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11

11.1

11.2

Functions and Relations

Graphs of Functions and Relations

Functions Working in a world of numbers, designers of racing boats blend art with science to

design attractive boats that are also fast and safe. If the sail area is increased, the

boat will go faster but will be less stable in open seas. If the displacement is

increased, the boat will be more stable but slower. Increasing length increases

speed but reduces stability. To make yacht racing both competitive and safe,

racing boats must satisfy complex systems of rules, many of which involve

mathematical formulas.

After the 1988 mismatch between Dennis Conner’s catamaran and New

Zealander Michael Fay’s 133-foot monohull, an international group of yacht design­

ers rewrote the America’s Cup rules to ensure the fairness of the race. In addition

to hundreds of pages of other rules, every yacht must satisfy the basic inequality

3 L + 1.25VsS – 9.8VDs – 24.000,

0.679

which balances the length L, the sail area S, and the displacement D.

11.3

11.4

In the 1979 Fastnet Race, 15 sailors lost

Transformations their lives. After Exide Challenger’s carbon-fiber of Graphs

Graphs of Polynomial Functions

keel snapped off, Tony Bullimore spent

4 days inside the overturned hull before

being rescued by the Australian navy. Yacht

11.5

11.6

11.7

racing is a dangerous sport. To determine the

Graphs of Rational general performance and safety of a yacht, Functions designers calculate the displacement-length

ratio, the sail area-displacement ratio, the Combining Functions

ballast-displacement ratio, and the capsize

screening value. Inverse Functions

D is

pl ac

em en

t- le

ng th

r at

io 800

0 25 30 35 40 45 50

Length at water line (ft)

d 25,000 lbs600

400

200

In Exercises 83 and 84 of Section 11.6 we will see how

composition of functions is used to define the displacement-

length ratio and the sail area-displacement ratio.

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690 Chapter 11 Functions 11-2

11.1 Functions and Relations

We have been using the language of functions and function notation since we first studied formulas in Chapter 2. In this section we will review what we have already studied about functions and delve further into this important concept.

In This Section

U1V The Concept of a Function

U2V Functions Expressed by Formulas

U3V Functions Expressed by Tables

U4V Functions Expressed by Ordered Pairs

U5V The Vertical-Line Test

U6V Domain and Range

U7V Function Notation

U1V The Concept of a Function If the value of the variable y is determined by the value of the variable x, then y is a function of x. So “is a function of” means “is uniquely determined by.” But what does uniquely determined mean? According to the dictionary “determine” means “to settle conclusively.” There can be no ambiguity. There is only one y for any x.

The x-value is thought of as input and the y-value as output. If y is a function of x, then there is only one output for any input. For example, after a shopper places an order on the Internet, the shopper is asked to input a ZIP code so that the shipping cost (output) can be determined. For that order the shipping cost is a function of ZIP code. Note that many different ZIP codes can correspond to the same output. If any ZIP code caused the computer to output more than one shipping cost, then shipping cost is not a function of ZIP code. The shopper is confused and probably cancels the order. See Fig. 11.1.

Input: ZIP code Output: Shipping cost Input: ZIP code Output: Shipping cost

Shipping cost is a function of ZIP code

70454 70402 02116 98431

\$5

\$8 \$10

Shipping cost is not a function of ZIP code

49858

32118 27886

\$9 \$12 \$6 \$7

Figure 11.1

E X A M P L E 1 Deciding if y is a function of x In each case determine whether y is a function of x.

a) Consider all possible circles. Let y represent the area of a circle and x represent its radius.

b) Consider all possible first-class letters mailed today in the United States. Let y represent the weight of a letter and x represent the amount of postage on the letter.

c) Consider all students at Pasadena City College. Let y represent the weight of a student to the nearest pound and x represent the height of the same student to the nearest inch.

d) Consider all possible rectangles. Let y represent the area of a rectangle and x represent the width.

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11-3 11.1 Functions and Relations 691

e) Consider all cars sold at Bill Hood Ford this year where the sales tax rate is 9%. Let y represent the amount of sales tax and x represent the selling price of the car.

Solution a) Can the area of a circle be determined from its radius? The well-known formula

A = 7r2 (or in this case y = 7×2) indicates exactly how to determine the area if the radius is known. So there is only one area for any given radius and y is a function of x.

b) Can the weight of a letter be determined if the amount of postage on the letter is known? There are certainly letters that have the same amount of postage and different weights. Since the weight cannot be determined conclusively from the postage, the weight is not a function of the postage and y is not a function of x.

c) Can the weight of a student be determined from the height of the student? Imagine that we have a list containing the weights and heights for all students. There will certainly be two 5 ft 9 in. students with different weights. So weight cannot be determined from the height and y is not a function of x.

d) Can the area of a rectangle be determined from the width? Among all possible rectangles there are infinitely many rectangles with width 1 ft and different areas. So the area is not determined by the width and y is not a function of x.

e) Can the amount of sales tax be determined from the price of the car? The formula y = 0.09x is used to determine the amount of tax. For example, the tax on every \$20,000 car is \$1800. So y is a function of x.

Now do Exercises 1–8

U2V Functions Expressed by Formulas If you get a speeding ticket in St. John’s Parish, Louisiana, there is a rule that is used to determine your fine. You can mail to the judge \$153 plus \$1 for every mile per hour over 80 miles per hour, but if your speed is over 90 miles per hour you must appear before the judge. Since there is no ambiguity, the amount of the fine is a function of your speed.

Function (as a Rule)

A function is a rule by which any allowable value of one variable (the indepen­ dent variable) determines a unique value of a second variable (the dependent variable).

There are many ways to express a rule. A rule can be expressed verbally (as in the speeding ticket), with a formula, a table, or a graph. Of course, in mathematics we prefer the preciseness that formulas or equations provide. Since a formula such as A = 7r2 gives us a rule for obtaining the value of the dependent variable A from the value of the independent variable r, we say that this formula is a function. Formulas are used to describe or model relationships between variables.

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692 Chapter 11 Functions 11-4

E X A M P L E 2 Writing a formula for a function A carpet layer charges \$25 plus \$4 per square yard for installing carpet. Write the total charge C as a function of the number n of square yards of carpet installed.

Solution At \$4 per square yard, n square yards installed cost 4n dollars. If we include the \$25 charge, then the total cost is 4n + 25 dollars. Thus, the equation

C = 4n + 25

expresses C as a function of n.

Now do Exercises 9–12

Any formula that has the form y = mx + b with m * 0 is a linear function. If m = 0, then the formula has the form y = b and is called a constant function. So in Example 2, the charge is a linear function of the number of square yards installed and C = 4n + 25 is a linear function.

E X A M P L E 3 A function in geometry Express the area of a circle as a function of its diameter.

Solution The area of a circle is given by A = 7r2. Because the radius of a circle is one-half of the diameter, we have r = d

2 . Now replace r by d

2 in the formula A = 7r2:

A = 7(d 2r 2

= 7

4 d2

So A = 7 4 d2 expresses the area of a circle as a function of its diameter.

Now do Exercises 13–18

U3V Functions Expressed by Tables Tables are often used to provide a rule for pairing the value of one variable with the value of another. For a table to define a function, each value of the independent variable must correspond to only one value of the dependent variable.

E X A M P L E 4 Functions defined by tables Determine whether each table expresses y as a function of x.

a) b) c) Weight (lbs) Cost (\$)

x y

0 to 10 4.60

Weight (lbs) Cost (\$) x y

0 to 15 4.60

10 to 30 12.75

31 to 79 32.90

80 to 99 55.82

11 to 30 12.75

31 to 79 32.90

80 to 99 55.82

x y

1 1

-1 1

2 2

-2 2

3 3

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11-5 11.1 Functions and Relations 693

Solution a) For each allowable weight, this table gives a unique cost. So the cost is a function

of the weight and y is a function of x.

b) Using this table a weight of say 12 pounds would correspond to a cost of \$4.60 and also to \$12.75. Either the table has an error or perhaps there is some other factor that is being used to determine cost. In any case the weight does not determine a unique cost and y is not a function of x.

c) In this table every allowable value for x corresponds to a unique y-value, so y is a function of x. Note that different values of x corresponding to the same y-value are permitted in a function.

In a function, every value for the independent variable determines conclusively a corresponding value for the dependent variable. If there is more than one possible value for the dependent variable, then the set of ordered pairs is not a function.

Now do Exercises 19–26

U4V Functions Expressed by Ordered Pairs A computer at your grocery store determines the price of each item by searching a long list of ordered pairs in which the first coordinate is the universal product code and the second coordinate is the price of the item with that code. For each product code there is a unique price. This process certainly satisfies the rule definition of a function. Since the set of ordered pairs is the essential part of this rule, we say that the set of ordered pairs is a function.

Function (as a Set of Ordered Pairs)

A function is a set of ordered pairs of real numbers such that no two ordered pairs have the same first coordinates and different second coordinates.

Note the importance of the phrase “no two ordered pairs have the same first coordinates and different second coordinates.” Imagine the problems at the grocery store if the com­ puter gave two different prices for the same universal product code. Note also that the product code is an identification number and it cannot be used in calculations. So the computer can use a function defined by a formula to determine the amount of tax, but it cannot use a formula to determine the price from the product code.

Any set of ordered pairs is called a relation. A function is a special relation.

E X A M P L E 5 Relations given as lists of ordered pairs Determine whether each relation is a function. (Determine whether y is a function of x.)

a) {(1, 2), (1, 5), (3, 7)} b) {(4, 5), (3, 5), (2, 6), (1, 7)}

Solution a) This relation is not a function because (1, 2) and (1, 5) have the same first coordi­

nate but different second coordinates.

b) This relation is a function. Note that the same second coordinate with different first coordinates is permitted in a function.

Now do Exercises 27–34

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694 Chapter 11 Functions 11-6

The solution set to any equation in x and y is the set of ordered pairs that satisfy the equation. For example, the solution set to x = y2 is expressed in set-builder nota­ tion as {(x, y) I x = y2}. Since any set of ordered pairs is a relation, this solution set is a relation. We can use the definition of a function to determine whether the solution set is a function. For simplicity we often refer to an equation in x and y as a relation or a function.

E X A M P L E 6 Relations given as equations Determine whether each relation is a function. (Determine whether y is a function of x.)

a) x = y2

b) y = 2x

c) x = I y I

Solution a) Is it possible to find two ordered pairs with the same first coordinate and different

second coordinates that satisfy x = y2? Since (1, 1) and (1, -1) both satisfy x = y2 , this relation is not a function.

b) The equation y = 2x indicates that the y-coordinate is always twice the x-coordinate. Ordered pairs such as (0, 0), (2, 4), and (3, 6) satisfy y = 2x. It is not possible to find two ordered pairs with the same first coordinate and different second coordinates. So y = 2x is a function.

c) The equation x = I y I is satisfied by ordered pairs such as (2, 2) and (2, -2) because 2 = I 2 I and 2 = I -2 I are both correct. So this relation is not a function.

To determine whether an equation expresses y as a function of x, always select a number for x (the indepen­ dent variable) and then see if there is more than one corresponding value for y (the dependent variable). If there is more than one corresponding y-value, then y is not a function of x.

Now do Exercises 35–62

U5V The Vertical-Line Test Since every graph illustrates a set of ordered pairs, every graph is a relation. To deter­ mine whether a graph is a function, we must see whether there are two (or more) ordered pairs on the graph that have the same first coordinate and different second coordinates. Two points with the same first coordinate lie on a vertical line that crosses the graph.

1

y

x

1

4

2

5

3

1 2

2

1

3

4

32

(4, 2)

(4, 2)

Figure 11.2

The Vertical-Line Test

A graph is the graph of a function if and only if there is no vertical line that crosses the graph more than once.

If there is a vertical line that crosses a graph twice (or more) as in Fig. 11.2, then we have two points with the same x-coordinate and different y-coordinates, and the graph is not the graph of a function. If you mentally consider every possible vertical line and none of them crosses the graph more than once, then you can conclude that the graph is the graph of a function.

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11-7 11.1 Functions and Relations 695

E X A M P L E 7 Using the vertical-line test Which of these graphs are graphs of functions?

a) b)

c)

Solution Neither (a) nor (c) is the graph of a function, since we can draw vertical lines that cross these graphs twice. The graph (b) is the graph of a function, since no vertical line crosses it more than once.

Now do Exercises 63–68

1

y

x

1

2

4

2

3 4

3

1

2

1

3

4

1

y

x

1

4

2

3

312

2

1

3

4

y

x 1

2

2

1

4

1

2

1

3

3

4

4 32 3

Domain (inputs) Range (outputs)

0 1

1 2

2

0

1

4

Figure 11.3

The vertical-line test illustrates the visual difference between a set of ordered pairs that is a function and one that is not. Because graphs are not precise and not always complete, the vertical-line test might be inconclusive.

U6V Domain and Range A relation (or function) is a set of ordered pairs. The set of all first coordinates of the ordered pairs is the domain of the relation (or function). The set of all second coor­ dinates of the ordered pairs is the range of the relation (or function). A function is a rule that pairs each member of the domain (the inputs) with a unique member of the range (the outputs). See Fig. 11.3. If a function is given as a table or a list of ordered pairs, then the domain and range are determined by simply reading them from the table or list. More often, a relation or function is given by an equation, with no domain stated. In this case, the domain consists of all real numbers that, when substituted for the independent variable, produce real numbers for the dependent variable.

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696 Chapter 11 Functions 11-8

E X A M P L E 8 Identifying the domain and range Determine the domain and range of each relation.

a) {(2, 5), (2, 7), (4, 3)} b) y = 2x c) y = Vx – 1s

Solution a) The domain is the set of first coordinates, {2, 4}. The range is the set of second

coordinates, {3, 5, 7}. b) Since any real number can be used in place of x in y = 2x, the domain is (-o, o).

Since any real number can be used in place of y in y = 2x, the range is also (-o, o).

c) Since the square root of a negative number is not a real number, we must have x – 1 : 0 or x : 1. So the domain is the interval [1, o). Since the square root of a nonnegative real number is a nonnegative real number, we must have y : 0. So the range is the interval [0, o).

Domain Range f

4 11

Figure 11.4

E X A M P L E 9

Now do Exercises 69–80

U7V Function Notation If y is a function of x, we can use the notation f (x) to represent y. The expression f (x) is read as “f of x.” The notation f (x) is called function notation. So if x is the inde­ pendent variable, then either y or f (x) is the dependent variable. For example, the func­ tion y = 2x + 3 can be written as

f (x) = 2x + 3.

We use y and f (x) interchangeably. We can think of f as the name of the function. We may use letters other than f. For example g(x) = 2x + 3 is the same function as f (x) = 2x + 3. The ordered pairs for each function are identical. Note that f (x) does not mean f times x. The expression f (x) represents the second coordinate when the first coordinate is x.

If f (x) = 2x + 3, then f(4) = 2(4) + 3 = 11. So the second coordinate is 11 if the first coordinate is 4. The ordered pair (4, 11) is an ordered pair in the function f. Figure 11.4 illustrates this situation.

Using function notation Let f (x) = 3x – 2 and g(x) = x2 – x. Evaluate each expression.

a) f (-5) b) g(-5) c) f (0) + g(3)

Solution a) Replace x by -5 in the equation defining the function f :

f (x) = 3x – 2

f (-5) = 3(-5) – 2

= -17

So f (-5) = -17.

b) Replace x by -5 in the equation defining the function g:

g(x) = x2 – x

g(-5) = (-5)2 – (-5) = 30

So g(-5) = 30.

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11-9 11.1 Functions and Relations 697

c) Since f (0) = 3(0) – 2 = -2 and g(3) = 32 – 3 = 6, we have f(0) + g(3) = -2 + 6 = 4.

Now do Exercises 81–96

CAUTION The notation f (x) does not mean f times x.

E X A M P L E 10 An application of function notation To determine the cost of an in-home repair, a computer technician uses the linear function C(n) = 40n + 30, where n is the time in hours and C(n) is the cost in dollars. Find C(2) and C(4).

Solution Replace n with 2 to get

C(2) = 40(2) + 30 = 110.

Replace n with 4 to get

C(4) = 40(4) + 30 = 190.

So for 2 hours the cost is \$110 and for 4 hours the cost is \$190.

U Calculator Close-Up V

A graphing calculator has function notation built in. To find C(2) and C(4) with a graphing calculator, enter y1 = 40x + 30 as shown here:

Now do Exercises 97–104

To find C(2) and C(4), enter y1(2) and y1(4) as shown here: In this section we studied functions of one variable. However, a variable can be a

function of another variable or a function of many other variables. For example, your grade on the next test is not a function of the number of hours that you study for it. Your grade is a function of many variables: study time, sleep time, work time, your mother’s IQ, and so on. Even though study time alone does not determine your grade, it is the variable that has the most influence on your grade.

Warm-Ups ▼

Fill in the blank. 1. A set of ordered pairs is a .

2. A is a set of ordered pairs in which no two have the same first coordinate and different second coordinates.

3. If m is a of w, then m is uniquely determined by w.

4. The of a relation is the set of all first coordinates of the ordered pairs.

5. The of a relation is the set of all second coordinates of the ordered pairs.

6. In function notation f (x) is used for the variable.

True or false? 7. The set {(1, 2), (3, 0), (9, 0)} is a function.

8. The set {(2, 1), (0, 3), (0, 9)} is a function.

9. The diameter of a circle is a function of the radius.

10. The equation y = x2 is a function.

11. Every relation is a function.

12. The domain of {(2, 1), (0, 3), (0, 9)} is {0, 2}.

13. The range of {(2, 1), (0, 3), (0, 9)} is {0, 1, 2, 3, 9}.

14. The domain of f(x) = Vxs is [0, o). 15. If h(x) = x2 – 3, the h(-2) = 1.

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1 1

.1 Exercises

U Study Tips V • Instructors love to help students who are eager to learn. • Show a genuine interest in the subject when you ask questions and you will get a good response from your instructor.

U1V The Concept of a Function

In each situation determine whether y is a function of x. Explain your answer. See Example 1.

1. Consider all gas stations in your area. Let x represent the price per gallon of regular unleaded gasoline and y represent the number of gallons that you can get for \$10.

2. Consider all items at Sears. Let x represent the universal product code for an item and y represent the price of that item.

3. Consider all students taking algebra at your school. Let x represent the number of hours (to the nearest hour) a student spent studying for the first test and y represent the student’s score on the test.

4. Consider all students taking algebra at your school. Let x represent a student’s height to the nearest inch and y represent the student’s IQ.

5. Consider the air temperature at noon today in every town in the United States. Let x represent the Celsius temperature for a town and y represent the Fahrenheit

10. A developer prices condominiums in Florida at \$20,000 plus \$40 per square foot of living area. Express the cost C as a function of the number of square feet of living area s.

11. The sales tax rate on groceries in Mayberry is 9%. Express the total cost T (including tax) as a function of the total price of the groceries S.

12. With a GM MasterCard, 5% of the amount charged is cred­ ited toward a rebate on the purchase of a new car. Express the rebate R as a function of the amount charged A.

13. Express the circumference of a circle as a function of its radius.

14. Express the circumference of a circle as a function of its diameter.

15. Express the perimeter P of a square as a function of the length s of a side.

16. Express the perimeter P of a rectangle with width 10 ft as a function of its length L.

temperature. 17. Express the area A of a triangle with a base of 10 m as a

6. Consider all first-class letters mailed within the United States today. Let x represent the weight of a letter and y represent the amount of postage on the letter.

7. Consider all items for sale at the nearest Wal-Mart. Let x represent the cost of an item and y represent the universal product code for the item.

8. Consider all packages shipped by UPS. Let x represent the weight of a package and y represent the cost of shipping that package.

U2V Functions Expressed by Formulas

Write a formula that describes the function. See Examples 2 and 3.

9. A small pizza costs \$5.00 plus 50 cents for each topping. Express the total cost C as a function of the number of toppings t.

function of its height h.

18. Express the area A of a trapezoid with bases 12 cm and 10 cm as a function of its height h.

U3V Functions Expressed by Tables

Determine whether each table expresses the second variable as a function of the first variable. See Example 4.

19. 20.x y

1 1

4 2

9 3

16 4

25 5

36 6

49 8

x y

2 4

3 9

4 16

5 25

8 36

9 49

10 100

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11-11

21. 22.t v

2 2

-2 2

3 3

-3 3

4 4

-4 4

5 5

s W

5 17

6 17

-1 17

-2 17

-3 17

7 17

8 17

23. a P

2 2

2 -2

3 3

3 -3

4 4

4 -4

5 5

24. n r

17 5

17 6

17 -1

17 -2

17 -3

17 -4

17 -5

25. b q

1970 0.14

1972 0.18

1974 0.18

1976 0.22

1978 0.25

1980 0.28

26. c h

345 0.3

350 0.4

355 0.5

360 0.6

365 0.7

370 0.8

380 0.9

U4V Functions Expressed by Ordered Pairs

Determine whether each relation is a function. See Example 5.

27. {(2, 4), (3, 4), (4, 5)} 28. {(2, -5), (2, 5), (3, 10)} 29. {(-2, 4), (-2, 6), (3, 6)} 30. {(3, 6), (6, 3)} 31. {(7, -1), (7, 1)} 32. {(-0.3, -0.3), (-0.2, 0), (-0.3, 1)}

33. �(1, 2 1r�2

1 34. �(1, 7r, (- , 7r, (1, 7r�3 3 6

11.1 Functions and Relations 699

Find two ordered pairs that satisfy each equation and have the same x-coordinate but different y-coordinates. Answers may vary. See Example 6.

235. x = 2y2 36. x = y2

37. x = I 2y I 38. I x I = I y I

2 239. x2 + y = 1 40. x2 + y = 4

4 4 441. x = y 42. x = y

43. x – 2 = I y I 44. x + 5 = I y I

Determine whether each relation is a function. See Example 6. 245. y = x 46. y = x2 + 3

47. x = I y I + 1 48. I x I = I y + 1 I 49. y = x 50. x = y + 4

251. x = y4 + 1 52. x4 = y 53. y = Vxs 54. x = Vys 55. I x I = I 2y I 56. I 4x I = I 2y I

2 457. x2 + y = 9 58. x2 + y = 1 59. x = 2Vy 60. ss y = Vx – 5 61. x + 5 = I y I 62. x – 2 = I y I

U5V The Vertical-Line Test

Use the vertical-line test to determine which of the graphs are graphs of functions. See Example 7.

63. 64.

y

x

2

1

3

32 1

2

2

3

3

y

x

1

2

1

3

32 1

2

2 1

3

3

65. 66.

y

x

2

3

32

1

1

2

2 1

3

3

y

x

2

3

32

1

2

2 1

3

3

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700 Chapter 11 Functions 11-12

67. 68. expresses its height h(t) in feet as a function of the time t in seconds.

a) Find h(2), the height of the ball 2 seconds after it is dropped.

b) Find h(4).

