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Exploratory Problems for Chapter 3Flipping, Shifting, Shrinking, and Stretching: Exercising Functions In these exercises you will experiment with altering the input and output of functio

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equation u2− 5u + 4 = 0 for u Once we have found u, we can solve7for x easily because

we know u = x2 Breaking the original (hard) problem into two (easier) subproblems makes solving the original problem possible

u2− 5u + 4 = 0 (u − 4)(u − 1) = 0

u = 4 or u = 1 But u = x2

x2= 4 or x2= 1

x = ±2 or x = ±1

The solutions to x4− 5x2+ 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

P R O B L E M S F O R S E C T I O N 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)) Do not let g(x) = x; do not let f (x) = x

(a) h(x) =√x2+ 3

(b) h(x) =√x +√5

x (c) h(x) =3x23

+2x (d) h(x) = 5(x2+ 3x3)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+ 2x2)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) = x2 If

h(t − 2) =

t − 2+ 1

2 , what is g(x)?

7 For more on solving quadratics, see Appendix A: Algebra.

8

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Decompose the functions in Problems 6 through 9 by finding functions f (x) and g(x),

6 h(x) = (x2+ 7x + 1)3

7 h(x) =x21+4

8 h(x) =√x2+ 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

are not unique

10 h(x) =√3

x+2

11 h(x) = 3x4+ 2x2+ 3

12 h(x) = 5(x − π)2+ 4(x − π) + 7

13 h(x) = 2|3x − 4|

14 h(x) =x+1x+2

15 h(x) =√5x2+ 3

16 h(x) = 32x+ 3x+ 1

17 h(x) = 4π2x+ 3πx+ 2

In Problems 18 through 20, find functions f , g, and h such that k(x) = f (g(h(x)))

18 k(x) = √3

x 2 +4

19 k(x) = (√1

x+1) 9

20 k(x) =(x2+ 1)3+ 5

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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) = x2 On the 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.

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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 =x−k1 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) = −12(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) = x2 Un-raveling a function is just like unUn-raveling a very complicated algebraic expression; you need to start from the innermost part and remember the order of operations rules

(b) Let q(x) = −x−42 + 3 Interpret the graph of q in terms of transformations of a familiar function Where are the vertical and horizontal asymptotes?

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(c) Below is the graph of f (x) Graph the following.

i y = f (2x)

ii y = 3f (x2) 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)

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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.9We’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) =x−1−1 + 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

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Suppose that a is a zero of f (x) This means that f (a) = 0 Therefore kf (a) =

k · 0 = 0 as well 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) + k is the graph of f (x) shifted up k units

If k < 0, the graph of f (x) + k is the 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) − 3 is not 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)

2 1 1 2

y

x

y = 2f(x)

2 1 1 2

–1 –2

–3

y

x

y = 2f(x) –3

2

1

–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.

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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

compressedhorizontally

2 1

1

–1 –2

y

x

y = f(x)

2 3 4 5 6 7 8 1

1 2

–1 –1 –2

–2

y

x

y = f(–2x)

y = f ( (x2

2 1 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) To alter the input is to alter x; to alter the output is to alter y Notice that y = f (x) + 2 is equivalent

to y − 2 = f (x) Similarly, y = 2f (x) is equivalent toy2= 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 y2 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 x2stretches 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

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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 = x2 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

2 1

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 –1 1 2 3 4

f

x

–3 –2 –1 1 2 3 4

|f |

x

Figure 3.12

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Reciprocals: f (x)1

Suppose a function f (t) gives the price of gasoline in dollarsgallon at time t We may be interested

in the functionf (t )1 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 minutesitem 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 =f (x)1 related to the graph of y = f (x)? We already have some experience with graphing reciprocals; we know how to graph f (x) = x and g(x) =f (x)1 = 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 =m(x)1 = 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) = x and g(x) =x1 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,f (x)1 is positive; where f (x) is negative,f (x)1 is negative Where f (x) is zero,f (x)1 is undefined

Where |f (x)| = 1 , |f (x)1 | = 1

Where |f (x)| > 1 , |f (x)1 | < 1

1 LARGE positive number= small positive number

Where |f (x)| < 1 , |f (x)1 | > 1

1 small positive number= LARGE positive number

Suppose that f (x) → ∞ as x → ∞ Then as x → ∞, f (x)1 → 0

Suppose that f (x) → L as x → ∞, where L is a nonzero constant Then as

x → ∞, 1 →L1

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