multiplying E 1 by (E 1) 3x1 + x2 + x4 = 6 (E 2) 2x1 + x2 − x3 = 4 (E 3) x − 3x − 2x = 0
2
3
4 1
3 to get a coeﬃcient of 1 on x1 . (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 1
Replace E 1 with 3 E 1
−−−−−−−
−−−−−−→
(E 2)
2x1 + x2 − x3 = 4 (E 3) x − 3x − 2x = 0
2
3
4 Next we eliminate x1 from E 2 (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 (E 2)
2x1 + x2 − x3 = 4 (E 3) x − 3x − 2x = 0
2
3
4 (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 Replace E 2
2
−− − − −→
−−−−−
(E 2) 1 x2 − x3 − 3 x4 = 0
3 with −2E 1 + E 2 (E 3) x − 3x − 2x = 0
2
3
4 We switch E 2 and E 3 to get a coeﬃcient of 1 for x2 . (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 (E 2) 1 x2 − x3 − 2 x4 = 0
3
3 (E 3) x − 3x − 2x = 0
2
3
4 Switch E 2 and E 3 −− − − − −
− − − − −→ (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 (E 2) (E 3) x2 − 3x3 − 2x4 = 0 1
3 x2 2
− x3 − 3 x4 = 0 8.1 Systems of Linear Equations: Gaussian Elimination 411 Finally, we eliminate x2 in E 3. (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 (E 2) x2 − 3x3 − 2x4 = 0 (E 3) 1 x − x − 2 x = 0
3
32
34 (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 Replace E 3
−−−−−→
−−−−−
(E 2) x2 − 3x3 − 2x4 = 0
1 with − 3 E 2 + E 3 (E 3)
0=0 Equation E 3 reduces to 0 = 0,which is always true. Since we have no equations with x3
or x4 as leading variables, they are both free, which means we have a consistent dependent
system. We parametrize the solution set by letting x3 = s and x4 = t and obtain from E 2
1
that x2 = 3s + 2t. Substituting this and x4 = t into E 1, we have x1 + 3 (3s + 2t) + 1 t = 2
3
which gives x1 = 2 − s − t. Our solution is the set {(2 − s − t, 2s + 3t, s, t) : −∞ < s, t < ∞}.13
We leave it to the reader to verify that the substitutions x1 = 2 − s − t, x2 = 3s + 2t, x3 = s
and x4 = t satisfy the equations in the original system.
Like all algorithms, Gaussian Elimination has the advantage of always producing what we
need, but it can also be ineﬃcient at times. For example, when solving 2 above, it is clear after we
eliminated the x’s in the second step to get the system 1 (E 1) x + 3 y − 1 z = 2
2
2 (E 2)
−15y + 4z = −3 (E 3)
−15y + 4z =
3
that equations E 2 and E 3 when taken together form a contradiction since we have identical left hand
sides and diﬀerent right hand sides. The algorithm takes two more steps to reach this contradiction.
We also note that substitution in Gaussian Elimination is delayed until all the elimination is done,
thus it gets called backsubstitution. This may also be ineﬃcient in many cases. Rest assured,
the technique of substitution as you may have learned it in Intermediate Algebra will once again
take center stage in Section 8.4. Lastly, we note that the system in 3 above is underdetermined,
and as it is consistent, we have free variables in our answer. We close this section with a standard
‘mixture’ type application of systems of linear equations.
Example 8.1.3. Lucas needs to create a 500 milliliters (mL) of a 40% acid solution. He has stock
solutions of 30% and 90% acid as well as all of the distilled water he wants. Setup and solve a
system of linear equations which determines all of the possible combinations of the stock solutions
and water which would produce the required solution.
Solution. We are after three unknowns, the amount (in mL) of the 30% stock solution (which
we’ll call x), the amount (in mL) of the 90% stock solution (which we’ll call y ) and the amount
13 Here, any choice of s and t will determine a solution which is a point in 4dimensional space. Yeah, we have
trouble visualizing that, too. 412 Systems of Equations (in mL) of water (which we’ll call w). We now need to determine some relationships between these
variables. Our goal is to produce 500 milliliters of a 40% acid solution. This product has two
deﬁning characteristics. First, it must be 500 mL; second, it must be 40% acid. We take each
of these qualities in turn. First, the total volume of 500 mL must be the sum of the contributed
volumes of the two stock solutions and the water. That is
amount of 30% stock solution + amount of 90% stock solution + amount of water = 500 mL
Using our deﬁned variables, this reduces to x + y + w = 500. Next, we need to make sure the ﬁnal
solution is 40% acid. Since water contains no acid, the acid will come from the stock solutions only.
We ﬁnd 40% of 500 mL to be 200 mL which means the ﬁnal solution must contain 200 mL of acid.
We have
amount of acid in 30% stock solution + amount of acid 90% stock solution = 200 mL
The amount of acid in x mL of 30% stock is 0.30x and the amount of acid in y mL of 90% solution
is 0.90y . We have 0.30x + 0.90y = 200. Converting to fractions,14 our system of equations becomes x + y + w = 500 3 x + 9 y = 200
10 10 We ﬁrst eliminate the x from the second equation (E 1) x + y + w = 500 (E 1) x + y + w = 500
3
Replace E 2 with − 10 E 1 + E 2
−−−−−−−−−→
−−−−−−−−− (E 2) 3 x + 9 y = 200 (E 2) 3 y − 3 w = 50
10
10
5
10
Next, we get a coeﬃcient of 1 on the leading variable in E 2 (E 1) x + y + w = 500 (E 1) x + y + w = 500
Replace E 2 with 5 E 2
− − − − − −3−
−−−−−−→ (E 2) 3 y − 3 w = 50 (E 2)
1
y − 2 w = 250
5
10
3
Notice that we have no equation to determine w, and as such, w is free. We set w = t and from E 2
1
1
3
get y = 2 t + 250 . Substituting into E 1 gives x + 2 t + 250 + t = 500 so that x = − 2 t + 1250 . This
3
3
3
3
1250 1
250
system is consistent, dependent and its solution set is { − 2 t + 3 , 2 t + 3 , t : −∞ < t < ∞}.
While this answer checks algebraically, we have neglected to take into account that x, y and w,
being amounts of acid and water, need to be nonnegative. That is, x ≥ 0, y ≥ 0 and w ≥ 0. The
3
1
constraint x ≥ 0 gives us − 2 t + 1250 ≥ 0, or t ≤ 2500 . From y ≥ 0, we get 2 t + 250 ≥ 0 or t ≥ − 500 .
3
9
3
3
The condition z ≥ 0 yields t ≥ 0, and we see that when we take the set theoretic intersection of
these intervals, we get 0 ≤ t ≤ 2500 . Our ﬁnal answer is { − 3 t + 1250 , 1 t + 250 , t : 0 ≤ t ≤ 2500 }.
9
2
3
2
3
9
Of what practical use is our answer? Suppose there is only 100 mL of the 90% solution remaining
and it is due to expire. Can we use all of it to make our required solution? We would have y = 100
14 We do this only because we believe students can use all of the practice with fractions they can get! 8.1 Systems of Linear Equations: Gaussian Elimination 413 1
so that 2 t + 250 = 100, and we get t = 100 . This means the amount of 30% solution required is
3
3
3
00
3
x = − 2 t + 1250 = − 2 13 + 1250 = 1100 mL, and for the water, w = t = 100 mL. The reader is
3
3
3
3
invited to check that mixing these three amounts of our constituent solutions produces the required
40% acid mix. 414 8.1.1 Systems of Equations Exercises 1. Put the following systems of linear equations into triangular form and then solve the system if possible. Classify each system as consistent independent, consistent dependent, or
inconsistent. −5x + y = 17 x − y + z = −4 (a) (i)
x+y = 5
−3x + 2y + 4z = −5 x − 5y + 2z = −18 x+y+z = 3 2x − 4y + z = −7
(b) 2x − y + z = 0 −3x + 5y + 7z = 7
(j)
x − 2y + 2z = −2 −x + 4y − 2z =
3 4x − y + z = 5 2x − y + z = 1
(c) 2y + 6z = 30 (k)
x+z = 5
2x + 2y − z = 1 3x + 6 y + 4 z = 9 4x − y + z = 5 (d) 2y + 6z = 30
x − 3y − 4z = 3 (l)
x+z = 6
3x + 4y − z = 13 2x − 19y − 19z = 2 x + y + z = −17 (e) x+y+z = y − 3z =
0
4 (m)
2x − 4y − z = −1 x − 2y + 3 z = 7 x−y =
2
(f)
−3x + y + 2z = −5 2x + 2 y + z = 3
x−y+z =
8 (n)
3x + 3y − 9z = −6 3x − 2y + z = −5 7x − 2y + 5z = 39
(g)
x + 3y − z = 12 x + y + 2z = 2x − 3y + z = −1 0 (o)
4x − 4y + 4z = −13 2x − y + z = −1 6x − 5y + 7z = −25
(h)
4x + 3 y + 5 z =
1 5y + 3z =
4 8.1 Systems of Linear Equations: Gaussian Elimination (p) 2x1 + x2 − 12x3 − x4 = 16 −x1 + x2 + 12x3 − 4x4 = −5 3x + 2x − 16x − 3x = 1
2
3
4 x1 + 2x2 − 5x4 = x1 − x3 = −2 2x2 − x4 =
0
(q) x1 − 2x2 + x3 =
0 −x3 + x4 =
1 25
11 (r) 415 x1 − x2 − 5x3 + 3x4 = −1
x1 + x2 + 5x3 − 3x4 = 0 x2 + 5x3 − 3x4 = 1 x1 − 2x2 − 10x3 + 6x4 = −1 2. Find two other forms of the parametric solution to Exercise 1c above by reorganizing the
equations so that x or y can be the free variable.
3. At The Old Home Fill’er Up and Keep on aTruckin’ Cafe, Mavis mixes two diﬀerent types
of coﬀee beans to produce a house blend. The ﬁrst type costs $3 per pound and the second
costs $8 per pound. How much of each type does Mavis use to make 50 pounds of a blend
which costs $6 per pound?
4. At The Crispy Critter’s Head Shop and Patchouli Emporium along with their dried up weeds,
sunﬂower seeds and astrological postcards they sell an herbal tea blend. By weight, Type I
herbal tea is 30% peppermint, 40% rose hips and 30% chamomile, Type II has percents 40%,
20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How
much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal
parts peppermint, rose hips and chamomile?
5. Discuss with your classmates how you would approach Exercise 4 above if they needed to use
up a pound of Type I tea to make room on the shelf for a new canister.
6. Discuss with your classmates why it is impossible to mix a 20% acid solution with a 40% acid
solution to produce a 60% acid solution. If you were to try to make 100 mL of a 60% acid
solution using stock solutions at 20% and 40%, respectively, what would the triangular form
of the resulting system look like? 416 8.1.2 Systems of Equations Answers 1. Because triangular form is not unique, we give only one possible answer to that part of the
question. Yours may be diﬀerent and still be correct. x+y = 5
Consistent independent
(a)
Solution (−2, 7) y=7 x − 5y − 7z = −7 3
3
3 5
(b)
y + 4z =
2 z=
0 x − 1y + 1z = 5 4
4
4 (c)
y + 3z = 15 0= 0 x − 1y + 1z = 5 4
4
4 (d)
y + 3z = 15 0= 1 Consistent independent
Solution (1, 2, 0) x + y + z = −17
(e) y − 3z =
0 Consistent dependent
Solution (−4t − 17, 3t, t)
for all real numbers t x − 2y + 3 z = 7 (f)
y − 11 z = − 16
5
5 z=
1 x + y + 2z = 0 (g)
y − 3z =
6
2 z = −2 x − 1y + 1z = −1 2
2
2 3
3
(h)
y + 5z =
5 0=
1 Consistent independent
Solution (2, −1, 1) Consistent dependent
Solution (−t + 5, −3t + 15, t)
for all real numbers t Inconsistent
No solution Consistent independent
Solution (1, 3, −2) Inconsistent
no solution 8.1 Systems of Linear Equations: Gaussian Elimination x − y + z = −4 (i)
y − 7z = 17 z = −2 x − 2y + 2z = −2 1
(j)
y=
2 z=
1 Consistent independent
Solution (1, 3, −2) x − 1y + 1z = 2
2 2
(k)
y − 3z = z= 417 Consistent independent
Solution 1 , 2 , 1
33 Consistent independent
1
Solution −3, 2 , 1 1
2 0
1 x − 3y − 4z = 3 4
(l)
y + 11 z = 13
13 0= 0 x+y+z = 4 1
(m)
y + 2z = 3
2 0=1 Consistent dependent
51
Solution 19 t + 13 , − 11 t +
13
13
for all real numbers t x−y+z = 8 (n)
y − 2z = −5 z=
1 x − 3y + 1z = −1 2
2
2 (o)
y + z = − 11
2 0=
0 x + 2 x − 16 x − x 1 32
4 33 x2 + 4x3 − 3x4
(p) 0 0 Consistent independent
Solution (4, −3, 1) 4
13 , t Inconsistent
no solution Consistent dependent
Solution −2t − 35 , −t −
4
for all real numbers t = 25
3 = 2 = 0 = 0 11
2 ,t Consistent dependent
Solution (8s − t + 7, −4s + 3t + 2, s, t)
for all real numbers s and t 418 Systems of Equations (q) x − x = −2 1
3 x2 − 1 x4 =
0
2 x3 − 1 x4 = 2 x4 = Consistent independent
Solution (1, 2, 3, 4) 1
4 x1 − x2 − 5x3 + 3 x4 x2 + 5x3 − 3x4
(r) 0 0 = −1
= 1
2 = 1 = Inconsistent
No solution 0 2. If x is the free variable then the solution is (t, 3t, −t + 5) and if y is the free variable then the
1
solution is 3 t, t, − 1 t + 5 .
3
3. Mavis needs 20 pounds of $3 per pound coﬀee and 30 pounds of $8 per pound coﬀee.
4. 4
3 − 1 t pounds of Type I,
2 2
3 − 1 t pounds of Type II and t pounds of Type III where 0 ≤ t ≤ 4 .
2
3 8.2 Systems of Linear Equations: Augmented Matrices* 8.2 419 Systems of Linear Equations: Augmented Matrices* In Section 8.1 we introduced Gaussian Elimination as a means of transforming a system of linear
equations into triangular form with the ultimate goal of producing an equivalent system of linear
equations which is easier to solve. If take a step back and study the process, we see that all of our
moves are determined entirely by the coeﬃcients of the variables involved, and not the variables
themselves. Much the same thing happened when we studied long division in Section 4.2. Just as
we developed synthetic division to streamline that process, in this section, we introduce a similar
bookkeeping device to help us solve systems of linear equations. To that end, we deﬁne a matrix
as a rectangular array of real numbers. We typically enclose matrices with square brackets, ‘[ ’ and
‘ ]’, and we size matrices by the number of rows and columns they have. For example, the size
(sometimes called the dimension) of 3 0 −1 2 −5 10 is 2 × 3 because it has 2 rows and 3 columns. The individual numbers in a matrix are called its
entries and are usually labeled with double subscripts: the ﬁrst tells which row the element is in
and the second tells which column it is in. The rows are numbered from top to bottom and the
columns are numbered from left to right. Matrices themselves are usually denoted by uppercase
letters (A, B , C , etc.) while their entries are usually denoted by the corresponding letter. So, for
instance, if we have 3
0 −1 A=
2 −5 10
then a11 = 3, a12 = 0, a13 = −1, a21 = 2, a22 = −5, and a23 = 10. We shall explore matrices as
mathematical objects with their own algebra in Section ?? and introduce them here solely as a
bookkeeping device. Consider the system of linear equations from number 2 in Example 8.1.2 (E 1) 2x + 3 y − z = 1 (E 2)
10x − z = 2 (E 3) 4x − 9y + 2z = 5
We encode this system into a matrix by assigning each equation to a corresponding row. Within
that row, each variable and the constant gets its own column, and to separate the variables on the
left hand side of the equation from the constants on the right hand side, we use a vertical bar, .
Note that in E 2, since y is not present, we record its coeﬃcient as 0. The matrix associated with 420 Systems of Equations this system is
x
(E 1) → y z c 2 3 −1 1 10
0 −1 2 (E 2) → (E 3) →
4 −9
25
This matrix is called an augmented matrix because the column containing the constants is
appended to the matrix containing the coeﬃcients.1 To solve this system, we can use the same
kind operations on the rows of the matrix that we performed on the equations of the system. More
speciﬁcally, we have the following analog of Theorem 8.1 below.
Theorem 8.2. Row Operations: Given an augmented matrix for a system of linear equations,
the following row operations produce an augmented matrix which corresponds to an equivalent
system of linear equations.
Interchange any two rows.
Replace a row with a nonzero multiple of itself.a
Replace a row with itself plus a nonzero multiple of another row.b
a
b That is, the row obtained by multiplying each entry in the row by the same nonzero number.
Where we add entries in corresponding columns. As a demonstration of the moves in Theorem 8.2, we revisit some of the steps that were used in
solving the systems of linear equations in Example 8.1.2 of Section 8.1. The reader is encouraged to
perform the indicated operations on the rows of the augmented matrix to see that the machinations
are identical to what is done to the coeﬃcients of the variables in the equations. We ﬁrst see a
demonstration of switching two rows using the ﬁrst step of part 1 in Example 8.1.2. (E 1) 3x − y + z = (E 2) 2x − 4y + 3z (E 3)
x−y+z 3 −1 2 −4 1 −1
1 3 = 16
=
1 3 5 3 16 15 Switch E 1 and E 3 −− − − − −
− − − − −→ (E 1) x−y+z = 5 (E 2) 2x − 4y + 3z = 16 (E 3)
3x − y + z = 3 1 −1
15 Switch R1 and R3 −− − − − −
− − − − − → 2 −4
3 16 3 −1
13 We shall study the coeﬃcient and constant matrices separately in Section ??. 8.2 Systems of Linear Equations: Augmented Matrices* 421 Next, we have a demonstration of replacing a row with a nonzero multiple of itself using the ﬁrst
step of part 3 in Example 8.1.2. (E 1) 3x1 + x2 + x4 = 6 (E 2) 2x1 + x2 − x3 = 4 (E 3) x − 3x − 2x = 0
2
3
4 3 2 0 1 0 16 (E 1) x1 + 1 x2 + 1 x4 = 2 3
3 1
Replace E 1 with 3 E 1
−−−−−−−
−−−−−−→
(E 2)
2x1 + x2 − x3 = 4 (E 3) x − 3x − 2x = 0
2
3
4 Replace R1 with 1 R1 3
− − − − − −→
1 −1
0 4 −− − − − − − 1 −3 −2 0 1 2 0 1
3 1
3 0 2 1 −1
0 4 1 −3 −2 0 Finally, we have an example of replacing a row with itself plus a multiple of another row using the
second step from part 2 in Example 8.1.2. (E 1) x + 3 y − 1 z = 2
2 (E 2)
10x − z = (E 3) 4x − 9y + 2z = 3
2 1 10
0 4 −9 −1
2
−1
2 1
2 2
5
1
2 1 (E 1) x + 3 y − 1 z = 2
2
2 Replace E 2 with −10E 1 + E 2
−−−−−−−−−→
−−−−−−−−−
(E 2)
−15y + 4z = −3 Replace E 3 with −4E 1 + E 3 (E 3)
−15y + 4z =
3 Replace R2 with −10R1 + R2
−−−−−−−−−
2 −− − − − − − − −→ Replace R3 with −4R1 + R3
5 3
2 1 0 −15 0 −15 1
−2 1
2 4 −3 4
3 The matrix equivalent of ‘triangular form’ is row echelon form. The reader is encouraged to
refer to Deﬁnition 8.3 for comparison. Note that the analog of ‘leading variable’ of an equation
is ‘leading entry’ of a row. Speciﬁcally, the ﬁrst nonzero entry (if it exists) in a row is called the
leading entry of that row.