98. Velocity. If a ball is dropped from a height of 256 ft, then the formula

v(t) = -32t

expresses its velocity v(t) in feet per second as a function of time t in seconds.

a) Find v(0), the velocity of the ball at time t = 0. b) Find v(4).

U6V Domain and Range 99. Area of a square. Find a formula that expresses the area Determine the domain and range of each relation. See of a square A as a function of the length of its side s. Example 8.

69. {(4, 1), (7, 1)} 100. Perimeter of a square. Find a formula that expresses the perimeter of a square P as a function of the length of its70. {(0, 2), (3, 5)} side s.

71. {(2, 3), (2, 5), (2, 7)} 72. {(3, 1), (5, 1), (4, 1)} 101. Cost of fabric. If a certain fabric is priced at \$3.98 per 73. y = x + 1 yard, express the cost C(x) as a function of the number of 74. y = 3x + 1 yards x. Find C(3).

75. y = 5 – x 102. Earned income. If Mildred earns \$14.50 per hour, 76. y = -2x + 1

express her total pay P(h) as a function of the number of 77. y = Vx – 2s hours worked h. Find P(40). 78. y = Vx + 4s

s79. y = V2x 103. Cost of pizza. A pizza parlor charges \$14.95 for a pizza 80. y = V2x – 4 plus \$0.50 for each topping. Express the total cost of as

pizza C(n) in dollars as a function of the number of toppings n. Find C(6).

U7V Function Notation

Let f(x) = 3x – 2, g(x) = -x2 + 3x – 2, and h(x) = I x + 2 I. 104. Cost of gravel. A gravel dealer charges \$50 plus \$30 per Evaluate each expression. See Example 9. cubic yard for delivering a truckload of gravel. Express the 81. f (0) 82. f (1) total cost C(n) in dollars as a function of the number of cubic 83. f (4) 84. f (100) yards delivered n. Find C(12). 85. g(-2) 86. g(-3) 87. h(-3) 88. h(-19) 89. h(-4.236) 90. h(-1.99)

Getting More Involved 91. f (2) + g(3) 92. f (1) – g(0) g(2) h(-10) 105. Writing

93. 94. h(-3) f (2) Consider y = x + 2 and y > x + 2. Explain why one

95. f (-1) . h(-4) 96. h(0) . g(0) of these relations is a function and the other is not.

Solve each problem. See Example 10. 106. Writing

97. Height. If a ball is dropped from the top of a 256-ft Consider the graphs of y = 2 and x = 3 in the building, then the formula rectangular coordinate system. Explain why one of

h(t) = 256 – 16t2 these relations is a function and the other is not.

y

x

2

3

3

1

1 1

2

2 1

3

y

x

2

3

3

1

21 1

2

3

3

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11-13 11.2 Graphs of Functions and Relations 701

In This Section

U1V Linear and Constant Functions

U2V Absolute Value Functions

U4V Square-Root Functions

11.2 Graphs of Functions and Relations

Functions were introduced in Section 11.1. In this section, we will study the graphs of several types of functions. We graphed linear functions in Chapter 3 and quadratic functions in Chapter 10, but for completeness we will review them here.

U1V Linear and Constant Functions U5V Piecewise Functions Linear functions get their name from the fact that their graphs are straight lines. U6V Graphing Relations

Linear Function

A linear function is a function of the form

f (x) = mx + b,

where m and b are real numbers with m * 0.

The graph of the linear function f (x) = mx + b is exactly the same as the graph of the linear equation y = mx + b. If m = 0, then we get f (x) = b, which is called a constant function. If m = 1 and b = 0, then we get the function f (x) = x, which is called the identity function. When we graph a function given in function notation, we usually label the vertical axis as f (x) rather than y.

Now do Exercises 1–2

x

The domain and range of a function can be determined from the formula or the graph. However, the graph is usually very helpful for understanding domain and range.

E X A M P L E 1 Graphing a constant function Graph f(x) = 3, and state the domain and range.

Solution The graph of f (x) = 3 is the same as the graph of y = 3, which is the horizontal line in Fig. 11.5. Since any real number can be used for x in f (x) = 3 and since the line in Fig. 11.5 extends without bounds to the left and right, the domain is the set of all real numbers, (-o, o). Since the only y-coordinate for f (x) = 3 is 3, the range is {3}.

f (x)

2

32112 3

4

5

1 4

f (x) 3

4

Domain ( , )

Figure 11.5

E X A M P L E 2 Graphing a linear function Graph the function f (x) = 3x – 4, and state the domain and range.

Solution The y-intercept is (0, -4) and the slope of the line is 3. We can use the y-intercept and the slope to draw the graph in Fig. 11.6 on the next page. Since any real number can be used for x in f(x) = 3x – 4, and since the line in Fig. 11.6 extends without bounds to the left and

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702 Chapter 11 Functions 11-14

right, the domain is the set of all real numbers, (-o, o). Since the graph extends without bounds upward and downward, the range is the set of all real numbers, (-o, o).

Now do Exercises 3–10

Figure 11.6

f (x)

x

Domain ( , )

R an

ge (

, )

f (x) 3x 4

1

2

3 2

1

2 3

4

5

1

2

4

3

3

5

4

U2V Absolute Value Functions The equation y = I x I defines a function because every value of x determines a unique value of y. We call this function the absolute value function.

Absolute Value Function

The absolute value function is the function defined by

f (x) = I x I.

To graph the absolute value function, we simply plot enough ordered pairs of the function to see what the graph looks like.

E X A M P L E 3 The absolute value function Graph f(x) = I x I, and state the domain and range.

Solution To graph this function, we find points that satisfy the equation f (x) = I x I.

x -2 -1 0 1 2

f (x) = I x I 2 1 0 1 2

Plotting these points, we see that they lie along the V-shaped graph shown in Fig. 11.7. Since any real number can be used for x in f (x) = I x I and since the graph extends without bounds to the left and right, the domain is (-o, o). Because I x I is never negative, the

The most important feature of an absolute value function is its V-shape. If we had plotted only points in the first quadrant, we would not have seen the V-shape. So for an absolute value function we always plot enough points to see the V-shape.

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11-15 11.2 Graphs of Functions and Relations 703

f (x)

x1 32

1

23

4

5

2

414

3

f (x) � �x �

Domain ( �, �)

R an

ge [

0, )

Figure 11.7

graph does not go below the x-axis. So the range is the set of nonnegative real numbers, [0, o).

Now do Exercises 11–12

Many functions involving absolute value have graphs that are V-shaped, as in Fig. 11.7. To graph functions involving absolute value, we must choose points that determine the correct shape and location of the V-shaped graph.

E X A M P L E 4 Other functions involving absolute value Graph each function, and state the domain and range.

a) f (x) = I x I – 2 b) g(x) = I 2x – 6 I

Solution a) Choose values for x and find f(x).

x -2 -1 0 1 2

f (x) = I x I – 2 0 -1 -2 -1 0

Plot these points and draw a V-shaped graph through them as shown in Fig. 11.8. The domain is (-o, o), and the range is [-2, o).

1

f (x)

x

2

32

1

23

4

5

3

4

2

1

f (x) x 2

Domain ( , )

R an

ge [

2, )

4

Figure 11.8

U Calculator Close-Up V

To check Example 4(a) set

y1 = abs(x) – 2

and then press GRAPH.

To check Example 4(b) set

y2 = abs(2x – 6)

and then press GRAPH.

10

10

10

10

10

10

10

10

g(x)

x

2

1

3

4

5

3

4 5 1

2 1 21 3

g(x) 2x 6

Domain ( , )

R an

ge [

0, )

Figure 11.9

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704 Chapter 11 Functions 11-16

b) Make a table of values for x and g(x).

x 1 2 3 4 5

g(x) = I 2x – 6 I 4 2 0 2 4

Draw the graph as shown in Fig. 11.9. The domain is (-o, o), and the range is [0, o).

Now do Exercises 13–20

U3V Quadratic Functions A function defined by a second-degree polynomial is a quadratic function.

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.

In Chapter 10 we learned that the graph of any quadratic function is a parabola, which opens upward or downward. The vertex of a parabola is the lowest point on a parabola that opens upward or the highest point on a parabola that opens downward. Parabolas will be discussed again when we study conic sections later in this text.

E X A M P L E 5 A quadratic function Graph the function g(x) = 4 – x2, and state the domain and range.

Solution We plot enough points to get the correct shape of the graph.

x -2 -1 0 1 2

g(x) = 4 – x2 0 3 4 3 0

See Fig. 11.10 for the graph. The domain is (-o, o). From the graph we see that the largest y-coordinate is 4. So the range is (-o, 4].

Now do Exercises 21–28

Figure 11.10

g(x)

x

2

1

3

5

3

4

2

4 1

1 1 3

3

4

5

g(x) 4 x2 Domain ( , )

R an

ge (

, 4 ]

U Calculator Close-Up V

You can find the vertex of a parabola with a calculator. For example, graph

y = -x2 – x + 2.

Then use the maximum feature, which is found in the CALC menu. For the left bound pick a point to the left of the vertex; for the right bound pick a point to the right of the vertex; and for the guess pick a point near the vertex.

4

6

6

4

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11-17 11.2 Graphs of Functions and Relations 705

U4V Square-Root Functions We define the square-root function as follows.

Square-Root Function

The square-root function is the function defined by

f (x) = Vxs.

Because squaring and square root are inverse operations, the graph of f(x) = Vsx is related to the graph of f(x) = x2. Recall that there are two square roots of every positive real number, but the radical symbol represents only the positive root. That is why we get only half of a parabola, as shown in Example 6.

E X A M P L E 6 Square-root functions Graph each equation, and state the domain and range.

a) y = Vxs b) y = Vx + 3s

Solution a) The graph of the equation y = Vxs and the graph of the function f (x) = Vxs are the

same. Because Vxs is a real number only if x : 0, the domain of this function is the set of nonnegative real numbers. The following ordered pairs are on the graph:

x 0 1 4 9

y = Vxs 0 1 2 3

The graph goes through these ordered pairs as shown in Fig. 11.11. Note that x is chosen from the nonnegative numbers. The domain is [0, o) and the range is [0, o).

b) Note that Vx + 3s is a real number only if x + 3 : 0, or x : -3. So we make a table of ordered pairs in which x : -3:

x -3 -2 1 6

y = Vx + 3s 0 1 2 3

The graph goes through these ordered pairs as shown in Fig. 11.12. The domain is [-3, o) and the range is [0, o).

Figure 11.12

y

x

Domain [–3, )

R an

ge [

0, )

1

3

123

2

4 532 61

√ —–––– x � 3y

y

x

2

1

3

4 5 1

1 32

4

6 7 8 9

2

1

y .x

Domain [0, )

R an

ge [

0, )

Figure 11.11

Now do Exercises 29–36

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706 Chapter 11 Functions 11-18

x-1 -1

3-2

1

2-3

4

5

6

7

2

41-4

3

y = x

y = x, x : 0

y = -x, x < 0

y

Figure 11.13

U5V Piecewise Functions Most of our functions are defined by a single formula, but functions can be defined by different formulas for different regions of the domain. Such functions are called piecewise functions. The simplest example of a piecewise function is the absolute value function. The graph of f(x) = Ix I is the straight line y = x to the right of the y-axis and the straight line y = -x to the left of the y-axis, as shown in Fig. 11.13. So f(x) = Ix I could be written as

x for x : 0 f(x) = � . -x for x < 0

In Example 7, we graph some piecewise functions.

E X A M P L E 7 Graphing piecewise functions Graph each function.

a) f(x) = � b) f(x) = � Solution

a) For x : 0, we graph the line y = 1 2x. For x < 0, we graph the line y = -3x. Make a table of ordered pairs for each.

x (x : 0) 0 2 4 6 x (x < 0) -0.1 -1 -2 -3

y = 1 2

x 0 1 2 3 y = -3x 0.3 3 6 9

Plot these ordered pairs and draw the lines as shown in Fig. 11.14. Note that both lines “start” at the origin and neither line extends below the x-axis.

b) For x : 0, we graph the curve y = Vxs. For x < 0, we graph the line y = -x + 2. Make a table of ordered pairs for each.

x (x : 0) 0 1 4 9 x (x < 0) -0.1 -1 -2 -3

y = Vxs 0 1 2 3 y = -x + 2 2.1 3 4 5

Plot these ordered pairs and sketch the graph, as shown in Fig. 11.15. Note that the line comes right up to the point (0, 2) but does not include it. So the point is shown with a hollow circle. The point (0, 0) is included on the curve. So it is shown with a solid circle.

Now do Exercises 37–44

Vxs for x : 0 -x + 2 for x < 0

1 2 x for x : 0

-3x for x < 0

x1 1

32 23

4

5

6

7

2

41

3

y

x, x 0 3x, x < 0�f (x)

5

1 2

Figure 11.14

x1 1

32

1

23

4

5

6

7

41

y

.Jx, x : 0 x + 2, x < 0�f(x)

5

3

Figure 11.15

U6V Graphing Relations A function is a set of ordered pairs in which no two have the same first coordinate and different second coordinates. A relation is any set of ordered pairs. The domain of a relation is the set of x-coordinates of the ordered pairs and the range of a relation is the set of y-coordinates of the ordered pairs. In Example 8, we graph the relation x = y2. Note that this relation is not a function because ordered pairs such as (4, 2) and (4, -2) satisfy x = y2.

– c

c

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11-19 11.2 Graphs of Functions and Relations 707

E X A M P L E 8 Graphing relations that are not functions Graph each relation, and state the domain and range.

a) x = y2

b) x = l y – 3 l

Solution a) Because the equation x = y2 expresses x in terms of y, it is easier to choose the

y-coordinate first and then find the x-coordinate:

x = y2 4 1 0 1 4

y -2 -1 0 1 2

Figure 11.16 shows the graph. The domain is [0, o) and the range is (-o, o).

b) Again we select values for y first and find the corresponding x-coordinates:

x = l y – 3 l 2 1 0 1 2

y 1 2 3 4 5

Plot these points as shown in Fig. 11.17. The domain is [0, o) and the range is (-o, o).

Now do Exercises 45–56

1

y

x

-2

1

3

4 5 -1

3

-3

-4

x = y2

2

4

6 7

2

R an

ge (

, )

Domain [0, c)

Figure 11.16

1

y

x

1

3

4 52 3

4

5

6 7

2

R an

ge (

, )

Domain [0, )

6

x y 3

Figure 11.17 Note that y = x2 is a function and x = y2 is not a function because we have agreed

that x is always the independent variable and y the dependent variable. You can deter­ 2mine the y-coordinate from x in y = x2, but y can’t be determined from x in x = y .

If we use variables other than x and y, then we must know which is the independent variable to decide if the equation is a function. For example, if W = a2 and a is the independent variable, then W is a function of a.

Warm-Ups ▼

Fill in the blank. 1. A function has the form f(x) = mx + b where

m � 0.

2. A function has the form f(x) = k where k is a real number.

3. The function is f(x) = x.

4. The graph of a quadratic function is a .

5. An function has a V-shaped graph.

True or false? 6. The graph of a function is a picture of all of the ordered

pairs of the function.

7. The domain of f(x) = 3 is (-o, o).

8. The range of a quadratic function is (-o, o).

9. The y-axis and the f(x)-axis are the same.

10. The domain of x = y2 is [0, o).

11. The domain of f(x) = �x – 1 is (1, o). 12. The domain of a quadratic function is (-o, o).

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1 1

.2 Exercises

U Study Tips V • Success in school depends on effective time management, which is all about goals. • Write down your long-term, short-term, and daily goals. Assess them, develop methods for meeting them, and reward yourself when you do.

U1V Linear and Constant Functions

Graph each function, and state its domain and range. See Examples 1 and 2.

1. h(x) = -2 2. f (x) = 4

3. f (x) = 2x – 1 4. g(x) = x + 2

1 2 5. g(x) = 33 x + 2 6. h(x) = 33 x – 4

2 3

9. y = -0.3x + 6.5 10. y = 0.25x – 0.5

U2V Absolute Value Functions

Graph each absolute value function and state its domain and range. See Examples 3 and 4.

11. f (x) = l x l + 1 12. g(x) = l x l – 3

13. h(x) = l x + 1 l 14. f (x) = l x – 2 l

15. g(x) = l 3x l 16. h(x) = l -2x l

2 3 37. y = – 3 x + 3 8. y = -3 x + 4 17. f (x) = l 2x – 1 l 18. y = l 2x – 3 l33 4

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11-21 11.2 Graphs of Functions and Relations 709

19. f (x) = I x – 2 I + 1 20. y = I x – 1 I + 2 U4V Square-Root Functions

Graph each square-root function, and state its domain and range. See Example 6.

29. g(x) = 2Vxs 30. g(x) = Vxs – 1

Graph each quadratic function, and state its domain and range. See Example 5.

21. y = x2 22. y = -x2

223. g(x) = x2 + 2 24. f (x) = x – 4

25. f (x) = 2×2 26. h(x) = -3×2

27. y = 6 – x2 28. y = -2×2 + 3

31. f (x) = Vsx – 1 32. f (x) = Vx + 1s

33. h(x) = -Vx 34. h(x) = -Vx – 1s s

35. y = Vxs + 2 36. y = 2Vsx + 1

U5V Piecewise Functions

Graph each piecewise function. See Example 7.

x for x : 0 for x : 0 37. f(x) = � 38. f(x) = �3x + 1 -4x for x < 0 -x + 1 for x < 0

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710 Chapter 11 Functions 11-22

2 for x > 1 3 for x > -2 47. x = -y2 48. x = 1 – y2 39. f(x) = � 40. f(x) = �-2 for x – 1 -4 for x – -2

41. f(x) = �Vxs for x + 3 for x > 1 x – 1 42. f(x) = �Vx – 2 s for x : 3

6 – x for x < 3 49. x = 5 50. x = -3

43. f(x) = �Vxs x – 4 for for 0 – x – 4 x > 4 51. x + 9 = y2 52. x + 3 = I y I

44. f(x) = �Vx + 1s x – 5 for for -1 – x – 3 x > 3 53. x = Vys 54. x = -Vys

U6V Graphing Relations

Graph each relation, and state its domain and range. See Example 8.

45. x = I y I 46. x = -I y I 55. x = (y – 1)2 56. x = (y + 2)2

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11-23 11.2 Graphs of Functions and Relations 711

Miscellaneous 67. y = -x2 + 4x – 4 68. y = -2 I x – 1 I + 4 Graph each function, and state the domain and range.

57. f (x) = 1 – I x I s58. h(x) = Vx – 3

259. y = (x – 3)2 – 1 60. y = x – 2x – 3

61. y = I x + 3 I + 1 62. f (x) = -2x + 4

63. y = Vxs – 3 64. y = 2 I x I

65. y = 3x – 5 66. g(x) = (x + 2)2

Classify each function as either a linear, constant, quadratic, square-root, or absolute value function.

s69. f (x) = Vx – 3 70. f (x) = ⏐x⏐ + 5 71. f (x) = 4 72. f (x) = 4x – 7 73. f (x) = 4×2 – 7 74. f (x) = -3 75. f (x) = 5 + Vxs 76. f (x) = ⏐x – 99⏐ 77. f (x) = 99x – 100 78. f (x) = -5×2 + 8x + 2

Graphing Calculator Exercises 279. Graph the function f (x) = Vxs, and explain what this

graph illustrates.

80. Graph the function f(x) = 1, and state the domain and range.x

1281. Graph y = x , y = 2x 2, and y = 2×2 on the same coordinate sys­

tem. What can you say about the graph of y = ax2 for a > 0?

2 282. Graph y = x , y = x2 + 2, and y = x – 3 on the same screen. What can you say about the position of y = x2 + k relative to y = x2?

283. Graph y = x , y = (x + 5)2, and y = (x – 2)2 on the same screen. What can you say about the position of y = (x – h)2 relative to y = x2?

84. You can graph the relation x = y2 by graphing the two functions y = Vsx and y = -Vsx. Try it and explain why this works.

85. Graph y = (x – 3)2, y = I x – 3 I, and y = Vx – 3 ons the same coordinate system. How does the graph of y = f (x – h) compare to the graph of y = f (x)?

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712 Chapter 11 Functions 11-24

In This Section

U1V Horizontal Translation

U2V Stretching and Shrinking

U3V Reflecting

U4V Vertical Translation

U5V Multiple Transformations

U Calculator Close-Up V

Note that for a translation of six units to the left, x + 6 must be written in parentheses on a graphing calculator.

7

8 6

7

11.3 Transformations of Graphs

If a, h, and k are real numbers with a * 0, then the graph of y = af(x – h) + k is a transformation of the graph of y = f(x). All of the transformations of a function form a family of functions. For example, all functions of the form y = a (x – h)2 + k form the square or quadratic family because they are transformations of y = x2 . The absolute-value family consists of functions of the form y = a|x – h| + k, and the square-root family consists of functions of the form y = aVx – hs + k. Understanding families of functions makes graphing easier because all of the functions in a family have similar graphs. The graph of any function in the square family is a parabola. We will now see what effect each of the numbers a, h, and k has on the graph of the original function y = f(x).

U1V Horizontal Translation According to the order of operations, to find y in y = af(x – h) + k, we first subtract h from x, evaluate f (x – h), multiply by a, and then add k. The order is important here, and we look at the effects of these numbers in the order h, a, and k.

Consider the graphs of f (x) = Vxs, g(x) = Vx – 2 s shown ins, and h(x) = Vx + 6 Fig. 11.18. In the expression Vx – 2s, subtracting 2 is the first operation to perform. So every point on the graph of g is exactly two units to the right of a corresponding point on the graph of f. (We must start with a larger value of x to get the same y-coordinate because we first subtract 2.) Every point on the graph of h is exactly six units to the left of a corresponding point on the graph of f.

h(

Figure 11.18

x

x) .x + 6 f(x) .x

g(x) .x 2

y

4 2 6 2 4 6 8 10 12

1

1 2

2

4 5 6

Translating to the Right or Left

If h > 0, then the graph of y = f (x – h) is a translation to the right of the graph of y = f (x).

If h < 0, then the graph of y = f(x – h) is a translation to the left of the graph of y = f (x).

E X A M P L E 1 Horizontal Translation Sketch the graph of each function and state the domain and range.

a) f (x) = (x – 2)2 b) f (x) = I x + 3 I

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11-25 11.3 Transformations of Graphs 713

Figure 11.19

2 1

y

x

2 3

1

1 2 3 4 5

f (x) (x 2)2

345 2 1

Figure 11.20

2 1

y

x

2

4 5

1

1 2

f (x) |x � 3| 345678 2 1

Now do Exercises 1–8

Solution a) The graph of f (x) = (x – 2)2 is a translation two units to the right of the familiar

graph of f (x) = x2. Calculate a few ordered pairs for accuracy. The points (2, 0), (0, 4), and (4, 4) are on the graph in Fig. 11.19. Since any real number can be used in place of x in (x – 2)2, the domain is (-o, o). Since the graph extends upward from (2, 0), the range is [0, o).

b) The graph of f (x) = I x + 3 I is a translation three units to the left of the familiar graph of f (x) = I x I. The points (0, 3), (-3, 0), and (-6, 3) are on the graph in Fig. 11.20. Since any real number can be used in place of x in I x + 3 I, the domain is (-o, o). Since the graph extends upward from (-3, 0), the range is [0, o).

