**Unformatted text preview: **This page intentionally left blank Student’s Solutions Manual
to accompany Jon Rogawski’s Single Variable CALCULUS
SECOND EDITION BRIAN BRADIE
Christopher Newport University ROGER LIPSETT W. H. FREEMAN AND COMPANY
NEW YORK © 2012 by W. H. Freeman and Company
ISBN-13: 978-1-4292-4290-5
ISBN-10: 1-4292-4290-6
All rights reserved
Printed in the United States of America First Printing W. H. Freeman and Company, 41 Madison Avenue, New York, NY 10010
Houndmills, Basingstoke RG21 6XS, England
CONTENTS
Chapter 1 PRECALCULUS REVIEW
1.1
1.2
1.3
1.4
1.5 Real Numbers, Functions, and Graphs
Linear and Quadratic Functions
The Basic Classes of Functions
Trigonometric Functions
Technology: Calculators and Computers
Chapter Review Exercises 1
1
8
13
16
23
27 Chapter 2 LIMITS 31 2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9 31
37
46
49
57
61
66
73
76
82 Limits, Rates of Change, and Tangent Lines
Limits: A Numerical and Graphical Approach
Basic Limit Laws
Limits and Continuity
Evaluating Limits Algebraically
Trigonometric Limits
Limits at Infinity
Intermediate Value Theorem
The Formal Definition of a Limit
Chapter Review Exercises Chapter 3 DIFFERENTIATION
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9 Definition of the Derivative
The Derivative as a Function
Product and Quotient Rules
Rates of Change
Higher Derivatives
Trigonometric Functions
The Chain Rule
Implicit Differentiation
Related Rates
Chapter Review Exercises 91
91
101
112
119
126
132
138
147
157
165 Chapter 4 APPLICATIONS OF THE DERIVATIVE 174
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8 Linear Approximation and Applications
Extreme Values
The Mean Value Theorem and Monotonicity
The Shape of a Graph
Graph Sketching and Asymptotes
Applied Optimization
Newton’s Method
Antiderivatives
Chapter Review Exercises Chapter 5 THE INTEGRAL
5.1
5.2
5.3 Approximating and Computing Area
The Definite Integral
The Fundamental Theorem of Calculus, Part I 174
181
191
198
206
220
236
242
250 260
260
274
284 5.4
5.5
5.6 The Fundamental Theorem of Calculus, Part II
Net Change as the Integral of a Rate
Substitution Method
Chapter Review Exercises Chapter 6 APPLICATIONS OF THE INTEGRAL
6.1
6.2
6.3
6.4
6.5 Derivative of f (x) = bx and the Number e
Inverse Functions
Logarithms and Their Derivatives
Exponential Growth and Decay
Compound Interest and Present Value
Models Involving y ! = k ( y − b)
L’Hôpital’s Rule
Inverse Trigonometric Functions
Hyperbolic Functions
Chapter Review Exercises Chapter 8 TECHNIQUES OF INTEGRATION
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8 370
370
378
383
393
398
401
407
415
424
431 446 Integration by Parts
446
Trigonometric Integrals
457
Trigonometric Substitution
467
Integrals Involving Hyperbolic and Inverse Hyperbolic
Functions
481
The Method of Partial Fractions
485
Improper Integrals
503
Probability and Integration
520
Numerical Integration
525
Chapter Review Exercises
537 Chapter 9 FURTHER APPLICATIONS OF THE
INTEGRAL AND TAYLOR
POLYNOMIALS
9.1
9.2
9.3
9.4 317 Area Between Two Curves
317
Setting Up Integrals: Volume, Density, Average Value 328
Volumes of Revolution
336
The Method of Cylindrical Shells
346
Work and Energy
355
Chapter Review Exercises
362 Chapter 7 EXPONENTIAL FUNCTIONS
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9 290
296
300
307 Arc Length and Surface Area
Fluid Pressure and Force
Center of Mass
Taylor Polynomials
Chapter Review Exercises 555
555
564
569
577
593
iii iv C A L C U L U S CON TEN T S Chapter 10 INTRODUCTION TO DIFFERENTIAL