Definition 8.4. A matrix is said to be in row echelon form provided all of the following
conditions hold:
1. The ﬁrst nonzero entry in each row is 1.
2. The leading 1 of a given row must be to the right of the leading 1 of the row above it.
3. Any row of all zeros cannot be placed above a row with nonzero entries. 422 Systems of Equations To solve a system of a linear equations using an augmented matrix, we encode the system into an
augmented matrix and apply Gaussian Elimination to the rows to get the matrix into rowechelon
form. We then decode the matrix and back substitute. The next example illustrates this nicely.
Example 8.2.1. Use an augmented matrix to transform
into triangular form. Solve the system. 3x − y + z = x + 2y − z = 2x + 3y − 4z = the following system of linear equations 8
4
10 Solution. We ﬁrst encode the system into an augmented matrix. 3x − y + z = 8 3 −1
18 Encode into the matrix −− − − − − −→ 1
−−−−−−−
2 −1 4
x + 2y − z = 4 2x + 3y − 4z = 10
2
3 −4 10 Thinking back to Gaussian Elimination at an equations level, our ﬁrst order of business is to
get x in E 1 with a coeﬃcient of 1. At the matrix level, this means getting a leading 1 in R1. This
is in accordance with the ﬁrst criteria in Deﬁnition 8.4. To that end, we interchange R1 and R2. 3 −1
18
1
2 −1 4 Switch R1 and R2 − − − − − → 3 −1
1
2 −1 4 − − − − − −
1 8 2
3 −4 10
2
3 −4 10
Our next step is to eliminate the x’s from E 2 and E 3. From a matrix standpoint, this means
we need 0’s below the leading 1 in R1. This guarantees the leading 1 in R2 will be to the right of
the leading 1 in R1 in accordance with the second requirement of Deﬁnition 8.4. 1
2 −1 4
1
2 −1
4 Replace R2 with −3R1 + R2 −−−−−−−−−
− − − − − − − − → 0 −7 3 −1
1 8
4 −4 Replace R3 with −2R1 + R3 2
3 −4 10
0 −1 −2
2
Now we repeat the above process for the variable y which means we need
entry in R2 to be 1. 1
2 −1
4
1
2 −1 4 Replace R2 with − 1 R2 7
4
−−−−−−− 0 −7
4 −4 − − − − − − − → 0
1 −7 4
7 0 −1 −2
2
0 −1 −2 2 to get the leading 8.2 Systems of Linear Equations: Augmented Matrices* 423 To guarantee the leading 1 in R3 is to the right of the leading 1 in R2, we get a 0 in the second
column of R3. 1
2 −1 0
1 −4
7 0 −1 −2 4 4
7 2 1 Replace R3 with R2 + R3
−− − − − − − −→ 0
−−−−−−−− 0 2 −1 1 −4
7 0 − 18
7 4 4
7 18
7 Finally, we get the leading entry in R3 to be 1. 1 0 0 2 −1 1 4
−7 0 − 18
7 4
7 Replace R3 with − 18 R3
4 −− − − − − − −
− − − − − − −→
7 18
7 1 0 0 2 −1 4 1 −4
7 4
7 0 1 −1 Decoding from the matrix gives a system in triangular form 1 0 0 −1 4 1 −4
7 4
7 2 0 1 −1 x + 2y − z = 4 Decode from the matrix
4
−−−−−−−→
−−−−−−−
y − 4z =
7
7 z = −1 4
4
We get z = −1, y = 4 z + 4 = 7 (−1) + 7 = 0 and x = −2y + z + 4 = −2(0) + (−1) + 4 = 3 for
7
7
a ﬁnal answer of (3, 0, −1). We leave it to the reader to check. As part of Gaussian Elimination, we used row operations to obtain 0’s beneath each leading 1
to put the matrix into row echelon form. If we also require that 0’s are the only numbers above a
leading 1, we have what is known as the reduced row echelon form of the matrix. Definition 8.5. A matrix is said to be in reduced row echelon form provided both of the
following conditions hold:
1. The matrix is in row echelon form.
2. The leading 1s are the only nonzero entry in their respective columns. Of what signiﬁcance is the reduced row echelon form of a matrix? To illustrate, let’s take the
row echelon form from Example 8.2.1 and perform the necessary steps to put into reduced row
echelon form. We start by using the leading 1 in R3 to zero out the numbers in the rows above it. 424 Systems of Equations 1 0 0 2 −1 4 1 −4
7 4
7 0 Finally, we take care 12 0 1 00 1 −1 Replace R1 with R3 + R1
− − − − − − − −→ − − − − − − 4− − − Replace R2 with 7 R3 + R2 0 1 0
0 0 0 1 −1 of the 2 in R1 above the leading 1 in 0
3 Replace R1 with −2R2 + R1
−−−−−−−−→
0
0 −−−−−−−−− 1 −1 3 120 R2. 100
3 0 1 0
0 0 0 1 −1 To our surprise and delight, when we decode this matrix, we
without having to deal with any backsubstitution at all. x 100
3 Decode from the matrix
−−−−−−→
−−−−−−−
0 1 0
0
y z
0 0 1 −1 obtain the solution instantly = 3 = 0 = −1 Note that in the previous discussion, we could have started with R2 and used it to get a zero
above its leading 1 and then done the same for the leading 1 in R3. By starting with R3, however,
we get more zeros ﬁrst, and the more zeros there are, the faster the remaining calculations will be.2
It is also worth noting that while a matrix has several3 row echelon forms, it has only one reduced
row echelon form. The process by which we have put a matrix into reduced row echelon form is
called GaussJordan Elimination.
Example 8.2.2. Solve the following system using an augmented matrix. Use GaussJordan Elimination to put the augmented matrix into reduced row echelon form. x2 − 3x1 + x4 = 2 2x1 + 4x3 = 5 4x − x = 3
2 4 Solution. We ﬁrst encode the system into a matrix. (Pay x2 − 3x1 + x4 = 2 Encode into the matrix
−−−−−−−
2x1 + 4x3 = 5 − − − − − − − → 4x2 − x4 = 3
2
3 attention to the subscripts!) −3
1
0
12 2
0
4
0 5 0
4
0 −1 3 Carl also ﬁnds starting with R3 to be more symmetric, in a purely poetic way.
inﬁnite, in fact 8.2 Systems of Linear Equations: Augmented Matrices*
Next, we get a leading −3
1 2
0 0
4 1 in the ﬁrst column of R1. 0
12 Replace R1 with − 1 R1 − − − − − − 3−
4
0 5 −−−−−−−→ 0 −1 3 425 1 −1
3 2 0 0 −1
3 0 4
0 5 3 0 4 −2
3 −1 Now we eliminate the nonzero entry below our leading 1. 1 2 0 −1
3 0 −1
3 0 4 0 4 0 −1 2
−3 1 0 0 Replace R2 with −2R1 + R2
−−−−−−−−→
5 −−−−−−−−− 3 We proceed to get a leading 1 in R2. 1
2
1
0 −3 −3 1 −3 Replace R2 with 3 R2 2
19 − − − − − −2−
2
0
− − − − − −→
4 3
3
3 0
4
0 −1
3 1 0 0 0 1
−3 −2
3 2
3 4 2
3 19
3 4 0 −1 3 −1
3 −1
3 0 1
−3 −2
3 1 6 1 19
2 4 0 −1 3 We now zero out the entry below the leading 1 in R2. 1 1 −3 0
1 0
4 2
−3 Replace R3 with −4R2 + R3
−−−−−−−−→
1 19 − − − − − − − − −
2 −1
3 1
0 −3 6
0 Next, it’s time for a leading 1 in R3. 1
1 −3
0 −1 −2 3
3 1 Replace R3 with − 24 R3 19 − − − − − − − −
0
1
6
1 − − − − − − −→ 2 0
0 −24 −5 −35
The matrix is now in row echelon form. To get the reduced
last leading 1 we produced and work to get 0’s above it. 1
1
2
0 −3 −3 1 −3 Replace R2 with −6R3 + R2
0
−−−−−−−−→
1
6
1 19 − − − − − − − − − 2 5
35
0
0
1 24
24 1
0 −1 −2
3
3 1 −3 19
0
1
6
1 2 0
0 −24 −5 −35 1 1 −3 0
1 0
0 −2 3 1 19 2 1
0 −3 6
1 5
24 35
24 row echelon form, we start with the 1 0 0 −1
3 0 1
−3 −2
3 1 0 1
−4 3
4 0 1 5
24 35
24 426 Systems of Equations Lastly, we get a 0 above the leading 1 of R2. 1 0 0 1
−3 0 −1
3 2
−3
3
4
35
24 1 0 −1
4 0 1 5
24 At last, we decode to 0
1 0
1 0
0 Replace R1 with 1 R2 + R1 − − − − − − 3− − − − − − − − − − −→ 1 0 0 0 5
− 12 5
− 12 1 0 −1
4 3
4 0 1 5
24 35
24 0 get
0 5
− 12 5
− 12 0 −1
4 3
4 1 5
24 35
24 x − 5x = −5 1 12 4 12 Decode from the matrix
1
3
−−−−−−−→
−−−−−−−
x− x =
4 2 44 35 x3 + 5 x4 =
24
24 5
We have that x4 is free and we assign it the parameter t. We obtain x3 = − 24 t + 35 , x2 = 1 t + 3 ,
24
4
4
5
5
5
51
3
5
35
and x1 = 12 t − 12 . Our solution is 12 t − 12 , 4 t + 4 , − 24 t + 24 , t : −∞ < t < ∞ and leave it to
the reader to check. Like all good algorithms, putting a matrix in row echelon or reduced row echelon form can easily
be programmed into a calculator, and, doubtless, your graphing calculator has such a feature. We
use this in our next example.
Example 8.2.3. Find the quadratic function which passes through the points (−1, 3), (2, 4), (5, −2).
Solution. According to Deﬁnition 3.4, a quadratic function has the form f (x) = ax2 + bx + c
where a = 0. Our goal is to ﬁnd a, b and c so that the three given points are on the graph of
f . If (−1, 3) is on the graph of f , then f (−1) = 3, or a(−1)2 + b(−1) + c = 3 which reduces to
a − b + c = 3, an honesttogoodness linear equation with the variables a, b and c. Since the point
(2, 4) is also on the graph of f , then f (2) = 4 which gives us the equation 4a + 2b + c = 4. Lastly,
the point (5, −2) is on the graph of f gives us 25a + 5b + c = −2. Putting these together, we obtain
a system of three linear equations. Encoding this into an augmented matrix produces 1 −1
1
a−b+c =
3
3 Encode into the matrix −− − − − − −→ 4
−−−−−−−
4a + 2 b + c =
4
2
1
4 25a + 5b + c = −2
25
5
1 −2
7
Using a calculator,4 we ﬁnd a = − 18 , b = 13 and c = 37 . Hence, the one and only quadratic
18
9
72
13
which ﬁts the bill is f (x) = − 18 x + 18 x + 37 . To verify this analytically, we see that f (−1) = 3,
9
f (2) = 4, and f (5) = −2. We can use the calculator to check our solution as well by plotting the
three data points and the function f .
4 We’ve tortured you enough already with fractions in this exposition! 8.2 Systems of Linear Equations: Augmented Matrices* 7
The graph of f (x) = − 18 x2 + 427 13
18 x + 37
9 with the points (−1, 3), (2, 4) and (5, −2) 428 8.2.1 Systems of Equations Exercises 1. State whether the given matrix is in reduced row echelon form, row echelon form
neither of those forms. 1143
1043
103 (a) (c) 0 1 3 6 (e) 0 1 3 6
013 0001
0000
3 −1
13 1000
(b) 2 −4
3 16 1143 (f) (d) 0 1 0 0 1 −1
15
0136 only or in 0 0 0 0001
2. The following matrices are in reduced row echelon form. Decode from each matrix the solution
of the corresponding system of linear equations or state that the system is inconsistent. 1 0 −2
10030
1
0
9 −3 (a) (d) 0 1 2 6 0 (f) 0
01
7
1 −4 20 00001
0
0
0
0
1 0 0 −3 (b) 0 1 0 20 1
0 −8
17
10000 0 0 1 19
0 0 1 4 3 0
1
4 −3 2 (g) (e) 0 0 0 0 0 1
0
0
00
4
1003 (c) 0 1 0 6 −6 0
0
0
00
00000 2
0010
3. Solve the following systems of linear equations using the techniques discussed in this section.
Compare and contrast these techniques with those you used to solve the systems in the
Exercises in Section 8.1. −5x + y = 17 4x − y + z = 5 (a) (c)
x+y = 5
2y + 6z = 30 x+z = 5 x+y+z = 3 (b)
2x − y + z = 0 −3x + 5y + 7z = 7 8.2 Systems of Linear Equations: Augmented Matrices* (d) x − 2y + 3 z = (j) 2x + 2 y + z =
3 3x − 2y + z = −5 (e)
x + 3y − z = 12 x + y + 2z =
0 2x − y + z = −1 (f)
4x + 3 y + 5 z =
1 5y + 3z =
4 x − y + z = −4 (g)
−3x + 2y + 4z = −5 x − 5y + 2z = −18 2x − 4y + z = −7 (h)
x − 2y + 2z = −2 −x + 4y − 2z =
3 2x − y + z = 1 (i)
2x + 2y − z = 1 3x + 6 y + 4 z = 9 (k) x − 3y − 4z = 7 −3x + y + 2z = −5 429 2x − 19y − 19z = x+y+z = (l) 3x + 4y − z = 13 (n) 2
4 2x − 4y − z = −1
x−y =
x−y+z = 2
8 3x + 3y − 9z = −6 7x − 2y + 5z 2x − 3y + z (m)
4x − 4y + 4z 6x − 5y + 7z 3 = 39 = −1 = −13
= −25 x1 − x3 = −2
2x2 − x4 = 0 x − 2x + x =
1
2
3 −x3 + x4 = 0
1 4. The price for admission into the StitzZeager Sasquatch Museum and Research Station is $15
for adults and $8 for kids 13 years old and younger. When the Zahlenreich family visits the
museum their bill is $38 and when the Nullsatz family visits their bill is $39. One day both
families went together and took an adult babysitter along to watch the kids and the total
admission charge was $92. Later that summer, the adults from both families went without
the kids and the bill was $45. Is that enough information to determine how many adults
and children are in each family? If not, state whether the resulting system is inconsistent or
consistent dependent. In the latter case, give at least two plausible solutions. 5. Use the technique in Example 8.2.3 to ﬁnd the line between the points (−3, 4) and (6, 1).
How does your answer compare to the slopeintercept form of the line in Equation 3.3? 430 Systems of Equations 6. With the help of your classmates, ﬁnd at least two diﬀerent row echelon forms for the matrix 1 23 4 12 8 8.2 Systems of Linear Equations: Augmented Matrices* 8.2.2 431 Answers 1. (a) Reduced row echelon form
(b) Neither
(c) Row echelon form only (d) Reduced row echelon form
(e) Reduced row echelon form
(f) Row echelon form only 2. (a) (−2, 7)
(b) (−3, 20, 19)
(c) (−3t + 4, −6t − 6, 2, t)
for all real numbers t
(d) (2, −1, 1)
(e) (1, 3, −2)
(f) Inconsistent
(g) (1, 3, −2)
1
(h) −3, 2 , 1
2
(i) 1 , 3 , 1
3
19
51
4
(j) 13 t + 13 , − 11 t + 13 , t
13
for all real numbers t (k) Inconsistent 3. (a) (−2, 7)
(b) (1, 2, 0) (l) (4, −3, 1)
(m) −2t − 35
4 , −t − 11
2 ,t (n) Inconsistent
(o) (8s − t + 7, −4s + 3t + 2, s, t)
for all real numbers s and t
(p) (−9t − 3, 4t + 20, t)
for all real numbers t
(q) Inconsistent (c) (−t + 5, −3t + 15, t)
for all real numbers t
(d) (1, 2, 3, 4) 4. Let x1 and x2 be the numbers of adults and children, respectively, in the Zahlenreich family
and let x3 and x4 be the numbers of adults and children, respectively, in the Nullsatz family.
The system of equations determined by the given information is 15x1 + 8x2 = 38 15x3 + 8x4 = 39 15x + 8x + 15x + 8x = 77 1
2
3
4 15x1 + 15x3 = 45
We subtracted the cost of the babysitter in E3 so the constant is 77, not 92. This system is
8
8
consistent dependent and its solution is 15 t + 2 , −t + 4, − 15 t + 13 , t . Our variables repre5
5
sent numbers of adults and children so they must be whole numbers. Running through the
values t = 0, 1, 2, 3, 4 yields only one solution where all four variables are whole numbers;
t = 3 gives us (2, 1, 1, 3). Thus there are 2 adults and 1 child in the Zahlenreichs and 1 adult
and 3 kids in the Nullsatzs. 432 Systems of Equations 8.3 Determinants and Cramer’s Rule* 8.3.1 Definition and Properties of the Determinant In this section we assign to each square matrix A a real number, called the determinant of A,
which will eventually lead us to yet another technique for solving consistent independent systems
of linear equations. The determinant is deﬁned recursively, that is, we deﬁne it for 1 × 1 matrices
and give a rule by which we can reduce determinants of n × n matrices to a sum of determinants
of (n − 1) × (n − 1) matrices.1 This means we will be able to evaluate the determinant of a 2 × 2
matrix as a sum of the determinants of 1 × 1 matrices; the determinant of a 3 × 3 matrix as a sum
of the determinants of 2 × 2 matrices, and so forth. To explain how we will take an n × n matrix
and distill from it an (n − 1) × (n − 1), we use the following notation.
Definition 8.6. Given an n × n matrix A where n > 1, the matrix Aij is the (n − 1) × (n − 1)
matrix formed by deleting the ith row of A and the j th column of A. For example, using the matrix A below, we ﬁnd the matrix A23 by deleting the second row and
third column of A. 3 12 31 Delete R2 and C 3 A = 0 −1 5 − − − − − −
− − − − − → A23 = 21
2
14
We are now in the position to deﬁne the determinant of a matrix.
Definition 8.7. Given an n × n matrix A the determinant of A, denoted det(A), is deﬁned
as follows
If n = 1, then A = [a11 ] and det(A) = det ([a11 ]) = a11 .