U Calculator Close-Up V

A typical graphing calculator can draw 10 curves on the same screen. On this screen there are the curves

2 2y = 0.1x , y = 0.2x , and so on, 2through y = x .

4

2 2

1

U2V Stretching and Shrinking 12Consider the graphs of f (x) = x , g(x) = 2×2, and h(x) = 2 x

2 shown in Fig. 11.21. Every point on g(x) = 2×2 corresponds to a point directly below on the graph of f (x) = x2. The y-coordinate on g is exactly twice as large as the corresponding y-coordinate on f. This situation occurs because, in the expression 2×2, multiplying by 2 is the last operation performed. Every point on h corresponds to a point directly above on f, where the y-coordinate on h is half as large as the y-coordinate on f. The factor 2 has stretched the graph of f to form the graph of g, and the factor

1 has shrunk2

the graph of f to form the graph of h.

2 1

y

x

2 3 4 5

1 2 3 4

f (x) x2

g(x) 2×2 h(x) x21 2 345 2 1 5

Figure 11.21

Stretching and Shrinking

If a > 1, then the graph of y = af(x) is obtained by stretching the graph of y = f (x). If 0 < a < 1, then the graph of y = af(x) is obtained by shrinking the graph of y = f(x).

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714 Chapter 11 Functions 11-26

Note that the last operation to be performed in stretching or shrinking is multi­ plication by a. Whereas the function g(x) = 2Vsx is obtained by stretching f (x) = Vsx by a factor of 2, h(x) = V2xs is not.

Stretching and shrinking Graph the functions f (x) = Vxs, g(x) = 2Vxs, and h(x) = 1 2Vxs on the same coordinate system.

Solution The graph of g is obtained by stretching the graph of f, and the graph of h is obtained by shrinking the graph of f. The graph of f includes the points (0, 0), (1, 1), and (4, 2). The graph of g includes the points (0, 0), (1, 2), and (4, 4). The graph of h includes the points (0, 0), (1, 0.5), and (4, 1). The graphs are shown in Fig. 11.22.

Figure 11.22

y

x2 31

2 1

3 4 5

4 5

f (x) �x

g(x) 2�x

h(x) �x1 2

1

2 1

E X A M P L E 2

U Calculator Close-Up V

The following calculator screen shows the curves y = Vxs, y = 2Vxs, y = 3Vxs, and so on, through y = 10Vxs.

9

30

5

1

Figure 11.23

5 4 3 2

y

x32

2 1

3 4 5

4 5

f (x) x2

g(x) x2

345 2

U Calculator Close-Up V

With a graphing calculator, you can quickly see the result of modifying the formula for a function. If you have a graphing calculator, use it to graph the functions in the examples. Experimenting with it will help you to understand the ideas in this section.

10

10 10

10

Now do Exercises 9–16

U3V Reflecting Consider the graphs of f (x) = x2 and g(x) = -x2 shown in Fig. 11.23. Notice that the graph of g is a mirror image of the graph of f. For any value of x we compute the y-coordinate of an ordered pair of f by squaring x. For an ordered pair of g we square first and then find the opposite because of the order of operations. This gives a correspondence between the ordered pairs of f and the ordered pairs of g. For every ordered pair on the graph of f there is a corresponding ordered pair directly below it on the graph of g, and these ordered pairs are the same distance from the x-axis. We say that the graph of g is obtained by reflecting the graph of f in the x-axis or that g is a reflection of the graph of f.

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11-27 11.3 Transformations of Graphs 715

Reflection

The graph of y = -f (x) is a reflection in the x-axis of the graph of y = f (x).

E X A M P L E 3 Reflection Sketch the graphs of each pair of functions on the same coordinate system.

a) f (x) = Vxs, g(x) = -Vxs b) f (x) = I x I, g(x) = -I x I

Solution In each case the graph of g is a reflection in the x-axis of the graph of f. Recall that we graphed the square-root function and the absolute value function in Section 11.2. Figures 11.24 and 11.25 show the graphs for these functions.

Now do Exercises 17–24

Figure 11.24

1

2

3

4

1

2

4

5

5

3

21 3 5 64

y

x

f(x) �x

g(x) �x

34 2 1

Figure 11.25

5 4 3 2

y

x32

2 3 4 5

4 5

f (x) |x |

g(x) |x |

345 2

U4V Vertical Translation Consider the graphs of the functions f (x) = Vxs, g(x) = Vxs + 2, and h(x) = Vsx – 6 shown in Fig. 11.26. In the expression Vxs + 2, adding 2 is the last operation to per­ form. So every point on the graph of g is exactly two units above a corresponding point on the graph of f, and g has the same shape as the graph of f. Every point on the graph of h is exactly six units below a corresponding point on the graph of f. The graph of g is an upward translation of the graph of f, and the graph of h is a downward trans­ lation of the graph of f.

4

6

2

2

y

x

2

4

h(x) .x 6

f(x) .x

g(x) .x + 2

2 4 6 8

Figure 11.26

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716 Chapter 11 Functions 11-28

Translating Upward or Downward

If k > 0, then the graph of y = f (x) + k is an upward translation of the graph of y = f (x).

If k < 0, then the graph of y = f (x) + k is a downward translation of the graph of y = f (x).

Vertical translation Graph the function f (x) = I x I – 6, and state the domain and range.

Solution The graph of f (x) = I x I – 6 is a translation six units downward of the familiar graph of f (x) = I x I. Calculate a few ordered pairs for accuracy. The ordered pairs (0, -6), (1, -5), and (-1, -5) are on the graph in Fig. 11.27. Since any real number can be used in place of x in I x I – 6, the domain is (-o, o). Since the graph extends upward from (0, -6), the range is [-6, o).

E X A M P L E 4

Figure 11.27

4 3 2 1

y

x321

2 1

3

4

f (x) � |x| 6

3 4 2 1

Now do Exercises 25–30

U5V Multiple Transformations When graphing a function containing more than one transformation, perform the transformations in the following order:

Strategy for graphing y = af(x – h) + k

To graph y = af(x – h) + k, start with the graph of y = f (x) and perform

1. Horizontal translation (right for h > 0 and left for h < 0)

2. Stretching/shrinking (stretch for a > 1 and shrink for 0 < a < 1)

3. Reflection (reflect in x-axis for a < 0 or y = -f (x))

4. Vertical translation (up for k > 0 and down for k < 0).

Note that the order in which you reflect, stretch, or shrink does not matter. It does mat­ ter that you do vertical translation last. For example, if y = x2 is reflected in the x-axis and then moved up two units, the equation is y = -x2 + 2. If it is moved up two units and then reflected in the x-axis, the equation is y = -(x2 + 2) or y = -x2 – 2. A change in the order produces different functions.

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5

11-29 11.3 Transformations of Graphs 717

E X A M P L E 5 A multiple transformation of y � Vxs Graph the function y = -2Vx – 3s, and state the domain and range.

Solution Start with the graph of y = Vxs through (0, 0), (1, 1), and (4, 2), as shown in Fig. 11.28. Translate it three units to the right to get the graph of y = Vx – 3s. Stretch this graph by a factor of two to get the graph of y = 2Vx – 3s shown in Fig. 11.28. Now reflect in the x-axis to get the graph of y = -2Vx – 3s. To get an accurate graph calculate a few points on the final graph as follows:

Since x – 3 must be nonnegative in the expression -2Vx – 3s, we must have x – 3 : 0 and x : 3. So the domain is [3, o). Since the graph extends downward from the point (3, 0), the range is (-o, 0].

x 3 4 7

y = -2Vx – 3s 0 -2 -4

U Calculator Close-Up V

5 4 3 2 1

y

x21

2 1

3 4 5

4 5 6 7 8

y 2�x 3

y 2�x 3

y �x 3

y �x

2 1

Figure 11.28

Now do Exercises 31–32 You can check Example 5 by graph­ ing y = -2Vx – 3s with a graphing calculator.

10 The graph of y = x2 is a parabola opening upward with vertex (0, 0). The graph

of a function of the form y = a(x – h)2 + k is a transformation of y = x2 and is also a parabola. It opens upward if a > 0 and downward if a < 0. Its vertex is (h, k). In10 Example 6, we graph a multiple transformation of y = x2.

10

E X A M P L E 6 A multiple transformation of the parabola y � x2 Graph the function y = -2(x + 3)2 + 4, and state the domain and range.

Solution Think of the parabola y = x2 through (-1, 1), (0, 0), and (1, 1). To get the graph of y = -2(x + 3)2 + 4, translate it three units to the left, stretch by a factor of two, reflect in the x-axis, and finally translate upward four units. The graph is a stretched parabola open­ ing downward from the vertex (-3, 4) as shown in Fig. 11.29. To get an accurate graph

Figure 11.29

1

2

3

1

2

4

5

3

1 2 3

y

x

( 3, 4)

( 4, 2)

( 2, 2)

54

y 2(x � 3)2 � 4

5

345 2 1

( 5, 4) ( 1, 4)

dug84356_ch11a.qxd 9/14/10 2:33 PM Page 718

718 Chapter 11 Functions 11-30

calculate a few points around the vertex as follows:

Since any real number can be used for x in -2(x + 3)2 + 4, the domain is (-o, o). Since the graph extends downward from (-3, 4), the range is (-o, 4].

x -5 -4 -3 -2 -1

y = -2(x + 3)2 + 4 -4 2 4 2 -4

Now do Exercises 33–34

Understanding transformations helps us to see the location of the graph of a func­ tion. To get an accurate graph we must still calculate ordered pairs that satisfy the equation. However, if we know where to expect the graph, it is easier to choose appro­ priate ordered pairs.

Now do Exercises 35–48

E X A M P L E 7 A multiple transformation of the absolute value function y � x Graph the function y = 1

2 I x – 4 I – 1, and state the domain and range.

Solution Think of the V-shaped graph of y = I x I through (-1, 1), (0, 0), and (1, 1). To get the graph of y = 1

2 I x – 4 I – 1, translate y = I x I to the right four units, shrink by a factor

of 1 2 , and finally translate downward one unit. The graph is shown in Fig. 11.30.

To get an accurate graph, calculate a few points around the lowest point on the V-shaped graph as follows:

Since any real number can be used for x in 1 2

I x – 4 I – 1, the domain is (-o, o). Since the graph extends upward from (4, -1), the range is [-1, o).

x 2 4 6

y = 1 2

I x – 4 I – 1 0 -1 0

y

x

3

4 1

1 2

4

5

6 7 8 9

2

1

y x 4 11 2

(4, 1)

(6, 0)(2, 0)

Figure 11.30

dug84356_ch11a.qxd 9/18/10 10:06 AM Page 719

U1V Horizontal Translation

Graph each function and state the domain and range. See Example 1.

1. y = x + 3 2. y = x – 1

3. f (x) = (x – 3)2 4. f (x) = (x + 1)2

11-31 11.3 Transformations of Graphs 719

Warm-Ups ▼

Fill in the blank. 1. The graph of y = -f(x) is a of the graph of

y = f (x).

2. The graph of y = f(x) + k for k > 0 is a(n) of the graph of y = f (x).

3. The graph of y = f(x) + k for k < 0 is a(n) of the graph of y = f(x).

4. The graph of y = f(x – h) for h > 0 is a translation to the of the graph of y = f(x).

5. The graph of y = f(x – h) for h < 0 is a translation to the of the graph of y = f(x).

6. The graph of y = af(x) is a of the graph of y = f(x) if a > 1.

7. The graph of y = af(x) is a of the graph of y = f(x) if 0 < a < 1.

True or false? 8. The graph of f(x) = -x2 is a reflection in the x-axis of

the graph of f(x) = x2 .

9. The graph of y = lx – 3 l lies 3 units to the left of the graph of y = lx l.

10. The graph of y = lx l – 3 lies 3 units below the graph of y = lx l.

11. The graph of f(x) = 2 is a reflection in the x-axis of the graph of f(x) = -2.

12. The graph of y = -2×2 can be obtained by stretching and reflecting the graph of y = x2 .

13. The graph of y = Vx – 3 + 5 has the same shape as the graph of y = Vx .

1 1

.3Exercises

U Study Tips V • When you take notes, leave space. Go back later and fill in details and make corrections. • You can even leave enough space to work another problem of the same type in your notes.

5. f (x) = V 6. x – 1 f (x) = Vx + 6

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720 Chapter 11 Functions 11-32

7. f (x) = � x + 2 � 8. f (x) = � x – 4 � U3V Reflecting

Sketch the graphs of each pair of functions on the same coordinate system. See Example 3.

17. f (x) = V2x 18. y = x, y = -xs,

g(x) = -V2xs

U2V Stretching and Shrinking

Use stretching and shrinking to graph each function, and state the domain and range. See Example 2.

1 9. f (x) = 3×2 10. f (x) = x2

3

19. f (x) = x2 + 1, 20. f (x) = I x I + 1, 2 I g(x) = -(x + 1) g(x) = – x I – 1

1 11. y = x 12. y = 5x

5

21. y = Vx – 2 22. y = I x – 1 I, s, y = -Vx – 2 y = -I Is x – 1

1 13. f (x) = 3Vxs 14. f (x) = Vsx

3

23. f (x) = x – 3, 24. f (x) = x2 – 2, 1 2 15. y = 4

I x I 16. y = 4I x I g(x) = 3 – x g(x) = 2 – x

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11-33 11.3 Transformations of Graphs 721

U4V Vertical Translation 33. f (x) = (x + 3)2 – 5 34. f (x) = -2×2

Graph each function, and state the domain and range. See Example 4.

25. y = Vxs + 1 26. y = Vxs – 3

227. f (x) = x – 4 28. f (x) = x2 + 2

29. y = � x � + 2 30. y = � x � – 4

U5V Multiple Transformations

Sketch the graph of each function, and state the domain and range. See Examples 5–7. See the Strategy for graphing y = af(x – h) + k on page 716.

31. y = Vxs- 2 + 1 32. y = -Vx + 3s

35. y = -� x + 3 � 36. y = � x – 2 � + 1

37. y = -Vxs+ 1 – 2 38. y = -3Vx + 4s + 6

39. y = -2� x – 3 � + 4 40. y = 3� x – 1 � + 2

41. y = -2x + 3 42. y = 3x – 1

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722 Chapter 11 Functions 11-34

43. y = 2(x + 3)2 + 1 44. y = 2(x + 1)2 – 2 c) d)

y

x321

2 1

3 4 5

44 3 2 1 1

y

x321

2 1

3 4 5

44 3 2 1 1 2 3 4 5

45. y = -2(x – 4)2 + 2 46. y = -2(x – 1)2 + 3

e) f )

y

x321

2 3 4 5

44 3 2 1 1

y

x321

2 1

3 4 5

44 3 2 1 1

47. y = -3(x – 1)2 + 6 48. y = 3(x + 2)2 – 6

g) h)

Match each function with its graph a–h.

49. y = 2 + Vxs y = V2 + x50. s

51. y = 2Vxs 52. y = �x 2 1

53. y = Vxs 54. sy = 2 + Vx – 2 2

55. y = -2Vxs 56. y = V-xs

a) b)

y

x321

2 1

3 4 5

4 5 62 1 1 2 3 4 5

y

x321

2 3 4 5

44 3 2 1 1 2 3 4 5

y

x321

2 1

3 4 5

4 53 2 1 1 2

y

x321

2 1

3 4 5

44 3 2 1 1 2

Getting More Involved

57. If the graph of y = x2 is translated eight units upward, then what is the equation of the curve at that location?

58. If the graph of y = x2 is translated six units to the right, then what is the equation of the curve at that location?

59. If the graph of y = Vxs is translated five units to the left, then what is the equation of the curve at that location?

60. If the graph of y = Vxs is translated four units downward, then what is the equation of the curve at that location?

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11-35

61. If the graph of y = I x I is translated three units to the left and then five units upward, then what is the equation of the curve at that location?

62. If the graph of y = I x I is translated four units downward and then nine units to the right, then what is the equation of the curve at that location?

Graphing Calculator Exercises

63. Graph f (x) = I x I and g(x) = I x – 20 I + 30 on the same screen of your calculator. What transformations will transform the graph of f into the graph of g?

11.3 Transformations of Graphs 723

64. Graph f (x) = (x + 3)2, g(x) = x2 + 32, and h(x) = x2 + 6x + 9 on the same screen of your calculator.

a) Which two of these functions has the same graph? Why are they the same?

b) Is it true that (x + 3)2 = x2 + 9 for all real numbers x?

c) Describe each graph in terms of a transformation of the graph of y = x2.

Math at Work Sailboat Design

Mention sailing and your mind drifts to exotic locations, azure seas with soothing tropical breezes, crystal-clear waters, and dazzling white sand. But sailboat designers live in a world of computers, numbers, and formulas. Some of the measurements and formulas used to describe the sailing characteristics and stability of sailboats are the maximum hull speed formula, the sail area-displacement ratio, and the motion-comfort ratio.

To estimate the theoretical maximum hull speed (M) in knots, designers use the formula M = 1.34VLWLs, where LWL is the loaded waterline length (the length of the hull at the waterline). See the accompanying figure.

Sail area-displacement ratio r indicates how fast the boat is in light wind. It is given by r =

D A 2�3, where A is the sail area in square feet and D is the displacement in cubic feet. Values

of r range from 10 to 15 for cruisers and above 24 for high-performance racers. The motion-comfort ratio MCR, created by boat designer Ted Brewer, predicts the speed

of the upward and downward motion of the boat as it encounters waves. The faster the motion, the more uncomfortable the passengers. If D is the displacement in pounds, LWL the loaded waterline length in feet, LOA the length overall, and B is the beam (width) in feet, then

MCR = .

As the displacement increases, MCR increases. As the length and beam increases, MCR decreases. MCR should be in the low 30’s for a boat with an LOA of 42 feet.

D 2 3

B3�4 (1 7 0

LWL + 1 3

LOAr

8

10

12

6

4

2

40 50 30

M ax

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h ul

l s pe

ed (

kn ot

s)

M 1.34.LWL

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724 Chapter 11 Functions 11-36

Mid-Chapter Quiz Sections 11.1 through 11.3 Chapter 11

Determine whether y is a function of x using the ordered pairs given in each table.

1. x -1 0 3 4 3

y 4 2 6 8 5

2. x 2 4 6 8 10

y 1 2 3 4 5

Determine whether each set of ordered pairs is a function.

3. {(99, 0), (76, 0), (44, 0)}

4. {(1/2, 6), (1/3, 12), (1/4, -9)}

Determine whether y is a function of x for each relation.

5. y = x2 + 100

6. x = y2 + 100

Find the domain and range of each relation.

7. y = Vx – 3

8. {(1, 2), (3, 4), (20, 30), (40, 30)}

Graph each relation, and state its domain and range.

9. f(x) = 2x – 4 10. g(x) = l2x – 4l

11. h(x) = -x2 + 3 – 512. y = Vx + 6

for x : 013. y = x + 1 14. x = y2 – 2 2 for x < 0

15. y = -2(x – 1)2 + 3

Miscellaneous.

16. Find f(3) if f(x) = -x2 + 9.

17. Find g(-4) if g(x) = -2lx + 4l – 9.

18. If the graph of y = x2 is translated 2 units to the left and 4 units upward, then what is the equation of the curve in its final position?

19. If the graph of y = lx l is stretched by a factor of 2, trans­ lated 3 units to the left, and reflected in the x-axis, then what is the equation of the curve in its final position?

1 20. If the graph of y = V x is shrunk by a factor of , trans­2

lated 9 units to the left and 5 units downward, then what is the equation of the curve in its final position?

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11-37 11.4 Graphs of Polynomial Functions 725

In This Section

U1V Cubic Functions

U2V Quartic Functions

U3V Symmetry

U4V Behavior at the x-Intercepts

11.4 Graphs of Polynomial Functions

We have already graphed constant functions, linear functions, and quadratic func- tions, which are polynomial functions of degree 0, 1, and 2, respectively. In this section we will graph some polynomial functions with degrees that are greater than 2.

U1V Cubic Functions U5V Transformations

U6V Solving Polynomial A third-degree polynomial function is called a cubic function. The most basic third-

Inequalities degree polynomial function is f (x) = x3, which is called the cubing function. We can graph it by plotting some ordered pairs that satisfy the equation f (x) = x3.

E X A M P L E 1 The cubing function Graph the function f (x) = x3, and identify the intercepts.

Solution Make a table of ordered pairs as follows:

x -2 -1 0 1 2

f (x) = x3 -8 -1 0 1 8

Plot these ordered pairs, and sketch a smooth curve through them as shown in Fig. 11.31. The x-intercept and the y-intercept are both at the origin, (0, 0).

Now do Exercises 1–2

Figure 11.31

2 1

3 4 5 6 7 8 9

y

x1 32 4

( 2, 8)

(2, 8)

4 3 2

4 5 6 7 8

f (x) x3

In general, the x-intercepts for a polynomial function can be difficult to find, but we will consider only polynomial functions for which the x-intercepts can be found by fac­ toring. Example 2 shows a third-degree polynomial function that has three x-intercepts.

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726 Chapter 11 Functions 11-38

Plot these ordered pairs, and sketch a smooth curve through them as shown in Fig. 11.32.

Now do Exercises 3–10

E X A M P L E 2 A cubic function with three x-intercepts Graph the function f (x) = x3 – 4x, and identify the intercepts.

Solution The y-intercept is found by replacing x with 0. Since f (0) = 03 – 4(0) = 0, the y-intercept is (0, 0). The x-intercepts are found by replacing y or f (x) with 0 and then solving for x:

x3 – 4x = 0

x(x2 – 4) = 0 x (x – 2)(x + 2) = 0 Factor completely.

x = 0 or x – 2 = 0 or x + 2 = 0 Zero factor property

x = 0 or x = 2 or x = -2

The x-intercepts are (-2, 0), (0, 0), and (2, 0). Now make a table that includes those val­ ues for x:

Figure 11.32

6 3

9 12 15

y

x4 5

f (x) x3 4x

4 3 2 1

6 9

12 15 18

3 35

x -3 -1 1 3

f (x) = x3 – 4x -15 3 -3 15

U2V Quartic Functions A fourth-degree polynomial function is called a quartic function. The most basic fourth-degree polynomial function is f (x) = x4. We can graph it by plotting some ordered pairs that satisfy the equation f (x) = x4.

E X A M P L E 3 The most basic fourth-degree polynomial function Graph f (x) = x4, and identify the intercepts.

Solution Since f (0) = 04 = 0, the y-intercept is (0, 0). Since x4 = 0 is satisfied only if x = 0, the only x-intercept is also (0, 0). Make a table of ordered pairs as follows:

x -2 -1 1 2

f (x) = x4 16 1 1 168 10 12 14 16 18

y

x21 4 54 3 2 1

4 6

2

44

35

f (x) x4

Plot these ordered pairs, and sketch a smooth curve through them as shown in Fig. 11.33. The shape of f (x) = x4 is similar to a parabola, except it is “flatter” on the bottom.

