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17

Chapter 1 Functions andGraphs

1.1 Modeling and Equation Solving

Alternate Example 3 Comparing Pizzas

A pizzeria sells a rectangular pizza 10″ by 12″ for the same price as its large round pizza(12″ diameter). If both pizzas are of the same thickness, which option gives the mostpizza for the money?

SOLUTION

We need to compare the areas of the pizzas. Fortunately, geometry has providedalgebraic models that allow us to compare the areas from the given information. For therectangular pizza:

Area = l × w = 10 × 12 = 120 square inches.

For the circular pizza:

Area r . = =

= ≈π π π2

2122

36 113 1 squarre inches.

The rectangular pizza is larger and therefore gives more for the money.

Exercises for Alternate Example 3

In Exercises 1–6, the two pizzas are of the same thickness. Which option gives the most pizza for the money?

1. square pizza 10 inches on a siderectangular pizza 10 inches long and 8 inches wide

2. square pizza 10 inches on a sidecircular pizza with a 6-inch diameter

3. rectangular pizza 12 inches long and 7 inches widerectangular pizza 10 inches long and 8 inches wide

4. circular pizza with an 8-inch diametercircular pizza with a 9-inch diameter

5. square pizza 15 inches on a sidesquare pizza 14 inches on a side

6. circular pizza with a 10-inch diameterrectangular pizza 10 inches long and 8 inches wide �

Alternate Example 6 Solving an Equation Algebraically

Find all real numbers x for which 2x3 = –3x2 + 2x.

SOLUTION

We begin by changing the form of the equation to 2x3 + 3x2 – 2x = 0.

We can then solve this equation algebraically by factoring.

2 3 2 02 3 2 0

3 2

2

x x xx x xx x

– ( – ) (

+ =+ =

+ 22 2 1 0)( – ) x =

x x x

x x

–

–

= + = =

= =

0 2 0 2 1 0

0 2

or or

or orr x = 12

18 Chapter 1 Functions and Graphs


In Exercises 7–12, solve the equation algebraically.

7. x3 = 4x2 – 3x 8. x3 + 2x2 = 8x 9. x3 + 16x = 8x2 10. 2x3 = 13x2 – 20x 11. 2x3 = 5x2

12. 6x3 = 17x2 – 5x �

Alternate Example 7 Solving an Equation: Comparing Methods

Solve the equation x2 = 3x – 1.

[–5.1, 5.1] by [–4.7, 4.7]

ZeroX=.381966 Y=0

Figure 1.1 The graph ofy = x2 – 3x + 1.

SOLUTION

Solve AlgebraicallyThe given equation is equivalent to x2 – 3x + 1 = 0. This quadratic equation has irrationalroots that can be found by the quadratic formula.

x

= + ≈3 5

22.61803

and

x

– = ≈3 5

20.381966

While the decimal answers are certainly accurate enough for all practical purposes, it isimportant to note that only the expressions found by the quadratic formula give the exactreal number answers. The tidiness of exact answers is a worthy mathematical goal.Realistically, however, exact answers are often impossible to obtain, even with the mostsophisticated mathematical tools.

Solve GraphicallyWe first find an equivalent equation with 0 on the right-hand side: x2 – 3x + 1 = 0. Wethen graph the equation y = x2 – 3x + 1, as shown in Figure 1.1.

We then use the grapher to locate the x-intercepts of the graph:

x ≈ 0.381966 and x ≈ 2.61803.


In Exercises 13–18, solve the equation.

13. x2 – 4x + 2 = 0 14. x2 – 5x + 2 = 0 15. x2 = x + 7 16. x2 + 3x = 1 17. x2 + 4x = 6

18. x2 = 4x + 7 �

Alternate Example 8 Applying the Problem-Solving Process

The engineers at an auto manufacturer pay students $0.10 per mile plus $20 per day toroad test their new vehicles.

(a) How much did the auto manufacturer pay Dennis to drive 200 miles in one day?

(b) Erma earned $30 test-driving a new car in one day. How far did she drive?

SOLUTION

ModelA picture of a car or of Dennis or Erma would not be helpful, so we go directly todesigning the model. Both Dennis and Erma earned $20 for one day, plus $0.10 per mile.Multiply dollars/miles to get dollars.

So, if p represents the pay for driving x miles in one day, our algebraic model is

p = 20 + 0.10x.