If n > 1, then A = [aij ]n×n and det(A) = det [aij ]n×n = a11 det (A11 ) − a12 det (A12 ) + − . . . + (−1)1+n a1n det (A1n ) There are two commonly used notations for the determinant of a matrix A: ‘det(A)’ and ‘A’
We have chosen to use the notation det(A) as opposed to A because we ﬁnd that the latter is
often confused with absolute value, especially in the context of a 1 × 1 matrix. In the expansion
1 We will talk more about the term ‘recursively’ in Section 9.1. 8.3 Determinants and Cramer’s Rule* 433 a11 det (A11 ) − a12 det (A12 )+ − . . . +(−1)1+n a1n det (A1n ), the notation ‘+ − . . . +(−1)1+n a1n ’ means
that the signs alternate and the ﬁnal sign is dictated by the sign of the quantity (−1)1+n . Since
the entries a11 , a12 and so forth up through a1n comprise the ﬁrst row of A, we say we are ﬁnding
the determinant of A by ‘expanding along the ﬁrst row’. Later in the section, we will develop a
formula for det(A) which allows us to ﬁnd it by expanding along any row. 4 −3 we get
Applying Deﬁnition 8.7 to the matrix A = 2
1 4 −3 det(A) = det 2
1
= 4 det (A11 ) − (−3) det (A12 )
= 4 det([1]) + 3 det([2])
= 4(1) + 3(2)
= 10 ab we get
For a generic 2 × 2 matrix A = cd ab det(A) = det cd = a det (A11 ) − b det (A12 )
= a det ([d]) − b det ([c])
= ad − bc
This formula is worth remembering
Equation 8.1. For a 2 × 2 matrix, det ab
cd = ad − bc 3
1 Applying Deﬁnition 8.7 to the 3 × 3 matrix A = 0 −1 2
1 2 5 we obtain 4 434 Systems of Equations 3
1 det(A) = det 0 −1 2
1 2 5 4 = 3 det (A11 ) − 1 det (A12 ) + 2 det (A13 ) = 3 det −1 5
14 − det 05
24 + 2 det 0 −1
2 1 = 3((−1)(4) − (5)(1)) − ((0)(4) − (5)(2)) + 2((0)(1) − (−1)(2))
= 3(−9) − (−10) + 2(2)
= −13
To evaluate the determinant of a 4 × 4 matrix, we would have to evaluate the determinants of
four 3 × 3 matrices, each of which involves the ﬁnding the determinants of three 2 × 2 matrices. 8.3.2 Cramer’s Rule In this section, we introduce a theorem which enables us to solve a system of linear equations by
means of determinants only. As usual, the theorem is stated in full generality, using numbered
unknowns x1 , x2 , etc., instead of the more familiar letters x, y , z , etc. The proof of the general
case is best left to a course in Linear Algebra.
Theorem 8.3. Cramer’s Rule: Suppose AX = B is the matrix form of a system of n linear
equations in n unknowns where A is the coeﬃcient matrix, X is the unknowns matrix, and
B is the constant matrix. If det(A) = 0, then the corresponding system is consistent and
independent and the solution for unknowns x1 , x2 , . . . xn is given by:
xj = det (Aj )
,
det(A) where Aj is the matrix A whose j th column has been replaced by the constants in B . In words, Cramer’s Rule tells us we can solve for each unknown, one at a time, by ﬁnding the
ratio of the determinant of Aj to that of the determinant of the coeﬃcient matrix. The matrix Aj
is found by replacing the column in the coeﬃcient matrix which holds the coeﬃcients of xj with
the constants of the system. The following example ﬂeshes out this method.
Example 8.3.1. Use Cramer’s Rule to solve for the indicated unknowns. 8.3 Determinants and Cramer’s Rule* 435 2x1 − 3x2 =
4
for x1 and x2
1. Solve 5x + x = −2
1
2 2x − 3y + z = −1 2. Solve
x−y+z =
1 for z . 3x − 4z =
0
Solution.
1. Writing this system in matrix form, we ﬁnd A= 2 −3
5 1 X= x1
x2 B= 4
−2 To ﬁnd the matrix A1 , we remove the column of the coeﬃcient matrix A which holds the
coeﬃcients of x1 and replace it with the corresponding entries in B . Likewise, we replace the
column of A which corresponds to the coeﬃcients of x2 with the constants to form the matrix
A2 . This yields A1 = 4 −3
−2 1 A2 = 2 4 5 −2 Computing determinants, we get det(A) = 17, det (A1 ) = −2 and det (A2 ) = −24, so that
x1 = det (A1 )
2
=−
det(A)
17 x2 = det (A2 )
24
=−
det(A)
17 2
The reader can check that the solution to the system is − 17 , − 24 .
17 2. To use Cramer’s Rule to ﬁnd z , we identify x3 as z . We have 2 −3 1 A = 1 −1
1 3
0 −4 x X= y B= z −1 2 −3 −1 A3 = Az = 1 −1
1 3
0
0 1 0 Expanding both det(A) and det (Az ) along the third rows (to take advantage of the 0’s) gives
z= det (Az )
−12
6
=
=
det(A)
−10
5 436 Systems of Equations
The reader is encouraged to solve this system for x and y similarly and check the answer. 8.3 Determinants and Cramer’s Rule* 8.3.3 437 Exercises 1. Compute the determinant of the following matrices. 12 −7 (a) B = (f) G = −5
3 6 15 (b) C = 14 35 2
(g) V = xx (c) Q =
1 2x (d) L = 1
x3 ln(x)
x3 3
− x4 1−3 ln(x)
x4 4 (e) F = 3 1 6 −3 4 −3 2
6 1 2 3 3 11 4 19 2
3
i j k −1
0
5 9 −4 −2
1 0 −3 0 2 −2
(h) H = −5
0 1
0 8 7 16 0 41 2. Use Cramer’s Rule to solve the system of linear equations. 3x + 7y = 26 x+y+z = 3 (a) 5x + 12y = 39
(b)
2x − y + z = 0 −3x + 5y + 7z = 7
3. Use Cramer’s Rule to solve for x4 in the following x1 − x3 2x2 − x4 x − 2x + x
1
2
3 −x3 + x4 system of linear equations.
= −2
= 0 = 0 = 1 438 Systems of Equations 8.3.4 Answers 1. (a)
(b)
(c)
(d) (e) det(F ) = −12 det(B ) = 1
det(C ) = 0
det(Q) = x2
1
det(L) = x7 (f) det(G) = 0
(g) det(V ) = 20i + 43j + 4k
(h) det(H ) = −2 2. (a) x = 39, y = −13 8.4 (b) x = 1, y = 2, z = 0 Systems of NonLinear Equations and Inequalities In this section, we study systems of nonlinear equations and inequalities. Unlike the systems of
linear equations for which we have developed several algorithmic solution techniques, there is no
general algorithm to solve systems of nonlinear equations. Moreover, all of the usual hazards of
nonlinear equations like extraneous solutions and unusual function domains are once again present.
Along with the tried and true techniques of substitution and elimination, we shall often need equal
parts tenacity and ingenuity to see a problem through to the end. You may ﬁnd it necessary to
review topics throughout the text which pertain to solving equations involving the various functions
we have studied thus far. To get the section rolling we begin with a fairly routine example.
Example 8.4.1. Solve the following systems of equations. Verify your answers algebraically and
graphically. x2 + y 2 = 4 x2 + y 2 = 4
1.
3. 4x2 + 9y 2 = 36 y − 2x = 0 2. x2 + y2 = 4 4x2 − 9y 2 = 36 x2 + y 2 = 4
4. y − x2 = 0 Solution:
1. Since both equations contain x2 and y 2 only, we can eliminate one of the variables as we did
in Section 8.1. (E 1)
x2 + y 2 = 4 (E 2) 4x2 + 9y 2 = 36 (E 1) x2 + y 2 = 4
Replace E 2 with
−− − − −→
−−−−− (E 2)
−4E 1 + E 2
5y 2 = 20 From 5y 2 = 20, we get y 2 = 4 or y = ±2. To ﬁnd the associated x values, we substitute each
value of y into one of the equations to ﬁnd the resulting value of x. Choosing x2 + y 2 = 4,
we ﬁnd that for both y = −2 and y = 2, we get x = 0. Our solution is thus {(0, 2), (0, −2)}. 8.4 Systems of NonLinear Equations and Inequalities 439 To check this algebraically, we need to show that both points satisfy both of the original
equations. We leave it to the reader to verify this. To check our answer graphically, we sketch
both equations and look for their points of intersection. The graph of x2 + y 2 = 4 is a circle
centered at (0, 0) with a radius of 2, whereas the graph of 4x2 + 9y 2 = 36, when written in the
2
2
standard form x + y4 = 1 is easily recognized as an ellipse centered at (0, 0) with a major
9
axis along the xaxis of length 6 and a minor axis along the y axis of length 4. We see from
the graph that the two curves intersect at their y intercepts only, (0, ±2).
2. We proceed as before to eliminate one of the variables (E 1)
x2 + y 2 = 4 (E 2) 4x2 − 9y 2 = 36 (E 1) x2 + y 2 = 4
Replace E 2 with
−− − − −→
−−−−− (E 2) −13y 2 = 20
−4E 1 + E 2 Since the equation −13y 2 = 20 admits no real solution, the system is inconsistent. To verify
this graphically, we note that x2 + y 2 = 4 is the same circle as before, but when writing the
2
2
second equation in standard form, x − y4 = 1, we ﬁnd a hyperbola centered at (0, 0) opening
9
to the left and right with a transverse axis of length 6 and a conjugate axis of length 4. We
see that the circle and the hyperbola have no points in common.
y y 1 −3 −2 −1 1 1 2 3 −3 x −2 −1 −1 Graphs for 1 2 3 x −1 x2 + y2 = 4 4x2 + 9y 2 = 36 Graphs for x2 + y 2 = 4 4x2 − 9y 2 = 36 3. Since there are no like terms among the two equations, elimination won’t do us any good.
We turn to substitution and from the equation y − 2x = 0, we get y = 2x. Substituting this
√
into x2 + y 2 = 4 gives x2 + (2x)2 = 4. Solving, we ﬁnd 5x2 = 4 or x = ± 2 5 5 . Returning
to the equation we used for the substitution, y = 2x, we ﬁnd y =
√ √ √
45
5 √
25
, so
√
√5
25
45
− 5 , − 5 . We when x = one solution is 2 5 5 , 4 5 5 . Similarly, we ﬁnd the other solution to be
leave it to the reader that both points satisfy both equations, so that our ﬁnal answer is
√
√
√
√
2545
, 5 , − 2 5 5 , − 4 5 5 . The graph of x2 + y 2 = 4 is our circle from before and the
5
graph of y − 2x = 0 is a line through the origin with slope 2. Though we cannot verify the
numerical values of the points of intersection from our sketch, we do see that we have two
solutions: one in Quadrant I and one in Quadrant III as required. 440 Systems of Equations 4. While it may be tempting to solve y − x2 = 0 as y = x2 and substitute, we note that this
system is set up for elimination.1 (E 1) x2 + y 2 = 4 (E 2) y − x2 = 0 (E 1) x2 + y 2 = 4
Replace E 2 with
−− − − −→
−−−−− (E 2) y 2 + y = 4
E1 + E2
√ From y 2 + y = 4 we get y 2 + y − 4 = 0 which gives y = −1± 17 . Due to the complicated
2
nature of these answers, it is worth our time to make a quick sketch of both equations to head
oﬀ any extraneous solutions we may encounter. We see that the circle x2 + y 2 = 4 intersects
the parabola y = x2 exactly twice,√and both of these points have a positive y value. Of the
two solutions for y , only y = −1+ 17 is positive, so to get our solution, we substitute this
2
√
√
√
−1+ 17
2 = 0 and solve for x. We get x = ±
= ± −2+2 17 . Our solution is
into y − x
2
2
√
√
√
√
√
√
−2+2 17 −1+ 17
−2+2 17 −1+ 17
,
,−
,
, which we leave to the reader to verify.
2
2
2
2
y y 1 −3 −2 −1 1 1 2 3 x −3 −2 −1 1 2 3 x −1 Graphs for x2 + y 2 = 4 y − 2x = 0 Graphs for x2 + y 2 = 4 y − x2 = 36 A couple of remarks about Example 8.4.1 are in order. First note that, unlike systems of linear
equations, it is possible for a system of nonlinear equations to have more than one solution without
having inﬁnitely many solutions. In fact, while we characterize systems of nonlinear equations as
being ‘consistent’ or ‘inconsistent,’ we generally don’t use the labels ‘dependent’ or ‘independent’.
Secondly, as we saw with number 4, sometimes making a quick sketch of the problem situation can
save a lot of time and eﬀort. While in general the curves in a system of nonlinear equations may
not be easily visualized, it sometimes pays to take advantage when they are. Our next example
provides some considerable review of many of the topics introduced in this text.
Example 8.4.2. Solve the following systems of equations. Verify your answers algebraically and
graphically, as appropriate.
1 We encourage the reader to solve the system using substitution to see that you get the same solution. 8.4 Systems of NonLinear Equations and Inequalities x2 + 2xy − 16 = 0
1. y 2 + 2xy − 16 = 0 441 y + 4e2x = 1
2. y 2 + 2 ex = 1 z (x − 2) = x 3. (x − 2)2 + y 2 = 1 yz = y Solution. 1. At ﬁrst glance, it doesn’t appear as though elimination will do us any good since it’s clear
that we cannot completely eliminate one of the variables. The alternative, solving one of
the equations for one variable and substituting it into the other, is full of unpleasantness.
Returning to elimination, we note that it is possible to eliminate the troublesome xy term,
and the constant term as well, by elimination and doing so we get a more tractable relationship
between x and y (E 1) x2 + 2xy − 16 = 0 (E 2) y 2 + 2xy − 16 = 0 (E 1) x2 + 2xy − 16 = 0
−− − − −→
−−−−− (E 2)
−E 1 + E 2
y 2 − x2 = 0
Replace E 2 with We get y 2 − x2 = 0 or √ = ±x. Substituting y = x into E 1 we get x2 + 2x2 − 16 = 0 so
y
2 = 16 or x = ± 4 3 . On the other hand, when we substitute y = −x into E 1, we get
that x
3
3
√ x2 − 2x2 − 16 = 0 or x2 = −16 which gives no real solutions. Substituting each of x = ± 4 3 3
√ √ √ √ 4343
, − 4 3 3 , − 4 3 3 . We
into the substitution equation y = x yields the solution
3,3
leave it to the reader to show that both points satisfy both equations and now turn to verifying
−2
our solution graphically. We begin by solving x2 +2xy − 16 = 0 for y to obtain y = 162xx . This
function is easily graphed using the techniques of Section 5.2. Solving the second equation,
y 2 + 2xy − 16 = √ for y , however, is more complicated. We use the quadratic formula to
0,
obtain y = −x ± x2 + 16 which would require the use of Calculus or a calculator to graph.
Believe it or not, we don’t need either because the equation y 2 + 2xy − 16 = 0 can be obtained
from the equation x2 + 2xy − 16 = 0 by interchanging y and x. Thinking back to Section
6.2, this means we can obtain the graph of y 2 + 2xy − 16 = 0 by reﬂecting the graph of
x2 + 2xy − 16 = 0 across the line y = x. Doing so conﬁrms that the two graphs intersect
twice: once in Quadrant I, and once in Quadrant III as required. 442 Systems of Equations
y
4
3
2
1 −4 −3 −2 −1 1 2 3 4 x −1
−2
−3
−4 The graphs of x2 + 2xy − 16 = 0 and y 2 + 2xy − 16 = 0 2. Unlike the previous problem, there seems to be no avoiding substitution and a bit of algebraic
unpleasantness. Solving y + 4e2x = 1 for y , we get y = 1 − 4e2x which, when substituted
2
into the second equation, yields 1 − 4e2x + 2ex = 1. After expanding and gathering like
terms, we get 16e4x − 8e2x + 2ex = 0. Factoring gives us 2ex 8e3x − 4ex + 1 = 0, and since
2ex = 0 for any real x, we are left with solving 8e3x − 4ex + 1 = 0. We have three terms,
and even though this is not a ‘quadratic in disguise’, we can beneﬁt from the substitution
u = ex . The equation becomes 8u3 − 4u + 1 = 0. We ﬁnd u = 1 is a zero and use synthetic
2
division to factor the left hand side as u − 1 8u2 + 4u − 2 . We use the quadratic formula
2
√
−1± 5
.
4
1
= ln 2 to solve 8u2 + 4u − 2 = 0 and ﬁnd u = √
1
ex = −1± 5 . From ex = 2 , we get x
4
√
√
that −1− 5 < 0, so ex = −1− 5 has no real
4
4
√
x = ln −1+ 5 . We now return to y = 1 −
4 1
2 and
√
−1± 5
, we ﬁrst note
4
√
x = −1+ 5 , so that
e
4 Since u = ex , we now must solve ex =
= − ln(2). As for ex = solutions. We are left with 4e2x to ﬁnd the accompanying y values for each
of our solutions for x. For x = − ln(2), we get y = 1 − 4e2x
= 1 − 4e−2 ln(2)
= 1 − 4eln(1/4)
= 1−4
=0 1
4 8.4 Systems of NonLinear Equations and Inequalities
For x = ln √
−1+ 5
4 443 , we have
y = 1 − 4e2x
2 ln = 1 − 4e ln = 1 − 4e
= 1−4
= 1−4
= √
−1+ 5
2
√ “ “ √”
−1+ 5
4 √ ”2
−1+ 5
4 √2
−1+ 5
4
√
3− 5
8 √ We get two solutions, (0, − ln(2)), ln −1+ 5 , −1+ 5 . It is a good review of the prop4
2
erties of logarithms to verify both solutions, so we leave that to the reader. We are able to
sketch y = 1 − 4e2x using transformations, but the second equation is more diﬃcult and we
resort to the calculator. We note that to graph y 2 + 2ex = 1, we need to graph both the
√
positive and negative roots, y = ± 1 − 2ex . After some careful zooming,2 we conﬁrm our
solutions. √
The graphs of y = 1 − 4e2x and y = ± 1 − 2ex .
3. Our last system involves three variables and gives some insight on how to keep such systems
organized. Labeling the equations as before, we have E1 E2 z (x − 2) = x
yz = y E 3 (x − 2)2 + y 2 = 1 The easiest equation to start with appears to be E 2. While it may be tempting to divide
both sides of E 2 by y , we caution against this practice because it presupposes y = 0. Instead,
2 The calculator has trouble conﬁrming the solution (− ln(2), 0) due to its issues in graphing square root functions.
If we mentally connect the two branches of the thicker curve, we see the intersection. 444 Systems of Equations
we take E 2 and rewrite it as yz − y = 0 so y (z − 1) = 0. From this, we get two cases: y = 0
or z = 1. We take each case in turn.
Case 1: y = 0. Substituting y = 0 into E 1 and E 3, we get E 1 z (x − 2) = x E 3 (x − 2)2 = 1
Solving E 3 for x gives x = 1 or x = 3. Substituting these values into E 1 gives z = −1 when
x = 1 and z = 3 when x = 3. We obtain two solutions, (1, 0, −1) and (3, 0, 3).
Case 2: z = 1. Substituting z = 1 into E 1 and E 3 gives us E1
(1)(x − 2) = x E 3 (1 − 2)2 + y 2 = 1
Equation E 1 gives us x − 2 = x or −2 = 0, which is a contradiction. This means we have
no solution to the system in this case, even though E 3 is solvable and gives y = 0. Hence,
our ﬁnal answer is {(1, 0, −1), (3, 0, 3)}. These points are easy enough to check algebraically
in our three original equations, so that is left to the reader. As for verifying these solutions
graphically, they require plotting surfaces in three dimensions and looking for intersection
points. While this is beyond the scope of this book, we provide a snapshot of the graphs of
our three equations near one of the solution points, (1, 0, −1). Example 8.4.2 showcases some of the ingenuity and tenacity mentioned at the beginning of
the section. Sometimes you just have to look at a system the right way to ﬁnd the most eﬃcient
method to solve it. Sometimes you just have to try something.