EQUATIONS
601
10.1
10.2
10.3
10.4 Solving Differential Equations
Graphical and Numerical Methods
The Logistic Equation
First-Order Linear Equations
Chapter Review Exercises Chapter 11 INFINITE SERIES
11.1
11.2
11.3
11.4 Sequences
Summing an Infinite Series
Convergence of Series with Positive Terms
Absolute and Conditional Convergence 601
614
621
626
637 646
646
658
669
683 11.5 The Ratio and Root Tests
11.6 Power Series
11.7 Taylor Series
Chapter Review Exercises 690
697
710
727 Chapter 12 PARAMETRIC EQUATIONS, POLAR
COORDINATES, AND CONIC
SECTIONS
742
12.1
12.2
12.3
12.4
12.5 Parametric Equations
Arc Length and Speed
Polar Coordinates
Area and Arc Length in Polar Coordinates
Conic Sections
Chapter Review Exercises 742
759
766
780
789
801 1 PRECALCULUS REVIEW
1.1 Real Numbers, Functions, and Graphs
Preliminary Questions
1. Give an example of numbers a and b such that a < b and |a| > |b|. solution Take a = −3 and b = 1. Then a < b but |a| = 3 > 1 = |b|. 2. Which numbers satisfy |a| = a? Which satisfy |a| = −a? What about |−a| = a? solution The numbers a ≥ 0 satisfy |a| = a and | − a| = a. The numbers a ≤ 0 satisfy |a| = −a.
3. Give an example of numbers a and b such that |a + b| < |a| + |b|. solution Take a = −3 and b = 1. Then |a + b| = | − 3 + 1| = | − 2| = 2, but |a| + |b| = | − 3| + |1| = 3 + 1 = 4. Thus, |a + b| < |a| + |b|. 4. What are the coordinates of the point lying at the intersection of the lines x = 9 and y = −4? solution The point (9, −4) lies at the intersection of the lines x = 9 and y = −4.
5. In which quadrant do the following points lie?
(a) (1, 4)
(b) (−3, 2) (c) (4, −3) (d) (−4, −1) solution
(a) Because both the x- and y-coordinates of the point (1, 4) are positive, the point (1, 4) lies in the first quadrant.
(b) Because the x-coordinate of the point (−3, 2) is negative but the y-coordinate is positive, the point (−3, 2) lies in
the second quadrant.
(c) Because the x-coordinate of the point (4, −3) is positive but the y-coordinate is negative, the point (4, −3) lies in
the fourth quadrant.
(d) Because both the x- and y-coordinates of the point (−4, −1) are negative, the point (−4, −1) lies in the third quadrant.
6. What is the radius of the circle with equation (x − 9)2 + (y − 9)2 = 9? solution The circle with equation (x − 9)2 + (y − 9)2 = 9 has radius 3.
7. The equation f (x) = 5 has a solution if (choose one):
(a) 5 belongs to the domain of f .
(b) 5 belongs to the range of f . solution The correct response is (b): the equation f (x) = 5 has a solution if 5 belongs to the range of f .
8. What kind of symmetry does the graph have if f (−x) = −f (x)? solution If f (−x) = −f (x), then the graph of f is symmetric with respect to the origin. Exercises
1. Use a calculator to find a rational number r such that |r − π 2 | < 10−4 . solution r must satisfy π 2 − 10−4 < r < π 2 + 10−4 , or 9.869504 < r < 9.869705. r = 9.8696 = 12337
1250 would
be one such number.
In Exercises
3–8,
expressare
thetrue
interval
of an
involving absolute value.
Which
of (a)–(f)
for a in
=terms
−3 and
b =inequality
2?
(a) a2]< b
3. [−2, solution
(d) 3a <|x|
3b≤ 2 (b) |a| < |b| (e) −4a < −4b (c) ab > 0
1
1
(f)
<
a
b 5. (0, 4)
(−4, 4)
solution The midpoint of the interval is c = (0 + 4)/2 = 2, and the radius is r = (4 − 0)/2 = 2; therefore, (0, 4)
can be expressed as |x − 2| < 2. 7. [1, 5]
[−4, 0]
solution The midpoint of the interval is c = (1 + 5)/2 = 3, and the radius is r = (5 − 1)/2 = 2; therefore, the
interval [1, 5] can be expressed as |x − 3| ≤ 2.