Section 1.2 Functions and Their Properties 19

Solve Algebraically(a) To get Dennis’s pay we let x = 200 and solve for p:

p . ( )

= +=

20 0 10 20040

(b) To get Erma’s mileage we let p = 30 and solve for x.

30 20 0 1010 0 10

100 10100

. .

.

= +=

=

=

xx

x

x

Support GraphicallyFigure 1.2a shows that the point (200, 40) is on the graph of y = 20 + 0.1x, supporting ouranswer to (a). Figure 1.2b shows that the point (100, 30) is on the graph of y = 20 + 0.1x,supporting out answer to (b). (We could also have supported our answer numerically bysimply substituting in for each x and confirming the value of p.)

InterpretDennis earned $40 for driving 200 miles in one day. Erma drove 300 miles in one day toearn $30.

[0, 250] by [0, 50]

(a)

X=200 Y=40

[0, 250] by [0, 50]

(b)

X=100 Y=30

Figure 1.2 Graphical support for the algebraic solution in Alternate Example 8.

Exercises for Alternate Example 8In Exercises 19–21, the engineers at an auto manufacturer paystudents $0.20 per mile plus $50 per day to road test their newvehicles. Answer each question.

19. How much did the auto manufacturer pay Jacob to drive 250miles in one day?

20. Maria earned $85 test-driving a new car in one day. How far didshe drive?

21. Kyle earned $75 test-driving a new car in one day and $90 on asecond day. How far did he drive altogether?

In Exercises 22–24, engineers have a tank that contains 50 gallons ofwater. They let in 10 gallons of water per minute. Answer eachquestion.

22. How many gallons of water will be in the tank after 12 minutes?

23. How many minutes will it take for the tank to contain 85 gallonsin all?

24. How many minutes will it take for the tank to contain 700gallons in all?

�

1.2 Functions and Their Properties

Alternate Example 3 Finding the Domain of a Function

Find the domain of each of these functions:

(a) f x x( ) = + 4

(b) g x

xx

( ) –

=3

(c) A x x( ) ,= 2 where A(x) is the area of a square with sides of length x.


SOLUTION

Solve Algebraically(a) The expression under a radical may not be negative. We set x + 4 0 and solve tofind x –4. The domain of f is the interval [–4, ∞).

(b) The expression under a radical may not be negative; therefore x 0. Also, thedenominator of a fraction may not be zero; therefore x 3. The domain of g is theinterval [0, ∞) with the number 3 removed, which we can write as the union of twointervals: [0, 3) ∪ (3, ∞).

(c) The algebraic expression has domain all real numbers, but the behavior beingmodeled restricts s from being negative. The domain of A is the interval [0, ∞).

Support GraphicallyWe can support our answers in (a) and (b) graphically, as the calculator should not plotpoints where the function is undefined.

(a) Notice that the graph of y x = + 4 (Figure 1.3a) shows points only for x –4, as

expected.

(b) The graph of y

xx

–

=3

(Figure 1.3b) shows points only for x 0 as expected, but

shows an unexpected line through the x-axis at x = 3. This line, a form of grapher failuredescribed on page 78 of the text, should not be there. Ignoring it, we see that 3, asexpected, is not in the domain.

(c) The graph of y x = 2 (Figure 1.3c) shows the unrestricted domain of the algebraic

expression; all real numbers. The calculator has no way of knowing that x is the length ofa side of a square.

[–5.1, 5.1] by [–4.7, 4.7]

(a)

[–5.1, 5.1] by [–4.7, 4.7]

(c)

[–5.1, 5.1] by [–4.7, 4.7]

(b)

Figure 1.3 Graphical support of the algebraic solutions in Alternate Example 3. The vertical line in (b) should be ignored because it resultsfrom grapher failure. The points in (c) with negative x-coordinates should be ignored because the calculator does not know that x is a length(but we do).


In Exercises 1–8, find the domain of each of these functions.

1. f x x( ) – = 2 1 2. g xx

x( )

=

+22

3. P s s( ) ,= 4 where P(s) is the perimeter of a square with sides of length s.

4. h xx x

x( )

( )

=+ 1

5. f x x( ) – = 2 1 6. V r r( ) ,= 4

33π V(r) the volume of a sphere with radius r.