We close this section discussing how nonlinear inequalities can be used to describe regions in
the plane which we ﬁrst introduced in Section 3.4. Before we embark on some examples, a little
motivation is in order. Suppose we wish to solve x2 < 4 − y 2 . If we mimic the algorithms for solving
nonlinear inequalities in one variable, we would gather all of the terms on one side and leave a 0 8.4 Systems of NonLinear Equations and Inequalities 445 on the other to obtain x2 + y 2 − 4 < 0. Then we would ﬁnd the zeros of the left hand side, that
is, where is x2 + y 2 − 4 = 0, or x2 + y 2 = 4. Instead of obtaining a few numbers which divide the
real number line into intervals, we get an equation of a curve, in this case, a circle, which divides
the plane into two regions  the ‘inside’ and ‘outside’ of the circle  with the circle itself as the
boundary between the two. Just like we used test values to determine whether or not an interval
belongs to the solution of the inequality, we use test points in the each of the regions to see which
of these belong to our solution set.3 We choose (0, 0) to represent the region inside the circle and
(0, 3) to represent the points outside of the circle. When we substitute (0, 0) into x2 + y 2 − 4 < 0,
we get −4 < 4 which is true. This means (0, 0) and all the other points inside the circle are part of
the solution. On the other hand, when we substitute (0, 3) into the same inequality, we get 5 < 0
which is false. This means (0, 3) along with all other points outside the circle are not part of the
solution. What about points on the circle itself? Choosing a point on the circle, say (0, 2), we get
0 < 0, which means the circle itself does not satisfy the inequality.4 As a result, we leave the circle
dashed in the ﬁnal diagram.
y
2 −2 2 x −2 The solution to x2 < 4 − y 2
We put this technique to good use in the following example.
Example 8.4.3. Sketch the solution to the following nonlinear inequalities in the plane. 1. y 2 − 4 ≤ x < y + 2
x2 + y 2 ≥ 4
2. x2 − 2x + y 2 − 2y ≤ 0
Solution.
1. The inequality y 2 − 4 ≤ x < y + 2 is a compound inequality. It translates as y 2 − 4 ≤ x
and x < y + 2. As usual, we solve each inequality and take the set theoretic intersection
to determine the region which satisﬁes both inequalities. To solve y 2 − 4 ≤ x, we write
y 2 − x − 4 ≤ 0. The curve y 2 − x − 4 = 0 describes a parabola since exactly one of the
variables is squared. Rewriting this in standard form, we get y 2 = x + 4 and we see that the
vertex is (−4, 0) and the parabola opens to the right. Using the test points (−5, 0) and (0, 0),
3
The theory behind why all this works is, surprisingly, the same theory which guarantees that sign diagrams work
the way they do  continuity and the Intermediate Value Theorem  but in this case, applied to functions of more
than one variable.
4
Another way to see this is that points on the circle satisfy x2 + y 2 − 4 = 0, so they do not satisfy x2 + y 2 − 4 < 0. 446 Systems of Equations
we ﬁnd that the solution to the inequality includes the region to the right of, or ‘inside’, the
parabola. The points on the parabola itself are also part of the solution, since the vertex
(−4, 0) satisﬁes the inequality. We now turn our attention to x < y + 2. Proceeding as before,
we write x − y − 2 < 0 and focus our attention on x − y − 2 = 0, which is the line y = x − 2.
Using the test points (0, 0) and (0, −4), we ﬁnd points in the region above the line y = x − 2
satisfy the inequality. The points on the line y = x − 2 do not satisfy the inequality, since
the y intercept (0, −2) does not. We see that these two regions do overlap, and to make the
graph more precise, we seek the intersection of these two curves. That is, we need to solve
the system of nonlinear equations (E 1) y 2 = x + 4 (E 2) y = x − 2
Solving E 1 for x, we get x = y 2 − 4. Substituting this into E 2 gives y = y 2 − 4 − 2, or
y 2 − y − 6 = 0. We ﬁnd y = −2 and y = 3 and since x = y 2 − 4, we get that the graphs
intersect at (0, −2) and (5, 3). Putting all of this together, we get our ﬁnal answer below.
y y y 3 x −5 4
− −3 y2 − 4 ≤ x 2 −3 x<y+2 3 4 5 x −5 4
− 2 3 4 5 x −3 y2 − 4 ≤ x < y + 2 2. To solve this system of inequalities, we need to ﬁnd all of the points (x, y ) which satisfy
both inequalities. To do this, we solve each inequality separately and take the set theoretic
intersection of the solution sets. We begin with the inequality x2 + y 2 ≥ 4 which we rewrite as
x2 + y 2 − 4 ≥ 0. The points which satisfy x2 + y 2 − 4 = 0 form our friendly circle x2 + y 2 = 4.
Using test points (0, 0) and (0, 3) we ﬁnd that our solution comprises the region outside the
circle. As far as the circle itself, the point (0, 2) satisﬁes the inequality, so the circle itself
is part of the solution set. Moving to the inequality x2 − 2x + y 2 − 2y ≤ 0, we start with
x2 − 2x + y 2 − 2y = 0. Completing the squares, we obtain (x − 1)2 + (y − 1)2 = 2, which is
√
a circle centered at (1, 1) with a radius of 2. Choosing (1, 1) to represent the inside of the
circle, (1, 3) as a point outside of the circle and (0, 0) as a point on the circle, we ﬁnd that
the solution to the inequality is the inside of the circle, including the circle itself. Our ﬁnal
answer, then, consists of the points on or outside of the circle x2 + y 2 = 4 which lie on or
inside the circle (x − 1)2 + (y − 1)2 = 2. To produce the most accurate graph, we need to ﬁnd
where these circles intersect. To that end, we solve the system 8.4 Systems of NonLinear Equations and Inequalities 447 (E 1)
x2 + y 2 = 4 (E 2) x2 − 2x + y 2 − 2y = 0
We can eliminate both the x2 and y 2 by replacing E 2 with −E 1 + E 2. Doing so produces
−2x − 2y = −4. Solving this for y , we get y = 2 − x. Substituting this into E 1 gives
x2 + (2 − x)2 = 4 which simpliﬁes to x2 + 4 − 4x + x2 = 4 or 2x2 − 4x = 0. Factoring yields
2x(x − 2) which gives x = 0 or x = 2. Substituting these values into y = 2 − x gives the
points (0, 2) and (2, 0). The intermediate graphs and ﬁnal solution are below.
y y y 3 3 2 2 1 1
−1 x −3 −2 −1 2 x −3 −2 −1 2
−1 −2 −2 −3 x2 + y 2 ≥ 4 −1 −3 x2 − 2x + y 2 − 2y ≤ 0 Solution to the system. x 448 Systems of Equations 8.4.1 Exercises 1. Solve the following systems of nonlinear equations.
the same set of axes to verify the solution set. x2 − y = 4 x2 + y 2
(a)
(c) x2 + y 2 = 4 16x2 + 4y 2 x2 + y 2 = 4 x2 + y 2
(b)
(d) x2 − y = 5 9x2 − 16y 2 Sketch the graph of both equations on = 16
= 64
= 16 = 144 (e) x2 + y 2 = 16 1 y 2 − 1 x2 =
9
16 x2 + y 2 = 16
(f) x−y = 2 1 2. Solve the following systems of nonlinear equations. Use a graph to help you avoid any potential
extraneous solutions. x2 − y 2 = 1 (x − 2)2 + y 2 = 1
y =
x3 + 8
(a)
(d)
(g) x2 + 4 y 2 = 4 y = 10x − x2
x2 + 4 y 2 = 4
√ x+1−y = 0 x2 + y 2 = 25 x2 + y 2 = 25 (b)
(e) y−x = 1
x2 + 4 y 2 = 4
(h)
4x2 − 9y = 0 3y 2 − 16x = 0 x + 2y 2 = 2 x2 + y 2 = 25
(c)
(f) x2 + 4 y 2 = 4 x2 + (y − 3)2 = 10
3. Consider the system of nonlinear equations below 4+3 = 1
x y
3 2
+ = −1
xy
If we let u = 1
x and v = 1
y then the system becomes 4u + 3 v =
1 3u + 2v = −1 This associated system of linear equations can then be solved using any of the techniques
1
presented earlier in the chapter to ﬁnd that u = −5 and v = 7. Thus x = u = − 1 and
5
1
1
y = v = 7.
We say that the original system is linear in form because its equations are not linear but a
few basic substitutions reveal a structure that we can treat like a system of linear equations.
Each system given below is linear in form. Make the appropriate substitutions to help you
solve for x and y . 8.4 Systems of NonLinear Equations and Inequalities 4x3 + 3 √y =
1
(a) 3x3 + 2√y = −1 449 4ex + 3e−y =
1
(b) 3ex + 2e−y = −1 4 ln(x) + 3y 2 =
1
(c) 3 ln(x) + 2y 2 = −1 4. Solve the following system x2 + √ y + log2 (z ) = 6 √
3x2 − 2 y + 2 log2 (z ) = 5 −4x2 + √y − 3 log (z ) = 11
2
5. Sketch the solution to each system of nonlinear inequalities in the plane. x2 − y 2 ≤ 1 y > 10x − x2
(a)
(d) x2 + 4 y 2 ≥ 4
y <
x3 + 8 x + 2y 2 > 2
x2 + y 2 < 25
(b)
(e) x2 + (y − 3)2 ≥ 10 x2 + 4 y 2 ≤ 4 (x − 2)2 + y 2 < 1 x2 + y 2 ≥ 25
(c)
(f) y−x ≤ 1
x2 + 4 y 2 < 4
6. Systems of nonlinear equations show up in third semester Calculus in the midst of some really
cool problems. The system below came from a problem in which we were asked to ﬁnd the
dimensions of a rectangular box with a volume of 1000 cubic inches that has minimal surface
area. The variables x, y and z are the dimensions of the box and λ is called a Lagrange
multiplier. With the help of your classmates, solve the system.5 2y + 2z = λyz 2x + 2z = λxz 2y + 2x = λxy xyz = 1000
7. According to Theorem 10.5 in Section 10.1, the polynomial p(x) = x4 + 4 can be factored into
the product linear and irreducible quadratic factors. In this exercise, we present a method
for obtaining that factorization.
(a) Show that p has no real zeros.
5 If using λ bothers you, change it to w when you solve the system. 450 Systems of Equations
(b) Because p has no real zeros, its factorization must be of the form (x2 + ax + b)(x2 + cx + d)
where each factor is an irreducible quadratic. Expand this quantity and gather like terms
together.
(c) Create and solve the system of nonlinear equations which results from equating the
coeﬃcients of the expansion found above with those of x4 + 4. You should get four
equations in the four unknowns a, b, c and d. Write p(x) in factored form. 8. Factor q (x) = x4 + 6x2 − 5x + 6. 8.4 Systems of NonLinear Equations and Inequalities 8.4.2 451 Answers √
1. (a) (±2, 0), ± 3, −1 (d) (±4, 0) y y
4
2 3
2 1 1
−2 −1 1 x 2 −6 −5 −4 −3 −2 −1
−1 −1 1 2 3 3 4 4 5 6 x x −2
−2 −3
−4 −3 √ √ −4 (e) ± 4 5 7 , ± 125 2 y (b) No solution 4
3 y 2 2 1 1 −4 −3 −2 −1
−1 1 2 −2
−2 −1 1 x 2 −3 −1 −4 −2 (f) 1 + −3 √
√
√
√
7, −1 + 7 , 1 − 7, −1 − 7
y −4 4
3
2
1 (c) (0, ±4) −4 −3 −2 −1
−1 y
4
3 2 3 4 x −3 2 1 −2
−4 1
−4 −3 −2 −1
−1 1 2 3 4 x −2
−3
−4 2. (a) ±2 √ √
10
15
5 ,± 5 (e) (3, 4), (−4, −3)
(f) (±3, 4) (b) (0, 1)
(c) (0, ±1), (2, 0) (g) (−4, −56), (1, 9), (2, 16) (d) (h) (3, 4) √
5
4
3, ± 9 452 Systems of Equations √
3. (a) − 3 5, 49 √
(c) e−5 , 7 (b) No solution 4. (1, 4, 8), (−1, 4, 8) y x2 − y 2 ≤ 1
5. (a) x2 + 4 y 2 ≥ 4 2
1 −2 −1 1 x 2 −1
−2 (b) x2 + y 2 < 25 y
4 x2 + (y − 3)2 ≥ 10 3
2
1
−5 −4 −3 −2 −1
−1 1 2 3 4 −2
−3
−4
−5 (x − 2)2 + y 2 < 1
(c) x2 + 4 y 2 < 4 y
1 1 −1 2 x 5 x 8.4 Systems of NonLinear Equations and Inequalities 453
y y > 10x − x2
(d)
y <
x3 + 8 16
9
−4 −3 −2 −1 1 2 x −56 y x + 2y 2 > 2
(e) x2 + 4 y 2 ≤ 4 1 1 2 x −1 y x2 + y 2 ≥ 25
(f) y−x ≤ 1 7
6
5
4
3
2
1
−6 5 4 3 2 −1
− − − − −1
−2
−3
−4
−5
−6 6. x = 10, y = 10, z = 10, λ = 2
5 7. (c) x4 + 4 = (x2 − 2x + 2)(x2 + 2x + 2)
8. x4 + 6x2 − 5x + 6 = (x2 − x + 1)(x2 + x + 6) 123456 x 454 Systems of Equations Chapter 9 Sequences and Series
9.1 Sequences When we ﬁrst introduced a function as a special type of relation in Section 2.1, we did not put
any restrictions on the domain of the function. All we said was that the set of xcoordinates of the
points in the function F is called the domain, and it turns out that any subset of the real numbers,
regardless of how weird that subset may be, can be the domain of a function. As our exploration
of functions continued beyond Section 2.1, we saw fewer and fewer functions with ‘weird’ domains.
It is worth your time to go back through the text to see that the domains of the polynomial,
rational, exponential, logarithmic and algebraic functions discussed thus far have fairly predictable
domains which almost always consist of just a collection of intervals on the real line. This may lead
some readers to believe that the only important functions in a College Algebra text have domains
which consist of intervals and everything else was just introductory nonsense. In this section, we
introduce sequences which are an important class of functions whose domains are the set of natural
numbers.1 Before we get to far ahead of ourselves, let’s look at what the term ‘sequence’ means
mathematically. Informally, we can think of a sequence as an inﬁnite list of numbers. For example,
consider the sequence 1 3 9 27
,− , ,− ,...
2 4 8 16 (1) As usual, the periods of ellipsis, . . ., indicate that the proposed pattern continues forever. Each
1
of the numbers in the list is called a term, and we call 2 the ‘ﬁrst term’, − 3 the ‘second term’,
4
9
the ‘third term’ and so forth. In numbering them this way, we are setting up a function, which
8
we’ll call a per tradition, between the natural numbers and the terms in the sequence.
1 Recall that this is the set {1, 2, 3, . . .}. 456 Sequences and Series n a(n) 1 1
2 2 −3
4 3 9
8 4 − 27
16 .
.
. .
.
. In other words, a(n) is the nth term in the sequence. We formalize these ideas in our deﬁnition
of a sequence and introduce some accompanying notation.
Definition 9.1. A sequence is a function a whose domain is the natural numbers. The value
a(n) is often written as an and is called the nth term of the sequence. The sequence itself is
usually denoted {an }∞ .
n=1 1
Applying the notation provided in Deﬁnition 9.1 to the sequence given (1), we have a1 = 2 ,
a2 = − 3 , a3 = 9 and so forth. Now suppose we wanted to know a117 , that is, the 117th term in
4
8
the sequence. While the pattern of the sequence is apparent, it would beneﬁt us greatly to have an
explicit formula for an . Unfortunately, there is no general algorithm that will produce a formula for
every sequence, so any formulas we do develop will come from that greatest of teachers, experience.
In other words, it is time for an example. Example 9.1.1. Write the ﬁrst four terms of the following sequences.
∞ 5n−1
1. an = n , n ≥ 1
3 4. (−1)k
2. bk =
,k≥0
2k + 1 5. a1 = 7, an+1 = 2 − an , n ≥ 1 3. {2n − 1}∞
n=1 6. f0 = 1, fn = n · fn−1 , n ≥ 1 1 + (−1)j
j j =2 Solution.
1. Since we are given n ≥ 1, the ﬁrst four terms of the sequence are a1 , a2 , a3 and a4 . Since
the notation a1 means the same thing as a(1), we obtain our ﬁrst term by replacing every
1−1
1
occurrence of n in the formula for an with n = 1 to get a1 = 5 31 = 3 . Proceeding similarly,
2−1
3−1
4−1
5
we get a2 = 5 32 = 9 , a3 = 5 33 = 25 and a4 = 5 34 = 125 .
27
81 9.1 Sequences 457 2. For this sequence we have k ≥ 0, so the ﬁrst four terms are b0 , b1 , b2 and b3 . Proceeding
as before, replacing in this case the variable k with the appropriate whole number, we get
(−1)1
(1)0
(−1)2
(−1)3
1
b0 = 2(0)+1 = 1, b1 = 2(1)+1 = − 1 , b2 = 2(2)+1 = 5 and b3 = 2(3)+1 = − 1 . (This sequence
3
7
is called an alternating sequence since the signs alternate between + and −. The reader is
encouraged to think what component of the formula is producing this eﬀect.)
3. From {2n − 1}∞ , we have that an = 2n − 1, n ≥ 1. We get a1 = 1, a2 = 3, a3 = 5 and
n=1
a4 = 7. (The ﬁrst four terms are the ﬁrst four odd natural numbers. The reader is encouraged
to examine whether or not this pattern continues indeﬁnitely.)
4. Proceeding as in the previous problem, we set aj =
1
a4 = 2 and a5 = 0. 1+(−1)j
,
j j ≥ 2. We ﬁnd a2 = 1, a3 = 0, 5. To obtain the terms of this sequence, we start with a1 = 7 and use the equation an+1 = 2 − an
for n ≥ 1 to generate successive terms. When n = 1, this equation becomes a1 + 1 = 2 − a1
which simpliﬁes to a2 = 2 − a1 = 2 − 7 = −5. When n = 2, the equation becomes a2 + 1 = 2 − a2
so we get a3 = 2 − a2 = 2 − (−5) = 7. Finally, when n = 3, we get a3 + 1 = 2 − a3 so
a4 = 2 − a3 = 2 − 7 = −5.
6. As with the problem above, we are given a place to start with f0 = 1 and given a formula
to build other terms of the sequence. Substituting n = 1 into the equation fn = n · fn−1 ,
we get f1 = 1 · f0 = 1 · 1 = 1. Advancing to n = 2, we get f2 = 2 · f1 = 2 · 1 = 2. Finally,
f3 = 3 · f2 = 3 · 2 = 6.
