3.3 Decomposition of Functions 121 equation u 2 − 5u + 4 = 0 for u. Once we have found u, we can solve 7 for x easily because we know u =x 2 . Breaking the original (hard) problem into two (easier) subproblems makes solving the original problem possible. u 2 − 5u + 4 = 0 (u − 4)(u −1) = 0 u = 4oru=1 But u = x 2 . x 2 = 4orx 2 =1 x=±2orx=±1. The solutions to x 4 − 5x 2 + 4 = 0 are x =−2, −1, 1, 2. ◆ Calculus involves the study of rates of change of functions. Later in this text, when we are looking for the rate of change of a composite function, we will use the same approach of decomposition to decompose the function and find its rate of change from our knowledge of the rates of change of these simpler functions. 8 PROBLEMS FOR SECTION 3.3 1. Let h(x) =f(g(x))and suppose that h(x) = 1 √ x 2 +6 . Write possible formulas for f(x) and g(x). 2. For each of the functions given below, give possible formulas for f(x)and g(x) such that h(x) =f(g(x)).Donot let g(x) =x; do not let f(x)=x. (a) h(x) = √ x 2 + 3 (b) h(x) = √ x + 5 √ x (c) h(x) = 3 3x 2 +2x (d) h(x) = 5(x 2 + 3x 3 ) 3 3. Let j(x)= 2 3 √ 4x 2 +3x . Suppose that j(x)=h(g(f (x))). Write possible formulas for f(x),g(x), and h(x). None of f , g, and h should be the identity function. 4. Let j(x) =10(x −2 + 2x 2 ) 3 . Give two possible decompositions of j(x) such that j(x)=f(g(h(x))). None of the functions f , g, and h should be the identity function. 5. Let h(x) =f(g(x)),where f(x)=x 2 .If h(t − 2) =  1 t − 2 + 1  2 , what is g(x)? 7 For more on solving quadratics, see Appendix A: Algebra. 8 The method of constructing the rate of change function of a composite function is called the Chain Rule. 122 CHAPTER 3 Functions Working Together Decompose the functions in Problems 6 through 9 by finding functions f(x)and g(x), f(x)=x and g(x) =x, such that h(x) =f(g(x)). 6. h(x) =(x 2 + 7x + 1) 3 7. h(x) = 1 x 2 +4 8. h(x) = √ x 2 + 1 9. h(x) =( √ x) 3 −2 √ x + 3 In Problems 10 through 17, find functions f and g such that h(x) = f(g(x))and neither f nor g is the identity function, i.e., f(x)=xand g(x) =x. Answers to these problems are not unique. 10. h(x) = 3 √ x+2 11. h(x) =3x 4 + 2x 2 + 3 12. h(x) =5(x − π) 2 + 4(x − π) + 7 13. h(x) =2|3x − 4| 14. h(x) = x+1 x+2 15. h(x) = √ 5x 2 + 3 16. h(x) =3 2x + 3 x + 1 17. h(x) =4π 2x + 3π x + 2 In Problems 18 through 20, find functions f , g, and h such that k(x) =f(g(h(x))) and f(x)=x,g(x) =x, and h(x) = x. 18. k(x) = 3 √ x 2 +4 19. k(x) = 1 ( √ x+1) 9 20. k(x) =  (x 2 + 1) 3 + 5 Exploratory Problems for Chapter 3 123 Exploratory Problems for Chapter 3 Flipping, Shifting, Shrinking, and Stretching: Exercising Functions In these exercises you will experiment with altering the input and output of functions and draw conclusions about the effects of these alterations on the graph of the function. Part of being a scientist is learning how to design an experiment. A graphing calculator or a computer will be an invaluable tool. We’ll suggest some experiments, but you are expected to come up with most of them on your own. Experiment until you have confidence in your conclusions. Look for patterns, test your discoveries, and then try to generalize. When you generalize, think about why these generalizations make sense. Com- pare your conclusions with those of your classmates and resolve any discrepancies. 1. Modifying the output of f (a) How does the graph of y = kf(x)(where k is a constant) relate to that of y = f(x)? You might want to break your answer into cases depending upon the sign of k and whether |k| is greater than, less than, or equal to 1. i. Does multiplying f(x)by a constant change the location of its zeros? If so, how? ii. Does multiplying f(x) by a constant affect where the graph has turning points (peaks and valleys)? Does it affect the value of the function at these peaks and valleys? (b) How does the graph of y = f(x)+k(where k is a constant) relate to that of y = f(x)? i. Does adding a constant to f(x)change the location of its zeros? ii. Does adding a constant to f(x) affect where the graph has turning points (peaks and valleys)? Does it affect the value of the function at these peaks and valleys? (c) How does the graph of y = cf(x)+k (where c and k are constants) relate to that of y = f(x)? In particular, let f(x)=x 2 .Onthe same set of axes sketch the graphs of f(x)and h(x), where h(x) =−f(x)+2. Basic order of operation rules tell us to multiply by −1 first and then add 2. Do we follow the same procedure when analyzing the function graphically? In other words, to obtain the graph of h do we first flip the graph of f across the x-axis and then shift it up 2 units? Demonstrate that if we first shift the graph of f up 2 units and then flip the result across the x-axis that we will obtain the graph of y =−[f(x)+2], not the graph of h. 124 CHAPTER 3 Functions Working Together 2. Altering the function’s input (a) How is the graph of y = f(x −k) (where k is a constant) related to that of y = f(x)? Distinguish between the cases where k is positive and where k is negative. i. How are the zeros of f(x −k) related to the zeros of f(x)? In particular, if x = r is a zero of f(x),what can we say about a zero of f(x−k)? ii. Does replacing x by (x − k) affect where the graph has turning points (peaks and valleys)? Does it affect the value of the function at these peaks and valleys? iii. Given the generalizations you made, how would you ex- pect the graph of y = 1 x−k to look? Will the position of the vertical asymptote be affected by k? Will the position of the horizontal asymptote be affected by k? Explain your reasoning. Check your expectations by looking at some concrete cases. (b) How is the graph of y = f(kx) (where k is a constant) related to that of y = f(x)?Try looking at a function such as f(x)= (x − 1)(x − 2)(x + 4) and experimenting with various values of k. Break your answer into cases: k > 1, k < 1, k =−1,k<−1, and −1 < k < 0. i. What is the effect on the x-intercepts? ii. Do the locations of the peaks and valleys change? iii. Do the heights of the peaks and valleys change? 3. Alterations to both input and output (a) Let p(x) =− 1 2 (x + 3) 2 − 1. Interpret the graph of p in terms of transformations of a familiar function. Note: p(x) can be viewed as a variation of the familiar function f(x)=x 2 .Un- raveling a function is just like unraveling a very complicated algebraic expression; you need to start from the innermost part and remember the order of operations rules. (b) Let q(x) =− 2 x−4 + 3. Interpret the graph of q in terms of transformations of a familiar function. Where are the vertical and horizontal asymptotes? Exploratory Problems for Chapter 3 125 (c) Below is the graph of f(x).Graph the following. i. y = f(2x) ii. y = 3f( x 2 ) iii. y =−f(x−1) iv. y = 2f(x −1)+3 – 4 – 4 –3 –3 –2 –2 –1 –1 1 1 3 3 2 2 4 4 f x (1, 3) (3, –1) (–2, 2) (– 4, –1) 126 CHAPTER 3 Functions Working Together 3.4 ALTERED FUNCTIONS, ALTERED GRAPHS: STRETCHING, SHRINKING, SHIFTING, AND FLIPPING Suppose we are quite familiar with a certain function f and know the shape of its graph. Now suppose we alter this function f by composing it with another function, g, where g is a very simple function. 9 We’ll create a new but related function. By constructing g(f (x)) we alter the output of f . By constructing f(g(x))we alter the input of f . Suppose, for instance, g(x) = 3x. Then g(f (x)) = 3f(x) and f(g(x))=f(3x). If g(x) = x − 2, then g(f (x)) = f(x)−2 and f(g(x))=f(x −2). If you are a good sleuth, you can determine a great deal about these new but related functions, such as 3f(x)or f(x −2)or f(3x) − 2. The exploratory problems preceding this section were designed to help you figure out how your knowledge of f can lead you to understand its close relatives. Being able to construct functions that stretch, shrink, shift, and flip graphs with which you are familiar is an extremely powerful tool. Not only can it brighten a rainy day, it can also help you model real-life phenomena by enabling you to alter functions you know. Conversely, being able to take what you know about f and apply it to other related functions can save you a great deal of time because you will not have to start from scratch with every new function you encounter. For example, by the time you are done with this section, you should be able to look at a function like r(x) = −1 x−1 + 2 and break it down into a sequence of operations performed on a function that you know well. This will help you determine what the graph of y = r(x) will look like. Variations on the Theme of y = f(x) Let k be a constant. 