1 (−2, 8) June 7, 2011 LTSV SSM Second Pass 2 CHAPTER 1 PRECALCULUS REVIEW In Exercises 9–12, write the inequality in the form a < x < b.
9. |x| < 8 solution −8 < x < 8 11. |2x + 1| < 5
|x − 12| < 8
solution −5 < 2x + 1 < 5 so −6 < 2x < 4 and −3 < x < 2
In Exercises
|3x −13–18,
4| < 2express the set of numbers x satisfying the given condition as an interval.
13. |x| < 4
solution (−4, 4) 15. |x − 4| < 2
|x| ≤ 9
solution The expression |x − 4| < 2 is equivalent to −2 < x − 4 < 2. Therefore, 2 < x < 6, which represents the
interval (2, 6).
17. |4x − 1| ≤ 8
|x + 7| < 2
solution The expression |4x − 1| ≤ 8 is equivalent to −8 ≤ 4x − 1 ≤ 8 or −7 ≤ 4x ≤ 9. Therefore, − 74 ≤ x ≤ 94 ,
which represents the interval [− 74 , 94 ].
In Exercises
|3x +19–22,
5| < 1describe the set as a union of finite or infinite intervals.
19. {x : |x − 4| > 2}
solution x − 4 > 2 or x − 4 < −2 ⇒ x > 6 or x < 2 ⇒ (−∞, 2) ∪ (6, ∞) 21. {x : |x 2 − 1| > 2} {x : |2x + 4| > 3}
√
√
solution x 2 − 1 > 2 or x 2 − 1 < −2 ⇒ x 2 > 3 or x 2 < −1 (this will never happen) ⇒ x > 3 or x < − 3 ⇒
√
√
(−∞, − 3) ∪ ( 3, ∞).
23. Match (a)–(f) with (i)–(vi).
{x : |x 2 + 2x| > 2}
(a) a > 3
!
!
!
1 !!
!
(c) !a − ! < 5
3
(e) |a − 4| < 3
(i)
(ii)
(iii)
(iv)
(v)
(vi) (b) |a − 5| < 1
3 (d) |a| > 5
(f) 1 ≤ a ≤ 5 a lies to the right of 3.
a lies between 1 and 7.
The distance from a to 5 is less than 13 .
The distance from a to 3 is at most 2.
a is less than 5 units from 13 .
a lies either to the left of −5 or to the right of 5. solution
(a) On the number line, numbers greater than 3 appear to the right; hence, a > 3 is equivalent to the numbers to the right
of 3: (i).
(b) |a − 5| measures the distance from a to 5; hence, |a − 5| < 13 is satisfied by those numbers less than 13 of a unit from
5: (iii).
(c) |a − 13 | measures the distance from a to 13 ; hence, |a − 13 | < 5 is satisfied by those numbers less than 5 units from
1 : (v).
3
(d) The inequality |a| > 5 is equivalent to a > 5 or a < −5; that is, either a lies to the right of 5 or to the left of −5: (vi).
(e) The interval described by the inequality |a − 4| < 3 has a center at 4 and a radius of 3; that is, the interval consists
of those numbers between 1 and 7: (ii).
(f) The interval described by the inequality 1 < x < 5 has a center at 3 and a radius of 2; that is, the interval consists of
those numbers less than 2 units from 3: (iv).
# an interval. Hint: Plot y = x 2 + 2x − 3.
25. Describe {x :"x 2 + 2x < 3} as
x
< 0 as an interval.
Describe x :
solution The inequality
x + 1x 2 + 2x < 3 is equivalent to x 2 + 2x − 3 < 0. In the figure below, we see that the graph of
2
y = x + 2x − 3 falls below the x-axis for −3 < x < 1. Thus, the set {x : x 2 + 2x < 3} corresponds to the interval
−3 < x < 1. June 7, 2011 LTSV SSM Second Pass S E C T I O N 1.1 y = x2 + 2x − 3 −4 −3 −2 Real Numbers, Functions, and Graphs 3 y
10
8
6
4
2
−2 1 2 x 27. Show that if a > b, then b−1 > a −1 , provided that a and b have the same sign. What happens if a > 0 and b < 0?
Describe the set of real numbers satisfying |x − 3| = |x − 2| + 1 as a half-infinite interval.
solution Case 1a: If a and b are both positive, then a > b ⇒ 1 > ab ⇒ b1 > a1 .
