76906943-MATH21-Module-Part-1 - 1 MODULE I FUNCTIONS...

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Unformatted text preview: 1 MODULE I FUNCTIONS CONTENTS: Lesson 1: Lesson 2: Lesson 3: Lesson 4: Lesson 5: Functions and Functional Notation Piecewise Defined Functions Domain and Range of Functions Graphs of Functions Operations on Functions OVERVIEW OF THE MODULE: One of the fundamental concepts that students deal with in calculus is functions. In this module, we will define and develop the concept of functions, their graphs, ways of transforming and combining functions and enumerate some important functions that plays very important role in calculus. GENERAL OBJECTIVES: After completing this module, the students are expected to accomplish the following:       define functions; evaluate functions; graph functions; determine the domain and range of functions; transform one function into another function and combine functions to obtain new functions; and enumerate some important functions and study their basic properties. 2 HISTORICAL NOTES: ³The Calculus´ is a phrase we use to denote that branch of mathematics which studies properties of functions (³curves´) which are associated with the limit process (continuity, differentiation, integration). Calculus is the introductory level of a more general branch of mathematics which is called ³analysis.´ Analysis deals generally with infinite processes and includes such areas as real analysis, complex analysis, and differential equations. Kline calls the calculus one of the two greatest creations in the history of mathematics. The term function ± Latin functio ± first appeared in a mathematical article in the Acta Eruditorem to denote various tasks that a straight line may accomplish with respect to a curve, such as forming a chord, tangent, or normal. The article was signed O.V.E. but is attributed to Gottfried von Leibniz, the German mathematician. In an article from 1694, also in Acta Eruditorem, Leibniz gave the term ³function´ a more specific meaning by letting it denote the slope of a curve, a definition that has very little in common with the present day mathematical definition of a function. The Swiss mathematician Leonhard Euler in 1749 defined a function as a variable quantity that is dependent upon another quantity, thereby approaching today¶s definition. The notation f(x) is attributed to him and he wrote one of the most influential algebra books of all time. Euler¶s definition was challenged when the French physicist and mathematician Joseph Fourier in 1822 presented his work on heat flow (Theorie analytique de la Chaleur). For his investigations, Fourier introduced series with sines and cosines as terms, which led to the concept that a given representation of a function may be valid for only a certain range of values. Based on Fourier¶s investigations, the German mathematician Lejeune Dirichlet in 1837 proposed that, from the mathematical point of view, a function is a correspondence that assigns a unique value of the dependent variable to every permitted value of an independent variable. There will be a reason to return to this definition many times in this text. Another milestone in the development of analysis was the appearance of The Theory of Analytic Functions, by Joseph Louis Lagrange, in 1797. This may be thought of as the beginning of the modern theory of functions of a real variable. (adapted from Mathematics From the Birth of Numbers by Jan Gullberg, 1997) 3 SUGGESTED READINGS: Goldstein, Lay and Schneider. Calculus and its Applications (7th edition). Prentice Hall, Singapore: 1998. Anton, Howard. Calculus Brief Edition (6th edition). W iley, Singapore: 1999. Stewart, James. Calculus ± Concepts and Contexts (2nd edition). Brooks/Cole, Singapore: 2001. 