Some remarks about Example 9.1.1 are in order. We ﬁrst note that since sequences are functions,
we can graph them in the same way we graph functions. For example, if we wish to graph the
sequence {bk }∞ from Example 9.1.1, we graph the equation y = b(k ) for the values k ≥ 0. That
k=0
is, we plot the points (k, b(k )) for the values of k in the domain, k = 0, 1, 2, . . .. The resulting
collection of points is the graph of the sequence. Note that we do not connect the dots in a pleasing
fashion as we are used to doing, because the domain is just the whole numbers in this case, not a
collection of intervals of real numbers. If you feel a sense of nostalgia, you should see Section 1.2.
y
3
2 1
1
2 −1
2 1 2 3 x −1
−3
2 Graphing y = bk = (−1)k
,k≥0
2k + 1 Speaking of {bk }∞ , the astute and mathematically minded reader will correctly note that
k=0
this technically isn’t a sequence, since according to Deﬁnition 9.1, sequences are functions whose
domains are the natural numbers, not the whole numbers, as is the case with {bk }∞ . In other
k=0 458 Sequences and Series words, to satisfy Deﬁnition 9.1, we need to shift the variable k so it starts at k = 1 instead of
k = 0. To see how we can do this, it helps to think of the problem graphically. What we want
is to shift the graph of y = b(k ) to the right one unit, and thinking back to Section 2.5, we can
accomplish this by replacing k with k − 1 in the deﬁnition of {bk }∞ . Speciﬁcally, let ck = bk−1
k=0
k−1 k−1 ( 1)
where k − 1 ≥ 0. We get ck = 2(−−1)+1 = (−1) 1 , where now k ≥ 1. We leave to the reader to
2k−
k
verify that {ck }∞ generates the same list of numbers as does {bk }∞ , but the former satisﬁes
k=0
k=1
Deﬁnition 9.1, while the latter does not. Like so many things in this text, we acknowledge that
this point is pedantic and join the vast majority of authors who adopt a more relaxed view of
Deﬁnition 9.1 to include any function which generates a list of numbers which can be then be
matched up with the natural numbers.2 Finally, we wish to note the sequences in parts 5 and 6 are
examples of sequences described recursively. In each instance, an initial value of the sequence is
given which is then followed by a recursion equation − a formula which enables us to use known
terms of the sequence to determine other terms. The terms of the sequence in part 6 are given a
special name: fn is called nfactorial. We will study factorials in greater detail in Section ??. The
world famous Fibonacci Numbers are deﬁned recursively and are explored in the exercises. While
none of the sequences worked out to be the sequence in (1), they do give us some insight into what
kinds of patterns to look for. Two patterns in particular are given in the next deﬁnition. Definition 9.2. Arithmetic and Geometric Sequences: Suppose {an }∞ k is a sequencea
n=
If there is a number d so that an+1 = an + d for all n ≥ k , then {an }∞ k is called an
n=
arithmetic sequence. The number d is called the common diﬀerence.
If there is a number r so that an+1 = ran for all n ≥ k , then {an }∞ k is called an
n=
geometric sequence. The number r is called the common ratio. a Note that we have adjusted for the fact that not all ‘sequences’ begin at n = 1. Both arithmetic and geometric sequences are deﬁned in terms of recursion equations. In English,
an arithmetic sequence is one in which we proceed from one term to the next by always adding
the ﬁxed number d. The name ‘common diﬀerence’ comes from a slight rewrite of the recursion
equation from an+1 = an + d to an+1 − an = d. Analogously, a geometric sequence is one in which
we proceed from one term to the next by always multiplying by the same ﬁxed number r. If r = 0,
we can rearrange the recursion equation to get an+1 = r, hence the name ‘common ratio.’ Some
an
sequences are arithmetic, some are geometric and some are neither as the next example illustrates.3
Example 9.1.2. Determine if the following sequences are arithmetic, geometric or neither. If
arithmetic, ﬁnd the common diﬀerence d; if geometric, ﬁnd the common ratio r.
2
Math fans will delight to know we are basically talking about the ‘countably inﬁnite’ subsets of the real number
line when we do this.
3
Sequences which are both arithmetic and geometric are discussed in the Exercises. 9.1 Sequences 459 1. an = 5n−1
,n≥1
3n 3. {2n − 1}∞
n=1 2. bk = (−1)k
,k≥0
2k + 1 4. 1 3 9 27
,− , ,− ,...
2 4 8 16 Solution. A good rule of thumb to keep in mind when working with sequences is “When in doubt,
write it out!” Writing out the ﬁrst several terms can help you identify the pattern of the sequence
should one exist.
25
1. From Example 9.1.1, we know that the ﬁrst four terms of this sequence are 1 , 5 , 27 and 125 .
39
81
To see if this is an arithmetic sequence, we look at the successive diﬀerences of terms. We
5
25
5
ﬁnd that a2 − a1 = 9 − 1 = 2 and a3 − a2 = 27 − 9 = 10 . Since we get diﬀerent numbers,
3
9
27
there is no ‘common diﬀerence’ and we have established that the sequence is not arithmetic.
To investigate whether or not it is geometric, we compute the ratios of successive terms. The
ﬁrst three ratios
5
25
125
a2
5 a3
5
a4
5
and
=9= ,
= 27 =
= 81 =
1
5
25
a1
3 a2
3
a3
3
3
9
27 suggest that the sequence is geometric. To prove it, we must show that an+1
an an+1
an = r for all n. 5(n+1)−1
n+1
5n
3n
5
= 3n−1 = n+1 · n−1 =
3
5
3
5
n
3 This sequence is geometric with common ratio r = 5 .
3
2. Again, we have Example 9.1.1 to thank for providing the ﬁrst four terms of this sequence:
1
8
1, − 1 , 1 and − 7 . We ﬁnd b1 − b0 = − 4 and b2 − b1 = 15 . Hence, the sequence is not
35
3
1
arithmetic. To see if it is geometric, we compute b1 = − 3 and b2 = − 3 . Since there is no
b0
b1
5
‘common ratio,’ we conclude the sequence is not geometric, either.
3. As we saw in Example 9.1.1, the sequence {2n − 1}∞ generates the odd numbers: 1, 3, 5, 7, . . ..
n=1
Computing the ﬁrst few diﬀerences, we ﬁnd a2 − a1 = 2, a3 − a2 = 2, and a4 − a3 = 2. This
suggests that the sequence is arithmetic. To verify this, we ﬁnd
an+1 − an = (2(n + 1) − 1) − (2n − 1) = 2n + 2 − 1 − 2n + 1 = 2
This establishes that the sequence is arithmetic with common diﬀerence d = 2. To see if it is
5
geometric, we compute a2 = 3 and a3 = 3 . Since these ratios are diﬀerent, we conclude the
a1
a2
sequence is not geometric.
4. We met our last sequence at the beginning of the section. Given that a2 − a1 = − 5 and
4
a3 − a2 = 15 , the sequence is not arithmetic. Computing the ﬁrst few ratios, however, gives us
8
a2
a4
3 a3
3
3
a1 = − 2 , a2 = − 2 and a3 = − 2 . Since these are the only terms given to us, we assume that
the pattern of ratios continue in this fashion and conclude that the sequence is geometric. 460 Sequences and Series We are now one step away from determining an explicit formula for the sequence given in (1).
We know that it is a geometric sequence and our next result gives us the explicit formula we require.
Equation 9.1. Formulas for Arithmetic and Geometric Sequences:
An arithmetic sequence with ﬁrst term a and common diﬀerence d is given by an = a + (n − 1)d, n≥1 A geometric sequence with ﬁrst term a and common ratio r = 0 is given by an = arn−1 , n≥1 While the formal proofs of the formulas in Equation 9.1 require the techniques set forth in
Section ??, we attempt to motivate them here. According to Deﬁnition 9.2, given an arithmetic
sequence with ﬁrst term a and common diﬀerence d, the way we get from one term to the next is by
adding d. Hence, the terms of the sequence are: a, a + d, a + 2d, a + 3d, . . . . We see that to reach
the nth term, we add d to a exactly (n − 1) times, which is what the formula says. The derivation of
the formula for geometric series follows similarly. Here, we start with a and go from one term to the
next by multiplying by r. We get a, ar, ar2 , ar3 and so forth. The nth term results from multiplying
a by r exactly (n − 1) times. We note here that the reason r = 0 is excluded from Equation 9.1 is
to avoid an instance of 00 which is an indeterminant form.4 With Equation 9.1 in place, we ﬁnally
have the tools required to ﬁnd an explicit formula for the nth term of the sequence given in (1).
3
We know from Example 9.1.2 that it is geometric with common ratio r = − 2 . The ﬁrst term is
n−1
3
a = 1 so by Equation 9.1 we get an = arn−1 = 1 − 2
for n ≥ 1. After a touch of simplifying,
2
2
n−1 we get an = (−3)n
for n ≥ 1. Note that we can easily check our answer by substituting in values
2
of n and seeing that the formula generates the sequence given in (1). We leave this to the reader.
Our next example gives us more practice ﬁnding patterns.
Example 9.1.3. Find an explicit formula for the nth term of the following sequences.
1. 0.9, 0.09, 0.009, 0.0009, . . .
2. 2
22
, 2, − , − , . . .
5
37 24
8
3. 1, − , , − , . . .
7 13 19
Solution.
4 See the footnotes on page 217 in Section 4.1 and page 333 of Section 7.1. 9.1 Sequences 461 1. Although this sequence may seem strange, the reader can verify it is actually a geometric
1
9
9
1 n−1
for
sequence with common ratio r = 0.1 = 10 . With a = 0.9 = 10 , we get an = 10 10
9
n ≥ 0. Simplifying, we get an = 10n , n ≥ 1. There is more to this sequence than meets the
eye and we shall return to this example in the next section.
2. As the reader can verify, this sequence is neither arithmetic nor geometric. In an attempt
to ﬁnd a pattern, we rewrite the second term with a denominator to make all the terms
appear as fractions. We have 2 , 2 , − 2 , − 2 , . . .. If we associate the negative ‘−’ of the last two
51
3
7
2
2
2
terms with the denominators we get 5 , 2 , −3 , −7 , . . .. This tells us that we can tentatively
1
sketch out the formula for the sequence as an = d2 where dn is the sequence of denominators.
n
Looking at the denominators 5, 1, −3, −7, . . ., we ﬁnd that they go from one term to the next
by subtracting 4 which is the same as adding −4. This means we have an arithmetic sequence
on our hands. Using Equation 9.1 with a = 5 and d = −4, we get the nth denominator by
2
the formula dn = 5 + (n − 1)(−4) = 9 − 4n for n ≥ 1. Our ﬁnal answer is an = 9−4n , n ≥ 1.
3. The sequence as given is neither arithmetic nor geometric, so we proceed as in the last problem
to try to get patterns individually for the numerator and denominator. Letting cn and dn
c
denote the sequence of numerators and denominators, respectively, we have an = dn . After
n
some experimentation,5 we choose to write the ﬁrst term as a fraction and associate the
4
negatives ‘−’ with the numerators. This yields 1 , −2 , 13 , −8 , . . .. The numerators form the
17
19
sequence 1, −2, 4, −8, . . . which is geometric with a = 1 and r = −2, so we get cn = (−2)n−1 ,
for n ≥ 1. The denominators 1, 7, 13, 19, . . . form an arithmetic sequence with a = 1 and
d = 6. Hence, we get dn = 1 + 6(n − 1) = 6n − 5, for n ≥ 1. We obtain our formula for
n−1
c
an = dn = (−2)−5 , for n ≥ 1. We leave it to the reader to show that this checks out.
6n
n
While the last problem in Example 9.1.3 was neither geometric nor arithmetic, it did resolve
into a combination of these two kinds of sequences. If handed the sequence 2, 5, 10, 17, . . ., we would
be hardpressed to ﬁnd a formula for an if we restrict our attention to these two archetypes. We
said before that there is no general algorithm for ﬁnding the explicit formula for the nth term of
a given sequence, and it is only through experience gained from evaluating sequences from explicit
formulas that we learn to begin to recognize number patterns. The pattern 1, 4, 9, 16, . . . is rather
recognizable as the squares, so the formula an = n2 , n ≥ 1 may not be too hard to determine.
With this in mind, it’s possible to see 2, 5, 10, 17, . . . as the sequence 1 + 1, 4 + 1, 9 + 1, 16 + 1, . . .,
so that an = n2 + 1, n ≥ 1. Of course, since we are given only a small sample of the sequence, we
shouldn’t be too disappointed to ﬁnd out this isn’t the only formula which generates this sequence.
1
5
For example, consider the sequence deﬁned by bn = − 4 n4 + 2 n3 − 31 n2 + 25 n − 5, n ≥ 1. The
4
2
reader is encouraged to verify that it also produces the terms 2, 5, 10, 17. In fact, it can be shown
that given any ﬁnite sample of a sequence, there are inﬁnitely many explicit formulas all of which
generate those same ﬁnite points. This means that there will be inﬁnitely many correct answers to
some of the exercises in this section.6 Just because your answer doesn’t match ours doesn’t mean
5 Here we take ‘experimentation’ to mean a frustrating guessandcheck session.
For more on this, see When Every Answer is Correct: Why Sequences and Number Patterns Fail the Test by
Professor Don White of Kent State University.
6 462 Sequences and Series it’s wrong. As always, when in doubt, write your answer out. As long as it produces the same
terms in the same order as what the problem wants, your answer is correct.
We would be remiss to close this section without mention of the utility of sequences in everyday
life. Indeed, sequences play a major role in the Mathematics of Finance, as we have already seen
with Equation 7.2 in Section 7.5. Recall that if we invest P dollars at an annual percentage rate r
and compound the interest n times per year, the formula for Ak , the amount in the account after k
rk
r
r k −1
compounding periods, is Ak = P 1 + n = P 1 + n
1+ n
, k ≥ 1. We now spot this as a
r
r
geometric sequence with ﬁrst term P 1 + n and common ratio 1 + n . In retirement planning,
it is seldom the case that an investor deposits a set amount of money into an account and waits for
it to grow. Usually, additional payments of principal are made at regular intervals and the value of
the investment grows accordingly. This kind of investment is called an Investment Retirement
Account and will be discussed in the next section once we have developed more mathematical
machinery. 9.1.1 Exercises 1. Write out the four terms of the following sequences.
(a) an = 2n − 1, n ≥ 0
j (j +1)
2 (b) dj = (−1)
(c) {5k − 2}∞
k=1
(d)
(e)
(f) (g) a1 = 3, an+1 = an − 1, n ≥ 1
(h) d0 = 12, dm = dm1 , m ≥ 1 ,j≥1 100 (i) b1 = 2, bk+1 = 3bk + 1, k ≥ 1
cj (j) c0 = −2, cj = (j +1)(1+2) , m ≥ 1
j ∞
n2 +1
n+1 n=0
xn ∞
n2 n=1
∞
ln(n)
n
n=1 1
an , n ≥ 1
xn+1 + sn , n (k) a1 = 117, an+1 =
(l) s0 = 1, sn+1 = ≥0 (m) F0 = 1, F1 = 1, Fn = Fn1 + Fn2 , n ≥ 2 (This is the famous Fibonacci Sequence )
2. Find an explicit formula for the nth term of the following sequences. Use the formulas in
Equation 9.1 as needed.
1
(d) 1, 2 , 3 ,
3 (a) 3, 5, 7, 9, . . .
(b) 1,
(c) 1, 1
1
−2, 1, −8,
4
248
3, 5, 7, . . . ... (e) 1,
(f) x, 11
4, 9,
3
−x ,
3 4
27 ,
1
16 ,
x5
5, ... (g) 0.9, 0.99, 0.999, 0.9999, . . . ... (h) 27, 64, 125, 216, . . . 7
−x ,
7 ... (i) 1, 0, 1, 0, . . . 3. Find a sequence which is both arithmetic and geometric. (Hint: Start with an = c for all n.)
4. Show that a geometric sequence can be transformed into an arithmetic sequence by taking
the natural logarithm of the terms. 9.2 Series and Summation Notation 463 5. Thomas Robert Malthus is credited with saying, “The power of population is indeﬁnitely
greater than the power in the earth to produce subsistence for man. Population, when
unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical
ratio. A slight acquaintance with numbers will show the immensity of the ﬁrst power in
comparison with the second.” (See this webpage for more information.) Discuss this quote
with your classmates from a sequences point of view.
6. This classic problem involving sequences shows the power of geometric sequences. Suppose
that a wealthy benefactor agrees to give you one penny today and then double the amount
she gives you each day for 30 days. So, for example, you get two pennies on the second day
and four pennies on the third day. How many pennies do you get on the 30th day? What is
the total dollar value of the gift you have received?
7. Research the terms ‘arithmetic mean’ and ‘geometric mean.’ With the help of your classmates,
show that a given term of a arithmetic sequence ak , k ≥ 2 is the arithmetic mean of the
term immediately preceding, ak−1 it and immediately following it, ak+1 . State and prove an
analogous result for geometric sequences.
8. Discuss with your classmates how the results of this section might change if we were to
examine sequences of other mathematical things like complex numbers or matrices. Find an
explicit formula for the nth term of the sequence i, −1, −i, 1, i, . . .. List out the ﬁrst four terms
of the matrix sequences we discussed in Exercise ?? in Section ??. 9.1.2 Answers 1. (a)
(b)
(c)
(d) 0, 1, 3, 7
−1, −1, 1, 1
3, 8, 13, 18
5
1, 1, 3 , 5
2
2 3 (h) 12, 0.12, 0.0012, 0.000012
(i) 2, 7, 22, 67
1
1
(j) −2, − 1 , − 36 , − 720
3
1
1
(k) 117, 117 , 117, 117 4 (e) x, x , x , x
234 (l) 1, x + 1, x2 + x + 1, x3 + x2 + x + 1 (f) 0, ln(2), ln(3) , ln(4)
3
4
(g) 3, 2, 1, 0
2. (a) an = 1 + 2n, n ≥ 1
(b) an =
(c) an = 9.2 n−1
−1
,
2
n−1
2
2n−1 , n ≥ (m) 1, 1, 2, 3
n
, n≥1
3n−1
1
an = n2 , n ≥ 1
(−1)n−1 x2n−1
, n≥
2n−1 10n −1
10n , n≥1 (d) an = (g) an = n≥1 (e) (h) an = (n + 2)3 , n ≥ 1 1 (f) 1 (i) an = 1+(−1)n−1
,
2 n≥1 Series and Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems
concerning the sum of terms of a sequence. We begin with a deﬁnition, which, while intimidating,
is meant to make our lives easier. 464 Sequences and Series Definition 9.3. Summation Notation: Given a sequence {an }∞ k and numbers m and p
n=
satisfying k ≤ m ≤ p, the summation from m to p of the sequence {an } is written
p an = am + am+1 + . . . + ap
n=m The variable n is called the index of summation. The number m is called the lower limit
of summation while the number p is called the upper limit of summation. In English, Deﬁnition 9.3 is simply deﬁning a shorthand notation for adding up the terms of
the sequence {an }∞ k from am through ap . The symbol Σ is the capital Greek letter sigma and is
n=
shorthand for ‘sum’. The index of summation tells us which term to start with and which term
to end with. For example, using the sequence an = 2n − 1 for n ≥ 1, we can write the sum
a3 + a4 + a5 + a6 as
6 (2n − 1) = (2(3) − 1) + (2(4) − 1) + (2(5) − 1) + (2(6) − 1)
n=3 = 5 + 7 + 9 + 11
= 32
The index variable is considered a ‘dummy variable’ in the sense that it may be changed to any
letter without aﬀecting the value of the summation. For instance,
6 6 (2n − 1) =
n=3 6 (2k − 1) = (2j − 1)
j =3 k=3 One place you may encounter summation notation is in mathematical deﬁnitions. For example,
summation notation allows us to deﬁne polynomials as functions of the form
n ak xk f (x) =
k=0 for real numbers ak , k = 0, 1, . . . n. The reader is invited to compare this with what is given in
Deﬁnition 4.1. Summation notation is particularly useful when talking about matrix operations.
For example, we can write the product of the ith row Ri of a matrix A = [aij ]m×n and the j th
column Cj of a matrix B = [bij ]n×r as
n Ri · Cj = aik bkj
k=1 Again, the reader is encouraged to write out the sum and compare it to Deﬁnition ??. Our next
example gives us practice with this new notation. 9.2 Series and Summation Notation 465 Example 9.2.1.