7.

g x x

x( ) – = 3 8. m x x( ) = �

Section 1.2 Functions and Their Properties 21

Alternate Example 6 Analyzing a Function for Increasing-DecreasingBehavior

For each function, tell the intervals on which it is increasing and the intervals on which itis decreasing.

(a) f(x) = (x – 2)2 (b) g x x

x( )

– =

2

2 4

SOLUTION

Solve Graphically(a) We can see from the graph in Figure 1.4 that f is decreasing on (–∞, 2] andincreasing on [2, ∞). (Notice that we include 2 in both intervals. Don’t worry that thissets up some contradiction about what happens at 2, because we only talk about functionsincreasing or decreasing on intervals, and 2 is not an interval.)

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.4 The function f(x) = (x – 2)2 decreases on (–∞, 2] and increases on [2, ∞).

(b) We see from the graph in Figure 1.5 that g is increasing on (–∞, –2), increasing againon (–2, 0], decreasing on [0, 2), and decreasing again on (2, ∞).

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.5 The function g x xx

( )

=−

2

2 4 increases on (–∞, –2) and (–2, 0]; the function decreases

on [0, 2) and (2, ∞).


In Exercises 9–14, identify the intervals on which the function is increasing and decreasing.

9. f(x) = 3(x + 2)2 10. f(x) = –(x + 3)2 11. g x xx

( )

=−

2

2 16 12. g x x

x( )

=

−

2

2 25

13. f( )x x= − +2 1 14. f xx

x( ) = − +3

2

31 �


Alternate Example 9 Checking Functions for Symmetry

Tell whether each of the following functions is odd, even, or neither.

(a) f(x) = 2x2 + 1 (b) g(x) = x2 – 2x + 3 (c) h x x

x( )

– =

3

2 1

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.6 This graph appears to besymmetric with respect to the y-axis, sowe conjecture that f is an even function.

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.7 This graph does notappear to be symmetric with respect toeither the y-axis or the origin, so weconjecture that g is neither even norodd.

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.8 This graph appears to besymmetric with respect to the origin, sowe conjecture that h is odd.

SOLUTION

(a) Solve GraphicallyThe graphical solution is shown in Figure 1.6.

Confirm AlgebraicallyWe need to verify that f(–x) = f(x) for all x in the domain of f.

f(–x) = 2(–x)2 + 1 = 2x2 + 1 = f(x)

Since this identity is true for all x, the function f is indeed even.

(b) Solve GraphicallyThe graphical solution is shown in Figure 1.7.

Confirm AlgebraicallyWe need to verify that g(–x) g(x) and g(–x) –g(x)

g x x x

x x

(– ) (– ) – (– )

= += + +

2

2

2 3

2 3

g x x x

g x x x

( ) –

– ( ) – –

= += +

2

2

2 3

2 3

So, g(–x) g(x) and g(–x) –g(x).

We conclude that g is neither odd nor even.

(c) Solve GraphicallyThe graphical solution is shown in Figure 1.8.

Confirm AlgebraicallyWe need to verify that

h(–x) = –h(x)

for all x in the domain of h.

h x xx

xxh x

(– ) (– )(– ) –

– –

– ( )

=

=

=

3

2

3

2

1

1

Since this identity is true for all x except ±1 (which is not in the domain of h), thefunction is odd.


In Exercises 15–20, state whether each of the following functions is odd, even, or neither.

15. f(x) = 3x2 + 2 16. f(x) = x2 – 2x 17. m x xx

( ) –

= 216

2

218. c x x

x( )

=

+

2

2 4

19. f(x) = (x – 5)2 – 2x 20. g x xx

( ) –

=3

22 9�

Section 1.3 Twelve Basic Functions 23

Alternate Example 10 Identifying the Asymptotes of a Graph

Identify any horizontal or vertical asymptotes of the graph of y xx x

–

=+2 4 3

.

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.9 The graph of

y xx x

–

=+2 4 3

has vertical

asymptotes of x = 1 and x = 3 and ahorizontal asymptote of y = 0.

SOLUTION

The quotient xx x

xx x2 4 3 1 3 –

( – )( – )+

= is undefined at x = 1 and x = 3, which makes

them likely sites for vertical asymptotes. The graph (Figure 1.9) provides support,showing vertical asymptotes of at x = 1 and x = 3.