4 LESSON 1: Functions and Functional Notation SPECIFIC OBJECTIVES: At the end of this lesson, the students are expected to accomplish the following: y define functions; y distinguish between dependent and independent variables; y represent functions in different ways; and y evaluate functions. INSTRUCTIONAL STRATEGIES: Class discussion using highly-expository instructional method is encouraged to be used in this topic. Students are expected to complete the exercises in this lesson before introducing the next lesson. The instructor will use a portion of the class time to discuss problems in Exercise 1.1 while other problems may be assigned as homework for mastery. This lesson may also be introduced using the illustration below: A software consulting company monitored the costs of different software development projects and summarized observations in a table: Project Name 001 002 003 004 005 006 007 Duration ( in months) 15 10 50 25 10 25 15 Cost ( in P1,000) 315 215 1,015 515 215 515 315 Let them think about this question: ³Did you notice that there is a simple relationship between duration and cost? Can you write a formula to compute the cost based on the value of the duration?´ If we look at the table, we find that Cost = 15 + 20 * Duration. This is a much compact and useful description of the data than the table. You can use this formula to describe any project in the table and also to estimate the cost of a new project. 5 In order to gauge the students¶ understanding of the concept of functions, ask your students to collect clippings that illustrate functions. Ask your students to illustrate the function using the different representations of functions. Identify the dependent and independent variables. DISCUSSION OF THE TOPICS: In many practical situations, the value of one quantity may depend on the value of a second quantity. For example, The area A of a square depends on the value of its edge s. The rule that associates A and s is given by A = s2, such that for each value of s there corresponds one value of A. The person¶s salary S may depend on the number of hours h worked. The fare F on public utility jeepney depends on distance d traveled. For instance, for the first 4 kilometers, F is equal to P4.00 and an additional P0.50 for the next succeeding kilometers, that is F = 4, 0ede4 F = 4 + 0.50d, d>4 The value of F depends on the value of d. Such relationships can often be represented mathematically as functions. A function is a correspondence from a set X of real numbers x to a set Y of real numbers y, where the number y is unique for a specific value of x. This definition is illustrated in Figure 1.1. X Y y y y y y Figure 1.1 A variable y is a function of x if a relationship between x and y produces exactly one value of y for each value of x. 6 Example 1.1. According to a certain function, a number is obtained by subtracting 1 from the square of a certain number. What number does this function assign to 2? Solution: If we let y as the number produced by applying the function to the certain number, say x, then the function is expressed as y = x2 ± 1. Thus, if x = 0, y = -1; if x = 1, y = 0; if x = -1, y = 0; if x = 2, y = 3; etc. Illustrating these values on a table, we have x 0 1 -1 2 y -1 0 0 3 yyy Therefore, the number assigned to 2 is 3. Furthermore, in example 1.1, the set of values of x with its corresponding values of y can be denoted by (x, y), called an ordered pair. Thus, the set of all (x, y), say set A, as described in the function is A ={(0, -1), (1, 0), (-1, 0), (2, 3), . . .}. The numerical value of the variable y is determined by a chosen value of the variable x. For this reason, y is sometimes referred to as the dependent variable and x as the independent variable. In order to have a function, there must be one value of the dependent variable for each value of the independent variable. Or, there could also be two or more independent variables for every dependent variable. These correspondences are called one-to-one correspondence and many-to-one correspondence, respectively. Therefore, a function is a set of ordered pairs of numbers (x, y) in which no two distinct ordered pairs have the same first number. 