Case 1b: If a and b are both negative, then a > b ⇒ 1 < ab (since a is negative) ⇒ b1 > a1 (again, since b is negative).
Case 2: If a > 0 and b < 0, then a1 > 0 and b1 < 0 so b1 < a1 . (See Exercise 2f for an example of this).
< 12|x
, then
29. Show
that xif satisfy
|a − 5|both
< 12 |xand
Which
− |b
3| −
< 8|
2 and
− 5||(a
<+
1?b) − 13| < 1. Hint: Use the triangle inequality.
solution
|a + b − 13| = |(a − 5) + (b − 8)|
≤ |a − 5| + |b − 8|
< (by the triangle inequality) 1 1
+ = 1.
2 2 Suppose that |a − 6| ≤ 2 and |b| ≤ 3.
Suppose that |x − 4| ≤ 1.
What is the largest possible value of |a + b|?
(a) What is the maximum possible value of |x + 4|?
What is the smallest
possible value of |a + b|?
(b) Show that |x 2 − 16| ≤ 9.
solution |a − 6| ≤ 2 guarantees that 4 ≤ a ≤ 8, while |b| ≤ 3 guarantees that −3 ≤ b ≤ 3. Therefore 1 ≤ a + b ≤ 11.
It follows that
(a) the largest possible value of |a + b| is 11; and
(b) the smallest possible value of |a + b| is 1.
31.
(a)
(b) 33. Express r1 = 0.27 as a fraction. Hint: 100r1 − r1 is an integer. Then express r2 = 0.2666 . . . as a fraction.
Prove that |x| − |y| ≤ |x − y|. Hint: Apply the triangle inequality to y and x − y.
solution Let r1 = 0.27. We observe that 100r1 = 27.27. Therefore, 100r1 − r1 = 27.27 − 0.27 = 27 and
r1 = 27
3
=
.
99
11 Now, let r2 = 0.2666. Then 10r2 = 2.666 and 100r2 = 26.666. Therefore, 100r2 − 10r2 = 26.666 − 2.666 = 24 and
r2 = 4
24
=
.
90
15 35. The text states: If the decimal expansions of numbers a and b agree to k places, then |a − b| ≤ 10−k . Show that the
Represent 1/7 and 4/27 as repeating decimals.
converse is false: For all k there are numbers a and b whose decimal expansions do not agree at all but |a − b| ≤ 10−k .
solution Let a = 1 and b = 0.9 (see the discussion before Example 1). The decimal expansions of a and b do not
agree, but |1 − 0.9| < 10−k for all k.
37.
(a)
(b) Find the equation of the circle with center (2, 4):
Plot each pair of points and compute the distance between them:
with radius r = 3.
(a) (1, 4) and (3, 2)
(b) (2, 1) and (2, 4)
that passes through (1, −1).
(c) (0, 0) and (−2, 3)
(d) (−3, −3) and (−2, 3)
solution (a) The equation of the indicated circle is (x − 2)2 + (y − 4)2 = 32 = 9.
(b) First determine the radius as the distance from the center to the indicated point on the circle:
$
√
r = (2 − 1)2 + (4 − (−1))2 = 26. Thus, the equation of the circle is (x − 2)2 + (y − 4)2 = 26. 39. Determine the domain and range of the function
Find all points with integer coordinates located at a distance 5 from the origin. Then find all points with integer
coordinates located at a distance 5 fromf (2,
3).s, t, u} → {A, B, C, D, E}
: {r,
defined by f (r) = A, f (s) = B, f (t) = B, f (u) = E. solution The domain is the set D = {r, s, t, u}; the range is the set R = {A, B, E}. June 7, 2011 LTSV SSM Second Pass 4 CHAPTER 1 PRECALCULUS REVIEW In Exercises
find the
and
range
of the D
function.
Give 41–48,
an example
of adomain
function
whose
domain
has three elements and whose range R has two elements. Does a
exist whose domain D has two elements and whose range R has three elements?