For large values of x, the numerator (a large number) is dwarfed by the denominator

(a product of two large numbers), suggesting that lim ( – )( – )

x

xx x→ ∞

=1 3

0 . This would

indicate a horizontal asymptote of y = 0 as x → ∞. Similar logic suggests that

lim ( – )( – )

– –x

xx x→ ∞

= =1 3

0 0 , indicating the same horizontal asymptote as x → –∞.

Again, the graph provides support for this.


In Exercises 21–26, find all horizontal or vertical asymptotes of the function.

21. yx

–

= 142

22. yx

=−5

5 23.

y

x

– = 3

5 24. y x

x x

– =

+2 4 4 25. y x

x

=

+2 1

26. y xx x

– –

= +2 3 4

2 �

1.3 Twelve Basic Functions

Alternate Example 5 Analyzing a Function Graphically

Graph the function y = (x – 3)2. Then answer the following questions:

(a) On what interval is the function increasing? On what interval is it decreasing?

(b) Is the function odd, even, or neither?

(c ) Does the function have any extrema?

(d) How does the graph relate to the graph of the basic function y = x2?

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.10 The graph of y = (x – 3)2

SOLUTION

The graph is shown in Figure 1.10.

(a) The function is increasing if its graph is headed upward as it moves from left toright. We see that it is increasing on the interval [3, ∞). The function is decreasing if itsgraph is headed downward as it moves from left to right. We see that it is decreasing onthe interval (–∞, 3].

(b) The graph is not symmetric with respect to the y-axis, nor is it symmetric withrespect to the origin. The function is neither.

(c ) Yes, we see that the function has a minimum value of 0 at x = 3. (This is easilyconfirmed by the algebraic fact that (x – 3)2 0 for all x.)

(d) We see that the graph of y = (x – 3)2 is just the graph of y = x2 moved three units tothe right.



For Exercises 1–6, answer the following questions: (a) On what interval is the function increasing? On what interval is it decreasing? (b) Isthe function odd, even, or neither? (c ) Does the function have any extrema? (d) How does the graph relate to the graph of the basic functiony = x2?

1. y = –x2 2. y = (x + 1)2 3. y = –(x – 3)2 4. y = x2 + 1 5. y = x2 – 1 6. y = x2 – 4x + 4 �

Alternate Example 7 Defining a Function Piecewise

Using basic functions from this section, construct a piecewise definition for the functionwhose graph is shown in Figure 1.11. Is your function continuous?

x

y

4

3

2

1

–11 2 3 4 5–1–2

Figure 1.11 A piecewise defined function.

SOLUTION

This appears to be the graph of y = x2 to the left of x = 2 and the graph of y = 4 to theright of 2. We can therefore define it piecewise as

f xx x

x( )

= ≤

≥

2 2

4 2

if

if

The function is continuous.


In Exercises 7–12, using basic functions from this section, construct a piecewise definition for the function whose graph is shown.

7. 8. 9.

x

y

4321

–1–2 1 2 3 4 5x

y

4321

–1–2 1 2 3 4 5

x

y

2

–2

–2 2

10. 11. 12.

x

y

2

–2

–2 2x

y

2

–2

–2 2x

y

2

–2 2

–2

�

Section 1.4 Building Functions from Functions 25

Alternate Example 9 Analyzing a Function

Give a complete analysis of the basic function f(x) = x.

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.12 The graph of f(x) = x.

SOLUTION

BASIC FUNCTION The Square Root Function

f(x) = x (Figure 1.12)

Domain: all nonnegative real numbersRange: [0, ∞)ContinuousIncreasing on [0, ∞)Not symmetric with respect to the y-axis or the origin: neither even nor oddBounded belowLocal minimum at (0, 0)No horizontal asymptotesNo vertical asymptotes

End behavior: lim x

x→ ∞

= ∞


In Exercises 13–18, give a complete analysis of each basic function.

13. f(x) = x3 14. f(x) = ln x 15. f(x) = ex 16. f(x) = int (x) 17. f(x) = 1x 18. f(x) = |x| �

1.4 Building Functions from Functions

Alternate Example 2 Composing Functions

Let f(x) = 2x and g(x) = x. Find ( )( )f g x� and ( )( )g f x� and verify that the functions

f g � and g f � are not the same.

[–5.1, 5.1] by [–4.7, 4.7]

Figure 1.13 The graphs of yx = 2

and yx = 2 are not the same.