1. Find the following sums.
4 (a)
k=1 5 13
100k (b)
n=1 (−1)n+1
(x − 1)n
n 2. Write the following sums using summation notation.
(a) 1 + 3 + 5 + . . . + 117
111
1
(b) 1 − + − + − . . . +
234
117
(c) 0.9 + 0.09 + 0.009 + . . . 0. 0 · · · 0 9
n − 1 zeros Solution.
1. (a) We substitute k = 1 into the formula
4
k=1 13
100k = 13
100k and add successive terms until we reach k = 4. 13
13
13
13
+
+
+
1
2
3
100
100
100
1004 = 0.13 + 0.0013 + 0.000013 + 0.00000013
= 0.13131313
(b) We proceed as in part (a), replacing the index n, but not the variable x, with the values
1 through 5 and adding the resulting terms.
5
n=1 (−1)n+1
(x − 1)n =
n (−1)1+1
(−1)2+1
(−1)3+1
(x − 1)1 +
(x − 1)2 +
(x − 1)3
1
2
3
(−1)1+4
(−1)1+5
+
(x − 1)4 +
(x − 1)5
4
5 = (x − 1) − (x − 1)2 (x − 1)3 (x − 1)4 (x − 1)5
+
−
+
2
3
4
5 2. The key to writing these sums with summation notation is to ﬁnd the pattern of the terms.
To that end, we make good use of the techniques presented in Section 9.1.
(a) The terms of the sum 1, 3, 5, etc., form an arithmetic sequence with ﬁrst term a = 1
and common diﬀerence d = 2. We get a formula for the nth term of the sequence using
Equation 9.1 to get an = 1 + (n − 1)2 = 2n − 1, n ≥ 1. At this stage, we have the formula 466 Sequences and Series
for the terms, namely 2n − 1, and the lower limit of the summation, n = 1. To ﬁnish
the problem, we need to determine the upper limit of the summation. In other words,
we need to determine which value of n produces the term 117. Setting an = 117, we get
2n − 1 = 117 or n = 59. Our ﬁnal answer is
59 (2n − 1) 1 + 3 + 5 + . . . + 117 = n=1 (b) We rewrite all of the terms as fractions, the subtraction as addition, and associate the
negatives ‘−’ with the numerators to get
1 −1 1 −1
1
+
++
+ ... +
1
2
3
4
117
The numerators, 1, −1, etc. can be described by the geometric sequence1 cn = (−1)n−1
for n ≥ 1, while the denominators are given by the arithmetic sequence2 dn = n for
n−1
n ≥ 1. Hence, we get the formula an = (−1)
for our terms, and we ﬁnd the lower and
n
upper limits of summation to be n = 1 and n = 117, respectively. Thus 1− 111
1
+ − + −... +
234
117 117 =
n=1 (−1)n−1
n 9
(c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = 10n for
n ≥ 1. This gives us a formula for the summation as well as a lower limit of summation.
To determine the upper limit of summation, we note that to produce the n − 1 zeros to
the right of the decimal point before the 9, we need a denominator of 10n . Hence, n is
the upper limit of summation. Since n is used in the limits of the summation, we need
to choose a diﬀerent letter for the index of summation.3 We choose k and get
n 0.9 + 0.09 + 0.009 + . . . 0. 0 · · · 0 9 =
n − 1 zeros k=1 9
10k The following theorem presents some general properties of summation notation. While we shall
not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover,
there is much to be learned by thinking about why the properties hold. We invite the reader to
prove these results. To get started, remember, “When in doubt, write it out!”
1 This is indeed a geometric sequence with ﬁrst term a = 1 and common ratio r = −1.
It is an arithmetic sequence with ﬁrst term a = 1 and common diﬀerence d = 1.
3
To see why, try writing the summation using ‘n’ as the index. 2 9.2 Series and Summation Notation 467 Theorem 9.1. Properties of Summation Notation: Suppose {an } and {bn } are sequences
so that the following sums are deﬁned.
p p p an ± (an ± bn ) =
p p an , for any real number c. c an = c
n=m n=m n=m n=m
p n=j +1 p+r an =
n=m an , for any natural number m ≤ j < j + 1 ≤ p. an + an = p j p n=m n=m n=m bn an−r , for any whole number r.
n=m+r We now turn our attention to the sums involving arithmetic and geometric sequences. Given
an arithmetic sequence ak = a + (k − 1)d for k ≥ 1, we let S denote the sum of the ﬁrst n terms.
To derive a formula for S , we write it out in two diﬀerent ways S= a + ( a + d) + . . . + (a + (n − 2)d) + (a + (n − 1)d) S = (a + (n − 1)d) + (a + (n − 2)d) + . . . + (a + d) + a If we add these two equations and combine the terms which are aligned vertically, we get
2S = (2a + (n − 1)d) + (2a + (n − 1)d) + . . . + (2a + (n − 1)d) + (2a + (n − 1)d)
The right hand side of this equation contains n terms, all of which are equal to (2a + (n − 1)d)
so we get 2S = n(2a + (n − 1)d). Dividing both sides of this equation by 2, we obtain the formula
n
(2a + (n − 1)d)
2
If we rewrite the quantity 2a + (n − 1)d as a + (a + (n − 1)d) = a1 + an , we get the formula
S= S=n a1 + an
2 A helpful way to remember this last formula is to recognize that we have expressed the sum as
the product of the number of terms n and the average of the ﬁrst and nth terms.
To derive the formula for the geometric sum, we start with a geometric sequence ak = ark−1 ,
k ≥ 1, and let S once again denote the sum of the ﬁrst n terms. Comparing S and rS , we get 468 Sequences and Series = a + ar + ar2 + . . . + arn−2 + arn−1 S ar + ar2 + . . . + arn−2 + arn−1 + arn rS = Subtracting the second equation from the ﬁrst forces all of the terms except a and arn to cancel
out and we get S − rS = a − arn . Factoring, we get S (1 − r) = a (1 − rn ). Assuming r = 1, we can
divide both sides by the quantity (1 − r) to obtain
S=a 1 − rn
1−r If we distribute a through the numerator, we get a − arn = a1 − an+1 which yields the formula
a1 − an+1
1−r S=
In the case when r = 1, we get the formula S = a + a + ... + a = na
n times Our results are summarized below.
Equation 9.2. Sums of Arithmetic and Geometric Sequences:
The sum S of the ﬁrst n terms of an arithmetic sequence ak = a + (k − 1)d for k ≥ 1 is
n S= ak = n
k=1 a1 + an
2 = n
(2a + (n − 1)d)
2 The sum S of the ﬁrst n terms of a geometric sequence ak = ark−1 for k ≥ 1 is
n 1. S = ak =
k=1
n 2. S = a1 − an+1
=a
1−r
n ak =
k=1 1 − rn
, if r = 1.
1−r a = na, if r = 1.
k=1 While we have made an honest eﬀort to derive the formulas in Equation 9.2, formal proofs
require the machinery in Section ??. An application of the arithmetic sum formula which proves
useful in Calculus results in formula for the sum of the ﬁrst n natural numbers. The natural 9.2 Series and Summation Notation 469 numbers themselves are a sequence4 1, 2, 3, . . . which is arithmetic with a = d = 1. Applying
Equation 9.2,
n(n + 1)
1 + 2 + 3 + ... + n =
2
So, for example, the sum of the ﬁrst 100 natural numbers is 9.2.1 100(101)
2 = 5050.5 Exercises 1. Find the following sums.
5 9 (a) (5g + 3) 4 2j (c) g =4 j =0 8 2 (b)
k=3 1
k i=1
100 (3k − 5)xk (d) 12
(i + 1)
4 (e) (−1)n (f) n=1 k=0 2. Rewrite the sum using summation notation.
(a) 8 + 11 + 14 + 17 + 20
(b) 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8
x3 x5 x7
+
−
(c) x −
3
5
7
(d) 1 + 2 + 4 + · · · + 229 (e) 2 + 3
2 + 4
3 + 5
4 + 6
5 (f) − ln(3) + ln(4) − ln(5) + · · · + ln(20)
1
1
25 − 36
1
1
21
31
4
2 (x − 5)+ 4 (x 5) + 6 (x − 5) + 8 (x − 5) (g) 1 −
(h) 1
4 + 1
9 − 1
16 + 3. Find the sum of the ﬁrst 10 terms of the following sequences.
(a) an = 3 + 5n (b) bn = 1n
2 (c) cn = −2n + 5n
3 4. Prove the properties listed in Theorem 9.1.
5. Discuss with your classmates the problems which arise in trying to ﬁnd the sum of the
following geometric series. When in doubt, write them out!
∞ ∞ 2k−1 (a)
k=1 9.2.2 ∞ (1.0001)k−1 (b)
k=1 (−1)k−1 (c)
k=1 Answers 1.
4 This is the identity function on the natural numbers!
There is an interesting anecdote which says that the famous mathematician Carl Friedrich Gauss was given this
problem in primary school and devised a very clever solution.
5 470 Sequences and Series (b) (e) (d) −5 − 2x + x2 341
280 (f) 0
5 5 2. (a) (e) (3k + 5) k=1 k=1
k −1 (−1) k=3 4
k −1 (−1)
k=1 6 x
2k − 1 (g)
k=1 29 4 2k−1 (d) (h) k=1 3. (a) 305 k=1 (b) 7
9
13
(b)
99 (−1)k−1
k2
1
(x − 5)k
2k 1023
1024 (c) 17771050
59049 3383
333
5809
(d) −
990 4. (a) 9.3 (−1)k ln(k ) (f) k k=1 (c) k+1
k 20 8 (b) 17
2 (c) 63 (a) 213 (c) IRAs and Mortgages An important application of the geometric sum formula is the investment plan called an Investment Retirement Account (IRA). These diﬀer from the kind of investments we studied in
Section 7.5 in that payments are deposited into the account on an ongoing basis, and this complicates the mathematics a little.6 Suppose you have an account with annual interest rate r which
r
is compounded n times per year. We let i = n denote the interest rate per period. You will
make ongoing deposits of P dollars at the end of each compounding period. Let Ak denote the
amount in the account after k compounding periods. Then A1 = P , because we have made our
ﬁrst deposit at the end of the ﬁrst compounding period and no interest has been earned. During
the second compounding period, we earn interest on A1 so that our initial investment has grown
to A1 (1 + i) = P (1 + i) in accordance with Equation 7.1. When we add our second payment at the
end of the second period, we get
A2 = A1 (1 + i) + P = P (1 + i) + P = P (1 + i) 1 + 1
1+i The reason for factoring out the P (1 + i) will become apparent in short order. During the third
compounding period, we earn interest on A2 which then grows to A2 (1 + i). We add our third
6 The reader may wish to reread the discussion on compound interest in Section 7.5 before proceeding. 9.3 IRAs and Mortgages 471 payment at the end of the third compounding period to obtain
A3 = A2 (1 + i) + P = P (1 + i) 1 + 1
1+i (1 + i) + P = P (1 + i)2 1 + 1
1
+
1 + i (1 + i)2 During the fourth compounding period, A3 grows to A3 (1+ i), and when we add the fourth payment,
we factor out P (1 + i)3 to get
A4 = P (1 + i)3 1 + 1
1
1
+
+
2
1 + i (1 + i)
(1 + i)3 This pattern continues so that at the end of the k th compounding, we get
Ak = P (1 + i)k−1 1 + 1
1
1
+
+ ... +
2
1 + i (1 + i)
(1 + i)k−1 The sum in the parentheses above is the sum of the ﬁrst k terms of a geometric sequence with
1
a = 1 and r = 1+i . Using Equation 9.2, we get 1
1−
−k 1
1
1
(1 + i)k = (1 + i) 1 − (1 + i)
1+
+
+ ... +
= 1 1
1 + i (1 + i)2
i
(1 + i)k−1
1−
1+i
Hence, we get
Ak = P (1 + i)k−1 (1 + i) 1 − (1 + i)−k
i = P (1 + i)k − 1
i If we let t be the number of years this investment strategy is followed, then k = nt, and we get the
formula for the value after t years.
Equation 9.3. Future Value of an IRA: Suppose an IRA oﬀers an annual interest rate r
r
compounded n times per year. Let i = n be the interest rate per compounding period. If a
deposit P is made at the end of each compounding period, the amount A in the account after
t years is given by
P (1 + i)nt − 1
A=
i r
The reader is encouraged to substitute i = n into Equation 9.3 and simplify. Some familiar
equations arise which are cause for pause and meditation. Example 9.3.1. An IRA oﬀers a 6% annual interest rate, compounded monthly.
1. If monthly payments of $50 are made, ﬁnd the value of the IRA in 30 years. 472 Sequences and Series 2. How many years will it take for the IRA to grow to $100,000?
Solution.
1. We have r = 0.06 and n = 12 so that i =
A= r
n = 0.06
12 = 0.005. With P = 50 and t = 30, 50 (1 + 0.005)(12)(30) − 1
≈ 50225.75
0.005 Our ﬁnal answer is $50,225.75.
2. To ﬁnd how long it will take for the IRA to grow to $100,000, we set A = 100000 and solve
for t. We isolate the exponential and take the natural logarithm of both sides of the equation.
100000 = 50 (1 + 0.005)12t − 1
0.005 10 = (1.005)12t − 1
(1.005)12t = 11
ln (1.005)12t = ln(11) 12t ln(1.005) = ln(11)
t= ln(11)
≈ 40.06
12 ln(1.005) This means that it takes just over 40 years for the investment to grow to $100,000. Comparing
this with our answer to part 1, we see that in just 10 additional years, the value of the IRA
nearly doubles. This is a lesson worth remembering.
A mortgage is like a backwards IRA. Here’s an example.
Example 9.3.2. A family takes out a 30 year $100,000 mortgage to buy a house. The interest
rate is 6% per year and they will repay the loan in 360 equal monthly payments. To compute the
monthly payment imagine that the family has taken 360 loans
100000 = L1 + L2 + · · · + L360
and they will repay the k th loan at the end of the k th month with the monthly payment of a dollars.
The monthly interest rate is 6%/12 = 0.005 so the amount owed on the k th loan after k months is
(1.005)k Lk . This is the amount of the k th payment so a = (1.005)k Lk so Lk = a(1.005)−k so
100000 = L1 + L2 + · · · + L360 = a(1.005)−1 + a(1.005)−2 + · · · + a(1.005)−360 . 9.4 Infinite sums and Repeating Decimals* 473 Using Sigma notation and Theorem ?? with r = (1.005)−1 this may be written
360 a(1.005)−k = a 100000 =
k=1 (1.005)−361 − (1.005)−1
(1.005)−1 − 1 To evaluate the expression on the right multiply top and bottom by 1.005 to get
(1.005)−361 − (1.005)−1
(1.005)−360 − 1
1 − (1.005)−360
=
=
= 166.79
(1.005)−1 − 1
1 − 1.005
0.005
so 100000 = 166.79a so the monthly payment is
a = 100000/166.79 = 599.56. 9.3.1 Exercises 1. If monthly payments of $300 are made to an IRA with an APR of 2.5% compounded monthly
what is the value of the IRA after 17 years?
2. Show that the formula for the future value of an IRA where the periodic payment is made at
the beginning of the compounding period (rather than the end) is
A = P (1 + i) 9.3.2 (1 + i)nt − 1
i Answers 1. $76,163.67 9.4 Infinite sums and Repeating Decimals* Definition 9.4. For a sequence an the notation
lim an = b n→∞ means that that the numbers an are arbitrarily close to the number b when n is suﬃciently large.
The expression on the left is called the the limit of an as n becomes inﬁnite. For a series we also
use the notation
∞ n ak = lim
k=1 n→∞ ak .
k=1 474 Sequences and Series Theorem 9.2. If the ratio r of successive terms in a geometric series is less than one in absolute
value, then the sum of the inﬁnite geometric series is
∞ ark =
k=1 ar
.
1−r By Theorem ??
n ark =
k=1 ar
ar − arn+1
=
+ crn
1−r
1−r where c = ar/(1 − r). In Section ?? we saw that for r < 1 the graph of the exponential function
y = rx decays exponentially as x becomes large positive. This implies that
lim rn = 0 n→∞ so
∞ n
k ark = ar = lim n→∞ k=1 k=1 ar
ar
+ c lim rn =
.
n→∞
1−r
1−r Example 9.4.1. (Zeno’s Paradox) To travel one mile I must ﬁrst travel the ﬁrst half mile, then
half of the remaining distance, then half of the remaining distance, and so on. How can I ever go
whole distance? The answer is that the sum of all the distances is one:
11
3
+=,
24
4 111
3
++=,
248
4 111
1
15
+++
=,
2 4 8 16
16 The numerator is only one less than the denominator:
n
k=1 1
2 k = 111
1
2n − 1
+ + + ··· + n =
= 1 − 2−n .
248
2
2n The ﬁnite sums are getting closer to one and the inﬁnite sum is
∞
k=1 1
2 k = 1
2 1− 1
2 = 1. .... 9.4 Infinite sums and Repeating Decimals* 475 Example 9.4.2. We use Theorem 9.2 to prove that the inﬁnite repeating decimal 0.1 36 36 36 36 36 36 . . ..
is equal to 3/22. The ﬁrst step is to make the decimal look like a geometric series.
1
10
1
=
10
1
=
10
1
=
10 0.1 36 36 36 36 36 36 . . . = Now by Theorem ?? ∞
k=1 1
100 + 36 0.0 01 01 01 01 01 . . .
36
0.01 01 01 01 01 . . .
10
36
10−2 + 10−4 + 10−6 + 10−8 + · · ·
+
10
∞
1k
36
+
10
100
+ k=1 k = 0.01
1
0.01
=
=
1 − 0.01
0.99
99 so
0.1 36 36 36 36 . . . = 1
36 1
+
·.
10 10 99 Now we do the arithmetic:
1
36 1
99 + 36
135
15
3
+
·
=
=
=
=.
10 10 99
990
990
110
22
The concept of an inﬁnite sum makes the deﬁnition of decimal expansion more precise. If x is
a real number between 0 and 1 it has a decimal expansion
∞ dk 10−k x=
k=1 where each dk is an integer between 0 and 9. For most real numbers the digits dk won’t follow any
pattern, but
Theorem 9.3. A real number x is rational if and only if has a repeating decimal expansion like
the one in Example 9.4.2. A rational number is a ratio p/q of two integers p and q . To see why the decimal expansion
eventually repeats periodically imagine computing the decimal expansion by long division. At each
step in the long division algorithm we compute the next digit in the quotient, multiply that digit by
q , subtract the product to get the next remainder, and bring down the next digit from the dividend.
The remainder is smaller than q , otherwise we would have used a larger digit in the quotient we are
computing. Once we are computing digits of the right of the decimal point the digit we bring down
from the dividend is always zero and since the remainder is always less than q we will eventually
ﬁnd ourselves redoing what we have already done. 476 Sequences and Series As an example we express 1202/99 as a repeating decimal. We do long division as in elementary
school.
12.141
99  1202.000
99
212
188
14 0 ←
99
4 10
3 96
140 ←
After we are to the right of the decimal point we always bring down a zero and the two rows marked
with an arrow have the same remainder. Hence the process must cycle (repeat periodically) at that
point (every other time we do the same division and subtraction) so
1202
= 12.14 := 12. 14 14 14 . . .