7 Example 1.2. Determine whether or not each of the following sets represents a function: a. A = {(-1, -1), (10, 0), (2, -3), (-4, -1)} b. B = {(2, a), (2, -a), (2, 2a), (3, a2)} c. C = {(a, b)| a and b are integers and a = b2) d. D = {(a, b)| a and b are positive integers and a = b2} e. E = {(x, y)| y2 = x  4 } Solutions: a. A is a function. There are more than one element as the first component of the ordered pair with the same second component namely (-1, -1) and (-4, -1), called a many-to-one correspondence. One-to-many correspondence is a not function but many-to-one correspondence is a function. b. B is a not a function. There exists one-to-many correspondence namely, (2, a), (2, -a) and (2, 2a). c. C is not a function. There exists a one-to-many correspondence in C such as (1, 1) and (1, -1), (4, 2) and (4, -2), (9, 3) and (9, -3), etc. d. D is a function. The ordered pairs with negative values in solution c above are no longer elements of C since a and b are given as positive integers. Therefore, one-to-many correspondence does not exist anymore in set D. e. E is not a function because for every value of x, y will have two values. As we have seen in the discussions above, functions can be represented in four basic ways: numerically by tables or set notations; geometrically by graphs (to be illustrated later); algebraically by formulas; and, verbally by descriptions. Function Notation. A versatile notation for function is widely used. A letter such as f is chosen to stand for the function itself, and the value that the function assigns to x is denoted by f(x) where x is called the argument of the function. The symbol f(x) is read as ³f of x.´ Other letters such as h, g, F, G and H may also be used to denote functions. 8 Example 1.3. If f(x) = x2 ±1, find: (a) f(3), (b) f(0), (c) f(a), (d) f(a2 +1), (e) [f(a)]2, and (f) [f(a + h)]2. Solutions: a. f(3) = 32 ± 1 =8 b. f(0) = 02 ± 1 = -1 c. f(a) = a2 ± 1 d. f(a2 + 1) = (a2 + 1)2 ± 1 = a4 + 2a2 + 1 ± 1 = a4 + 2a2 e. [f(a)]2 = (a2 ±1)2 f. [f(a + h)]2 = [(a + h)2 ±1]2 = [(a2 + 2ah + h2) ± 1]2 = (a2 + 2ah + h2 ± 1)2 Example 1.4. Find (a) g(2 + h), (b) g(x + h), (c) = g( x  h)  g( x ) where h { 0 if g(x) h x . x 1 Solutions: 2h 2h 2h = = . ( 2  h)  1 2  h  1 3  h xh b. g( x  h) ! x  h 1 (x  1)(x  h)  x(x  h  1) xh x  g(x  h)  g(x) x  h  1 x  1 (x  h  1)(x  1) c. ! ! h h h a. g(2 + h) = h x 2  hx  x  h  x 2  hx  x ! h(x  1)(x  h  1) h(x  h  1)(x  1) 1 . ! (x  1)(x  h  1) ! 9 Example 1.5. Given that F(x) = 1/x, show that F(x + k) ± F(x) = k . x  kx 2 Solution: 1 , then xk 1 1  F(x + k) ± F(x) = x k x x  x k ! x( x  k ) k !2 x  kx F(x) = 1/x and F(x + k) = Example 1.6. Given H(y) = 4y , show that H(y + 1) ± H(y) = 3H(y). Solution: H(y) = 4y; H(y + 1) = 4y + 1, then H(y + 1) ± H(y) = 4y + 1 ± 4y = 4y (4 ± 1) = 3(4y); but H(y) = 4y. Therefore: H(y + 1) ± H(y) = 3 H(y). Example 1.7. If f(x, y) = x3 + 4xy2 + y3 , show that f(ax, ay) = a3f(x, y). Solution: Since f(x, y) = x3 + 4xy2 + y3 , then f(ax, ay) = (ax)3 + 4(ax)(ay)2 + (ay)3 = a3x3 + 4a3xy2 + a3y3 = a3(x3 + 4xy2 + y3); but f(x, y) = x3 + 4xy2 + y3 Therefore: f(ax, ay) = a3f(x, y). 10 Example 1.8. If f(u, v) = uv , uv ¨ 1 1¸ find f © , ¹  f(u, v). ªu vº Solution: 1 1 v u  uv 1 1¸ u v v -u ¨ ! uv ! , and f © , ¹ ! , then f(u, v) = uv ªu vº 1  1 v u u v uv uv v -u u- v v -u-u v 1 1¸ ¨ f © , ¹  f(u, v). !  ! u v uv uv ªu v º ! uv - 2u  2v ¨u v¸ ! 2© ¹ ; but f(u, v) ! uv u v ªu vº ¨ 1 1¸ Therefore, f © , ¹  f(u, v) ! -2f(u, v). ªu v º 11 EXERCISES: 1. Which of the following represents a function? a. A = {(2, -3), (1, 0), (0, 0), (-1, -1)} b. B = {(a, b)|b = ea} c. C = {(x, y)| y = 2x + 1} d. e. f. g. h. i. D = {(a, b)|b = 1  a2 } E = {(x, y)|y = (x -1)2 + 2} F = {(x, y)|x = (y+1)3 ± 2} G = {(x, y)|x2 + y2 = 1} H = {(x, y)|x e y} I = {(x, y)| |x| + |y| = 1} j. J = {(x, y)|x is positive integer and x = 1 7} y3 2. Given the function f defined by f(x) = 2x2 + 3x ± 1, find: a. f(0) f. f(3 ± x2) b. f(1/2) g. f(2x3) c. f(-3) h. f(x) + f(h) d. f(k + 1) i. [f(x)]2 ± [f(2)]2 f ( x  h)  f ( x ) e. f(h ± 1) j. ;h { 0 h 3. Given F(x) = a. F(-1) b. F(4) 2x  3 , find: d. F(1/2) F( x  h)  F( x ) ;h { 0 e. h c. F(2x + 3) 4. If J(x) = cos x, find J(0), J(T/2), J(T), J(-x) and J(T - y). 