Now do Exercises 11–12Figure 11.33

Example 4 shows a fourth-degree polynomial function that has four x-intercepts.

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11-39 11.4 Graphs of Polynomial Functions 727

E X A M P L E 4 A fourth-degree polynomial function with four x-intercepts Graph f (x) = x4 – 10×2 + 9, and identify the intercepts.

Solution To find the y-intercept replace x with 0. Since f (0) = 04 – 10(02) + 9 = 9, the y-intercept is (0, 9). To find the x-intercepts replace y or f (x) with 0 and then solve for x:

x4 – 10×2 + 9 = 0

(x2 -1)(x2 – 9) = 0 (x – 1)(x + 1)(x – 3)(x + 3) = 0 Factor completely.

x – 1 = 0 or x + 1 = 0 or x – 3 = 0 or x + 3 = 0

x = 1 or x = -1 or x = 3 or x = -3

The four x-intercepts are (�1, 0) and (�3, 0). Now make a table that includes those values for x:

x -4 -2 0 2 4

x4 – 10×2 + 9 105 -15 9 -15 105

Figure 11.34

100

x2 4 54 2

25

5

75

50

25

f (x) x4 10×2 + 9

f (x)

Plot these ordered pairs, and sketch a smooth curve through them as shown in Fig. 11.34.

Now do Exercises 13–20

U3V Symmetry Consider the graph of the quadratic function f (x) = x2 shown in Fig. 11.35. Notice that both (2, 4) and (-2, 4) are on the graph. In fact, f (x) = f (-x) for any value of x. We get the same y-coordinate whether we evaluate the function at a number or its opposite. This fact causes the graph to be symmetric about the y-axis. If we folded the paper along the y-axis, the two halves of the graph would coincide.

Figure 11.36

2 1

3 4 5 6 7 8 9

y

x1 32 4

( 2, 8)

(2, 8)

4 3 2

4 5 6 7 8

f (x) x3

2 1

3 4 5

y

x4 3 1 32 4

f (x) x2

2

2

1 1

Figure 11.35

If f (x) is a function such that f (x) = f (-x) for any value of x in its domain, then the graph of the function is said to be symmetric about the y-axis.

Consider the graph of f (x) = x3 shown in Fig. 11.36. It is not symmetric about the y-axis like the graph of f (x) = x2, but it has a different kind of symmetry. On the graph

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728 Chapter 11 Functions 11-40

of f (x) = x3 we find the points (2, 8) and (-2, -8). In this case f (x) and f (-x) are not equal, but f (-x) = -f (x). Notice that the points (2, 8) and (-2, -8) are the same dis­ tance from the origin and lie on a line through the origin.

If f (x) is a function such that f (-x) = -f (x) for any value of x in its domain, then the graph of the function is said to be symmetric about the origin.

Determining the symmetry of a graph Discuss the symmetry of the graph of each polynomial function.

a) f (x) = 5×3 – x b) f (x) = 2×4 – 3×2 c) f (x) = x2 – 3x + 6

Solution a) Since f (-x) = 5(-x)3 – (-x) = -5×3 + x, we have f (-x) = -f (x). So the graph

b) Since f (-x) = 2(-x)4 – 3(-x)2 = 2×4 – 3×2, we have f (x) = f (-x). So the graph is symmetric about the y-axis.

c) In this case f (-x) = (-x)2 – 3(-x) + 6 = x2 + 3x + 6. So f (-x) * f (x) and f (-x) * -f (x). This graph has neither type of symmetry.

E X A M P L E 5

Now do Exercises 21–38

U Calculator Close-Up V

We can use graphs to check the conclusions The graph of f (x) = 2×4 – 3×2 appears to be The graph of f (x) = x2 – 3x + 6 does not about symmetry that were arrived at symmetric about the y-axis. appear to have either type of symmetry. algebraically in Example 5. The graph of

3f (x) = 5x – x appears to be symmetric about the origin.

2 5 20

3 31 1

3 6

5 52

U4V Behavior at the x-Intercepts 2The graphs of y = x, y = x , y = x3, and y = x4 all have the same x-intercept (0, 0).

But they have two different types of behavior at that x-intercept. The graphs of y = x and y = x3 cross the x-axis at (0, 0), whereas the graphs of y = x2 and y = x4 touch but do not cross the x-axis at (0, 0). The reason for this behavior is the power of the fac­ tor x. If a nonzero number is raised to an odd power, the result has the same sign as the original number. But if the power is even, the result is positive. Since y = x and y = x3 have odd powers, the y-coordinates are positive to the right of (0, 0) and neg­ ative to the left of (0, 0), and the graph crosses the x-axis at (0, 0). Since the exponents in y = x2 and y = x4 are even, the y-coordinates are positive on either side of (0, 0), and the graphs touch but do not cross the x-axis at (0, 0). In general, we have the fol­ lowing theorem.

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11-41 11.4 Graphs of Polynomial Functions 729

Behavior at the x-Intercepts

Suppose that x – c is a factor of a polynomial function. The graph of the function crosses the x-axis at (c, 0) if x – c occurs an odd number of times and touches but does not cross the x-axis if x – c occurs an even number of times.

Since factoring can get difficult for higher-degree polynomials, we will often dis­ cuss functions that are given in factored form as in Example 6.

E X A M P L E 6 Behavior at the x-intercepts Find the x-intercepts, and discuss the behavior of the graph of each polynomial function at its x-intercepts.

a) f (x) = (x – 1)2(x – 3) b) y = x3 + 2×2 – x – 2

Solution a) Replace f (x) with 0 to find the x-intercepts:

(x – 1)2(x – 3) = 0 (x – 1)2 = 0 or x – 3 = 0

x – 1 = 0 or x = 3 x = 1

The x-intercepts are (1, 0) and (3, 0). Since the factor corresponding to (1, 0) is x – 1 and its power is even, the graph touches but does not cross the x-axis at (1, 0). Since the factor corresponding to (3, 0) is x – 3 and its power is odd, the graph crosses the x-axis at (3, 0).

b) Replace y with 0 to find the x-intercepts:

x3 + 2×2 – x – 2 = 0

x2(x + 2) – 1(x + 2) = 0 Factor by grouping.

(x2 – 1)(x + 2) = 0 Factor out x + 2. (x – 1)(x + 1)(x + 2) = 0 Factor completely.

x – 1 = 0 or x + 1 = 0 or x + 2 = 0

x = 1 or x = -1 or x = -2

The x-intercepts are (1, 0), (-1, 0), and (-2, 0). Since each factor occurs with the power of one and one is odd, the graph crosses the x-axis at each of the three x-intercepts.

Now do Exercises 39–52

U Calculator Close-Up V

The graphs of the functions in Example 6 support the conclusions that were made about the behavior at the x-intercepts.

8 8

44 4 4

8 8

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730 Chapter 11 Functions 11-42

U5V Transformations In Section 11.3 we learned how changes in the formula defining a function can trans­ form the graph of the function. In Example 7, we perform some transformations on

4f (x) = x3 and f (x) = x .

E X A M P L E 7 Transformations of graphs Write the equation of each curve in its final position.

a) The graph of f(x) = x3 is translated 3 units to the right and 2 units downward.

b) The graph of f (x) = x4 is translated 4 units to the left and reflected in the x-axis.

Solution a) To move the graph 3 units to the right, replace x with x – 3 to get f(x) = (x – 3)3 .

To move the graph 2 units downward, subtract 2. So f (x) = (x – 3)3 – 2 is the equation for the graph in its final position.

b) To move the graph 4 units to the left, replace x with x + 4 to get f(x) = (x + 4)4 . To reflect in the x-axis, multiply by -1. So f (x) = -(x + 4)4 is the equation for the graph in its final position.

Now do Exercises 53–60

U Calculator Close-Up V

The graph of f (x) = (x – 3)3 – 2 shows that it is a The graph of f (x) = -(x + 4)4 shows that it is a translation 3 units to the right and 2 units down- translation 4 units to the left and a reflection of

3 4ward of f (x) = x . f (x) = x .

8 2

6 1

1 6

8 8

U6V Solving Polynomial Inequalities An inequality such as x3 – 3x > 0 is a polynomial inequality. The graphical method that we used for quadratic inequalities in Section 10.5 can also be used with polyno­ mial inequalities. We can read the solution to the inequality from the graph of y = x3 – 3x provided we know all of the x-intercepts. Any value of x for which y > 0 on the graph is a solution to the inequality.

E X A M P L E 8 Solving a polynomial inequality with the graphical method Solve each polynomial inequality. Write the solution set in interval notation, and graph it.

a) x3 – 3x > 0 b) (x – 1)(x + 2)(x – 3) – 0

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11-43 11.4 Graphs of Polynomial Functions 731

Solution a) To solve x3 – 3x > 0, graph y = x3 – 3x. We can determine the solution set to the

inequality from the graph if we know the x-intercepts. So first find them by solving x3 – 3x = 0:

x3 – 3x = 0

x(x2 – 3) = 0

x = 0 or x2 – 3 = 0

x = 0 or x2 = 3

x = 0 or x = ±\33

The x-intercepts are (-\33, 0), (0, 0), and (\33, 0). The graph in Fig. 11.37 crosses the x-axis at each intercept. The inequality is satisfied for any x that corresponds to a positive y-coordinate on this graph. So the solution set to the inequality is (-\33, 0) U (\33, o) and its graph is shown in Fig. 11.38.

b) To solve (x – 1)(x + 2)(x – 3) – 0, graph y = (x – 1)(x + 2)(x – 3). To find the x-intercepts we solve (x – 1)(x + 2)(x – 3) = 0. The x-intercepts are (1, 0), (-2, 0), and (3, 0). The graph in Fig. 11.39 crosses the x-axis at each intercept.

The inequality is satisfied for any x that corresponds to a negative or zero y-coordinate on this graph. So the solution set is (-o, -2] U [1, 3] and its graph is shown in Fig. 11.40.

Now do Exercises 87–96

Figure 11.39

y

x

6

�4

2

4

1 �2

�4

�6

�8

�10

2 3

8

10

4�1�2�3

y � (x � 1) (x � 2) (x � 3)

(�2, 0) (1, 0) (3, 0)

y

x

6

-4

2

4

1 -2

-4

-6

-8

2 3

8

4

y = x2 – 3x

-1-2-3

(-[33, 0) ([33, 0)

-2-1 0 1 2

-3-2-1 0 1 2 3

Figure 11.37

Figure 11.38

Figure 11.40

The test-point method that we used for quadratic inequalities in Section 10.5 can be used also with polynomial inequalities. For this method we first find all of the roots to the polynomial, as in the graphical method. Instead of graphing the polynomial function, we plot the roots on a number line and then test a point in each interval deter­ mined by the roots.

E X A M P L E 9 Solving a polynomial inequality with the test-point method Solve each polynomial inequality. Write the solution set in interval notation and graph it.

a) x4 – 16 – 0 b) x4 – 3×3 – 18×2 0

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732 Chapter 11 Functions 11-44

Solution a) Find the roots to x4 – 16 = 0:

x4 -16 = 0

(x2 – 4)(x2 + 4) = 0

x2 – 4 = 0 or x2 + 4 = 0

x2 = 4 x2 = -4

x = �2

The only real solutions to the equation are -2 and 2. Locate these numbers on a number line as in Fig. 11.41.

Select the test points -3, 0, and 3. Test them in the original inequality x4 – 16 – 0.

(-3)4 – 16 – 0 Incorrect

(0)4 – 16 – 0 Correct

(3)4 – 16 – 0 Incorrect

Since 0 is the only test point that satisfies the inequality, the interval containing 0 is the solution set to the inequality. Because of the – symbol we include the endpoints. The solution set is [-2, 2], and its graph is shown in Fig. 11.42.

b) Find the roots to x4 – 3×3 – 18×2 = 0:

x2(x2 – 3x – 18) = 0

x2(x – 6)(x + 3) = 0

x2 = 0 or x – 6 = 0 or x + 3 = 0

x = 0 or x = 6 or x = -3

Now locate -3, 0, and 6 on a number line as in Fig. 11.43.

Select the test points -5, -1, 2 and 7. Use a calculator to test them in the original inequality x4 – 3×3 – 18×2 : 0:

(-5)4 – 3(-5)3 – 18(-5)2 – 0 Incorrect

(-1)4 – 3(-1)3 – 18(-1)2 – 0 Correct

(2)4 – 3(2)3 – 18(2)2 – 0 Correct

(7)4 – 3(7)3 – 18(7)2 – 0 Incorrect

The inequality is satisfied on the intervals containing -1 and 2, which are [-3, 0] and [0, 6]. Since the symbol is -, the endpoints of the intervals are included. The solution set is [-3, 0] U [0, 6], which is simplified to [-3, 6]. The graph of the solution set is shown in Fig. 11.44.

Figure 11.43

10-1-2-3-4-5 2 3 4 5 6 7

Test points

Figure 11.41

10-1-2-3-4 2 3 4

Test points

-3-2-1 0 1 2 3

Figure 11.42

Figure 11.44

-3 0 3 6

Now do Exercises 97–104

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U1V Cubic Functions

Graph each function, and identify the x- and y-intercepts. See Examples 1 and 2.

1. f x � x3 � 1 2. f x � x3 � 1

3. f (x) � x3 � 9x 4. f (x) � x3 � x

11-45 11.4 Graphs of Polynomial Functions 733

Warm-Ups ▼

Fill in the blank. 1. The graph of y � f (x) is symmetric about the if

f (x) � f (�x) for all x in the domain of f.

2. The graph of y � f (x) is symmetric about the if f (�x) � �f (x) for all x in the domain of f.

3. The graph of a polynomial function P(x) crosses the x-axis at c if x � c occurs an number of times in the prime factorization of P.

4. The graph of a polynomial function P(x) touches but does not cross the x-axis at c if x � c occurs an number of times in the prime factorization of P.

True or false? 5. The graph of f(x) � x3 � x is symmetric about the

y-axis.

6. The graph of y � 2x � 1 is symmetric about the origin.

7. If f(x) � 3x, then f(x) � f(�x) for any real number x.

8. If f(x) � 3×4 � 5×3 � 2×2 � 6x � 7, then f(�x) � 3×4 � 5×3 � 2×2 � 6x � 7.

9. There is only one x-intercept for the graph of f(x) � x2 � 4x � 4.

10. The graph of y � (x � 1)2(x � 4)4 does not cross the x-axis at either of its intercepts.

1 1

.4Exercises

U Study Tips V • Always study math with a pencil and paper. Just sitting back and reading the text rarely works. • A good way to study the examples in the text is to cover the solution with a piece of paper and see how much of the solution you can

( ) ( )

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734 Chapter 11 Functions 11-46

3 2 2 4 2 4 25. f (x) = -x – 4x 6. f (x) = -x3 + 3x 13. f (x) = x – 4x 14. f (x) = x – 9x

4 2 4 215. f (x) = x – 5x + 4 16. f (x) = x – 20x + 64 3 + x2 27. f (x) = x – 4x – 4 8. f (x) = x3 + 2x – 9x – 18

3 217. f (x) = x4 + x – 4x – 4x

3 2 3 29. f(x) = x – 3x – 9x + 27 10. f(x) = x – 2x – 4x + 8

3 218. f (x) = x4 + 2x – 9x -18x

U2V Quartic Functions

Graph each function, and identify the x- and y-intercepts. See Examples 3 and 4.

411. f (x) = x – 1 12. f (x) = x4 + 3

4 319. f (x) = x – 3x – 9×2 + 27x

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11-47 11.4 Graphs of Polynomial Functions 735

4 320. f (x) = x – 2x – 4×2 + 8x

U3V Symmetry

Discuss the symmetry of the graph of each polynomial function. See Example 5.

21. f (x) = 2x

22. f (x) = -x

23. f (x) = -2×2

24. f (x) = x2 + 1

25. f (x) = 2×3

26. f (x) = -x3

27. f (x) = x4

28. f (x) = x4 – x

29. f (x) = x3 – 5x + 1

30. f (x) = 5×3 + 7x

31. f (x) = 6×6 – 3×2 – x 6 4 232. f (x) = x – x + x – 8

33. f (x) = (x – 3)2

34. f (x) = 3(x + 2)2

235. f (x) = (x – 5)3

236. f (x) = (x + 1)2

37. f (x) = x

38. f (x) = -3x

U4V Behavior at the x-Intercepts

Find the x-intercepts, and discuss the behavior of the graph of each polynomial function at its x-intercepts. Check your answers with a graphing calculator if you have one. See Example 6.

39. f (x) = (x – 2)2(x – 8)

40. f (x) = (x + 3)2(x – 5)

41. f (x) = (x – 1)2(x + 4)2

42. f (x) = (x + 4)2(x + 6)2

43. f (x) = (x – 1)(x + 4)(x – 7)2

44. f (x) = (x + 1)(x – 3)(x + 9)2

245. f (x) = x3 + 6x – x – 6

246. f (x) = x3 + 5x – 4x – 20

247. f (x) = -x3 + 5x

48. f (x) = -x3 – 9×2

49. f (x) = x4 – 5×3

350. f (x) = x4 + x

251. f (x) = x4 + 6×3 + 9x

4 252. f (x) = x – 4×3 + 4x

U5V Transformations

Write the equation of each curve in its final position. See Example 7.

53. The graph of f (x) = x3 is translated 5 units to the right and 4 units downward.

54. The graph of f (x) = x3 is translated 2 units to the right and 1 unit upward.

55. The graph of f (x) = x3 is translated 6 units to the left and 3 units upward.

56. The graph of f (x) = x3 is translated 4 units to the left and 7 units downward.

57. The graph of f(x) = x3 is reflected in the x-axis.

58. The graph of f (x) = x3 is reflected in the x-axis and then translated 1 unit upward.

59. The graph of f (x) = x4 is translated 3 units to the right and then reflected in the x-axis.

60. The graph of f (x) = x4 is translated 5 units to the left and then reflected in the x-axis.

Miscellaneous

Match each polynomial function with its graph a–h.

61. f (x) = -2x + 3 62. f (x) = -2×2 + 3 63. f (x) = -2×3 + 3 64. f (x) = -2×2 + 4x + 3 65. f (x) = -x4 + 3 66. f (x) = x3 – 3×2

267. f (x) = x3 + 3x – x – 3 1

68. f (x) = x4 – 3 2

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736 Chapter 11 Functions 11-48

a) b) Sketch the graph of each polynomial function.

69. f (x) = 2x – 6 70. f (x) = -3x + 3

2 1

4 5

y

x2 434 3 2

2 3 4 5

1 1

2 1

4 5

y

x2 43

3

14 3 2

2

4 5

1

c) d) 71. f (x) = -x2 72. f (x) = x2 – 3

2 1

4 5

y

x21 434 3 2

2 3 4 5

1

2 1

5

y

x1 434 3 2 1

2 3 4 5

1

3 2 373. f (x) = x – 2x 74. f (x) = x – 4x

e) f )

1

5 4

y

x4324 3 2

2 3 4 5

1

2 1

5 4 3

y

x4324 2

4 5

75. f (x) = (x – 1)2(x + 1)2 76. f (x) = (x + 2)2(x – 1)

g) h)

1

5 4 3 2

y

x421

5

4 3 2 1

3 277. f (x) = (x – 1)2(x – 3) 78. f (x) = x + 2x – 3x

5 4

y

x43214 3 2 1

5

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11-49 11.4 Graphs of Polynomial Functions 737

4 2 3 2 279. f (x) = x – 4×3 + 4x 80. f (x) = -x4 + 6x – 9x 86. f (x) = (x – 20)2(x + 30)2x

Graphing Calculator Exercises Sketch the graph of each polynomial function. First graph the function on a calculator and use the calculator graph as a guide.

81. f (x) = x – 20 82. f (x) = (x – 20)2

83. f (x) = (x – 20)2(x + 30)

84. f (x) = (x – 20)2(x + 30)2

85. f (x) = (x – 20)2(x + 30)2x

U6V Solving Polynomial Inequalities

Solve each polynomial inequality using the graphical method. State the solution set using interval notation and graph it. See Example 8.

387. x – 4x > 0

388. x – 16x < 0

89. (x – 3)(x – 5)(x + 2) – 0

90. (x + 4)(x + 1)(x – 6) : 0

3 291. x – 2x : 0

92. -x3 + 5×2 – 0

93. x4 – 4×3 + 4×2 : 0

94. -x4 + 6×3 – 9×2 : 0

95. (x – 1)2(x + 1)2 – 0

96. (x + 2)2(x – 1) – 0

Solve each polynomial inequality using the test-point method. State the solution set using interval notation, and graph it. See Example 9.

497. x – 81 > 0

498. x – 1 < 0

4 3 299. x – x – 6x – 0

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738 Chapter 11 Functions 11-50

100. x4 + 2×3 – 8×2 : 0

101. x3 + 6×2 – 4x – 24 > 0

3 + 5×2102. x – 9x – 45 < 0

103. x4 – 10×2 + 9 – 0

104. x4 – 18×2 + 32 : 0

Getting More Involved In each case, find a polynomial function f(x) whose graph behaves in the required manner. Answers may vary.

105. The graph has only one x-intercept at (3, 0) and crosses the x-axis there.

106. The graph has only one x-intercept at (3, 0) but does not cross the x-axis there.

107. The graph has only two x-intercepts at (-2, 0) and (1, 0). It crosses the x-axis at (-2, 0) but does not cross at (1, 0).

108. The graph has only two x-intercepts at (5, 0) and (-6, 0). It does not cross the x-axis at either x-intercept.

In This Section

U1V Rational Functions

U2V Asymptotes

U3V Sketching the Graphs

U4V Rational Inequalities U1V Rational Functions

11.5 Graphs of Rational Functions

We first studied rational expressions in Chapter 6. In this section we will study functions that are defined by rational expressions.

A rational expression was defined in Chapter 6 as a ratio of two polynomials. If a ratio of two polynomials is used to define a function, then the function is called a rational function.

Rational Function P(x)If P(x) and Q(x) are polynomials with no common factor and f (x) = forQ(x)

Q(x) * 0, then f (x) is called a rational function.

The domain of a rational function is the set of all real numbers except those that cause the denominator to have a value of 0.

E X A M P L E 1 Domain of a rational function Find the domain of each rational function.

a) f (x) = x

x

3 1

b) g(x) = 2 x

x 2 –

4 3

Solution a) Since x – 1 = 0 only for x = 1, the domain of f is the set of all real numbers

except 1, (-o, 1) U (1, o).

b) Since x2 – 4 = 0 for x = �2, the domain of g is the set of all real numbers excluding 2 and -2, (-o, -2) U (-2, 2) U (2, o).