99
The proof that a real number with a repeating decimal expansion is rational is just like this
last computation. We ﬁrst write the number as the sum of a ﬁnite decimal and a negative power
of ten times a geometric series. We then use Theorem 9.2 and do the arithmetic. 9.4.1 Exercises 1. Express the following repeating decimals as a fraction of integers.
(a) 0.7 (b) 0.13 (c) 10.159 (d) −5.867 Chapter 10 Complex Numbers and the
Fundamental Theorem of Algebra
10.1 Complex Numbers Consider the polynomial p(x) = x2 + 1. The zeros of p are the solutions to x2 + 1 = 0, or x2 = −1.
This equation has no real solutions, but you may recall from Intermediate Algebra that we can
√
√
formally extract the square roots of both sides to get x = ± −1. The quantity −1 is usually relabeled i, the socalled imaginary unit.1 The number i, while not a real number, plays along well
with real numbers, and acts very much like any other radical expression. For instance, 3(2i) = 6i,
7i − 3i = 4i, (2 − 7i) + (3 + 4i) = 5 − 3i, and so forth. The key properties which distinguish i from
the real numbers are listed below.
Definition 10.1. The imaginary unit i satisﬁes the two following properties
1. i2 = −1
2. If c is a real number with c ≥ 0 then √
√
−c = i c Property 1 in Deﬁnition 10.1 establishes that i does act as a square root2 of −1, and property
2 establishes what we mean by the ‘principal square root’ of a negative real number. In property
2, it is important to remember the restriction on c. For example, it is perfectly acceptable to say
√
√
√
−4 = i 4 = i(2) = 2i. However, −(−4) = i −4, otherwise, we’d get
2=
1
2 √ 4= √
−(−4) = i −4 = i(2i) = 2i2 = 2(−1) = −2, Some technical mathematics textbooks label it ‘j ’.
Note the use of the indeﬁnite article ‘a’. Whatever beast is chosen to be i, −i is the other square root of −1. 478 Complex Numbers and the Fundamental Theorem of Algebra which is unacceptable.3 We are now in the position to deﬁne the complex numbers.
Definition 10.2. A complex number is a number of the form a + bi, where a and b are real
numbers and i is the imaginary unit.
√
2
Complex numbers include things you’d normally expect, like 3+2i and 5 − i 3. However, don’t
forget that a or b could be zero, which means numbers like 3i and 6 are also complex numbers. In
√
other words, don’t forget that the complex numbers include the real numbers, so 0 and π − 21
are both considered complex numbers. The arithmetic of complex numbers is as you would expect.
The only thing you need to remember are the two properties in Deﬁnition 10.1. The next example
should help recall how these animals behave.
Example 10.1.1. Perform the indicated operations and simplify. Write your ﬁnal answer in the
form4 a + bi.
√√
1. (1 − 2i) − (3 + 4i)
4. −3 −12
2. (1 − 2i)(3 + 4i)
3. 1 − 2i
3 − 4i 5. (−3)(−12) 6. (x − [1 + 2i])(x − [1 − 2i]) Solution.
1. As mentioned earlier, we treat expressions involving i as we would any other radical. We
combine like terms to get (1 − 2i) − (3 + 4i) = 1 − 2i − 3 − 4i = −2 − 6i.
2. Using the distributive property, we get (1 − 2i)(3 + 4i) = (1)(3) + (1)(4i) − (2i)(3) − (2i)(4i) =
3 + 4i − 6i − 8i2 . Recalling i2 = −1, we get 3 + 4i − 6i − 8i2 = 3 − 2i − (−8) = 11 − 2i.
3. How in the world are we supposed to simplify 1−2i ? Well, we deal with the denominator
3−4i
3 − 4i as we would any other denominator containing a radical, and multiply both numerator
and denominator by 3 + 4i (the conjugate of 3 − 4i).5 Doing so produces
1 − 2i 3 + 4 i
(1 − 2i)(3 + 4i)
11 − 2i
11
2
·
=
=
=
−
i
3 − 4i 3 + 4 i
(3 − 4i)(3 + 4i)
25
25 25
4. We use property 2 of Deﬁnition 10.1 √
ﬁrst, then√
apply the rules of radicals applicable to real
√
√
√√
radicals to get −3 −12 = i 3 i 12 = i2 3 · 12 = − 36 = −6.
3 We want to enlarge the number system so we can solve things like x2 = −1, but not at the cost of the established
rules already set in place. For that reason, the general properties of radicals simply do not apply for even roots of
negative quantities.
4
We’ll accept an answer of say 3 − 2i, although, technically, we should write this as 3 + (−2)i. Even we pedants
have our limits.
5
We will talk more about this in a moment. 10.1 Complex Numbers 479 5. We adhere to the order of operations here and perform the multiplication before the radical
√
to get (−3)(−12) = 36 = 6.
6. We can brute force multiply using the distributive property and see that
(x − [1 + 2i])(x − [1 − 2i]) = x2 − x[1 − 2i] − x[1 + 2i] + [1 − 2i][1 + 2i]
= x2 − x + 2ix − x − 2ix + 1 − 2i + 2i − 4i2
= x2 − 2x + 5 A couple of remarks about the last example are in order. First, the conjugate of a complex
number a + bi is the number a − bi. The notation commonly used for conjugation is a ‘bar’:
√
√
a + bi = a − bi. For example, 3 + 2i = 3 − 2i, 3 − 2i = 3+2i, 6 = 6, 4i = −4i, and 3 + 5 = 3+ 5.
The properties of the conjugate are summarized in the following theorem.
Theorem 10.1. Suppose z and w are complex numbers.
z=z
z+w =z+w
z w = zw
(z )n = z n , for any natural number n = 1, 2, 3, . . .
z is a real number if and only if z = z . Essentially, Theorem 10.1 says that complex conjugation works well with addition, multiplication, and powers. The proof of these properties can best be achieved by writing out z = a + bi
and w = c + di for real numbers a, b, c, and d. Next, we compute the left and right hand side
of each equation and check to see that they are the same. The proof of the ﬁrst property is
a very quick exercise.6 To prove the second property, we compare z + w and z + w. We have
z + w = a + bi + c + di = a − bi + c − di. To ﬁnd z + w, we ﬁrst compute
z + w = (a + bi) + (c + di) = (a + c) + (b + d)i
so
z + w = (a + c) + (b + d)i = (a + c) − (b + d)i = a − bi + c − di
6 Trust us on this. 480 Complex Numbers and the Fundamental Theorem of Algebra As such, we have established z + w = z + w. The proof for multiplication works similarly. The proof
that the conjugate works well with powers can be viewed as a repeated application of the product
rule, and is best proved using a technique called Mathematical Induction.7 The last property is a
characterization of real numbers. If z is real, then z = a + 0i, so z = a − 0i = a = z . On the other
hand, if z = z , then a + bi = a − bi which means b = −b so b = 0. Hence, z = a + 0i = a and is real. 10.2 The Fundamental Theorem of Algebra We now return to the business of zeros. Suppose we wish to ﬁnd the zeros of f (x) = x2 − 2x + 5.
To solve the equation x2 − 2x + 5 = 0, we note the quadratic doesn’t factor nicely, so we resort to
the Quadratic Formula, Equation 3.5 and obtain
x= −(−2) ± √
(−2)2 − 4(1)(5)
2 ± −16
2 ± 4i
=
=
= 1 ± 2i.
2(1)
2
2 Two things are important to note. First, the zeros, 1 + 2i and 1 − 2i are complex conjugates.
If ever we obtain nonreal zeros to a quadratic function with real coeﬃcients, the zeros will be a
complex conjugate pair. (Do you see why?) Next, we note that in Example 10.1.1, part 6, we
found (x − [1 + 2i])(x − [1 − 2i]) = x2 − 2x + 5. This demonstrates that the factor theorem holds
even for nonreal zeros, i.e, x = 1 + 2i is a zero of f , and, sure enough, (x − [1 + 2i]) is a factor of
f (x). It turns out that polynomial division works the same way for all complex numbers, real and
nonreal alike, and so the Factor and Remainder Theorems hold as well. But how do we know if a
general polynomial has any complex zeros at all? We have many examples of polynomials with no
real zeros. Can there be polynomials with no zeros whatsoever? The answer to that last question
is “No.” and the theorem which provides that answer is The Fundamental Theorem of Algebra.
Theorem 10.2. The Fundamental Theorem of Algebra: Suppose f is a polynomial function with complex number coeﬃcients of degree n ≥ 1, then f has least one complex zero. The Fundamental Theorem of Algebra is an example of an ‘existence’ theorem in mathematics.
Like the Intermediate Value Theorem, Theorem 4.1, the Fundamental Theorem of Algebra guarantees the existence of at least one zero, but gives us no algorithm to use in ﬁnding it. In fact,
there are polynomials whose real zeros, though they exist, cannot be expressed using the ‘usual’
combinations of arithmetic symbols, and must be approximated. The authors are fully aware that
the full impact and profound nature of the Fundamental Theorem of Algebra is lost on most students this level, and that’s ﬁne. It took mathematicians literally hundreds of years to prove the
theorem in its full generality, and some of that history is recorded here. Note that the Fundamental
Theorem of Algebra applies to polynomial functions with not only real coeﬃcients, but, those with
complex number coeﬃcients as well.
7 See Section ??. 10.2 The Fundamental Theorem of Algebra 481 Suppose f is a polynomial of degree n with n ≥ 1. The Fundamental Theorem of Algebra
guarantees us at least one complex zero, z1 , and, as such, the Factor Theorem guarantees that f (x)
factors as f (x) = (x − z1 ) q1 (x) for a polynomial function q1 , of degree exactly n − 1. If n − 1 ≥ 1,
then the Fundamental Theorem of Algebra guarantees a complex zero of q1 as well, say z2 , and so
the Factor Theorem gives us q1 (x) = (x − z2 ) q2 (x), and hence f (x) = (x − z1 ) (x − z2 ) q2 (x). We
can continue this process exactly n times, at which point our quotient polynomial qn has degree 0
so it’s a constant. This argument gives us the following factorization theorem.
Theorem 10.3. Complex Factorization Theorem: Suppose f is a polynomial function with
complex number coeﬃcients. If the degree of f is n and n ≥ 1, then f has exactly n complex
zeros, counting multiplicity. If z1 , z2 , . . . , zk are the distinct zeros of f , with multiplicities m1 ,
m2 , . . . , mk , respectively, then f (x) = a (x − z1 )m1 (x − z2 )m2 · · · (x − zk )mk . Note that the value a in Theorem 10.3 is the leading coeﬃcient of f (x) (Can you see why?)
and as such, we see that a polynomial is completely determined by its zeros, their multiplicities,
and its leading coeﬃcient. We put this theorem to good use in the next example.
Example 10.2.1. Let f (x) = 12x5 − 20x4 + 19x3 − 6x2 − 2x + 1.
1. Find all complex zeros of f and state their multiplicities.
2. Factor f (x) using Theorem 10.3
Solution.
1. Since f is a ﬁfth degree polynomial, we know we need to perform at least three successful
divisions to get the quotient down to a quadratic function. At that point, we can ﬁnd the
remaining zeros using the Quadratic Formula, if necessary. We get
1
2 12 −20
↓ 1
2 −6 −2 1 6 −7 6 0 −1 12 0 −2 6 −4 4 2 12 −14
↓ 1
−3 19 12 −8 8 4 ↓ −4 4 −4 12 −12 12 0 0 0 482 Complex Numbers and the Fundamental Theorem of Algebra
√ i
Our quotient is 12x2 − 12x + 12, whose zeros we ﬁnd to be 1±2 3 . From Theorem 10.3, we
know f has exactly 5 zeros, counting multiplicities, and as such we have the zero 1 with
2
1
multiplicity 2, and the zeros − 3 , √
1+i 3
2 and √
1−i 3
2, each of multiplicity 1. 2. Applying Theorem 10.3, we are guaranteed that f factors as 1
f (x) = 12 x −
2 2 1
x+
3 √
1+i 3
x−
2 √
1−i 3
x−
2 A true test of Theorem 10.3 (and a student’s mettle!) would be to take the factored form of f (x)
in the previous example and multiply it out8 to see that it really does reduce to the formula f (x) =
12x5 − 20x4 + 19x3 − 6x2 − 2x + 1. When factoring a polynomial using Theorem 10.3, we say that
it is factored completely over the complex numbers, meaning that it is impossible to factor
the polynomial any further using complex numbers. If we wanted to completely factor f (x) over
the real numbers then we would have stopped short of ﬁnding the nonreal zeros of f and factored
12
1
f using our work from the synthetic division to write f (x) = x − 2
x + 3 12x2 − 12x + 12 ,
2
x + 1 x2 − x + 1 . Since the zeros of x2 − x + 1 are nonreal, we call
or f (x) = 12 x − 1
2
3
2 − x + 1 an irreducible quadratic meaning it is impossible to break it down any further using
x
real numbers. The last two results of the section show us that, at least in theory, if we have a
polynomial function with real coeﬃcients, we can always factor it down enough so that any nonreal
zeros come from irreducible quadratics. Theorem 10.4. Conjugate Pairs Theorem: If f is a polynomial function with real number
coeﬃcients and z is a zero of f , then so is z . To prove the theorem, suppose f is a polynomial with real number coeﬃcients. Speciﬁcally, let
f (x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 . If z is a zero of f , then f (z ) = 0, which means
an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 = 0. Next, we consider f (z ) and apply Theorem 10.1 below.
8 You really should do this once in your life to convince yourself that all of the theory actually does work! 10.2 The Fundamental Theorem of Algebra 483 f (z ) = an (z )n + an−1 (z )n−1 + . . . + a2 (z )2 + a1 z + a0
= an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 since (z )n = z n = an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 since the coeﬃcients are real = an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 since z w = zw = an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0 since z + w = z + w = f (z )
=0
=0
This shows that z is a zero of f . So, if f is a polynomial function with real number coeﬃcients,
Theorem 10.4 tells us if a + bi is a nonreal zero of f , then so is a − bi. In other words, nonreal
zeros of f come in conjugate pairs. The Factor Theorem kicks in to give us both (x − [a + bi]) and
(x − [a − bi]) as factors of f (x) which means (x − [a + bi])(x − [a − bi]) = x2 + 2ax + a2 + b2 is
an irreducible quadratic factor of f . As a result, we have our last result of the section.
Theorem 10.5. Real Factorization Theorem: Suppose f is a polynomial function with real
number coeﬃcients. Then f (x) can be factored into a product of linear factors corresponding
to the real zeros of f and irreducible quadratic factors which give the nonreal zeros of f . We now present an example which pulls together all of the major ideas of this section.
Example 10.2.2. Let f (x) = x4 + 64.
1. Use synthetic division to show x = 2 + 2i is a zero of f .
2. Find the remaining complex zeros of f .
3. Completely factor f (x) over the complex numbers.
4. Completely factor f (x) over the real numbers.
Solution.
1. Remembering to insert the 0’s in the synthetic division tableau we have 484 Complex Numbers and the Fundamental Theorem of Algebra 1 2 + 2i 0 0 0 64 ↓ 2 + 2i 8i −16 + 16i −64 1 2 + 2i 8i −16 + 16i 0 2. Since f is a fourth degree polynomial, we need to make two successful divisions to get a
quadratic quotient. Since 2 + 2i is a zero, we know from Theorem 10.4 that 2 − 2i is also a
zero. We continue our synthetic division tableau.
2 + 2i 0 0 64 ↓ 2 + 2i
2 − 2i 1 0 8i −16 + 16i −64 1 2 + 2i 8i −16 + 16i 0 ↓ 2 − 2i 8 − 8i
1 4 8 16 − 16i
0 Our quotient polynomial is x2 + 4x + 8. Using the quadratic formula, we obtain the remaining
zeros −2 + 2i and −2 − 2i.
3. Using Theorem 10.3, we get f (x) = (x − [2 − 2i])(x − [2 + 2i])(x − [−2 + 2i])(x − [−2 − 2i]).
4. We multiply the linear factors of f (x) which correspond to complex conjugate pairs. We ﬁnd
(x − [2 − 2i])(x − [2 + 2i]) = x2 − 4x + 8, and (x − [−2 + 2i])(x − [−2 − 2i]) = x2 + 4x + 8.
Our ﬁnal answer f (x) = x2 − 4x + 8 x2 + 4x + 8 .
Our last example turns the tables and asks us to manufacture a polynomial with certain properties of its graph and zeros.
Example 10.2.3. Find a polynomial p of lowest degree that has integer coeﬃcients and satisﬁes
all of the following criteria:
the graph of y = p(x) touches the xaxis at 1
3, 0 x = 3i is a zero of p.
as x → −∞, p(x) → −∞
as x → ∞, p(x) → −∞ Solution. To solve this problem, we will need a good understanding of the relationship between
the xintercepts of the graph of a function and the zeros of a function, the Factor Theorem, the
role of multiplicity, complex conjugates, the Complex Factorization Theorem, and end behavior of 10.2 The Fundamental Theorem of Algebra 485 polynomial functions. (In short, you’ll need most of the major concepts of this chapter.) Since the
graph of p touches the xaxis at 1 , 0 , we know x = 1 is a zero of even multiplicity. Since we
3
3
1
are after a polynomial of lowest degree, we need x = 3 to have multiplicity exactly 2. The Factor
2
Theorem now tells us x − 1 is a factor of p(x). Since x = 3i is a zero and our ﬁnal answer is to
3
have integer (real) coeﬃcients, x = −3i is also a zero. The Factor Theorem kicks in again to give us
(x − 3i) and (x +3i) as factors of p(x). We are given no further information about zeros or intercepts
12
so we conclude, by the Complex Factorization Theorem that p(x) = a x − 3 (x − 3i)(x + 3i) for
a
some real number a. Expanding this, we get p(x) = ax4 − 23 x3 + 82a x2 − 6ax + a. In order to obtain
9
integer coeﬃcients, we know a must be an integer multiple of 9. Our last concern is end behavior.
Since the leading term of p(x) is ax4 , we need a < 0 to get p(x) → −∞ as x → ±∞. Hence, if we
choose x = −9, we get p(x) = −9x4 + 6x3 − 82x2 + 54x − 9. We can verify our handiwork using
the techniques developed in this chapter.
This example concludes our study of polynomial functions.9 The last few sections have contained
what is considered by many to be ‘heavy’ mathematics. Like a heavy meal, heavy mathematics
takes time to digest. Don’t be overly concerned if it doesn’t seem to sink in all at once, and pace
yourself on the exercises or you’re liable to get mental cramps. But before we get to the exercises,
we’d like to oﬀer a bit of an epilogue.
Our main goal in presenting the material on the complex zeros of a polynomial was to give the
chapter a sense of completeness. Given that it can be shown that some polynomials have real zeros
which cannot be expressed using the usual algebraic operations, and still others have no real zeros
at all, it was nice to discover that every polynomial of degree n ≥ 1 has n complex zeros. So like
we said, it gives us a sense of closure. But the observant reader will note that we did not give any
examples of applications which involve complex numbers. Students often wonder when complex
numbers will be used in ‘realworld’ applications. After all, didn’t we call i the imaginary unit?
How can imaginary things be used in reality? It turns out that complex numbers are very useful in
many applied ﬁelds such as ﬂuid dynamics, electromagnetism and quantum mechanics, but most
of the applications require Mathematics well beyond College Algebra to fully understand them.