41. function
f (x) = −x
solution D : all reals; R : all reals 43. f (x) = x 3 4
g(t) = t
solution D : all reals; R : all reals 45. f (x) = |x|√
g(t) = 2 − t
solution D : all reals; R : {y : y ≥ 0} 1
47. f (x) = 2 1
h(s) =
x
s
solution D : {x : x ) = 0}; R : {y : y > 0}
In Exercises 49–52,1determine where f (x) is increasing.
g(t) = cos
49. f (x) = |x + 1|t solution A graph of the function y = |x + 1| is shown below. From the graph, we see that the function is increasing
on the interval (−1, ∞).
y
2
1 −3 −2 −1 1 x 51. f (x) = x 4 3
f (x) = x
solution A graph of the function y = x 4 is shown below. From the graph, we see that the function is increasing on
the interval (0, ∞).
y
12
8
4
−2 −1 1 2 x In Exercises 53–58, find
1 the zeros of f (x) and sketch its graph by plotting points. Use symmetry and increase/decrease
f (x)where
= 4 appropriate.
information
x + x2 + 1
2
53. f (x) = x − 4
solution Zeros: ±2
Increasing: x > 0
Decreasing: x < 0
Symmetry: f (−x) = f (x) (even function). So, y-axis symmetry.
y
4
2
−2 −1 1 −2 2 x −4 55. f (x) = x 3 − 4x
f (x) = 2x 2 − 4
solution Zeros: 0, ±2; Symmetry: f (−x) = −f (x) (odd function). So origin symmetry.
y
10
5
−2 −1 −5
−10 June 7, 2011 1 2 x LTSV SSM Second Pass Real Numbers, Functions, and Graphs S E C T I O N 1.1 5 57. f (x) = 2 − x33
f (x) = x
√
solution This is an x-axis reflection of x 3 translated up 2 units. There is one zero at x = 3 2.
y
20
10
−2 −1−10
−20 1 x 2 59. Which of the curves in Figure 26 is the graph of a function?
1
f (x) =
(x − 1)2 + 1
y
y
x
x
(A) (B)
y y x x (C) (D) FIGURE 26 solution (B) is the graph of a function. (A), (C), and (D) all fail the vertical line test. 61. Determine whether the function is even, odd, or neither.
Determine1 whether the 1function is even, odd, or neither.
3−
2−t
(a) f(a)
(t)f=(x)4= x 5
−
(b)(b)g(t)
==
2t t−
g(t)
t2
t + t + 1 t4 − t + 1
1
(c) G(θ)
(d) H (θ) = sin(θ 2 )
(c) F=
(t)sin
= θ 4+ cos2 θ
t +t
solution
(a) This function is odd because
f (−t) = 1
1
−
(−t)4 + (−t) + 1 (−t)4 − (−t) + 1 1
1
= 4
− 4
= −f (t).
t −t +1 t +t +1
(b) g(−t) = 2−t − 2−(−t) = 2−t − 2t = −g(t), so this function is odd. (c) G(−θ ) = sin(−θ ) + cos(−θ ) = − sin θ + cos θ which is equal to neither G(θ) nor −G(θ), so this function is
neither odd nor even.
(d) H (−θ ) = sin((−θ )2 ) = sin(θ 2 ) = H (θ), so this function is even. 1
3 + 12x
2 −=3x + 4 as
63. Determine
f (x)
is increasing or decreasing.
Write f the
(x) interval
= 2x 4 −on5xwhich
x − 4 the sum of an even and an odd function. solution A graph of the function is shown below. From this graph we can see that f (x) is decreasing on (−∞, 4) and
also decreasing on (4, ∞).
y
6
4
2
0 x
2 −2 4 6 8 10 −2
−4
−6 June 7, 2011 LTSV SSM Second Pass 6 CHAPTER 1 PRECALCULUS REVIEW In Exercises
f (x)
be theisfunction
shown
in Figureor27.
State 65–70,
whetherletthe
function
increasing,
decreasing,
neither.
(a)
(b)
(c)
(d) Surface area of a sphere as a function of its radius
y
Temperature at a point on the equator as a function
of time
4 price of oil
Price of an airline ticket as a function of the
3 of volume
Pressure of the gas in a piston as a function
2
1
0 1 2 3 4 x FIGURE 27 65. Find the domain and range of f (x)?
solution D : [0, 4]; R : [0, 4] %1 &
67. Sketch
the the
graphs
of fof(2x),
, and
2f (x).