Now do Exercises 1–6

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11-51 11.5 Graphs of Rational Functions 739

E X A M P L E 2 Horizontal and vertical asymptotes Find the horizontal and vertical asymptotes for each rational function.

a) f (x) = x2 –

3 1

b) g(x) = x2 –

x

4

c) h(x) = 2 x

x

+

+

3 1

Solution a) The denominator x2 – 1 has a value of 0 if x = �1. So the lines x = 1 and

x = -1 are vertical asymptotes. If I x I is very large, the value of x2 –

3 1

is approximately 0. So the x-axis is a horizontal asymptote.

b) The denominator x2 – 4 has a value of 0 if x = �2. So the lines x = 2 and x = -2 are vertical asymptotes. If I x I is very large, the value of

x2 – x

4 is

approximately 0. So the x-axis is a horizontal asymptote.

c) The denominator x + 3 has a value of 0 if x = -3. So the line x = -3 is a vertical asymptote. If I x I is very large, the value of h(x) is not approximately 0.

U Calculator Close-Up V

The graph for Example 2(a) should consist of three separate pieces, but in connected mode the calculator connects the separate pieces. Even though the calculator does not draw a very good graph of this function, it does support the conclusion that the horizontal asymptote is the x-axis and the vertical asymptotes are x = -1 and x = 1.

5

40

40

5

U Calculator Close-Up V

If the viewing window is too large, a rational function will appear to touch its asymptotes.

8

8 8

8

Because the asymptotes are an important feature of a rational function, we should draw it so that it approaches but does not touch its asymptotes.

Figure 11.45

1

5 4 3 2

y

x1 4 52 31

f(x) 1 x

U2V Asymptotes Consider the simplest rational function f (x) = 1�x. Its domain does not include 0, but 0 is an important number for the graph of this function. The behavior of the graph of f when x is very close to 0 is what interests us. For this function the y-coordinate is the reciprocal of the x-coordinate. When the x-coordinate is close to 0, the y-coordinate is far from 0. Consider the following tables of ordered pairs that satisfy f (x) = 1�x:

x > 0 x < 0

x y

0.1 10

0.01 100

0.001 1000

0.0001 10,000

x y

-0.1 -10

-0.01 -100

-0.001 -1000

-0.0001 -10,000

As x gets closer and closer to 0 from above 0, the value of y gets larger and larger. We say that y goes to positive infinity. As x gets closer and closer to 0 from below 0, the values of y are negative but I y I gets larger and larger. We say that y goes to negative infinity. The graph of f gets closer and closer to the vertical line x = 0, and so x = 0 is called a vertical asymptote. On the other hand, as I x I gets larger and larger, y gets closer and closer to 0. The graph approaches the x-axis as x goes to infinity, and so the x-axis is a horizontal asymptote for the graph of f. See Fig. 11.45 for the graph of f (x) = 1�x.

In general, a rational function has a vertical asymptote for every number excluded from the domain of the function. The horizontal asymptotes are determined by the behavior of the function when I x I is large.

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740 Chapter 11 Functions 11-52

To understand the value of h(x), we change the form of the rational expression by using long division:

2 x + 3�2sxs+s 1s

2x + 6

-5

Writing the rational expression as quotient + re d m iv a i i s n o d r er, we get h(x) = 2 x

x +

+

3 1

=

2 + x –

+

5 3. If I x I is very large, x

+

5 3 is approximately 0, and so the y-coordinate

is approximately 2. The line y = 2 is a horizontal asymptote.

Now do Exercises 7–12

Example 2 illustrates two important facts about horizontal asymptotes. If the degree of the numerator is less than the degree of the denominator, then the x-axis

x – 4 is the horizontal asymptote. For example, y = x2 – 7 has the x-axis as a horizontal asymptote. If the degree of the numerator is equal to the degree of the denominator, then the ratio of the leading coefficients determines the horizontal asymptote. For

2x – 7 2 example, y = has y = as its horizontal asymptote. The remaining case is 3x – 5 3 when the degree of the numerator is greater than the degree of the denominator. This case is discussed next.

Each rational function of Example 2 had one horizontal asymptote and a vertical asymptote for each number that caused the denominator to be 0. The horizontal asymptote y = 0 occurs because, as I x I gets larger and larger, the y-coordinate gets closer and closer to 0. Some rational functions have a nonhorizontal line for an asymptote. An asymptote that is neither horizontal nor vertical is called an oblique asymptote or slant asymptote.

E X A M P L E 3 Finding an oblique asymptote Determine all of the asymptotes for

g(x) = 2×2 +

x +

3x 2 – 5

.

Solution If x + 2 = 0, then x = -2. So the line x = -2 is a vertical asymptote. Use long division to rewrite the function as quotient +

re d m iv a i i s n o d r er:

g(x) = = 2x – 1 + x

+

3 2

If I x I is large, the value of x – +

3 2

is approximately 0. So when I x I is large, the value of g(x) is approximately 2x – 1. The line y = 2x – 1 is an oblique asymptote for the graph of g.

2×2 + 3x – 5

x + 2

Now do Exercises 13–14

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11-53 11.5 Graphs of Rational Functions 741

We can summarize this discussion of asymptotes with the following strategy for finding asymptotes for a rational function.

Strategy for Finding Asymptotes for a Rational Function

Suppose f (x) = Q P(

( x x ) )

is a rational function with the degree of Q(x) at least 1.

1. Solve the equation Q(x) = 0. The graph of f has a vertical asymptote corre­ sponding to each solution to the equation.

2. If the degree of P(x) is less than the degree of Q(x), then the x-axis is a horizontal asymptote.

3. If the degree of P(x) is equal to the degree of Q(x), then find the ratio of the leading coefficients. The horizontal line through that ratio is the horizontal asymptote.

4. If the degree of P(x) is one larger than the degree of Q(x), then use division to rewrite the function as

quotient + re

d m iv a i i s n o d r er

.

The equation formed by setting y equal to the quotient gives us an oblique asymptote.

U3V Sketching the Graphs We now use asymptotes to help us sketch the graphs of some rational functions.

E X A M P L E 4 Graphing a rational function Sketch the graph of each rational function.

a) f (x) = x2 –

3 1

b) g(x) = x2 –

x

4

Solution a) From Example 2(a), the lines x = 1 and x = -1 are vertical asymptotes and the

x-axis is a horizontal asymptote. The vertical asymptotes are drawn with dashed lines as shown in Fig. 11.46 on page 742. Next, we find some ordered pairs that

satisfy f (x) = x2 –

3 1 . The graph goes through the points (0, -3), (�0.9, -15.789),

(�1.1, 14.286), (�2, 1), and (�3, 3 8r as it approaches its asymptotes in Fig. 11.46. b) From Example 2(b), the lines x = 2 and x = -2 are vertical asymptotes and the

x-axis is a horizontal asymptote. The vertical asymptotes are drawn with dashed lines

as shown in Fig. 11.47. Next, we find some ordered pairs that satisfy g(x) = x2 –

x 4 .

dug84356_ch11a.qxd 9/14/10 2:34 PM Page 742

742 Chapter 11 Functions 11-54

Now do Exercises 23–26

The graph goes through the points (0, 0), (1, -1 3r, (1.9, -4.872), (2.1, 5.122), (3, 3 5r,

and (4, 1 3r as it approaches its asymptotes in Fig. 11.47.

Figure 11.47 Figure 11.46

1

3 2

y

x24 4 52 3

f(x) 3

x2 1

2 1

U Calculator Close-Up V

This calculator graph supports the graph drawn in Fig. 11.47. Remember that the calculator graph can be misleading. The vertical lines drawn by the calculator are not part of the graph of the function.

4

10

10

4

E X A M P L E 5 Graphing a rational function Sketch the graph of each rational function.

a) h(x) = 2 x x + +

3 1 b) g(x) =

2×2 + x +

3x 2 – 5

Solution a) Draw the vertical asymptote x = -3 and the horizontal asymptote y = 2 from

Example 2(c) as dashed lines. The points (-2, -3), (0, 1 3r, (-1 2, 0r, (7, 1.5), (-4, 7), and (-13, 2.5) are on the graph shown in Fig. 11.48.

Figure 11.48 Figure 11.49

U Calculator Close-Up V

This calculator graph supports the graph drawn in Fig. 11.48. Note that if x is -3, there is no y-coordinate because x = -3 is the vertical asymptote.

4

20

10

10

b) Draw the vertical asymptote x = -2 and the oblique asymptote y = 2x – 1 from

Example 3 as dashed lines. The points (-1, -6), (0, -5 2r, (1, 0), (4, 6.5), and (-2.5, 0) are on the graph shown in Fig. 11.49.

x

x2 4 g(x)

2

1

y

x1 4 53

1

2

2x + 1 x + 3f(x)

10 8 6 4

y

x48 8642

4 6 8

10

12 x3 4 53215

2x 2 + 3x 5 x + 2g(x)

10 8 6 4 2

y

8 10

Now do Exercises 27–32

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11-55 11.5 Graphs of Rational Functions 743

U4V Rational Inequalities Inequalities involving rational expressions are rational inequalities. We can solve rational inequalities using the graphical method or the test-point method as we did for quadratic inequalities in Section 10.5 and polynomial inequalities in Section 11.4.

Solving a rational inequality graphically Solve each rational inequality. Write the solution set in interval notation, and graph it.

a) x x

+

1 2

> 0 b) x x

+

– 2 4

– 3

Solution a) To solve the rational inequality, graph y = x

x –

+

1 2 . The vertical asymptote is x = -2.

The x-intercept is (1, 0), and the y-intercept is �0, -1 2 �. The graph is shown in

Fig. 11.50. The values of x that satisfy the inequality are the same values of x for which y > 0 on the graph in the figure. The y-coordinates in the figure are positive for x in the interval (-o, -2) and also in the interval (1, o). So the solution set is (-o, -2) U (1, o) and the graph of the solution set is shown in Fig. 11.51.

b) First rewrite the inequality so that 0 is on the right:

x x

+

4 2

– 3

x x

+

4 2

– 3 – 0 Subtract 3 from each side.

x x

+

4 2

– 3(

x x –

2 2)

– 0 Get a common denominator.

-2 x x –

+

2 10

– 0 Get a single rational expression.

Now graph y = -2 x x –

+

2 10 . The vertical asymptote is x = 2. Find the x-intercept by

solving -2x + 10 = 0. The x-intercept is (5, 0). The y-intercept is (0, -5). The graph is shown in Fig. 11.52. From the graph we see that the y-coordinates are less than 0 for x in the interval (-o, 2) and less than or equal to zero for x in the inter­ val [5, o). So the solution set is (-o, 2) U [5, o). The graph of the solution set is shown in Fig. 11.53.

Now do Exercises 47–58

E X A M P L E 6

Figure 11.50

Figure 11.51

y

x

6

456

2

4

1 2

4

6

8

2 3

8

13

x 1 x+2y

210-1-2

y

x

6

2

4

6

8

4 6246

2x+10 x 2y

When solving an equation involving rational expressions, we multiply each side Figure 11.52

by the least common denominator. When solving a rational inequality, we do not use that technique. If we multiply each side of an inequality by a negative number, then the inequality symbol is reversed. But it is not reversed if we multiply by a positive

0 1 2 3 4 5 6 7 number. If the LCD involves a variable, then the value of the LCD could be positive Figure 11.53 or negative. We won’t know what to do with the inequality symbol if we multiply by

an LCD containing a variable. Example 7 illustrates the test-point method. The key fact here is that the

y-coordinates on the graph of a rational function can change sign only at an x-intercept or a vertical asymptote. The x-intercepts are found by setting the numerator equal to zero, and the vertical asymptotes are found by setting the denominator equal to zero. We locate these x-values on a number line and then test a point from each interval that is determined by them.

dug84356_ch11a.qxd 9/14/10 2:34 PM Page 744

744 Chapter 11 Functions 11-56

Solving a rational inequality with test points Solve each rational inequality. Write the solution set in interval notation, and graph it.

a) 2 4 x –

x 1

– 0

b) x x

+

1 2

> x –

x 4

c) x2 +

x2

2 +

x 2 + 4

: 0

Solution a) First solve 4 – x = 0 to get x = 4. Then solve 2x – 1 = 0 to get x = 1

2 . Plot 1

2 and 4 on a number line as shown in Fig. 11.54. We put a 0 above 4 and a U above 1 2

because the value of the rational expression is 0 when x = 4 and undefined if

x = 1 2 . Select three test points, say 0, 2, and 6. Now try each point in the original

inequality 2 4 x –

x 1

– 0:

2( 4 0 –

) – 0

1 – 0 Correct

2( 4 2 –

) – 2

1 – 0 Incorrect

2( 4 6 –

) – 6

1 – 0 Correct

Since 0 and 6 satisfy the inequality, the solution set consists of the intervals con­ taining 0 and 6. Since the inequality symbol is -, we include 4 in the solution set. Note that 1

2 does not satisfy the inequality. The solution set is the interval

(-o, 1 2 r U [4, o). The graph of the solution set is shown in Fig. 11.55. b) First rewrite the inequality with 0 on the right.

x x

+

1 2

> x –

x 4

x x

+

1 2

– x –

x 4

> 0

( ( x x

+

1 2 ) ) ( ( x x

4 4 ) )

– (x –

x(x 4)

+

(x 2 +

) 2)

> 0

> 0

(x + -7

2 x )(

+

x – 4

4) > 0

Now solve -7x + 4 = 0 to get x = 4 7 . The denominator is zero if x = -2 or x = 4.

Plot -2, 4 7 , and 4 on a number line as in Fig. 11.56. Select a test point in each

interval determined by these three numbers. We have chosen -4, 0, 3, and 5. Evaluate the original inequality at the test points:

4 4

+

1 2

> -4

4 4

Correct

0 0

+

1 2

> 0 –

0 4

Incorrect

x2 – 5x + 4 – (x2 + 2x)

(x + 2)(x – 4)

E X A M P L E 7

U 0

6543210

Test points

1 2

Figure 11.54

6543210 1 2

Figure 11.55

1

0U

0-1-2-3-4 2 3 4 5

U

Test points

4 7

Figure 11.56

dug84356_ch11a.qxd 9/14/10 2:34 PM Page 745

11-57 11.5 Graphs of Rational Functions 745

3 3

+

1 2

> 3 –

3 4

Correct

5 5

+

1 2

> 5 –

5 4

Incorrect

Since -4 and 3 satisfy the inequality, the solution set consists of the intervals con­ taining -4 and 3. Since the inequality symbol is >, no endpoints are included in the intervals. The solution set is the interval (-o, -2) U (4 7 , 4r. The graph of the solution set is shown in Fig. 11.57.

c) If x2 + 2 = 0, then x2 = -2 and there is no real solution to this equation. If x2 + 2x + 4 = 0, then

x = = -2 �

2 V-12s

and again there is no real solution. So the solution set is either all real numbers or

the empty set. To decide, test 0 in x2 +

x2

2 +

x 2 + 4

: 0:

02 + 0 2

2

( +

0)x 2

+ 4 : 0 Correct

Since the inequality is satisfied at the test point, it is satisfied for all real numbers. The solution set is (-o, o), and the graph of the solution set is shown in Fig. 11.58.

-2 � V22 – 4s(1)(4)s 2(1)

10-1-2-3-4 2 3 4 54 7

Figure 11.57

10-1-2 2

Figure 11.58

Now do Exercises 59–74

Warm-Ups ▼

Fill in the blank. 1. A function has the form f(x) = P(x)�Q(x) where

P(x) and Q(x) are polynomials with no common factor and Q(x) � 0.

2. The of a rational function is all real numbers except those that cause the denominator to be 0.

3. A asymptote is a vertical line that is approached by the graph of a rational function.

4. A asymptote is a horizontal line approached by the graph of a rational function.

5. An or asymptote is a nonvertical nonhorizontal line approached by the graph of a rational function.

True or false? 6. The domain of f(x) =

x – 1

9 is x = 9.

7. The domain of f(x) = x x

1 2

is (-o, -2) U (-2, 1) U (1, o).

8. The line x = 2 is the only vertical asymptote for the

graph of f(x) = x2 –

1 4 .

9. The x-axis is a horizontal asymptote for the graph of

f(x) = x2

x – 3 –

3x 9 +

x 5 .

10. The line y = 2x – 5 is an asymptote for the graph of

f(x) = 2x – 5 + 1 x .

11. The graph of f(x) = x2

x –

2

y-axis.

12. The solution set to x –

4 3

: 0 is (3, o).

dug84356_ch11a.qxd 9/18/10 10:08 AM Page 746

1 1

.5 Exercises

U Study Tips V • If your instructor does not tell you what is coming tomorrow, ask. • Read the material before it is discussed in class and your instructor’s explanation will make a lot more sense.

U1V Rational Functions

Find the domain of each rational function. See Example 1. 2

1. f (x) � �� x � 1 �2

2. f (x) � �� x � 3

x 1 2 � 3. f (x) � ��

x �2x � 3

4. f (x) � ��2x 5

5. f (x) � ��2 � x � 12

x 16

6. f (x) � ��2x � x � 6

U2V Asymptotes

Determine all asymptotes for the graph of each rational function. See Examples 2 and 3 and the Strategy for Finding Asymptotes for a Rational Function on page 741.

7 7. f (x) � ��

x � 4 �8

8. f (x) � �� x � 9

1 9. f (x) � ��2 �

�2 x 16

10. f (x) � ��2 �x 5x � 6

5x 11. f (x) � ��

x � 7

3x � 8 12. f (x) � ��

x � 2 22x

13. f (x) � �� x � 3

23x � 2 14. f (x) � ��

x � 1

U3V Sketching the Graphs

Match each rational function with its graph a–h. 2 1

15. f (x) � ��� 16. f (x) � ��� x x � 2

x 17. f (x) � ��

x � 2 1

19. f (x) � ��2 � x � 4

x 2x

21. f (x) � ��� 2

a)

3 4 5

y

x1 32 4�4�3

�5

c)

2 1

y

x1 2�4�3�2�1

�2 �3 �4 �5

�1

e) y

x1 3

3 2

4 5

1

�2�1

�2 �3 �4 �5

�1 6

x � 2 18. f (x) � ��

x 2x

20. f (x) � ��2x � 4 x2 � 2x � 1

22. f (x) � �� x

b)

2 3 4 5

y

x1 3 4 5 6�2�1

�2 �3 �4 �5

�1

1

d)

3 2

�4�3�2�1 �1

1 43

�2 �3 �4 �5

y

f)

3 2

4 5

y

x1 2 3

1

�5 �2�1

�3 �4 �5

�1

x

dug84356_ch11a.qxd 9/14/10 2:34 PM Page 747

11-59 11.5 Graphs of Rational Functions 747

g) 27. f (x) =

2 x

x

+

3 1

28. f (x) = 5

x

2

2

x

h)

29. f (x) = x2 – 3

x

x + 1 30. f (x) =

x3

x

+ 2

1

y

x1 31

2

3

1

y

x3

2 3 4 5

4 545 3

2 3 4 5

1

2 23x – 2x -x + 5x – 5 31. f (x) = 32. f (x) =

x – 1 x – 3

Determine all asymptotes, and sketch the graph of each function. See Examples 4 and 5.

2 -3 23. f (x) = 24. f (x) = Find all asymptotes, x-intercepts, and y-intercepts for the graph x + 4 x – 1

of each rational function, and sketch the graph of the function. 1 2

33. f (x) = 2 34. f (x) = 2x x – 4x + 4

x -2 25. f (x) = 2 26. f (x) = 2 +x – 9 x x – 2 2x – 3

35. f (x) = 2x + x – 6

dug84356_ch11a.qxd 9/14/10 2:34 PM Page 748

748 Chapter 11 Functions 11-60

x x 36. f (x) = 42. f (x) =

x2 + 4x + 4 x2 + x – 2

37. f (x) =

38. f (x) =

39. f (x) =

40. f (x) =

41. f (x) =

x + 1 2 x

x – 1 2 x

2x – 1 3 x – 9x

2 2x + 1 3 x – x

x 2 x – 1

2 43. f (x) =

x2 + 1

x 44. f (x) = 2 x + 1

2

45. f (x) = x

x + 1

2 x 46. f (x) =

x – 1

U4V Rational Inequalities

Solve each rational inequality using the graphical method. State the solution set using interval notation, and graph it. See Example 6.

1 47. > 0

x

dug84356_ch11a.qxd 9/22/10 11:16 AM Page 749

11-61

1 48. — – 0

x

x 49. — > 0

x – 3

a 50. — > 0

a + 2

x + 2 51. — – 0

x

w – 6 52. — – 0

w

t – 3 53. — > 0

t + 6

x – 2 54. — < 0

2x + 5

x 55. — > -1

x + 2

x + 3 56. — – -2

x

x – 3 57. — 2 2

x + 5

x + 2 58. — – 3

x – 6

Solve each rational inequality using the test-point method. State the solution set using interval notation, and graph it. See Example 7.

x – 4 59. — 2 0

x + 2

x – 3 60. — < 0

x + 5

11.5 Graphs of Rational Functions 749

x – 2 61. — < 1

x + 3

x – 3 62. — > 2

x + 4

3 1 63. — > —

x + 2 x + 1

1 1 64. — < —

x + 1 x – 1

2 1 65. — > —

x – 5 x + 4

3 2 66. — > —

x + 2 x – 1

m 3 67. — + — > 0

m – 5 m – 1

p 2 68. — + — – 0

p – 16 p – 6

x -8 69. — – —

x – 3 x – 6

x 2 70. — > —

x + 20 x + 8

2 x 71. — 2 02 x + 4

x2 + 2x + 3 72. — > 02 x

2 x 73. — < 02 x + 9

x2 + 4x + 5 74. — – 02 x + 1

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750 Chapter 11 Functions 11-62

Applications

Solve each problem.

75. Oscillating modulators. The number of oscillating modulators produced by a factory in t hours is given by the polynomial function n(t) = t2 + 6t for t : 1. The cost in dollars of operating the factory for t hours is given by the function c(t) = 36t + 500 for t : 1. The average cost per modulator is given by the rational function

36t + 500f (t) = 2 for t : 1. Graph the function f. What is t + 6t the average cost per modulator at time t = 20 and time t = 30? What can you conclude about the average cost per modulator after a long period of time?

76. Nonoscillating modulators. The number of nonoscillating modulators produced by a factory in t hours is given by the polynomial function n(t) = 16t for t : 1. The cost in dollars of operating the factory for t hours is given by the function c(t) = 64t + 500 for t : 1. The average cost per

64t + 500modulator is given by the rational function f (t) = 16t

for t : 1. Graph the function f. What is the average cost per modulator at time t = 10 and t = 20? What can you conclude about the average cost per modulator after a long period of time?

77. Average cost of an SUV. Mercedes-Benz spent \$700 million to design its new SUV (Motor Trend, www.motortrend.com). If it costs \$25,000 to manufacture each SUV, then the average cost per vehicle in dollars when x vehicles are manufactured is given by the rational function

25,000x + 700,000,000 . x

A(x) =

a) What is the horizontal asymptote for the graph of this function?

b) What is the average cost per vehicle when 50,000 vehicles are made?

c) For what number of vehicles is the average cost \$30,000?

d) Graph this function for x ranging from 0 to 100,000.