5. If G(x) = tan x, find G(T/6), G(x - T/2), G(-x) and express G(2x) as function of G(x). 6. Find all real values of x such that f(x) = 0 for f ( x ) ! 3 4  . x 1 x  2 7. Find the values of x for which f(x) = g(x) where f ( x ) ! 2x  1 and g(x) = x ± 1. 8. Let f(x) = ax such that f(1) = 6. What is the value of a? What is the value of a if f(1/2) = 1/8? 12 9. If h(x) = sin x, show that h(2x) = 2 h(x) h(T/2 ± x). 10. If g(y) = y 1 , show that [g( y )  g( y )] ! g( y 2 ). 1 y 2 11. If H(x) = logb 1/x, show that H(b-1/z) = 1/z. 13 LESSON 2: Piecewise Defined Functions OBJECTIVES: At the end of this lesson, the students are expected to accomplish the following: y define piecewise defined functions; y evaluate piecewise defined functions, and y enumerate some of the common functions that can be defined as piecewise defined functions. INSTRUCTIONAL STRATEGIES: The concept of a piecewise defined function is extremely important but it is given less attention and where problems involving this matter occurs without the introduction of its concepts to students. This approach is not quite bad but a more effective approach can be done to improve the students understanding about functions. Focus on transformation of absolute value functions into piecewise defined functions by maybe, discussing example 2.2 and example 2.3 and allowing students to solve exercise 2-1 as a group by focusing on one problem per group. Give them time to explain briefly this solution in class. DISCUSSION OF THE TOPICS: Sometimes a function is defined by more than one rule or by different formulas. This function is called a piecewise define function. A piecewise defined function is a function which is defined symbolically by using two or more functions. 14 if x 5 ® x3 ± 2 Example 2.1. A function f is defined by f ( x ) ! ¯ 25  x if  5 e x e 5 ± 3x if 5x ° Evaluate: f(-6), f(0), and f(10). Solution: To evaluate the function at the given value of x, consider the formula based on the value of x to which the function is to be evaluated. For instance, to evaluate f(-6), consider the first formula, since x = ±6 is less than 5, f(x) = x + 3 if x < -5. Then, f(-6) = -6 + 3 = -3. To evaluate f(0), consider f(x) = ±5 e x e 5. 25  x 2 which is the second formula since f (0) ! 25  02 = 5. [ why f(0) = 5 and not s5 considering 25 ! s5 ? ]. Evaluating f(10), we shall consider f(x) = 3 ± x, the third formula, since 5 < 10, f(10) = 3 ± x = 3 ± 10 = -7. Example 2.2. Define g(x) = |x| as a piecewise defined function and evaluate g(2), g(0) and g(2). Solution: From the definition of |x|, ®x if x u 0 . g( x ) ! ¯  ° x if x 0 Therefore, g(-2) = -(-2) = 2 g(0) = 0 g(2) = 2. 15 Example 2.3. Define F(x) = |x ± 1| - |x + 1| without absolute value bars, piecewise in the following intervals: (-g, -1), [-1, 1), [1, +g). Solution: From the definition of absolute value function, ® x  1 if x  1 u 0 | x  1 |! ¯  ° ( x  1) if x  1 0 ®x  1 if x u 1 or | x  1 |! ¯  ° x  1 if x 1. ® x  1 if x  1 u 0 Also, | x  1 |! ¯  ° ( x  1) if x  1 0 ®x  1 if x u 1 or | x  1 |! ¯  ° x  1 if x 1. If x  ( g,1) , |x ± 1| = -x + 1 and | x + 1| = -x ± 1 and |x ±1| - |x + 1| = -x + 1 ± (-x ±1) =-x+1+x+1 = 2. If x  [ 1,1) , |x ± 1| = -x + 1 and | x + 1| = x + 1 and |x ±1| - |x + 1| = -x + 1 ± (x +1) = - x + 1 - x -1 = -2x. If x  [1,g) , |x ± 1| = x - 1 and | x + 1| = x + 1 and |x ±1| - |x + 1| = x - 1 ± (x +1) = x-1-x-1 = -2. Therefore, the required piecewise defined function F(x) is defined as x 1 ® 2 if ± F( x ) ! ¯ 2x if  1 e x 1 ± 2 if  x u 1. ° Aside from absolute value function, there are other functions that can be defined as piecewise defined functions such as signum function and unit-step function. See exercise 1.2 numbers 2 and 4 for definition of these functions. 