10 y = kf(x):vertical stretching and shrinking plus flipping across the x-axis If k > 1, the graph of kf(x) is the graph of f(x) stretched vertically away from the x-axis, stretched by a factor of k. If 0 < k < 1, the graph of kf(x) is the graph of f(x) shrunk vertically toward the x-axis, rescaled by a factor of k. If −1 < k < 0, the graph of kf(x)is the graph of f(x)flipped across the x-axis and rescaled vertically by a factor of |k|. If k =−1, the graph of kf(x)is the graph of f(x)flipped across the x-axis. If −1 < k, the graph of kf(x) is the graph of f(x) flipped across the x-axis and stretched by a factor of |k|. 9 We will begin by looking at g where g(x) is of the form ax + b, a linear function. 10 A constant, such as k, that can vary from equation to equation but can be used to represent a family of related equations is often referred to as a parameter. 3.4 Altered Functions, Altered Graphs: Stretching, Shrinking, Shifting, and Flipping 127 Suppose that a is a zero of f(x). This means that f(a)=0.Therefore kf(a)= k·0=0aswell. kf(x)and f(x)have the same roots. Multiplying f(x)by a constant does not affect the location of its zeros. y = f(x)+k: vertical shifts If k > 0, the graph of f(x)+kisthe graph of f(x)shifted up k units. If k < 0, the graph of f(x)+kisthe graph of f(x)shifted down |k| units. y = cf(x)+k, where c and k are constants Just as 2a − 3 is not the same as 2(a − 3) = 2a − 6, the function 2f(x)−3isnot the same as 2(f (x) − 3). Following order of operations rules, we do the multiplication (i.e., the stretching or shrinking and possible flipping around the x-axis) before we do the addition or subtraction (i.e., the shifting up or down). Aside: Notice that because 2(f (x) − 3) = 2f(x)−6, we can say that shifting down 3 units then stretching vertically by a factor of 2 is the same as stretching vertically by a factor of 2 then shifting down 6 units. 2 1 y x y = f(x) 21 1 2 y x y = 2f(x) 21 1 2 –1 –2 –3 y x y = 2f(x) –3 21 –1 –2 –3 – 4 –5 – 6 y x y = f(x) – 3 y = 2[ f(x) – 3] Figure 3.9 y = f(x −k):horizontal shifts If k > 0, the graph of f(x −k)is the graph of f(x)shifted k units to the right. For example, replacing x by x − 2 shifts the graph of f to the right 2 units. If k > 0, the graph of f(x +k)is the graph of f(x)shifted k units to the left. For example, replacing x by x + 3 shifts the graph of f to the left 3 units. Be careful here. This is easy to get mixed up; although f(x+3)has a plus sign, the graph is shifted to the left. y = f(kx): horizontal shrinking and stretching plus flipping across the y-axis If k > 1, then the graph of f(kx) is the graph of f(x)compressed horizontally. If 0 < k < 1, then the graph of f(kx) is the graph of f(x)stretched horizontally. 128 CHAPTER 3 Functions Working Together If −1 < k < 0, then the graph of f(kx) is the graph of f(x)flipped across the y-axis and stretched horizontally. If k =−1, then the graph of f(kx) is the graph of f(x)flipped across the y-axis. If k < −1, then the graph of f(kx) is the graph of f(x)flipped across the y-axis and compressed horizontally. 21–134 1 –1 –2 y x y = f(x) 23456781 1 2 –1 –1–2 –2 y x y = f(–2x) y = f x 2 ( ( 21 1 –1 –1–2 –2 y x Figure 3.10 While the effects on the graph of f(x) of altering the output, for instance, y = 2f(x)+3, strike most people as natural, the effects of altering the input of f , such as f(2x) or f(x +1), often provoke consternation. This tends to be the case even if you discover these effects on your own. Below we will rewrite the altering of the output so that the symmetry between the two cases is more transparent. We begin with y = f(x).Toalter the input is to alter x; to alter the output is to alter y. Notice that y = f(x)+2isequivalent to y − 2 = f(x).Similarly, y = 2f(x)is equivalent to y 2 = f(x). Replacing y by y − 2 shifts the graph up 2 units; replacing y by y + 3 shifts the graph down 3 units. Replacing x by x − 2 shifts the graph right 2 units; replacing x by x + 3 shifts the graph left 3 units. Replacing y by y 2 stretches the graph vertically by a factor of 2; replacing y by 2y compresses the graph vertically. The new height is half the old one. Replacing x by x 2 stretches the graph horizontally by a factor of 2; replacing x by 2x compresses the graph horizontally. The x-coordinates