SOLUTION

( )( ) ( ( )) ( ) f g x f g x f x x� = = = 2

( )( ) ( ( )) ( ) g f x g f x g x x� = = =2 2

One verification that these functions are not the same is that they have different domains:

f g � is defined only for x 0, while g f � is defined for all real numbers. We couldalso consider their graphs (Figure 1.13), which agree only at x = 0 and x = 4.


In Exercises 1–6, find ( )( )f g x� and ( )( ).g f x�

1. f(x) = 2x2 and g(x) = 1x 2. f(x) = 2x and g(x) = x2 3. f(x) = x + 1 and g(x) = x

4. f(x) = x3 and g(x) = x – 1 5. f(x) = x2 and g(x) =

11x +

6. f(x) = 2x and g(x) = x 2 �


Alternate Example 4 Decomposing Functions

For each function h, find functions f and g such that h(x) = f(g(x)).

(a) h(x) = 2(x – 5)2 + 2(x – 5)

(b) h x x( ) – = 23 3

SOLUTION

(a) We can see that h is quadratic in x – 5. Let f(x) = 2x2 + 2 and let g(x) = x – 5. Then

h(x) = f(g(x)) = f(x – 5) = 2(x – 5)2 + 2(x – 5).

(b) We can see that h is the cube root of the function x2 – 3. Let f x x( ) = 3 and let

g(x) = x2 – 3. Then

h(x) = f(g(x)) = f(x2 – 3) = – x 23 3 .


In Exercises 7–12, for each function h, find functions f and g such that h(x) = f(g(x)). (Note: There may be more than one possibledecomposition.)

7. h(x) = 2(x2 + 5)2 8. h(x) = 112x +

9. h(x) = 1

1

3

2x +

10. h(x) = –2(x2 – 1)3 + x2 – 1

11. h(x) = 192x −

+ 3(x2 – 9)3 12. h(x) = x x2 23 + �

Alternate Example 7 Using Implicitly Defined Functions

Describe the graph of the relation 4x2 + 4xy + y2 = 4.

SOLUTION

This looks like a difficult task at first, but notice that the expression on the left of theequal sign is a factorable trinomial. This enables us to split the relation into twoimplicitly defined functions as follows.

4 4 4

2 42

2 2

2

x xy y

x yx

( )

+ + =+ =

+Factor.

yyx y x

= ±+ = +

22 2 2

Extract square roots.

or – – – – .

yy x y x

== + =

22 2 2 2or Solve for y

The graph consists of two parallel lines (Figure 1.14), each the graph of one of theimplicitly defined functions.

x

y

3

2

1

–11 2 3 4 5–1–2

–2

Figure 1.14 The graph of the relation4x2 + 4xy + y2 = 4


In Exercises 13–18, describe the graph of the relation.

13. (x – y)(x – y) = 9 14. x2 – y2 = 0 15. x2 – 3xy – 4y2 = 0 16. x2 – 5xy + 6y2 = 0

17. 9x2 – 12xy + 4y2 = 16 18. 3x2 + 2xy – y2 = 0 �

Section 1.5 Parametric Relations and Inverses 27

1.5 Parametric Relations and Inverses

Alternate Example 1 Defining a Function Parametrically

Consider the set of all ordered pairs (x, y) defined by the equations

x = t – 1

y = t2 – 1

where t is any real number.

(a) Find the points determined by t = –3, –2, –1, 0, 1, 2, and 3.

(b) Find an algebraic relationship between x and y. Is y a function of x?

(c) Graph the relation in the (x, y) plane.

x

y

3

3

t=3

t=2

t=1t=0

t=–1

t=–2

t=–3

Figure 1.15 (Alternate Example 1)

SOLUTION

(a) Substitute each value of t into the formulas for x and y to find the point that itdetermines parametrically:

t x = t – 1 y = t2 – 1 (x, y)

–3 –4 8 (–4, 8)

–2 –3 3 (–3, 3)

–1 –2 0 (–2, 0)

0 –1 –1 (–1, –1)

1 0 0 (0, 0)

2 1 3 (1, 3)

3 2 8 (2, 8)

(b) We can find the relationship between x and y algebraically by the method ofsubstitution. First we solve for t in terms of x to obtain x + 1.

y ty x

x

– ( ) –

== +=

2

2

11 1

Given

= + 1t x22

2

2 1 12

– .

+ += +

xx x

Expand

Simplify.