That does not mean you’ll never be be able to understand them; in fact, it is the authors’ sincere
hope that all of you will reach a point in your studies when the glory, awe and splendor of complex
numbers are revealed to you. For now, however, the really good stuﬀ is beyond the scope of this
text. We invite you and your classmates to ﬁnd a few examples of complex number applications
and see what you can make of them. A simple Internet search with the phrase ‘complex numbers in
real life’ should get you started. Basic electronics classes are another place to look, but remember,
they might use the letter j where we have used i.
For the remainder of the text, we will restrict our attention to real numbers. We do this
primarily because the ﬁrst Calculus sequence you will take, ostensibly the one that this text is
preparing you for, studies only functions of real variables. Also, lots of really cool scientiﬁc things
don’t require any deep understanding of complex numbers to study them, but they do need more
Mathematics like exponential, logarithmic and trigonometric functions. We believe it makes more
9 With the exception of the exercises on the next page, of course. 486 Complex Numbers and the Fundamental Theorem of Algebra sense pedagogically for you to learn about those functions now and then, after you’ve completed the
Calculus sequence, take a course in Complex Function Theory in your junior or senior year. It is in
that course that the true power of the complex numbers is released. But for now, in order to fully
1
prepare you for life immediately after College Algebra, we will say that functions like f (x) = 2
x +1
have a domain of all real numbers, even though we know x2 + 1 = 0 has two complex solutions,
namely x = ±i. Because x2 + 1 > 0 for all real numbers x, the fraction x21 is never undeﬁned in
+1
our real variable setting. 10.2 The Fundamental Theorem of Algebra 10.2.1 487 Exercises 1. We know i2 = −1 which means i3 = i2 · i = (−1) · i = −i and i4 = i2 · i2 = (−1)(−1) = 1.
Use this information to simplify the following.
(a) i5 (b) i304 (c) (2i)3 (d) (−i)23 2. Let z = 3 + 4i and w = 2 − i. Compute the following and express your answer in a + bi form.
(a) z + w
(b) w − z w
z
(f) w3 (c) z · w
z
(d)
w 3. Simplify the following.
√
(a) −49
√√
(b) −9 −16 (e) (c)
(d) √ (−9)(−16)
√
49 −4 4. Find the complex solutions of the following quadratic equations.
(a) x2 − 4x + 13 = 0 (b) 3x2 + 2x + 10 = 0 5. For each polynomial given below ﬁnd all of its zeros, completely factor it over the real numbers
and completely factor it over the complex numbers.
(a) x2 − 2x + 5 (i) 4x3 − 6x2 − 8x + 15 (b) x3 − 2x2 + 9x − 18 (j) x4 + x3 + 7x2 + 9x − 18 (c) x3 + 6x2 + 6x + 5 (k) 6x4 + 17x3 − 55x2 + 16x + 12 (d) 3x3 − 13x2 + 43x − 13 (l) x5 − x4 + 7x3 − 7x2 + 12x − 12 (e) x4 + 9x2 + 20 (m) x3 + 7x2 + 9x − 2 (f) 4x4 − 4x3 + 13x2 − 12x + 3 (n) −3x4 − 8x3 − 12x2 − 12x − 5 (g) x3 + 3x2 + 4x + 12 (o) 8x4 + 50x3 + 43x2 + 2x − 4 (h) 2x4 − 7x3 + 14x2 − 15x + 6 (p) 9x3 + 2x + 1 (q) x4 − 2x3 + 27x2 − 2x + 26 (Hint: x = i is one of the zeros.)
(r) 2x4 + 5x3 + 13x2 + 7x + 5 (Hint: x = −1 + 2i is a zero.)
6. Let z and w be arbitrary complex numbers. Show that z w = zw and z = z .
7. With the help of your classmates, build a polynomial p with integer coeﬃcients such that
x = −2 − i is a zero of p, p has a local maximum at the point (4, 0) and p(x) → −∞ as
x→∞ 488 Complex Numbers and the Fundamental Theorem of Algebra 10.2.2 Answers 1. (a) i5 = i4 · i = 1 · i = i
(b) i304 = (i4 )76 = 176 = 1
2. (a) z + w = 5 + 3i
(b) w − z = −1 − 5i
(c) z · w = 10 + 5i
√ −49 = 7i
√√
(b) −9 −16 = (3i) · (4i) = 12i2 = −12 3. (a) 4. (a) x = 2 ± 3i (c) (2i)3 = 8i3 = −8i
(d) (−i)23 = −i23 = −i20 · i3 = (−1)(−i) = i
z
2 11
=+i
w
5
5
w
2
11
(e)
=
−i
z
25 25
(f) w3 = 2 − 11i (d) √
(−9)(−16) = 144 = 12
√√
(d) 49 −4 = 7 · 2i = 14i
(c) √
1
29
(b) x = − ±
i
3
3 5. (a) x2 − 2x + 5 = (x − (1 + 2i))(x − (1 − 2i))
Zeros: x = 1 ± 2i
(b) x3 − 2x2 + 9x − 18 = (x − 2) x2 + 9 = (x − 2)(x − 3i)(x + 3i)
Zeros: x = 2, ±3i
√
√
1
3
3
3 +6x2 +6x+5 = (x+5)(x2 +x+1) = (x+5) x − − 1 +
x− − −
(c) x
i
i
2
2
2
2
√
1
3
Zeros: x = −5, x = − ±
i
2
2
(d) 3x3 − 13x2 + 43x − 13 = (3x − 1)(x2 − 4x + 13) = (3x − 1)(x − (2 + 3i))(x − (2 − 3i))
1
Zeros: x = , x = 2 ± 3i
3
√
√
(e) x4 + 9x2 + 20 = x2 + 4 x2 + 5 = (x − 2i)(x + 2i) x − i 5 x + i 5
√
Zeros: x = ±2i, ±i 5
√
√
(f) 4x4 − 4x3 + 13x2 − 12x + 3 = (2x − 1)2 (x2 + 3) = (2x − 1)2 (x + 3i)(x − 3i)
√
1
Zeros: x = , x = ± 3i
2
3 + 3x2 + 4x + 12 = (x + 3) x2 + 4 = (x + 3)(x + 2i)(x − 2i)
(g) x
Zeros: x = −3, ±2i
(h) 2x4 − 7x3 + 14x2 − 15x + 6 = (x − 1)2 2x2 − 3x + 6
√
√
3
39
3
39
2 x−
= 2(x − 1)
+
i
x−
−
i
4
4
4
4
√
3
39
Zeros: x = 1, x = ±
i
4
4 10.2 The Fundamental Theorem of Algebra
(i) 4x3 − 6x2 − 8x + 15 = (k) (l) (m) (n) (o) (p) (q)
(r) 3
2 4x2 − 12x + 10 = 31
31
+i
x−
−i
22
22
3
31
Zeros: x = − , x = ± i
2
22
x4 + x3 + 7x2 + 9x − 18 = (x + 2)(x − 1) x2 + 9 = (x + 2)(x − 1)(x + 3i)(x − 3i)
Zeros: x = −2, 1, ±3i
√
√
3
1
x−
x − −2 + 2 2
x − −2 − 2 2
6x4 +17x3 − 55x2 +16x +12 = 6 x +
3
2
√
1
3
Zeros: x = − , x = , x = −2 ± 2 2
3
2
x5 − x4 + 7x3 − 7x2 + 12x − 12 = (x − 1) x2 + 3 x2 + 4
√
√
3)(
= (x − 1)(x − i √ x + i 3)(x − 2i)(x + 2i)
Zeros: x = 1, ± 3i, ±2i
√
√
29
5
29
3 + 7x2 + 9x − 2 = (x + 2) x − − 5 +
x
x− − −
2
2
2
2
√
5
29
Zeros: x = −2, x = − ±
2
2
−3x4 − 8x3 − 12x2 − 12x − 5 = (x + 1)2 −3x2 − 2x − 5
√
√
1
14
14
2 x − −1 +
x− − −
= −3(x + 1)
i
i
3
3
3
3
√
1
14
Zeros: x = −1, x = − ±
i
3
3
√
√
1
1
8x4 + 50x3 + 43x2 + 2x − 4 = 8 x +
x−
(x − (−3 + 5))(x − (−3 − 5))
2
4
√
11
Zeros: x = − , , x = −3 ± 5
24
1
9x3 + 2x + 1 = x +
9x2 − 3x + 3
3
√
√
1
11
1
11
1
=9 x+
x−
+
i
x−
−
i
3
6
6
6
6
√
1
1
11
Zeros: x = − , x = ±
i
3
6
6
x4 − 2x3 +27x2 − 2x +26 = (x2 − 2x +26)(x2 +1) = (x − (1+5i))(x − (1 − 5i))(x + i)(x − i)
Zeros: x = 1 ± 5i, x = ±i
2x4 + 5x3 + 13x2 + 7x + 5 = x2 + 2x + 5 2x2 + x + 1 =
√
√
1
7
1
7
2(x − (−1 + 2i))(x − (−1 − 2i)) x − − + i
x− − −i
4
4
4
4
√
1
7
Zeros: x = −1 ± 2i, − ± i
4
4
4 x+ (j) x+ 489 3
2 x− 490 Complex Numbers and the Fundamental Theorem of Algebra Appendix A The Laws of Algebra Proved
A.1 The Laws of Algebra The Laws of Algebra.
All the rules of calculation that you learned in elementary school follow from the above fundamental laws. In particular, the Commutative and Associative Laws say that you can add a bunch
of numbers in any order and similarly you can multiply a bunch of numbers in any order. For
example,
(A + B ) + (C + D) = (A + C ) + (B + D), A.2 (A · B ) · (C · D) = (A · C ) · (B · D). The Analogy between Addition and Multiplication Because both addition and multiplication satisfy the commutative, associative, identity, and inverse
laws, there are other analogies:
Theorem A.1. Inverses are unique: b + a = 0 =⇒ b = −a; similarly b · a = 1 =⇒ b = a−1 .
Proof. Assume that b + a = 0. Then
step by with b=b+0 A=A+0 A=b = b + a + (−a) A + (−A) = 0 A=a = b + a + (−a) A + (B + C ) = (A + B ) + C A = b, B = a, C = −a = 0 + (−a) b+a=0 by hypothesis = −a 0+A=A A = −a Similarly assume that b · a−1 = 1. Then 492 The Laws of Algebra Proved step by with b=b·1 A=A·1 A=b = b · a · a−1 A · A−1 = 1 A=a = b · a · a−1 A · (B · C ) = (A · B ) · C A = b, B = a, C = a−1 = 1 · a−1 b·a=1 by hypothesis = a−1 1·A=A A = a−1 Here are the proofs of the properties (ivi) above. Notice how each proof of an addition rule is
immediately followed by the proof of the corresponding multiplication rule.
Theorem A.2. −(−a) = a and similarly (a−1 )−1 = a. Proof.
step by with a + (−a) = 0 A + (−A) = 0 A=a (−a) + a = 0 A+B =B+A A = a, B = −a a = −(−a) A + B = 0 =⇒ B = −A A = −a, B = a step by with a · a−1 = A A · A−1 A=a a−1 · a = 1 A·B =B·A A = a, B = a−1 a = −(−a) A · B = 1 =⇒ B = A−1 A = a−1 , B = a Theorem A.3. −(a + b) = −a − b and similarly (ab)−1 = a−1 b−1 . Proof. A.2 The Analogy between Addition and Multiplication
step 493 by with = (a + b) + (−a) + (−b) A − B ) := A + (−B ) A = −a, B = b = a + (−a) + b + (−b) (A + B ) + ( C + D ) = A = a, B = b, (a + b) + (−a − b) C = −a, D = −b = (A + C ) + (B + D)
= a + (−a) + 0 A + (−A) = 0 A=b =0+0 A + (−A) = 0 A=a =0 A+0=A A=0 −a − b = −(a + b) A + B = 0 =⇒ B = −A A = (a + b), B = −a − b step by with (A · B ) · (C · D ) = A = a, B = b, = (A · C ) · (B · D) C = a−1 , D = b−1 = a · a−1 · 1 A · A−1 A=b =1·1 A · A−1 A=a =1 A·1=A A=1 A · B = 1 =⇒ B = A−1 A = a−1 · b−1 , B = a · b (a · b) · (a−1 ) · (b−1 ) =
= a · a−1 · b · b−1 (a · b)−1 = a−1 · b−1 494 The Laws of Algebra Proved Theorem A.4. −(a − b) = b − a and similarly a
b −1 = b
.
a Proof.
step by with −(a − b) = − a + (−b) A − B := A + (−B ) A = a, B = b = −a − (−b) −(A + B ) = −A − B A = a, B = −b = −a + (−(−b)) A − B := A + (−B ) A = a, B = −b = −a + b −(−A) = A A=b = b + (−a) A+B =B+A A = −a, B = b =b−a A + (−B ) =: A − B A = b, B = a by with A/B := A · B −1 A = a, B = b = a−1 · (b−1 )−1 (A · B )−1 = A−1 · B −1 A = a, B = b−1 = a−1 · b (A−1 )−1 = A A=b = b · a−1 A·B =B·A A = a−1 , B = b = b/a A · B −1 =: A/B A = b, B = a step
(a/b)−1 = a · b−1 −1 Theorem A.5. (a − b) + (c − d) = (a + c) − (b + d) and similarly ac
ac
·=.
bd
bd Proof.
step by with = a + (−b) + (c − d) A − B := A + (−B ) A = a, B = b = a + (−b) + c + (−d) A − B := A + (−B ) A = c, B = d = a + c + (−b) + (−d) (A + B ) + ( C + D ) = A = a, B = −b, (a − b) + (c − d) = (A + C ) + (B + D) C = c, D = −d = a + c + (−b − d) A + (−B ) =: A − B A = −b, B = d = a + c + −(b + d) −A − B = −(A + B ) A = b, B = d = (a + c) − (b + d) A + (−B ) =: A − B A = a + c, B = b + d A.2 The Analogy between Addition and Multiplication 495 step by with (a/b) · (cd) = a · b−1 · (c/d) A/B := A · B −1 A = a, B = b A/B := A · B −1 A = c, B = d (A · B ) · (C · D ) = A = a, B = b−1 , = a · b−1 + c · d−1
= a · c · b−1 · d−1
= a · c · b−1 · d−1 C = c, D = d−1 = (A · C ) · (B · D) = a · c · (b−1 · d−1 ) A · B −1 =: A/B A = b−1 , B = d = a · c · (b · d)−1 A−1 · B −1 = (A · B )−1 A = b, B = d = (a · c)/(b · d) A · B −1 =: A/B A = a · c, B = b · d Theorem A.6. a − b = (a + c) − (b + c) and similarly a
ac
=.
b
bc Proof.
step by with a − b = (a − b) + 0 A=A+0 A=a−b = (a − b) − (c − c) 0=A−A A=c = (a + c) − (b + c) (A − B ) + ( C − D ) = A = a, B = b, = (A + C ) − (B + D) C=D=c step by with a/b = (a/b) · 1 A=A·1 A = a/b = (a/b)/(c/c) 1 = A/A A=c = (a · c)/(b · c) (A/B ) · (C/D) = A = a, B = b, = (A · C )/(B · D) C=D=c Theorem A.7. (a − b) − (c − d) = (a − b) + (d − c) and similarly a/b
ad
=·.
c/d
bc Proof.
step by with = a − b + −(c − d) A − B := A + (−B ) A = a − b, B = c − d = a−b + d−c −(A − B ) = B − A A = c, B = d (a − b) − (c − d) = 496 The Laws of Algebra Proved step by (a/b)/(c/d) = a/b · c/d −1 = a/b · d/c A.3 with A/B := A · B −1 A = a/b, B = c/d (A/B )−1 = B/A A = c, B = d Consequences of the Distributive Law step by with (a + b)c = c(a + b) AB = BA A = (a + b), B = c = ca + cb A(B + C ) = AB + AC A = c, B = a, C = b = ac + cb AB = BA A = c, B = a = ac + bc AB = BA A = c, B = b Theorem A.8. a · 0 = 0 step with 0 = −a + a 0 = −A + A A=a = −a + a · 1 A=A·1 A=a = −a + a · (1 + 0) A=A+0 A=1 = −a + (a · 1 + a · 0) A · (B + C ) = A · B + A · C A = a, B = a · 1, C = 0 = (−a + a · 1) + a · 0 A + (B + C ) = A · (A + B ) + C A = −a, B = a · 1, C = a · 0 = (−a + a) + a · 0 A·1=A A=a =0+a·0 −A + A = 0 A=a =a·0 Proof. by 0+A=A A=a·0 Theorem A.9. −a = (−1)a A.3 Consequences of the Distributive Law 497 step with 0=0·a 0=A·0=0·A A=a = (1 + (−1)) · a A + (−A) = 0 A=a = 1 · a + (−1) · a (A + B ) · C = A · C + B · C A = 1, B = −1, C = a = a + (−1) · a 1·A=A A=a (−1) · a = −a A + B = 0 =⇒ B = −A B = (−1)a, A = a step by with a · (−b) = a · [(−1) · b] −A = (−1)A A=b = [a · (−1)] · b A · (B · C ) = ( A · B ) · C A = a, B = −1, C = b = [(−1) · a] · b A·B =B·A A = a, B = −1 = (−1) · [a · b] A · B ) · C ) = (A · (B · C A = −1, B = a, C = b −(a · b) (−1)A = −A A=a·b step by with (−a) · (−b) = −[(−a) · b] A · (−B ) = −(A · B ) A = −a, B = b = −[b · (−a)] A·B =B·A A = −a, B = b = −[−(b · a)] A · (−B ) = −(A · B ) A = b, B = a = −[−(a · b)] A·B =B·A A = b, B = a =a·b Proof. by −(−A) = A A=a·b Theorem A.10. a(−b) = −ab Proof. Theorem A.11. (−a)(−b) = ab Proof. Theorem A.12. ac
ad + cb
+=
bd
bd 498 The Laws of Algebra Proved
step with (a + b)(c + d) = (a + b)c + (a +
b)d (A(B + C ) = AB + AC A = (a + b), B = c, C = d = ac + bc + (a + b)d (A + B )C = AC + BC A = a, B = b, C = c = ac + bc + ad + bd Proof. by (A + B )C = AC + BC A = a, B = b, C = d Theorem A.13. (a + b)2 = a2 + 2ab + b2
step with (a + b)2 = (a + b)(a + b) A2 = A · A A = (a + b) = a2 + ab + ba + b2 A + B )(C + D) = AB + AD +
BC + CD A = C = a, B = D = b = a2 + ab + ab + b2 AB = BA A = a, B = b = a2 + 1 · ab + 1 · ab + b2 A=1·A A = ab = a2 + (1 + 1)ab + b2 AC + BC = (A + B )C A = B = 1, C = ab = a2 + 2ab + b2 Proof. by 1+1=2 2 := 1 + 1 Theorem A.14. (a + b)(a − b) = a2 − b2
step with (a − b)(a + b) = a2 + ab +
(−b)a + (−b)b Proof. by
(A + B )(C + D) = AC + AD +
BC + BD A = C = a, B = −b, D = b (a − b)(a + b) = a2 + ab +
(−b)a − b2 (−A)A = −a2 A=b = a2 + ab + a(−b) − b2 AB = BA A = −b, B = a = a2 + ab − ab − b2 A(−B ) = −AB A = a, B = b = a2 + 0 − b2 A−A=0 A = ab = a2 − b2 A+0=A A = a2 ...
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This note was uploaded on 11/11/2011 for the course MATH 110 taught by Professor Staff during the Winter '08 term at BYU.
 Winter '08
 Staff
 Algebra, The Waves

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