are half the old ones. Taking Control: A Portable Problem-Solving Strategy Suppose you forget what happens when x is replaced by x + 2. Do you just say, “I used to know this but I’ve forgotten. If I study it I’ll remember”? No! If you want your knowledge to be portable you’ve got to find ways of reconstructing it so the knowledge that you’ve worked hard to gain doesn’t fly off in the wind or wear thin with the passing of time. 3.4 Altered Functions, Altered Graphs: Stretching, Shrinking, Shifting, and Flipping 129 Even though the details may fade, chances are you’ll remember some broad outlines. For instance, you might recall that alterations of the input variable x will be manifested horizontally. Or maybe you’ll remember that this is a shift, not a stretch. Probably you’ll have some question like “Is this a shift right or a shift left?” Make a hypothesis (for instance, that it is a shift right) and test the hypothesis out on a simple function like y = x 2 . Does the point (2, 0) satisfy the equation y = (x + 2) 2 ? No. Does the point (−2, 0) satisfy the equation y = (x + 2) 2 ? Yes. 21 y x –2 y x Figure 3.11 Learning to ask and answer questions that help you retain information is a large part of being a mathematician or scientist. More Transformations of Output In addition to shifting, stretching, and flipping, we can do a few other interesting things to functions to obtain new functions. Absolute Values: |f(x)| Suppose that v(t) is a function that gives velocity as a function of time. Then the speed function is given by |v(t)|. Similarly, if s(t) is a position function, then |s(t)| gives the distance from a benchmark position. How is the graph of y =|f(x)|related to the graph of y = f(x)?Where the graph of f(x)lies above or on the x-axis the graph of y =|f(x)|is identical; where the graph of f lies below the x-axis the graph of y =|f(x)|is obtained by flipping the graph of f over the x-axis, corresponding to multiplying the y values by −1. Notice that sharp corners are created. –3 –2 –11324 f x –3 –2 –11 324 |f| x Figure 3.12 130 CHAPTER 3 Functions Working Together Reciprocals: 1 f(x) Suppose a function f(t)gives the price of gasoline in dollars gallon at time t. We may be interested in the function 1 f(t) that gives the number of gallons of gas that can be purchased for a dollar. Or suppose a function tracking an employee’s training gives the amount of time it takes her to produce an item, with units of minutes item . We may be interested in the reciprocal function that gives us the number of items she can produce per minute. How is the graph of y = 1 f(x) related to the graph of y = f(x)?Wealready have some experience with graphing reciprocals; we know how to graph f(x)=x and g(x) = 1 f(x) = 1 x . Try the following exercise. EXERCISE 3.12 On the same set of axes, graph y = m(x) = x − 3 and its reciprocal, y = 1 m(x) = 1 x − 3 . Label all intercepts and asymptotes. Notice that you can graph these two functions by simply applying shifting principles to the functions f(x)=xand g(x) = 1 x . Instead of starting out that way, try to think in terms of reciprocals and use the shifts to check your work. General Principles: Sign, Magnitude, and Asymptotes (See Figure 3.13 for examples.) Where f(x)is positive, 1 f(x) is positive; where f(x)is negative, 1 f(x) is negative. Where f(x)is zero, 1 f(x) is undefined. Where |f(x)|=1,| 1 f(x) |=1. Where |f(x)|>1,| 1 f(x) |<1. 1 LARGE positive number = small positive number. Where |f(x)|<1,| 1 f(x) |>1. 1 small positive number = LARGE positive number. Suppose that f(x)→∞as x →∞.Then as x →∞, 1 f(x) →0. Suppose that f(x) →Lasx→∞, where L is a nonzero constant. Then as x →∞, 1 f(x) → 1 L . . function, we will use the same approach of decomposition to decompose the function and find its rate of change from our knowledge of the rates of change of these simpler functions. 8 PROBLEMS FOR SECTION. peaks and valleys? (c) How does the graph of y = cf(x)+k (where c and k are constants) relate to that of y = f(x)? In particular, let f(x)=x 2 .Onthe same set of axes sketch the graphs of f(x )and. the output of f (a) How does the graph of y = kf(x)(where k is a constant) relate to that of y = f(x)? You might want to break your answer into cases depending upon the sign of k and whether