78. Average cost of a pill. Assuming Pfizer spent a typical \$350 million to develop its latest miracle drug and \$0.10 each to make the pills, then the average cost per pill in dollars when x pills are made is given by the rational function

0.10x + 350,000,000 A(x) = .

x

a) What is the horizontal asymptote for the graph of this function?

b) What is the average cost per pill when 100 million pills are made?

c) For what number of pills is the average cost per pill \$2?

Photo for Exercise 78

http:www.motortrend.com

dug84356_ch11a.qxd 9/14/10 2:34 PM Page 751

11-63

d) Graph this function for x ranging from 0 to 100 million.

Graphing Calculator Exercises

Sketch the graph of each pair of functions in the same coordinate system. What do you observe in each case?

279. f (x) = x , g(x) = x2 + 1�x 2 280. f (x) = x , g(x) = x2 + 1�x

81. f (x) = I x I, g(x) = I x I + 1�x 282. f (x) = I x I, g(x) = I x I + 1�x

83. f (x) = Vxs, g(x) = Vxs + 1�x

11.6 Combining Functions 751

3 284. f (x) = x , g(x) = x3 + 1�x

Getting More Involved

In each case find a rational function whose graph has the required asymptotes. Answers may vary.

85. The graph has the x-axis as a horizontal asymptote and the y-axis as a vertical asymptote.

86. The graph has the x-axis as a horizontal asymptote and the line x = 2 as a vertical asymptote.

87. The graph has the x-axis as a horizontal asymptote and lines x = 3 and x = -1 as vertical asymptotes.

88. The graph has the line y = 2 as a horizontal asymptote and the line x = 1 as a vertical asymptote.

11.6 Combining Functions

In this section you will learn how to combine functions to obtain new functions. In This Section

U1V Basic Operations with Functions

U2V Composition U1V Basic Operations with Functions An entrepreneur plans to rent a stand at a farmers market for \$25 per day to sell straw­ berries. If she buys x flats of berries for \$5 per flat and sells them for \$9 per flat, then her daily cost in dollars can be written as a function of x:

C(x) = 5x + 25

Assuming she sells as many flats as she buys, her revenue in dollars is also a function of x:

R(x) = 9x

Because profit is revenue minus cost, we can find a function for the profit by subtract­ ing the functions for cost and revenue:

P(x) = R(x) – C(x) = 9x – (5x + 25) = 4x – 25

The function P(x) = 4x – 25 expresses the daily profit as a function of x. Since P(6) = -1 and P(7) = 3, the profit is negative if 6 or fewer flats are sold and positive if 7 or more flats are sold.

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752 Chapter 11 Functions 11-64

In the example of the entrepreneur we subtracted two functions to find a new func­ tion. In other cases we may use addition, multiplication, or division to combine two functions. For any two given functions we can define the sum, difference, product, and quotient functions as follows.

Sum, Difference, Product, and Quotient Functions

Given two functions f and g, the functions f + g, f – g, f . g, and � f

� are defined as g follows:

Sum function: ( f + g)(x) = f (x) + g(x) Difference function: ( f – g)(x) = f (x) – g(x) Product function: ( f . g)(x) = f (x) . g(x)

Quotient function: �� g f ��(x) = � g f( ( x x ) )� provided that g(x) � 0 The domain of the function f + g , f – g , f . g , or �

f � is the intersection of the domain g

of f and the domain of g. For the function � f

� we also rule out any values of x for which g(x) = 0.

g

E X A M P L E 2 Evaluating a sum function Let f (x) = 4x – 12 and g(x) = x – 3. Find ( f + g)(2).

Solution In Example 1(a) we found a general formula for the function f + g, namely, ( f + g)(x) = 5x – 15. If we replace x by 2, we get

( f + g)(2) = 5(2) – 15 = -5.

E X A M P L E 1 Operations with functions Let f (x) = 4x – 12 and g(x) = x – 3. Find the following.

a) ( f + g)(x) b) ( f – g)(x)

c) ( f . g)(x) d) �� g f ��(x) Solution

a) ( f + g)(x) = f (x) + g(x) = 4x – 12 + x – 3 = 5x – 15

b) ( f – g)(x) = f (x) – g(x) = 4x – 12 – (x – 3) = 3x – 9

c) ( f . g)(x) = f (x) . g(x) = (4x – 12)(x – 3) = 4×2 – 24x + 36

d) �� g f ��(x) = � g f( ( x x ) ) � = �4x x – – 1 3 2� = �4( x x – – 3 3)� = 4 for x � 3.

Note that we use f + g, f – g, f . g, and f�g to name these functions only because there is no application in mind here. We generally use a single letter to name functions after they are combined as we did when using P for the profit function rather than R – C.

Now do Exercises 1–4

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11-65 11.6 Combining Functions 753

We can also find ( f + g)(2) by evaluating each function separately and then adding the results. Because f(2) = -4 and g(2) = -1, we get

( f + g)(2) = f (2) + g(2) = -4 + (-1) = -5.

The difference between the first four operations with functions and composition is like the difference between parallel and series in electrical connections. Components connected in parallel operate simul­ taneously and separately. If compo­ nents are connected in series, then electricity must pass through the first component to get to the second component.

Now do Exercises 5–12

U2V Composition A salesperson’s monthly salary is a function of the number of cars he sells: \$1000 plus \$50 for each car sold. If we let S be his salary and n be the number of cars sold, then S in dollars is a function of n:

S = 1000 + 50n

Each month the dealer contributes \$100 plus 5% of his salary to a profit-sharing plan. If P represents the amount put into profit sharing, then P (in dollars) is a function of S:

P = 100 + 0.05S

Now P is a function of S, and S is a function of n. Is P a function of n? The value of n certainly determines the value of P. In fact, we can write a formula for P in terms of n by substituting one formula into the other:

P = 100 + 0.05S

= 100 + 0.05(1000 + 50n) Substitute S = 1000 + 50n.

= 100 + 50 + 2.5n Distributive property

= 150 + 2.5n

Now P is written as a function of n, bypassing S. We call this idea composition of functions.

E X A M P L E 3 The composition of two functions Given that y = x2 – 2x + 3 and z = 2y – 5, write z as a function of x.

Solution Replace y in z = 2y – 5 by x2 – 2x + 3:

z = 2y – 5

= 2(x2 – 2x + 3) – 5 Replace y by x2 – 2x + 3. = 2×2 – 4x + 1

The equation z = 2×2 – 4x + 1 expresses z as a function of x.

Now do Exercises 13–22

A composition of functions is simply one function followed by another. The out­ put of the first function is the input for the second. For example, let f (x) = x – 3 and g(x) = x2. If we start with 5, then f (5) = 5 – 3 = 2. Now use 2 as the input for g, g(2) = 22 = 4. So g( f (5)) = 4. The function that pairs 5 with 4 is called the composi­ tion of g and f, and we write (g o f )(5) = 4. Since we subtracted 3 first and then squared, a formula for g o f is (g o f )(x) = (x – 3)2. If we apply g first and then f, we get a different function, ( f o g)(x) = x2 – 3, the composition of f and g.

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754 Chapter 11 Functions 11-66

A composition of functions can be viewed as two function machines where the output of the first is the input of the second.

5

f(5) 2f(x) x 3

2

g(x) x2 g(2) 4

4

Composition of Functions

The composition of f and g is denoted f o g and is defined by the equation

( f o g)(x) = f(g(x)),

provided that g(x) is in the domain of f.

The notation f o g is read as “the composition of f and g” or “f compose g.” The diagram in Fig. 11.59 shows a function g pairing numbers in its domain with numbers in its range. If the range of g is contained in or equal to the domain of f, then f pairs the second coordinates of g with numbers in the range of f. The composition function f o g is a rule for pairing numbers in the domain of g directly with numbers in the range of f, bypassing the middle set. The domain of the function f o g is the domain of g (or a subset of it), and the range of f o g is the range of f (or a subset of it).

f o g

x g f

g(x)

Domain of g Range of g Domain of f

f (g(x))

CAUTION

Range of f

Figure 11.59

The order in which functions are written is important in composition. For the function f o g the function f is applied to g(x). For the function g o f the function g is applied to f (x). The function closest to the vari­ able x is applied first.

E X A M P L E 4 Evaluating compositions Let f(x) = 3x – 2 and g(x) = x2 + 2x. Evaluate each of the following expressions.

a) g( f (3)) b) f (g(-4)) c) (g o f )(2) d) ( f o g)(2)

Solution a) Because f (3) = 3(3) – 2 = 7, we have

g( f (3)) = g(7) = 72 + 2 . 7 = 63.

So g( f (3)) = 63.

b) Because g(-4) = (-4)2 + 2(-4) = 8, we have

f (g(-4)) = f (8) = 3(8) – 2 = 22.

So f (g(-4)) = 22.

c) Because (g o f )(2) = g( f (2)) we first find f (2):

f (2) = 3(2) – 2 = 4

Because f (2) = 4, we have

(g o f )(2) = g( f (2)) = g(4) = 42 + 2(4) = 24.

So (g o f )(2) = 24.

U Calculator Close-Up V

Set y1 = 3x – 2 and y2 = x2 + 2x. You can find the composition for Examples 4(c) and 4(d) by evaluating y2(y1(2)) and y1(y2(2)). Note that the order in which you evaluate the func­ tions is critical.

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11-67 11.6 Combining Functions 755

d) Because ( f o g)(2) = f(g(2)), we first find g(2):

g(2) = 22 + 2(2) = 8

Because g(2) = 8, we have

( f o g)(2) = f (g(2)) = f (8) = 3(8) – 2 = 22.

So ( f o g)(2) = 22.

Now do Exercises 23–36

In Example 4, we found specific values of compositions of two functions. In Example 5, we find a general formula for the two functions from Example 4.

E X A M P L E 5 Finding formulas for compositions Let f (x) = 3x – 2 and g(x) = x2 + 2x. Find the following.

a) (g o f )(x) b) ( f o g)(x)

Solution a) Since f (x) = 3x – 2, we replace f (x) with 3x – 2:

(g o f )(x) = g( f (x))

= g(3x – 2) Replace f (x) with 3x – 2.

= (3x – 2)2 + 2(3x – 2) Replace x in g(x) = x2 + 2x with 3x – 2.

= 9×2 – 12x + 4 + 6x – 4 Simplify.

= 9×2 – 6x

So (g o f )(x) = 9×2 – 6x.

b) Since g(x) = x2 + 2x, we replace g(x) with x2 + 2x:

( f o g)(x) = f (g(x)) Definition of composition

= f (x2 + 2x) Replace g(x) with x2 + 2x. = 3(x2 + 2x) – 2 Replace x in f (x) = 3x – 2 with x2 + 2x. = 3×2 + 6x – 2 Simplify.

So ( f o g)(x) = 3×2 + 6x – 2.

Now do Exercises 37–46

Notice that in Example 4(c) and (d), (g o f )(2) � ( f o g)(2). In Example 5(a) and (b) we see that (g o f )(x) and ( f o g)(x) have different formulas defining them. In gen­ eral, f o g � g o f. However, in Section 11.7 we will see some functions for which the composition in either order results in the same function.

It is often useful to view a complicated function as a composition of simpler func­ tions. For example, the function Q(x) = (x – 3)2 consists of two operations, subtract­ ing 3 and squaring. So Q can be described as a composition of the functions f (x) = x – 3 and g(x) = x2. To check this, we find (g o f )(x):

(g o f )(x) = g( f (x))

= g(x – 3)

= (x – 3)2

We can express the fact that Q is the same as the composition function g o f by writing Q = g o f or Q(x) = (g o f )(x).

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756 Chapter 11 Functions 11-68

E X A M P L E 6 Expressing a function as a composition of simpler functions Let f (x) = x – 2, g(x) = 3x, and h(x) = �x�. Write each of the following functions as a composition, using f, g, and h.

a) F(x) = �x – 2� b) H(x) = x – 4 c) K(x) = 3x – 6

Solution a) The function F consists of first subtracting 2 from x and then taking the square root

of that result. So F = h o f. Check this result by finding (h o f )(x):

(h o f )(x) = h( f (x)) = h(x – 2) = �x – 2�

b) Subtracting 4 from x can be accomplished by subtracting 2 from x and then subtracting 2 from that result. So H = f o f. Check by finding ( f o f )(x):

( f o f )(x) = f ( f (x)) = f (x – 2) = x – 2 – 2 = x – 4

c) Notice that K(x) = 3(x – 2). The function K consists of subtracting 2 from x and then multiplying the result by 3. So K = g o f. Check by finding (g o f )(x):

(g o f )(x) = g( f (x)) = g(x – 2) = 3(x – 2) = 3x – 6

Now do Exercises 47–56

In Example 6(a) we have F = h o f because in F we subtract 2 before tak­ ing the square root. If we had the function G(x) = �x� – 2, we would take the square root before subtracting 2. So G = f o h. Notice how important the order of operations is here.

CAUTION

In Example 7, we see functions for which the composition is the identity function. Each function undoes what the other function does. We will study functions of this type further in Section 11.7.

E X A M P L E 7 Composition of functions Show that ( f o g)(x) = x for each pair of functions.

a) f (x) = 2x – 1 and g(x) =

b) f(x) = x3 + 5 and g(x) = (x – 5)1�3

Solution a) ( f o g)(x) = f(g(x)) = f �� x + 2 1 ��

= 2�� x + 2 1 �� – 1 = x + 1 – 1 = x

b) ( f o g)(x) = f(g(x)) = f((x – 5)1�3) = ((x – 5)1�3)3 + 5 = x – 5 + 5 = x

x + 1 �

2

Now do Exercises 57–64

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11-69 11.6 Combining Functions 757

Warm-Ups ▼

Fill in the blank. 1. The function ( f + g)(x) is the of the functions f(x)

and g(x).

2. The function ( f – g)(x) is the of the functions f(x) and g(x).

3. The function ( f · g)(x) is the of the functions f(x) and g(x).

4. The function ( ffg)(x) is the of the functions f(x) and g(x).

5. The function ( f o g)(x) is the of the functions f(x) and g(x).

True or false? 6. For the composition of f and g, the function f is evaluated

after g.

7. If f(x) = x – 2 and g(x) = x + 3, then ( f – g)(x) = -5.

8. If f(x) = x2 – 2x and g(x) = 3x + 9, then ( f + g)(x) = x2 + x + 9.

9. If f(x) = x2 and g(x) = x + 2, then ( f o g)(x) = x2 + 2.

10. If f(x) = 3x and g(x) = – 3 x

-, then ( f o g)(x) = x.

11. The function f o g and g o f are always the same.

12. If F(x) = (x – 1)2 , h(x) = x – 1, and g(x) = x2, then F = g o h.

1 1

.6Exercises

U Study Tips V • Stay alert for the entire class period.The first 20 minutes are the easiest, and the last 20 minutes the hardest. • Think of how much time you will have to spend outside of class figuring out what happened during the last 20 minutes in which you

were daydreaming.

U1V Basic Operations with Functions

Let f(x) = 4x – 3 and g(x) = x2 – 2x. Find the following. See Examples 1 and 2.

1. ( f + g)(x)

3. ( f · g)(x)

5. ( f + g)(3)

7. ( f – g)(-3)

9. ( f · g)(-1)

2. ( f – g)(x)

f 4. ( -)(x)-g

6. ( f + g)(2)

8. ( f – g)(-2)

10. ( f · g)(-2)

11. – (4)-(g f )

U2V Composition

12. – (-2)-(g f )

Use the two functions to write y as a function of x. See Example 3.

13. y = 2a, a = 3x

14. y = w2, w = 5x

15. y = 3a – 2, a = 2x – 6

16. y = 2c + 3, c = -3x + 4 x + 1

17. y = 2d + 1, d = — 2

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� �

758 Chapter 11 Functions 11-70

2 – x 18. y = -3d + 2, d = ��

3

19. y = m2 – 1, m = x + 1

20. y = n2 – 3n + 1, n = x + 2 a – 3 2x + 3

21. y = ��, a = �� a + 2 1 – x w + 2 5x + 2

22. y = ��, w = �� w – 5 x – 1

x + 3Let f(x) = 2x – 3, g(x) = x2 + 3x, and h(x) = ��. Find the 2

following. See Examples 4 and 5.

23. (g o f )(1) 24. ( f o g)(-2)

25. ( f o g)(1) 26. (g o f )(-2)

27. ( f o f )(4) 28. (h o h)(3)

29. (h o f )(5) 30. ( f o h)(0)

31. ( f o h)(5) 32. (h o f )(0)

33. (g o h)(-1) 34. (h o g)(-1)

35. ( f o g)(2.36) 36. (h o f )(23.761)

37. (g o f )(x) 38. (g o h)(x)

39. ( f o g)(x) 40. (h o g)(x)

41. (h o f )(x) 42. ( f o h)(x)

43. ( f o f )(x) 44. (g o g)(x)

45. (h o h)(x) 46. ( f o f o f )(x)

Let f(x) = �x�, g(x) = x2, and h(x) = x – 3. Write each of the following functions as a composition using f, g, or h. See Example 6.

47. F(x) = �x – 3 48. N(x) = �x – 3

49. G(x) = x2 – 6x + 9 50. P(x) = x for x 2 0

2 1�451. H(x) = x – 3 52. M(x) = x

253. J(x) = x – 6 54. R(x) = �x�- 3

4 2 – 6�55. K(x) = x 56. Q(x) = �x�x + 9

Show that ( f o g)(x) = x and (g o f )(x) = x for each given pair of functions. See Example 7.

x – 5 57. f (x) = 3x + 5, g(x) = ��

3 x + 7

58. f (x) = 3x – 7, g(x) = �� 3

59. f (x) = x3 3 x + 9- 9, g(x) = �� 60. f (x) = x 3 x – 13 + 1, g(x) = ��

x – 1 x + 1 61. f (x) = ��, g(x) = ��

x + 1 1 – x x + 1 3x + 1

62. f (x) = ��, g(x) = �� x – 3 x – 1 1 1

63. f (x) = ��, g(x) = �� x x

1�3×364. f (x) = 2x , g(x) = ����2 Miscellaneous

Let f(x) = x2 and g(x) = x + 5. Determine whether each of these statements is true or false.

65. f (3) = 9

66. g(3) = 8

67. ( f + g)(4) = 21

68. ( f – g)(0) = 5

69. ( f . g)(3) = 72

70. ( f�g)(0) = 5 71. ( f o g)(2) = 14

72. (g o f )(7) = 54

73. f(g(x)) = x2 + 25

74. (g o f )(x) = x2 + 5

75. If h(x) = x2 + 10x + 25, then h = f o g.

76. If p(x) = x2 + 5, then p = g o f.

Applications

Solve each problem.

77. Area. A square gate in a wood fence has a diagonal brace with a length of 10 feet.

a) Find the area of the square gate. b) Write a formula for the area of a square as a function

of the length of its diagonal.

78. Perimeter. Write a formula for the perimeter of a square as a function of its area.

79. Profit function. A plastic bag manufacturer has deter­ mined that the company can sell as many bags as it can produce each month. If it produces x thousand bags in a month, the revenue is R(x) = x2 – 10x + 30 dollars, and the cost is C(x) = 2×2 – 30x + 200 dollars. Use the fact that profit is revenue minus cost to write the profit as a function of x.

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11-71

80. Area of a sign. A sign is in the shape of a square with a semicircle of radius x adjoining one side and a semicircle of diameter x removed from the opposite side. If the sides of the square are length 2x, then write the area of the sign as a function of x.

Figure for Exercise 80

x 2x

2x

x

81. Junk food expenditures. Suppose the average family spends 25% of its income on food, F = 0.25I, and 10% of each food dollar on junk food, J = 0.10F. Write J as a function of I.

82. Area of an inscribed circle. A pipe of radius r must pass through a square hole of area M as shown in the figure. Write the cross-sectional area of the pipe A as a function of M.

r

Figure for Exercise 82

83. Displacement-length ratio. To find the displacement- length ratio D for a sailboat, first find x, where x = (L�100)3 and L is the length at the water line in feet (www.sailing.com). Next find D, where D = (d�2240)�x and d is the displacement in pounds. a) For the Pacific Seacraft 40, L = 30 ft 3 in. and

d = 24,665 pounds. Find D. b) For a boat with a displacement of 25,000 pounds,

write D as a function of L.

11.6 Combining Functions 759

D is

pl ac

em en

t- le

ng th

r at

io

0

200

400

600

800

d 25,000 lbs

25 30 35 40 45 50

Length at water line (ft)

Figure for Exercise 83

c) The graph for the function in part (b) is shown in the accompanying figure. For a fixed displacement, does the displacement-length ratio increase or decrease as the length increases?

84. Sail area-displacement ratio. To find the sail area- displacement ratio S, first find y, where y = (d�64)2�3 and d is the displacement in pounds. Next find S, where S = A�y and A is the sail area in square feet.

a) For the Pacific Seacraft 40, A = 846 square feet (ft2) and d = 24,665 pounds. Find S.

b) For a boat with a sail area of 900 ft2, write S as a function of d.

c) For a fixed sail area, does S increase or decrease as the displacement increases?

Getting More Involved 85. Discussion

Let f (x) = �x� – 4 and g(x) = ��x. Find the domains of f, g, and g o f.

86. Discussion

Let f (x) = �x – 4 �. Find the� and g(x) = �x – 8 domains of f, g, and f + g.

Graphing Calculator Exercises

87. Graph y1 = x, y2 = �x� , and y3 = x + ��x in the same screen. Find the domain and range of y3 = x + �x� by examining its graph. (On some graphing calculators you can enter y3 as y3 = y1 + y2.)

88. Graph y1 = � x �, y2 = � x – 3 �, and y3 = � x � + � x – 3 �. Find the domain and range of y3 = � x � + � x – 3 � by examining its graph.

http:www.sailing.com

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760 Chapter 11 Functions 11-72

11.7 Inverse Functions

In Section 11.6, we introduced the idea of a pair of functions such that ( f o g)(x) = x and (g o f )(x) = x. Each function reverses what the other function does. In this section we explore that idea further.

In This Section

U1V Inverse of a Function

U2V Identifying Inverse Functions

U3V Switch-and-Solve Strategy

U4V Even Roots or Even Powers

U5V Graphs of f and f -1

Domain of f Range of f

6

7

8

f

1f

299

329

349

Range of f 1 Domain of f 1

Figure 11.60

Consider the universal product codes (UPC) and the prices for all of the items in your favorite grocery store. The price of an item is a function of the UPC because every UPC deter­ mines a price. This function is not invertible because you cannot deter­ mine the UPC from a given price.