16 EXERCISES: 1. Compute the indicated values of the given functions: t 4 ®3 if ± a. f ( x ) ! ¯t  1 if  4 e t e 4 , ± t if t"4 ° f(-6), f(-4) and f(16)  x 2 ® 4 if ± b. h( x ) ! ¯  1 if  2 e x e 2 , ±3 if 2"x ° h(-3), h(-2), h(T/2), h(e2) and h(2). ® 2  4 if x { 3 x , c. F( x ) ! ¯  2 if x ! 3 ° f(4), f(0), f(-3) and f(-2/3) 2. A unit step function of x denoted by U, is defined as 0 ® if x 0 U( x ) ! ¯ . 1 ° if 0 e x Formulate a piecewise defined funct ion for U(x ± 1) and evaluate U(-1), U(0) and U(1). 3. Define H(x) as a piecewise defined function and evaluate H(1), H(2), H(3), H(0) and H(-2) given by, H(x) = x - |x ± 2|. 4. Given a function h(x) function, is defined as  ® 1 if ± sgn( x ) ! ¯ 0 if ± if 1 ° = x ± 2 sgn(x) where sgn(x), called signum x0 x ! 0. 0x Evaluate: h(-2), h(0) and h(2). 17 5. For a given outside temperature T and wind speed v, the windchill index (WCI) is the equivalent temperature that exposed skin would feel with a wind speed of 4 mi/hr. An empirical formula for the WCI based on experience and observation is T, 0eve4 ® ± WCI ! ¯91.4  (91.4  T )(0.0203 v  0.304 v  0.474 ), 4 v 45 ± 1.6T  55, v u 45 ° where T is the air temperature in oF, v is the wind speed in mi/hr, and WCI is the equivalent temperature in oF. Find the WCI if the air temperature is 25oF and a) v = 3 mi/hr, b) v = 46 mi/hr and c) v= 15 mi/hr. 6. The number F(t) of fast food chains and related enterprises in the Philippines can be approximated by the following functions of time t in years since 1995: if 0 e t e 5 ® 50t  800 C(t) = ¯ 1, 5t °300 t  2,400 if a) Evaluate C(0), C(5) and C(7) and interpret the results. b) How fast was the number of fast food chains in the Philippines growing in 1997 and 2001? c) Use the model to estimate when there were 5,400 fast food chains in the Philippines. 18 LESSON 3: Domain and Range of Functions OBJECTIVES: At the end of this lesson, the students are expected to accomplish the following: y define domain and range of a function and y determine the domain and range of a function. INSTRUCTIONAL STRATEGIES: In order for the students to understand the concept of domain, it is suggested that you discuss the following topic: A calculator is designed to give the user an answer when the user entered a value. The calculator processes the input to produce the desired output. The processing that the calculator performs corresponds to the ³function.´ When the calculator produced an error message, it simply means that the input is not within the domain of the function. Enumerated below are some calculator keys, the equivalent function, input giving error message and domain. KEY x-1 x Sin-1x FUNCTION INPUT GIVING ERROR MESSAGE DOMAIN f(x) = 1/x x=0 all real numbers except 0 f(x) = x<0 all nonnegative integers x f(x) = Sin-1x x > 1 or x < 1 [-1, 1] Ask your students to experiment on the following keys and construct a table similar to the above: ln x, 3 x , Sin-1x, x2, Tan-1x 19 DISCUSSION OF THE TOPICS: A function has three elements, namely: the domain, the range and the rule of correspondence. Definition If y = f(x), then its domain is the set of all possible inputs (x-values) and its range is the set of outputs (y-values) that result when x varies over the domain of f. In the definition, the x variable represents the independent variable while the y variable represents the dependent. Hence the above definition may be restated as follows: In a function, the domain is the set of values that the independent variable is allowed to assume and the range is the set of all values that the dependent variable assumes. The rule of correspondence, or simply rule, is the formula which assigns to each element of the first set one and only one element in the second set. In the given figure below (Figure 1.1), set X refers to the domain and set Y refers to the range of the function defined. X Y y y y y y Figure 1.1 Example 3.1. Consider the set of ordered pairs {(2, 5), (3, 7), (4, 9), (5, 11)}. W hat are its do...
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