2 xand
Sketch
graphs
f (x f+ 2)
f (x)
+ 2. solution The graph of y = f (2x) is obtained by compressing the graph of y = f (x) horizontally by a factor of 2 (see
the graph below on the left). The graph of y = f ( 12 x) is obtained by stretching the graph of y = f (x) horizontally by a
factor of 2 (see the graph below in the middle). The graph of y = 2f (x) is obtained by stretching the graph of y = f (x)
vertically by a factor of 2 (see the graph below on the right).
y y y 4 4 8 3 3 6 2 2 4 1 1
1 2 3 4 x 2
2 f (2x) 4 6 8 x f (x/2) 1 2 3 4 x 2 f(x) 69. Extend the graph of f (x) to [−4, 4] so that it is an even function.
Sketch the graphs of f (−x) and −f (−x).
solution To continue the graph of f (x) to the interval [−4, 4] as an even function, reflect the graph of f (x) across
the y-axis (see the graph below).
y
4
3
2
1
−4 −2 2 4 x 71. Suppose that f (x) has domain [4, 8] and range [2, 6]. Find the domain and range of:
Extend the graph of f (x) to [−4, 4] so that it is an odd function.
(a) f (x) + 3
(b) f (x + 3)
(c) f (3x) (d) 3f (x) solution
(a) f (x) + 3 is obtained by shifting f (x) upward three units. Therefore, the domain remains [4, 8], while the range
becomes [5, 9].
(b) f (x + 3) is obtained by shifting f (x) left three units. Therefore, the domain becomes [1, 5], while the range remains
[2, 6]. (c) f (3x) is obtained by compressing f (x) horizontally by a factor of three. Therefore, the domain becomes [ 43 , 83 ],
while the range remains [2, 6]. (d) 3f (x) is obtained by stretching f (x) vertically by a factor of three. Therefore, the domain remains [4, 8], while the
range becomes [6, 18].
73. Suppose that the graph of f (x) = sin x is compressed horizontally by a factor of 2 and then shifted 5 units to the
2
right. Let f (x) = x . Sketch the graph over [−2, 2] of:
(a) f is
(xthe
+ 1)
(b) f (x) + 1
(a) What
equation for the new graph?
(c)
f
(5x)
(d)by5f
(b) What is the equation if you first shift by 5 and then compress
2?(x)
(c) Verify your answers by plotting your equations. June 7, 2011 LTSV SSM Second Pass Real Numbers, Functions, and Graphs S E C T I O N 1.1 7 solution
(a) Let f (x) = sin x. After compressing the graph of f horizontally by a factor of 2, we obtain the function g(x) =
f (2x) = sin 2x. Shifting the graph 5 units to the right then yields
h(x) = g(x − 5) = sin 2(x − 5) = sin(2x − 10).
(b) Let f (x) = sin x. After shifting the graph 5 units to the right, we obtain the function g(x) = f (x − 5) = sin(x − 5).
Compressing the graph horizontally by a factor of 2 then yields
h(x) = g(2x) = sin(2x − 5).
(c) The figure below at the top left shows the graphs of y = sin x (the dashed curve), the sine graph compressed
horizontally by a factor of 2 (the dash, double dot curve) and then shifted right 5 units (the solid curve). Compare this last
graph with the graph of y = sin(2x − 10) shown at the bottom left.
The figure below at the top right shows the graphs of y = sin x (the dashed curve), the sine graph shifted to the right
5 units (the dash, double dot curve) and then compressed horizontally by a factor of 2 (the solid curve). Compare this last
graph with the graph of y = sin(2x − 5) shown at the bottom right.
y y 1 1 −6 −4 −2 2 4 x 6 −6 −4 −2 −1 2 4 6 2 4 6 x −1
y y
1 1 −6 −4 −2 2 4 x 6 −6 −4 −2 x −1 −1 % &
75. Sketch
the28
graph
of fthe
(2x)
andof
f f12(x)
x ,=
where
(x)Match
= |x|the
+ 1functions
(Figure 28).
Figure
shows
graph
|x| +f1.
(a)–(e) with their graphs (i)–(v).
solution
by a factor of 2
(a) f (x The
− 1)graph of y = f (2x) is obta...

View
Full Document