This is consistent with the ordered pairs we had found in the table. As t varies over all thereal numbers, we will get all the ordered pairs in the relation y = x2 + 2x, which doesindeed define y as a function of x.

(c) Since the parametrically defined relation consists of all ordered pairs in the relationy = x2 + 2x, we can get the graph by simply graphing the parabola y = x2 + 2x. SeeFigure 1.15.


In Exercises 1–6, consider the set of all ordered pairs (x, y) defined by the equations. (a) Find the points determined by t = –3, –2, –1, 0, 1, 2,and 3. (b) Find an algebraic relationship between x and y. Is y a function of x? (c) Graph the relation in the (x, y) plane.

1. x = 2t and y = t – 1 2. x = t + 2 and y = 2t – 1 3. x = t – 1and y = t2 4. x = 2t and y = t2 – 1

5. x = 2t and y = t2 + 2 6. x = t + 1 and y = t2 – 3 �


Alternate Example 4 Finding an Inverse Function Algebraically

Find an equation for f –1(x) if f x xx

( ) –

.= 22 1

[–5.1, 5.1] by [–4.7, 4.7]


f x x

x( ) .

– = 2

2 1

SOLUTION

The graph of f in Figure 1.16 suggests that f is one-to-one. The original function satisfies

the equation y xx

–

.= 22 1

If f is truly one-to-one, the inverse function f –1 will satisfy the

equation x yy

–

.= 22 1

(Note that we just switch the x and y.)

x yy

x y y

–

( – )

=

=

22 1

2 1 2 2Multiply by y – 1..

2 22 2

xy x yxy y

– –

==

Distributive property

( – )

xy x x

Isolate the

Factor ou

y .terms

2 2 = tt .

Divide by 2 – 2.

y

xy xx

–

=2 2

Therefore f x xx

– ( ) –

12 2

= .


In Exercises 7–12, find an equation for f –1(x).

7. f x x

x( )

=

+ 1 8.

f x x

x( )

– =

1 9.

f x x

x( )

– = 2

3 4 10.

f x x

x( )

– = + 1

1 11.

f x x

x( )

– = +2 3

3

12. f x x( ) = − 23 �

Alternate Example 7 Finding an Inverse Function

Show that f x x( ) – = 2 3 has an inverse function and find a rule for f –1(x). State any

restrictions on the domains of f and f –1.

[–5.1, 5.1] by [–4.7, 4.7]


f x x( ) – = 2 3 and its inverse, a

restricted y x = +1

232

2 .

SOLUTION

Solve AlgebraicallyThe graph of f passes the horizontal line test, so f has an inverse function (Figure 1.17).Note that f has domain [1.5, ∞) and range [0, ∞).

To find f –1 we write

y x x y

x y

– . ,

–

= ≥ ≥

=

2 3 1 5 0

2 3

where

wheree . ,

–

.y x

x y

≥ ≥

=

1 5 0

2 2

Interchange andx y

3312

2 32

Square.

Solve for .y x = + y

Thus, f x x– ( ) 1 212

32

= + , with an inherited domain restriction of x 0. Figure 1.17 shows

the two functions. Note the domain restriction of x 0 imposed on the parabola

y x = +12

32

2 .

Section 1.6 Graphical Transformations 29

Support GraphicallyUse a grapher in parametric mode and compare the graphs of the two sets of parametricequations with Figure 1.17:

x t x x

y x y t

–

–

= =

= =

and 2 3

2 3


In Exercises 13–18, find an equation for f –1(x). State any restrictions on the domains of f and f –1.

13. f(x) = 2x + 3 14. f x

x( )

=

+1

2 5 15.

f x

x( )

– =

+1

3 2 16. f x x( ) = +5 1

17. f x x( ) – = 4 8 18. f x x( ) – = 2 1 �

1.6 Graphical Transformations

Alternate Example 3 Finding Equations for Reflections

Find an equation for the reflection of f x xx

( )

= ++

3 212

across each axis.

SOLUTION

Solve Algebraically

Across the x-axis: y f x xx

xx

– ( ) –

– –

= = ++

=+

3 21

3 212 2

Across the y-axis: y f x x

xx

x (– ) (– )

(– ) –

= = +

+= +3 2

13 2

2 2 ++

1

Support GraphicallyThe graphs in Figure 1.18 support our algebraic work.