U1V Inverse of a Function You can buy a 6-, 7-, or 8-foot conference table in the K-LOG Catalog for \$299, \$329, or \$349, respectively. The set

f = �(6, 299), (7, 329), (8, 349)�

gives the price as a function of the length. We use the letter f as a name for this set or function, just as we use the letter f as a name for a function in the function notation. In the function f, lengths in the domain �6, 7, 8� are paired with prices in the range �299, 329, 349�. The inverse of the function f, denoted f -1, is a function whose ordered pairs are obtained from f by interchanging the x- and y-coordinates:

-1f = �(299, 6), (329, 7), (349, 8)� -1We read f -1 as “f inverse.” The domain of f -1 is �299, 329, 349�, and the range of f

is �6, 7, 8�. The inverse function reverses what the function does: it pairs prices in the range of f with lengths in the domain of f. For example, to find the cost of a 7-foot table, we use the function f to get f (7) = 329. To find the length of a table costing \$349, we use the function f -1 to get f -1(349) = 8. Of course, we could find the length of a \$349 table by looking at the function f, but f -1 is a function whose input is price and whose output is length. In general, the domain of f -1 is the range of f, and the range of f -1 is the domain of f. See Fig. 11.60.

1CAUTION The -1 in f -1 is not read as an exponent. It does not mean ��. f

The cost per ink cartridge is a function of the number of boxes of ink cartridges purchased:

g = �(1, 4.85), (2, 4.60), (3, 4.60), (4, 4.35)�

If we interchange the first and second coordinates in the ordered pairs of this function, we get

�(4.85, 1), (4.60, 2), (4.60, 3), (4.35, 4)�.

This set of ordered pairs is not a function because it contains ordered pairs with the same first coordinates and different second coordinates. So g does not have an inverse function. A function is invertible if you obtain a function when the coordinates of all ordered pairs are reversed. So f is invertible and g is not invertible.

Any function that pairs more than one number in the domain with the same num­ ber in the range is not invertible, because the set is not a function when the ordered pairs are reversed. So we turn our attention to functions where each member of the domain corresponds to one member of the range and vice versa.

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11-73 11.7 Inverse Functions 761

One-to-One Function

If a function is such that no two ordered pairs have different x-coordinates and the same y-coordinate, then the function is called a one-to-one function.

In a one-to-one function each member of the domain corresponds to just one member of the range, and each member of the range corresponds to just one member of the domain. Functions that are one-to-one are invertible functions.

Inverse Function

The inverse of a one-to-one function f is the function f -1, which is obtained from f by interchanging the coordinates in each ordered pair of f.

E X A M P L E 1 Identifying invertible functions Determine whether each function is invertible. If it is invertible, then find the inverse function.

a) f = �(2, 4), (-2, 4), (3, 9)�

b) g = ��2, �1 2 ��, �5, � 1 5

��, �7, �1 7 ��� c) h = �(3, 5), (7, 9)�

Solution a) Since (2, 4) and (-2, 4) have the same y-coordinate, this function is not one­

to-one, and it is not invertible.

b) This function is one-to-one, and so it is invertible.

g -1 = ��� 1 2�, 2�, �� 1 5

�, 5�, �� 1 7�, 7�� c) This function is invertible, and h-1 = �(5, 3), (9, 7)�.

Now do Exercises 1–10

Figure 11.61

y

x

2

3

3

3

1

1 1 2123

You learned to use the vertical-line test in Section 11.1 to determine whether a graph is the graph of a function. The horizontal-line test is a similar visual test for determining whether a function is invertible. If a horizontal line crosses a graph two (or more) times, as in Fig. 11.61, then there are two points on the graph, say (x1, y) and (x2, y), that have different x-coordinates and the same y-coordinate. So the function is not one-to-one, and the function is not invertible.

Horizontal-Line Test

A function is invertible if and only if no horizontal line crosses its graph more than once.

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762 Chapter 11 Functions 11-74

E X A M P L E 2

Tests such as the vertical-line test and the horizontal-line test are certainly not accurate in all cases. We discuss these tests to get a visual idea of what graphs of functions and invertible functions look like.

Using the horizontal-line test Determine whether each function is invertible by examining its graph.

a) b)

Solution a) This function is not invertible because a horizontal line can be drawn so that it

crosses the graph at (2, 4) and (-2, 4).

b) This function is invertible because every horizontal line that crosses the graph crosses it only once.

Now do Exercises 11–14

y

x

3

2

4

3 4

1

1 21 1

y

x

3

2

4

3

1

1 2 1

123

U2V Identifying Inverse Functions Consider the one-to-one function f (x) = 3x. The inverse function must reverse the ordered pairs of the function. Because division by 3 undoes multiplication by 3, we

xcould guess that g(x) = �� is the inverse function. To verify our guess, we can use the fol­ 3

lowing rule for determining whether two given functions are inverses of each other.

Identifying Inverse Functions

Functions f and g are inverses of each other if and only if

(g o f )(x) = x for every number x in the domain of f and

( f o g)(x) = x for every number x in the domain of g.

In Example 3, we verify that f (x) = 3x and g(x) = � 3 x

� are inverses.

E X A M P L E 3 Identifying inverse functions Determine whether the functions f and g are inverses of each other.

a) f (x) = 3x and g(x) = � 3 x

� b) f (x) = 2x – 1 and g(x) = � 1 2

� x + 1

c) f (x) = x2 and g(x) = �x�

Solution a) Find g o f and f o g:

(g o f )(x) = g( f(x)) = g(3x) = � 3 3 x � = x

( f o g)(x) = f (g(x)) = f �� 3 x

�� = 3 . � 3 x

� = x

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11-75 11.7 Inverse Functions 763

Because each of these equations is true for any real number x, f and g are inverses of each other. We write g = f -1 or f -1(x) = �3

x �.

b) Find the composition of g and f :

(g o f )(x) = g( f (x))

= g(2x – 1) = � 1 2

� (2x – 1) + 1 = x + � 1 2

So f and g are not inverses of each other.

c) If x is any real number, we can write

(g o f )(x) = g( f (x))

= g(x2) = �x2� = � x �.

The domain of f is (-o, o), and � x � � x if x is negative. So g and f are not inverses of each other. Note that f (x) = x2 is not a one-to-one function, since both (3, 9) and (-3, 9) are ordered pairs of this function. Thus, f (x) = x2 does not have an inverse.

Now do Exercises 15–22

U3V Switch-and-Solve Strategy If an invertible function is defined by a list of ordered pairs, as in Example 1, then the inverse function is found by simply interchanging the coordinates in the ordered pairs. If an invertible function is defined by a formula, then the inverse function must reverse or undo what the function does. Because the inverse function interchanges the roles of x and y, we interchange x and y in the formula and then solve the new formula for y to undo what the original function did. The steps to follow in this switch-and-solve strategy are given in the following box and illustrated in Examples 4 and 5.

E X A M P L E 4 The switch-and-solve strategy Find the inverse of h(x) = 2x + 1.

Solution First write the function as y = 2x + 1, and then interchange x and y:

y = 2x + 1

x = 2y + 1 Interchange x and y.

x – 1 = 2y Solve for y.

� x –

2 1

� = y

h-1(x) = � x –

2 1

� Replace y by h-1(x).

Strategy for Finding f -1 by Switch-and-Solve

1. Replace f (x) by y.

2. Interchange x and y.

3. Solve the equation for y.

4. Replace y by f -1(x).

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764 Chapter 11 Functions 11-76

We can verify that h and h-1 are inverses by using composition:

(h-1 o h)(x) = h-1(h(x)) = h-1(2x + 1) = � 2x + 2 1 – 1 � = �

2 2 x � = x

(h o h-1)(x) = h(h-1(x)) = h�� x -2 1 �� = 2 . � x -2 1 � + 1 = x – 1 + 1 = x Now do Exercises 23–36

E X A M P L E 5 The switch-and-solve strategy If f (x) = �

x

x

+

1 3

�, find f -1(x).

Solution Replace f (x) by y, interchange x and y, and then solve for y:

y = � x

x

+

1 3

� Use y in place of f (x).

x = � y

y

+

1 3

� Switch x and y.

x(y – 3) = y + 1 Multiply each side by y – 3.

xy – 3x = y + 1 Distributive property

xy – y = 3x + 1

y(x – 1) = 3x + 1 Factor out y.

y = � 3 x x –

+

1 1

� Divide each side by x – 1.

f -1(x) = � 3 x x –

+

1 1

� Replace y by f -1(x).

To check, compute ( f o f -1)(x):

( f o f -1)(x) = f ��3 x x -+ 1 1�� = =

= = � 4 4 x � = x

You should check that ( f -1 o f )(x) = x.

3x + 1 + 1(x – 1) ��� 3x + 1 – 3(x – 1)

(x – 1)��3 x x -+ 1 1� + 1� —— (x – 1)��3 x x -+ 1 1� – 3�

� 3 x x – +

1 1

� + 1 —— � 3 x x – +

1 1

� – 3

Now do Exercises 37–40

If we use the switch-and-solve strategy to find the inverse of f (x) = x3, then we xget f -1(x) = x1�3. For h(x) = 6x we have h-1(x) = ��. The inverse of k(x) = x – 9 6

is k-1(x) = x + 9. For each of these functions there is an appropriate operation of arithmetic that undoes what the function does.

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11-77 11.7 Inverse Functions 765

Domain of g Range of g [0, o) [0, o)

3

g(x) [ x

2g 1(x) x for x : 0

Range of g 1 Domain of g 1

[ — 3

Figure 11.62

If a function involves two operations, the inverse function undoes those oper­ ations in the opposite order from which the function does them. For example, the func­ tion g(x) = 3x – 5 multiplies x by 3 and then subtracts 5 from that result. To undo these operations, we add 5 and then divide the result by 3. So,

x + 5 g -1(x) = ��.

3 xNote that g -1(x) � �� + 5. 3

U4V Even Roots or Even Powers We need to use special care in finding inverses for functions that involve even roots or even powers. We saw in Example 3(c) that f (x) = x2 is not the inverse of g(x) = �x �x�. However, because g(x) = � is a one-to-one function, it has an inverse. The domain of g is [0, o), and the range is [0, o). So the inverse of g must have domain [0, o) and range [0, o). See Fig. 11.62. The only reason that f (x) = x2 is not the inverse of g is that it has the wrong domain. So to write the inverse function, we must use the appropriate domain:

2g -1(x) = x for x 2 0

Note that by restricting the domain of g -1 to [0, o), g -1 is one-to-one. With this 1restriction it is true that (g o g -1)(x) = x and (g – o g)(x) = x for every nonnegative

number x.

E X A M P L E 6 Inverse of a function with an even exponent Find the inverse of the function f (x) = (x – 3)2 for x 2 3.

Solution Because of the restriction x 2 3, f is a one-to-one function with domain [3, o) and range [0, o). The domain of the inverse function is [0, o), and its range is [3, o). Use the switch­ and-solve strategy to find the formula for the inverse:

y = (x – 3)2

x = (y – 3)2

y – 3 = ±�x� y = 3 ± �x�

Because the inverse function must have range [3, o), we use the formula f -1(x) = 3 + �x�. Because the domain of f -1 is assumed to be [0, o), no restriction is required on x.

Now do Exercises 41–48

U5V Graphs of f and f -1 2Consider f (x) = x for x 2 0 and f -1(x) = �x�. Their graphs are shown in

Fig. 11.63 on page 766. Notice the symmetry. If we folded the paper along the line y = x, the two graphs would coincide.

If a point (a, b) is on the graph of the function f, then (b, a) must be on the graph -of f 1(x). See Fig. 11.64 on page 766. The points (a, b) and (b, a) lie on opposite sides

dug84356_ch11b.qxd 9/14/10 2:51 PM Page 766

1

3

2

4

766 Chapter 11 Functions 11-78

y

x

1

3

4

2

51 3

4

2

y xf 1

f

(b, a)

(a, b)

5

2

3

1 12345

y

x4 51 32

y x f (x) x2

x : 0

f 1(x)

2

3

1 12345

[ — x

Figure 11.63 Figure 11.64

of the diagonal line y = x and are the same distance from it. For this reason the graphs -of f and f 1 are symmetric with respect to the line y = x.

E X A M P L E 7

y

x

1

4

2

51 3

4

2

f (x)

f 1(x) x2 � 1 x : 0

x 1[ ———

2

3

1 123

Inverses and their graphs Find the inverse of the function f (x) = �x – 1�, and graph f and f -1 on the same pair of axes.

Solution To find f -1, first switch x and y in the formula y = �x – 1�:

x = �y – 1� x2 = y – 1 Square both sides.

x2 + 1 = y

Because the range of f is the set of nonnegative real numbers [0, o), we must restrict the domain of f -1 to be [0, o). Thus, f -1(x) = x2 + 1 for x 2 0. The two graphs are shown in Fig. 11.65.

Now do Exercises 49–58

Figure 11.65

Warm-Ups ▼

Fill in the blank. 1. The of a function is a function with the same

ordered pairs except that the coordinates are reversed. 2. The domain of f-1 is the of f. 3. The range of f-1 is the of f. 4. A function is if no two ordered pairs have the

same second coordinates with different first coordinates. 5. The graphs of f and f-1 are with respect to

the line y = x. 6. If a line can be drawn so that it crosses the

graph of a function more than once, then the function is not one-to-one.

True or false? 7. The inverse of {(1, 3), (2, 5)} is {(3, 1), (2, 5)}. 8. The function f(x) = 3 is one-to-one.

9. Only one-to-one functions are invertible.

10. The function f(x) = x4 is invertible.

11. If f(x) = -x, then f-1(x) = -x.

12. If h is invertible and h(7) = -95, then h-1(-95) = 7.

13. If f(x) = 4x + 5, then f-1(x) = � x –

4 5

�.

14. If g(x) = 3x – 6, then g -1(x) = � 1 3

�x + 2.

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4

6

8

2 2

– – –

Exercises

1 1

.7

U Study Tips V • When your mind starts to wander, don’t give in to it. • Recognize when you are losing it, and force yourself to stay alert.

U1V Inverse of a Function

Determine whether each function is invertible. If it is invertible, then find the inverse. See Example 1.

1. {(1, 3), (2, 9)} 2. {(0, 5), (-2, 0)} 3. {(-3, 3), (-2, 2), (0, 0), (2, 2)} 4. {(1, 1), (2, 8), (3, 27)} 5. {(16, 4), (9, 3), (0, 0)} 6. {(-1, 1), (-3, 81), (3, 81)} 7. {(0, 5), (5, 0), (6, 0)} 8. {(3, -3), (-2, 2), (1, -1)} 9. {(0, 0), (2, 2), (9, 9)}

10. {(9, 1), (2, 1), (7, 1), (0, 1)}

Determine whether each function is invertible by examining the graph of the function. See Example 2.

11. 12. y

x

6

4

1 3

8

1 3

4

6

8

2

y

x

2

6

4

1 3

8

1 3

4

6

8

2

13. 14. y

x

2

6

4

1 3

8

2 1 2 3

4

6

8

2

y

x

2

6

4

1 3

8

2 3

U2V Identifying Inverse Functions

Determine whether each pair of functions f and g are inverses of each other. See Example 3.

15. f(x) = 2x and g(x) = 0.5x

16. f(x) = 3x and g(x) = 0.33x 1

17. f(x) = 2x – 10 and g(x) = – x + 5- 2

x – 7 18. f(x) = 3x + 7 and g(x) = —

3 19. f(x) = -x and g(x) = -x

1 1 20. f(x) = — and g(x) = —

x x 11421. f(x) = x4 and g(x) = x

x 22. f(x) = l 2x l and g(x) = -�2 -�

U3V Switch-and-Solve Strategy

Find f 1. Check that ( f o f 1)(x) = x and ( f 1 o f )(x) = x. See Examples 4 and 5. See the Strategy for Finding f -1 by Switch-and-Solve box on page 763.

23. f(x) = 5x 24. h(x) = -3x

25. g(x) = x – 9 26. j(x) = x + 7

27. k(x) = 5x – 9 28. r(x) = 2x – 8

2 -1 29. m(x) = – 30. s(x) = —

x x

331. f(x) = V 3

x – 4 32. f(x) = Vx + 2

3 2 33. f(x) = — 34. f (x) = —

x – 4 x + 1

3

3 35. f(x) = V3x + 7 36. f(x) = V7 – 5x

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768 Chapter 11 Functions 11-80

x + 1 1 – x 51. f(x) = x2 – 1 for x 2 037. f(x) = �� 38. f(x) = �� x – 2 x + 3

x + 1 3x + 5 39. f(x) = �� 40. g(x) = ��

3x – 4 2x – 3

52. f(x) = x2 + 3 for x 2 0

U4V Even Roots or Even Powers

Find the inverse of each function. See Example 6. 4

41. p(x) = �x� 642. v(x) = ��x

43. f(x) = (x – 2)2 for x 2 2

44. g(x) = (x + 5)2 for x 2 -5

45. f(x) = x2 + 3 for x 2 0 53. f(x) = 5x

46. f(x) = x2 – 5 for x 2 0

47. f(x) = �x + 2

48. f(x) = �x – 4

-1 U5V Graphs of f and f

-Find the inverse of each function, and graph f and f 1 on the same pair of axes. See Example 7.

49. f(x) = 2x + 3 54. f(x) = � x

� 4

50. f(x) = -3x + 2 355. f(x) = x

dug84356_ch11b.qxd 9/14/10 2:51 PM Page 769

2�

11-81

56. f(x) = 2×3

57. f(x) = �x –

58. f (x) = �x + 3

Miscellaneous

Find the inverse of each function.

59. f(x) = 2x

60. f(x) = x – 1

61. f(x) = 2x – 1

62. f(x) = 2(x – 1) 363. f(x) = �x�

364. f(x) = 2�x� 3 �65. f(x) = �x – 1

3 �66. f(x) = �2x – 1

367. f(x) = 2��x – 1

3 �68. f(x) = 2�x – 1

11.7 Inverse Functions 769

For each pair of functions, find ( f -1 o f )(x).

69. f(x) = x -1(x) = �x + 13 – 1 and f 3 � x – 1370. f(x) = 2×3 + 1 and f -1(x) = ���2

1 71. f(x) = �� x – 3 and f -1(x) = 2x + 6

2 1 72. f(x) = 3x – 9 and f -1(x) = �� x + 3

3 1 -1(x) = �

1 73. f(x) = �� + 2 and f �

x x – 2 1 1

74. f(x) = 4 – �� and f -1(x) = �� x 4 – x

x + 1 2x + 1 75. f(x) = �� and f -1(x) = ��

x – 2 x – 1

3x – 2 2x + 2 76. f(x) = �� and f -1(x) = ��

x + 2 3 – x

Applications

Solve each problem.

77. Accident reconstruction. The distance that it takes a car to stop is a function of the speed and the drag factor. The drag factor is a measure of the resistance between the tire and the road surface. The formula S = �30LD� is used to determine the minimum speed S [in miles per hour (mph)] for a car that has left skid marks of length L feet (ft) on a surface with drag factor D.

a) Find the minimum speed for a car that has left skid marks of length 50 ft where the drag factor is 0.75.

b) Does the drag factor increase or decrease for a road surface when it gets wet?

c) Write L as a function of S for a road surface with drag factor 1 and graph the function.

20

40

0

60

M in

im um

s pe

ed (

m ph

)

0 20 40 60 80 100

Length of skid marks (ft)

D � 1

D � 0.75

D � 0.50

Figure for Exercise 77

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770 Chapter 11 Functions 11-82

78. Area of a circle. Let x be the radius of a circle and h(x) be the area of the circle. Write a formula for h(x) in terms of x. What does x represent in the notation h-1(x)? Write a formula for h-1(x).

79. Vehicle cost. At Bill Hood Ford in Hammond a sales tax of 9% of the selling price x and a \$125 title and license fee are added to the selling price to get the total cost of a vehicle. Find the function T(x) that the dealer uses to get the total cost as a function of the selling price x. Citizens National Bank will not include sales tax or fees in a loan. Find the function T-1(x) that the bank can use to get the selling price as a function of the total cost x.

80. Carpeting cost. At the Windrush Trace apartment complex all living rooms are square, but the length of x feet may vary. The cost of carpeting a living room is \$18 per square yard plus a \$50 installation fee. Find the function C(x) that gives the total cost of carpeting a living room of length x. The manager has an invoice for the total cost of a living room carpeting job but does not know in which apartment it was done. Find the function C-1(x) that gives the length of a living room as a function of the total cost of the carpeting job x.

Getting More Involved

81. Discussion

Let f(x) = xn where n is a positive integer. For which values of n is f an invertible function? Explain.

82. Discussion

Suppose f is a function with range (-o, o) and g is a function with domain (0, o). Is it possible that g and f are inverse functions? Explain.

Graphing Calculator Exercises

83. Most graphing calculators can form compositions of functions. Let f(x) = x2 and g(x) = �x�. To graph the composition g o f, let y1 = x2 and y2 = �y�1. The graph of y2 is the graph of g o f. Use the graph of y2 to determine whether f and g are inverse functions.

3 384. Let y1 = x3 = �x + 4 y1 + 4- 4, y2 �, and y3 = ��. The function y3 is the composition of the first two functions. Graph all three functions on the same screen. What do the graphs indicate about the relationship between y1 and y2?

C h

a p

t e

r

dug84356_ch11b.qxd 9/14/10 2:51 PM Page 771

� �

11-83 Chapter 11 Summary 771

11 Wrap-Up Summary

Relations and Functions Examples

Relation Any set of ordered pairs of real numbers �(1, 2), (1, 3)�

Function A relation in which no two ordered pairs have �(1, 2), (3, 5), (4, 5)� the same first coordinate and different second coordinates.

If y is a function of x, then y is uniquely determined by x. A function may be defined by a table, a listing of ordered pairs, or an equation.

Domain The set of first coordinates of the ordered pairs Function: y = x2, Domain: (-o, o)

Range The set of second coordinates of the ordered pairs. Function: y = x2, Range: [0, o)

Function notation If y is a function of x, the expression y = 2x + 3 f (x) is used in place of y. f (x) = 2x + 3

Vertical-line test If a graph can be crossed more than once by a vertical line, then it is not the graph of a function.

Linear function A function of the form f (x) = 3x – 7 f (x) = mx + b with m � 0 f (x) = -2x + 5

Constant function A function of the form f (x) = 2 f (x) = b, where b is a real number

Types of Functions Examples

Linear function y = mx + b or f (x) = mx + b for m � 0 f (x) = 2x – 3 Domain (-o, o), range (-o, o) If m = 0, y = b is a constant function. Domain (-o, o), range �b�

Absolute value y = �x � or f (x) = �x � f (x) = �x + 5 � function Domain (-o, o), range [0, o)

2Quadratic function f (x) = ax2 + bx + c for a � 0 f (x) = x – 4x + 3

Square-root function f (x) = �x f (x) = �x – 4 Domain [0, o), range [0, o)

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772 Chapter 11 Functions 11-84

Transformations of Graphs

Reflecting

Translating

Stretching and shrinking

Polynomial Functions

Polynomial function

Behavior at the x-intercepts

Polynomial inequality

Methods for solving

Rational Functions

Rational function

The graph of y = -f (x) is a reflection in the x-axis of the graph of y = f (x).