[–5.1, 5.1] by [–4.7, 4.7]

(a)

[–5.1, 5.1] by [–4.7, 4.7]

(b)

Figure 1.18 Reflections of f x x

x( )

= +

+3 2

12 across (a) the x-axis and (b) the y-axis.


In Exercises 1–6, find an equation for the reflection of each function across each axis.

1. f x x

x( )

– = +3 2

3 522.

f x

x( )

=

+1

2 5 3.

f x

x( )

– =

+1

3 2 4. f x x( ) = − +2 1

5. f x x

x( )

=

+2 76. f x x

x( )

– = 3

4 1

2

2�


Alternate Example 5 Finding Equations for Stretches and Shrinks

Let C1 be the curve defined by y1 = f(x) = x3 – x. Find equations for the following non-rigid transformations of C1:

(a) C2 is a vertical stretch of C1 by a factor of 2.

(b) C3 is a horizontal shrink of C1 by a factor of 1 4/ .

[–5.1, 5.1] by [–4.7, 4.7]

(a)

[–5.1, 5.1] by [–4.7, 4.7]

(b)

Figure 1.19 The graph of y1 = x3 – xshown with (a) a vertical stretch and(b) a horizontal shrink.

SOLUTION

Solve Algebraically(a) Denote the equation for C2 by y2. Then

y y

x x

x x

2 13

3

2

2

2 2

( – )

–

=

==

i

(b) Denote the equation for C3 by y3. Then

y f x

f x

x x

x

3

3

3

1 4

4

4 4

64

/

( )

( ) –

–

=

=== 4x

Support GraphicallyThe graphs in Figure 1.19 support our algebraic work.


In Exercises 7–12, find equations for the following non-rigid transformations of the given function.

7. f(x) = x2 + 2x; vertical stretch by a factor of 3

8. f(x) = 2x2 – 3x; horizontal shrink by a factor of 0.5

9. f(x) = 2x3 + 4x2; vertical shrink by a factor of 0.5

10. f(x) = –x3 + 5x; horizontal stretch by a factor of 2

11. f(x) = –2x3 + x2; vertical stretch by a factor of 3

12. f(x) = x2 – x; horizontal stretch by a factor of 2 �

Alternate Example 6 Combining Transformations in Order

(a) The graph of y = x2 undergoes the following transformations, in order. Find theequation of the graph that results.

• a horizontal shift 3 units to the left

• a vertical stretch by a factor of 2

• a vertical translation 4 units down.

(b) Apply the transformations in (a) in the opposite order and find the equation of thegraph that results.

SOLUTION

(a) Applying the transformations in order, we have

x x x x2 2 2 23 2 3 2 3 4 ( ) ( ) ( ) – ⇒ + ⇒ + ⇒ +

Expanding the final expression, we get the function y = 2x2 + 12x – 14.

Section 1.6 Graphical Transformations 31

(b) Applying the transformations in the opposite order, we have

x x x x2 2 2 24 2 4 2 3 4 – ( – ) (( ) – )⇒ ⇒ ⇒ +

Expanding the final expression, we get the function y = 2x2 + 12x + 10.

The second graph is 24 units higher than the first because the vertical stretch lengthensthe vertical translation when the translation occurs first. Order often matters whenstretches, shrinks, or reflections are involved.


In Exercises 13–18, the graph of y = x2 undergoes the following transformations, in order. (a) Find the equation of the graph that results.(b) Apply the transformations in (a) in the opposite order and find the equation of the graph that results.

13. a horizontal shift 2 units to the righta vertical shrink by a factor of 1 2/a vertical translation 2 units up.

14. a horizontal shift 1 unit to the righta vertical stretch by a factor of 3a vertical translation 1 unit down.

15. no horizontal shifta horizontal stretch by a factor of 2a vertical translation 3 units up.

16. a horizontal shift 3 units to the leftno horizontal or vertical stretch or shrinka vertical translation 1 unit down.

17. a horizontal shift 4 units to the righta vertical stretch by a factor of 3a vertical translation 4 units up.

18. a horizontal shift 2 units to the lefta horizontal shrink by a factor of 1 4/a vertical translation 5 units down. �

Alternate Example 7 Transforming a Graph Geometrically

The graph of y = f(x) is shown in Figure 1.20. Determine the graph of the compositefunction y = 2f(x – 1) – 2 by showing the effect of a sequence of transformations on thegraph of y = f(x).

x

y

4

3

2

1

–11 2 3 4 5–1–2

–2

Figure 1.20 The graph of thefunction y = f(x) in AlternateExample 7.