The graph of y = f(x) + k is k units above y = f(x) if k � 0 or �k � units below y = f(x) if k � 0.

The graph of y = f(x – h) is h units to the right of y = f(x) if h � 0 or �h � units to the left of y = f(x) if h � 0.

The graph of y = af (x) is obtained by stretching (if a � 1) or shrinking (if 0 � a � 1) the graph of y = f (x).

A function defined by a polynomial

If f (x) is a function such that f (x) = f (-x) for any value of x in its domain, then the graph of f is symmetric about the y-axis.

If f (x) is a function such that f (-x) = -f (x) for any value of x in its domain, then the graph of f is symmetric about the origin.

The graph of a polynomial function crosses the x-axis at (c, 0) if (x – c) has an odd exponent. The graph touches but does not cross the x-axis if (x – c) has an even exponent.

An inequality involving a polynomial

Use either the graphical method or the test-point method.

If P(x) and Q(x) are polynomials with no P(x)

common factor and f (x) = �� for Q(x) � 0, Q(x)

then f (x) is a rational function.

The graph of y = -x2 is a reflection of the graph of y = x2.

The graph of y = x2 + 3 is three units above y = x2, and y = x2 – 3 is three units below y = x2.

The graph of y = (x – 3)2 is three units to the right of y = x2, and y = (x + 3)2 is three units to the left.

The graph of y = 5×2 is obtained by stretching y = x2, and y = 0.1×2 is obtained by shrinking y = x2.

Examples

P(x) = x3 – x2 – 12x + 5

4If f (x) = x – x2, then 4 2f (-x) = (-x)4 – (-x)2 = x – x ,

and f (x) = f (-x).

If f (x) = x3 + x, then 3f (-x) = (-x)3 + (-x) = -x – x,

and f (-x) = -f (x).

Graph of f (x) = (x – 3)2(x + 5) touches but does not cross x-axis at (3, 0) and crosses x-axis at (-5, 0).

3x – x � 0

3x – x � 0 Solution set: (-1, 0) � (1, o)

Examples 2x – 1 1

f (x) = ��, f (x) = �� 3x – 2 x – 3

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11-85

Finding asymptotes for a rational function

P(x) f (x) = ��

Q(x)

Rational inequality

Methods for solving

Combining Functions

Sum

Difference

Product

Quotient

Composition of functions

Inverse Functions

One-to-one function

Inverse function

1. The graph of f has a vertical asymptote for each solution to the equation Q(x) = 0.

2. If the degree of P(x) is less than the degree of Q(x), then the x-axis is a horizontal asymptote.

3. If the degree of P(x) is equal to the degree of Q(x), then the horizontal asymptote is determined by the ratio of the leading coefficients.

4. If the degree of P(x) is one larger than the degree of Q(x), then use long division to find the quotient of P(x) and Q(x).

An inequality involving a rational expression

Use either the graphical method or the test-point method.

( f + g)(x) = f(x) + g(x)

( f – g)(x) = f (x) – g(x)

( f . g)(x) = f(x) . g(x)

f f (x) � (x) = ��� �g g(x)

(g o f )(x) = g( f (x)) ( f o g)(x) = f (g(x))

A function in which no two ordered pairs have different x-coordinates and the same y-coordinate

The inverse of a one-to-one function f is the function f -1, which is obtained from f by interchanging the coordinates in each ordered pair of f. The domain of f -1 is the range of f, and the range of f -1 is the domain of f.

Chapter 11 Summary 773

1 f (x) = ��

x – 2

Vertical: x = 2 Horizontal: x-axis

f (x) = � x �

x – 2

Vertical: x = 2 Horizontal: y = 1

2×2 + 3x – 5 -3 f (x) = �� = 2x – 1 + ��

x + 2 x + 2 Vertical: x = -2 Oblique: y = 2x -1

1 x + 3 2 x �� � 0, �� � 1, �� � �� x – 2 x – 9 x 2

1 �� � 0 x – 2 Solution set: (-o, 2)

Examples

For f (x) = x2 and g(x) = x + 1 ( f + g)(x) = x2 + x + 1

( f – g)(x) = x2 – x – 1

2( f . g)(x) = x3 + x

f x � (x) = ��� �g x +

2

1

(g o f )(x) = g(x2) = x2 + 1 ( f o g)(x) = f (x + 1)

= x2 + 2x + 1

Examples

f = �(2, 20), (3, 30)�

f -1 = �(20, 2), (30, 3)�

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774 Chapter 11 Functions 11-86

Horizontal-line test If there is a horizontal line that crosses the graph of a function more than once, then the function is not invertible.

Function notation Two functions f and g are inverses of each for inverse other if and only if both of the following

conditions are met. 1. (g o f )(x) = x for every number x in the f (x) = x3 + 1

domain of f. – 32. ( f o g)(x) = x for every number x in the 1(x ) = �x – 1f �

domain of g.

Switch-and-solve 1. Replace f (x) by y. y = x3 + 1 1strategy for finding f – 2. Interchange x and y. x = y3 + 1

3. Solve for y. x – 1 = y3 – 34. Replace y by f 1(x). y = �x – 1

3 �-f 1(x) = �x – 1

Graphs of f and f -1 Graphs of inverse functions are symmetric with respect to the line y = x.

Fill in the blank.

1. Any set of ordered pairs is a . 2. A is a set of ordered pairs in which no two have

the same first coordinate and different second coordinates. 3. The set of first coordinates of a relation is the . 4. The set of second coordinates of a relation is the . 5. The notation in which f(x) is used as the dependent

variable is notation. 6. A line that is approached by a curve is an . 7. An asymptote is neither horizontal nor vertical.

8. The of f and g is the function f o g where

( f o g)(x) = f(g(x)). 9. A function in which no two ordered pairs have the same

second coordinate and different first coordinates is a function.

Review Exercises

11.1 Functions and Relations Determine whether each relation is a function.

1. �(5, 7), (5, 10), (5, 3)�

2. �(1, 3), (4, 7), (1, 6)�

3. �(1, 1), (2, 1), (3, 3)�

10. The line test is a visual method for determining whether a graph is the graph of a function.

11. The line test is a visual method for determining whether a function is one-to-one.

12. The graph of y = -f(x) is a in the x-axis of the graph of y = f(x).

13. The graph of y = f(x) + c for c � 0 is an upward

of the graph of y = f(x).

14. If f(-x) = f (x), then the graph of f is about the y-axis.

15. If f(-x) = -f (x), then the graph of f is symmetric about the .

16. A ratio of two polynomial functions is a function.

4. �(2, 4), (4, 6), (6, 8)� 5. y = x2

6. x2 = 1 + y2

7. x = y4

8. y = �x – 1

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� �

11-87 Chapter 11 Review Exercises 775

Determine the domain and range of each relation. 225. y = x – 2x + 1 9. �(3, 5), (4, 9), (5, 1)�

10. �(2, 6), (6, 7), (8, 9)� 11. y = x + 1

12. y = 2x – 3

13. y = �x + 5 14. y = �x – 1

2 Let f(x) = 2x – 5 and g(x) = x2 + x – 6. Evaluate each 26. g(x) = x – 2x – 15

expression.

15. f (0) 16. f (-3)

17. g(0) 18. g(-2)

1 11 19. g���� 20. g�-����2 2 11.2 Graphs of Functions and Relations Graph each function, and state the domain and range. 27. k(x) = �x� + 2

21. f (x) = 3x – 4

28. y = �x – 2

22. y = 0.3x

29. y = 30 – x2

23. h(x) = � x � – 2

224. y = � x – 2 � 30. y = 4 – x

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x + 1� for 1 1 for x �

776 Chapter 11 Functions 11-88

11.3 Transformations of Graphs 31. f(x) = ��x + 4 for -4 � x � 0 Sketch the graph of each function, and state the domain and range.x + 2 for x � 0 37. y = �x�

38. y = -�x�

32. f(x) = �� – � x � 3 x – 3

39. y = -2�x�

Graph each relation, and state its domain and range.

33. x = 2

40. y = 2�x�

34. x = y2 – 1

41. y = �x – 2

35. x = � y � + 1

42. y = �x + 2

36. x = �y – 1

1 43. y = ���x�

2

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11-89 Chapter 11 Review Exercises 777

� 244. y = �x – 1 + 2 50. f (x) = (x – 3x – 4)(x + 3)

45. y = -�x + 1 + 3

46. y = 3�x + 4 – 5

11.4 Graphs of Polynomial Functions Graph each function, and identify the x- and y-intercepts.

47. f (x) = x3 – 25x

248. f (x) = x3 + 2x – 4x – 8

249. f (x) = (x – 4)(x – 1)

4 2 + 951. f (x) = x – 10x

52. f (x) = x4 – 4×3

Find the x-intercepts, and discuss the behavior of the graph of each polynomial function at its x-intercepts.

53. f (x) = x2 – 6x + 9

54. f (x) = x2 – 3x – 18

55. f (x) = (x – 3)(x + 5)(x – 4)2

56. f (x) = (x – 1)2(x + 7)

57. f (x) = x3 – 8×2 – 9x + 72

58. f (x) = x4 – 29×2 + 100

Solve each polynomial inequality. State the solution set using interval notation and graph it.

159. ��x3 – 2x � 0 2

60. x3 – 6×2 + 8x 2 0

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778 Chapter 11 Functions 11-90

4 2 2 061. x – 2x

4 262. -x + 2x – 1 2 0

63. x4 – 2×2 � 0

64. -x4 + 2×2 -1 � 0

3 265. x – x – 9x + 9 � 0

3 266. x – 4x – 16x + 64 � 0

3 267. x – x – 9x + 9 2 0

3 268. x – 4x – 16x + 64 � 0

11.5 Graphs of Rational Functions Find the domain of each rational function.

2x – 1 69. f (x) = ��

2x + 3 3x + 2

70. f (x) = ��2x – x – 12 1

71. f (x) = �� x2 + 9 x – 4

72. f (x) = ��2x – 9

Find all asymptotes for each rational function, and sketch the graph of the function.

2 -1 73. f (x) = �� 74. f (x) = ��

x – 3 x + 1

2

75. f (x) = � x � 76. f (x) = �

x �2 2x – 4 x – 4

2x – 1 -x – 1 77. f (x) = �� 78. f (x) = ��

x – 1 x

2 2x – 2x + 1 -x + x + 2 79. f (x) = �� 80. f (x) = ��

x – 2 x – 1

Solve each rational inequality. State the solution set using interval notation, and graph it.

x + 381. �� 2 0 x – 4

x + 5 82. �� � 0 x – 7

2x – 4 83. �� � 0

x

2 84. �

x � � 02x – 16

85. � x � � 2

x + 5

x + 486. �� � 4 x – 1

x2 + 4 87. �� � 0 x

2 88. �

x � 2 02x + 8

2x x 89. �� � �� x + 5 x – 2

90. � x � 2 �

x �

x – 6 x + 3

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11-91 Chapter 11 Review Exercises 779

11.6 Combining Functions Let f(x) = 3x + 5, g(x) = x2 – 2x, and h(x) = � x –

3

5 �. Find the

following.

120. f (x) = 2 – x2 for x 2 0

91. f (-3) 92. h(-4)

93. (h o f )(�2�) 94. ( f o h)(�) 95. (g o f )(2) 96. (g o f )(x)

97. ( f + g)(3) 98. ( f – g)(x)

99. ( f . g)(x)

101. ( f o f )(0)

100. � �(1) 102. ( f o f )(x)

f � g 121. f (x) = �

x

2

3

Let f(x) = � x �, g(x) = x + 2, and h(x) = x2. Write each of the following functions as a composition of functions, using f, g, or h.

103. F(x) = � x + 2 � 104. G(x) = � x � + 2

105. H(x) = x2 + 2

107. I(x) = x + 4

106.

108.

K(x) = x2 + 4x + 4

J(x) = x4 + 2 122. f (x) = -�

1

4 � x

11.7 Inverse Functions Determine whether each function is invertible. If it is invertible, find the inverse.

109. �(-2, 4), (2, 4)� 110. �(1, 1), (3, 3)�

Miscellaneous 111. f (x) = 8x 112. i(x) = -�

3

x � Sketch the graph of each function.

123. f (x) = 3 124. f (x) = 2x – 3

113. g(x) = 13x – 6 114. h(x) = �3 x – 6�

115. j(x) = � x

x

+

1

1 � 116. k(x) = � x � + 7

117. m(x) = (x – 1)2 118. n(x) = � 3

� 125. f (x) = x2 – 3 126. f (x) = 3 – x2 x

Find the inverse of each function, and graph f and f -1 on the same pair of axes.

119. f (x) = 3x – 1

127. f (x) = � x2 –

1 3

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780 Chapter 11 Functions

128. f (x) =� x �

(x – 1)(x + 2)

129. f (x) = x(x – 1)(x + 2)

3130. f (x) = x – 4×2 + 4x

Solve each inequality. State the solution set using interval notation.

131. 4 – 2x � 0

132. 2x – 3 � 0

2133. x – 3 � 0

134. 3 – x2 � 0 1

135. �� � 02x – 3 x

136. �� 2 0 (x – 1)(x + 2)

137. x(x – 1)(x + 2) � 0

138. x3 – 4×2 + 4x 2 0 x – 4

139. �� � 0 x + 3 2x – 1

140. �� 2 0 x + 5

141. (x – 2)(x + 1)(x – 5) 2 0

142. (x – 1)(x + 2)(2x – 5) � 0

143. x3 + 3×2 – x – 3 � 0

144. x3 + 5×2 – 4x – 20 2 0

Graphing Calculator Exercises Match the given inequalities with their solution sets (a through d) by examining a table or a graph.

145. x2 – 2x – 8 � 0 a. (-2, 2) � (8, o)

146. x2 – 3x � 54 b. (2, 4)

11-92

147. � x –

x 2

� � 2 c. (-2, 4)

148. � x –

3 2

� � � x +

5 2

� d. (-o, -6) � (9, o)

Solve each problem.

149. Inscribed square. Given that B is the area of a square inscribed in a circle of radius r and area A, write B as a function of A.

150. Area of a window. A window is in the shape of a square of side s, with a semicircle of diameter s above it. Write a function that expresses the total area of the window as a function of s.

s

s

Figure for Exercise 150

151. Composition of functions. Given that a = 3k + 2 and k = 5w – 6, write a as a function of w.

152. Volume of a cylinder. The volume of a cylinder with a fixed height of 10 centimeters (cm) is given by V = 10�r2, where r is the radius of the circular base. Write the volume as a function of the area of the base, A.

153. Square formulas. Write the area of a square A as a function of the length of a side of the square s. Write the length of a side of a square as a function of the area.

154. Circle formulas. Write the area of a circle A as a function of the radius of the circle r. Write the radius of a circle as a function of the area of the circle. Write the area as a function of the diameter d.

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11-93

Chapter 11 Test

Solve each problem.

1. Determine whether �(0, 5), (9, 5), (4, 5)� is a function.

2. Let f (x) = -2x + 5. Find f (-3).

3. Find the domain and range of the function y = �x – 7�.

4. A mail-order firm charges its customers a shipping and handling fee of \$3.00 plus \$0.50 per pound for each order shipped. Express the shipping and handling fee S as a function of the weight of the order n.

5. If a ball is tossed into the air from a height of 6 feet with a velocity of 32 feet per second, then its altitude at time t (in seconds) can be described by the function

A(t) = -16t2 + 32t + 6.

Find the altitude of the ball at 2 seconds.

Sketch the graph of each function or relation, and state the domain and range.

2 6. f (x) = -��x + 1

3

7. y = � x � – 4

8. g(x) = x2 + 2x – 8 9. x = y2

Chapter 11 Test 781

for x 2 0 11. y = -� x – 2 � 10. f (x) = ��x� -x – 3 for x � 0

12. y = �x + 5 – 2

Graph each function. Identify all intercepts.

13. f (x) = (x + 2)(x – 2)2

1 14. f (x) = ��2x – 4x + 4

2x – 3 15. f (x) = ��

x – 2

16. f (x) = x3 – x2 – 4x + 4

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– –

11-94 782 Chapter 11 Functions

Solve each inequality. State the solution set using interval notation.

3 + 4×217. x – 32x � 0 x

18. �� � 02x – 25

Let f(x) = -2x + 5 and g(x) = x2 + 4. Find the following.

19. f (-3) 20. (g o f )(-3)

21. f 1(11) 22. f 1(x)

23. (g + f )(x) 24. ( f . g)(1)

25. ( f -1 o f )(1776) 26. ( f�g)(2)

27. ( f o g)(x) 28. (g o f )(x)

Let f(x) = x – 7 and g(x) = x2. Write each of the following functions as a composition of functions using f and g.

29. H(x) = x2 – 7

30. W(x) = x2 – 14x + 49

Determine whether each function is invertible. If it is invertible, find the inverse.

31. {(2, 3), (4, 3), (1, 5)}

32. {(2, 3), (3, 4), (4, 5)}

Find the inverse of each function.

33. f(x) = x – 5 34. f(x) = 3x – 5

3 2x + 1 35. f(x) = �x� + 9 36. f(x) = �� x – 1

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11-95 Chapter 11 Making Connections 783

MakingConnections A Review of Chapters 1–11

Simplify each expression.

1. 125-2�3 2. �� 2 8

7 ��

-1�3

3. �18� – �8� 4. x5 . x3

12

5. 161�4 6. � x

� 3x

Find the real solution set to each equation. 2 27. x = 9 8. x = 8

2 29. x = x 10. x – 4x – 6 = 0

1�4 1�611. x = 3 12. x = -2

13. �x � = 8 14. � 5x – 4 � = 21

315. x = 8 16. (3x – 2)3 = 27

17. �2�x – 3 = 9 �18. �x – 2 = x – 8

Sketch the graph of each set.

19. �(x, y) � y = 5� 20. �(x, y) � y = 2x – 5�

21. �(x, y) � x = 5� 22. �(x, y) � 3y = x�

23. �(x, y) � y = 5×2� 24. �(x, y) � y = -2×2�

Find the missing coordinates in each ordered pair so that the ordered pair satisfies the given equation.

25. (2, ), (3, ), ( , 2), ( , 16), 2x = y

1 4x

2 26. ���, �, (-1, ), ( , 16), ( , 1), = y

Find the domain of each expression.

27. �x�

28. �6 – 2x

5x – 3 29. ��2x + 1

x – 3 30. ��2x – 10x + 9

Solve each system of equations, and state whether the system is independent, dependent, or inconsistent. 31. 4x – 9y = -1

2x + 12y = 5

32. 10x + 20y = 143 y = x + 5.2

33. 3x – 9y = 6 1 2

y = ��x – �� 3 3

34. x = 5y + 12 1

y = ��x – 7 5

35. x + y – z = -5 x – 2y + z = 11 3x – y – z = 3

36. 2x + y + 3z = 2 x + y + z = 5 3x + 2y + 4z = 6

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784 Chapter 11 Functions 11-96

Perform the indicated operations. 11 5

37. �� – �� 15 12 0.25

0.20

O pe

ra tin

g co

st (i

n do

lla rs

p er

m ile

)

12 10 38. �� . ��

35 21

4 14 39. �� � ��

9 15

0.15

0.10

0.05

2 1 40. �� + ��2 3x 6x

3 3 7 5xy 2a b 41. �� . ��

6a3b 3 5xy5

2 12a – 48 9a + 18 42. �� � ��2 2 a + a – 6 a – 9

Solve each problem. 43. Capital cost and operating cost. To decide when to replace

company cars, an accountant looks at two cost components: capital cost and operating cost. The capital cost C (the difference between the original cost and the salvage value) for a certain car is \$3000 plus \$0.12 for each mile that the car is driven.

a) Write the capital cost C as a linear function of x, the number of miles that the car is driven.

0 0 50 100

Miles (in thousands)

Figure for Exercise 43(b)

b) The operating cost P is \$0.15 per mile initially and increases linearly to \$0.25 per mile when the car reaches 100,000 miles. Write P as a function of x, the number of miles that the car is driven.

44. Total cost. The accountant in Exercise 43 uses the function CT = �� + P to find the total cost per mile.x

a) Find T for x = 20,000, 30,000, and 90,000.

b) Sketch a graph of the total cost function.

0 0

15

10

5

50 100

C ap

ita l c

os t

(i n

th ou

sa nd

s of

d ol

la rs

)

c) The accountant has decided to replace the car when T reaches \$0.38 for the second time. At what mileage will the car be replaced?Miles (in thousands)

d) For what values of x is T less than or equal to \$0.38? Figure for Exercise 43(a)

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11-97 Chapter 11 Critical Thinking 785

CriticalThinking For Individual or Group Work Chapter 11

These exercises can be solved by a variety of techniques, which may or may not require algebra. So be creative and think critically. Explain all answers. Answers are in the Instructor’s Edition of this text.

1. Knight moves. Draw a 3 by 3 chess board on paper, and place two pennies (P) and two nickels (N) in the corners as shown in (a) of the figure. Move the Ns to the positions of the Ps and the Ps to the position of the Ns using the moves that a knight can make in chess (one space verti­ cally followed by two spaces horizontally or one space horizontally followed by two spaces vertically). If you allow a P or an N to make more than one move on a given turn, then it takes six turns. Try it. Find the minimum number of turns required to interchange the coins starting with the arrangement in (b).

(a) (b)

Figure for Exercise 1

2. Friedman numbers. A Friedman number is a positive integer that can be written in some nontrivial way using its own digits together with the elementary operations (+, -, ., �, exponents, and grouping symbols). For example, 25 = 52, and 126 = 21 . 6. The only two-digit Friedman number is 25. Show that 121 and 125 are

Friedman numbers. There are 13 three-digit Friedman numbers. Find the other 10 three-digit Friedman numbers.

3. Large Friedman numbers. Show that 123,456,789 and 987,654,321 are Friedman numbers.

4. Year numbers. Using all four of the digits in the current year and only those digits, write expressions for the integers from 1 through 100. You may use grouping symbols, and the operations of addition, subtraction, multiplication, division, powers, roots, and factorial, but no two-digit numbers or decimal points. For example if the year is 2005, then 50 + 0 . 2 = 1, 50 + 20 = 2, and so on. See how far you can go. Vary the problem by trying another year (say 1776), or allowing decimal points, or two-digit numbers.

5. Real numbers. Two real numbers have a sum of 200 and a product of 50. What is the sum of their reciprocals?

6. Telling time. Find the first time after 11 A.M. for which the minute hand and hour hand of a clock form a perfect right angle. Find the time to the nearest tenth of a second. The answer is not 11:10.

7. Identity crisis. Determine the value of a that will make this equation an identity.

1 2 3 4 5 6 �� + �� + �� + �� + �� + �� x – 1 1 – x x – 1 1 – x x – 1 1 – x

7 8 9 10 a + �� + �� + �� + �� = ��

x – 1 1 – x x – 1 1 – x x – 1

8. Difference of two squares. Let a = 76006 + 7-6006 and b = 76006 2- 7-6006. Find a – b2.

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