SOLUTION

The graph of y = 2f(x – 1) + 2 can be obtained from the graph of y = f(x) by the followingsequence of transformations:

(a) a vertical stretch by a factor of 2 to get y = 2f(x) (Figure 1.21a)

(b) a horizontal translation 1 unit to the right to get y = 2f(x – 1) (Figure 1.21b)

(c) a vertical translation 2 units down to get y = 2f(x – 1) – 2 (Figure 1.21c)

(The order of the first two transformations can be reversed without changing the finalgraph.)

x

y

4

3

2

1

–11 2 3 4 5–1–2

–2

x

y

4

3

2

1

–11 2 3 4 5–1–2

–2

x

y

4

3

2

1

–11 2 3 4 5–1–2

–2

(a) (b) (c)

Figure 1.21 The graph of the function y = f(x) in Alternate Example 7.



In Exercises 19–24, the graph of y = f(x) is shown. Sketch the graph of the composite function.

x

y

3

3

19. y = 2f(x) – 1 20. y = 0.5f(x) + 1 21. y = 2f(x – 2) 22. y = 0.5f(x + 1) 23. y = 2f(x – 1) – 1

24. y = 1.5f(x – 1) + 1 �

1.7 Modeling with Functions

Alternate Example 1 Finding Functions from Formulas

A right circular cylinder has height 8 inches. Write the volume V of the cylinder as afunction of its base

(a) radius (b) diameter (c) circumference

SOLUTION

(a) The familiar formula from geometry gives V as a function of r with h = 8:

V = 8πr2

(b) This formula is not so familiar. However we know that r d ,=2

so we can substitute

that expression for r in the volume formula:

V r dd .= =

=8 8 222

22

π π π

(c) Since C = 2πr, we can solve for r to get r C .=2π

Then we substitute to get V:

V r C C = =

=8 822

22 2

π ππ π


In Exercises 1–6, write a function for each situation using known formulas.

1. A right rectangular prism has height 10 cm. Its base is asquare. Write functions for the volume V in terms of thelength s of a side of the base and in terms of the perimeter Pof the base.

2. A right circular cylinder has height is 10 cm and base withradius r. Write a functions for the volume V in terms of theradius r, the diameter d of the base, and circumference C ofthe base.

3. A circle is inscribed in a square. Write a function for the areaof the circle in terms of the side of the square.

4. A cube is contained in a sphere. Write a function for thesurface area of the sphere in terms of the side s ofthe cube.

5. A rectangle has length 4 and width w. Write functions forthe area A in terms of length and width and in terms of theperimeter.

6. A right circular cone has height 9. Write functionsfor the volume V in terms of the radius r, the diameter d ofthe base, and circumference C of the base.

�

Section 1.7 Modeling with Functions 33

Alternate Example 3 Protecting an Antenna

A small satellite dish is packaged with a cardboard cylinder for protection. The parabolicdish is 16 in. in diameter and 4 in. deep, and the diameter of the cardboard cylinder is8 in. How tall must the cylinder be to fit in the middle of the dish and be flush with thetop of the dish? (See Figure 1.22.)

SOLUTION

The diagram in Figure 1.22a showing the cross section of this 3-dimensional problem isalso a 2-dimensional graph of a quadratic function. We can transform our basic functiony = x2 with a vertical shrink so that it goes through the points (–8, 4) and (8, 4), therebyproducing a graph of the parabola in the coordinate plane (Figure 1.22b).

(a)

(b)

x

y

24

6

810

–2–4–6–8–10 2 4 6 8 10

(–8, 4) (8, 4)

Figure 1.22 (a) A parabolic satellite dish with a protective cardboard cylinder in the middle forpackaging. (b) The parabola in the coordinate plane.

y kx

k

( )

== ±

2

24 8

Vertical shrink

Substitute x == 8, = 4.y

k

±

= =k .464

116

Solve for

Thus, y x .= 116

2

To find the height of the cardboard cylinder, we first find the y-coordinate of the parabola4 inches from the center, that is, when x = 4:

y ( ) = =116

4 12

From that point to the top of the dish is 4 – 1 = 3 in.


In Exercises 7–12, a small satellite dish is packaged with a cardboard cylinder for protection. How tall must the cylinder be to fit in themiddle of the dish and be flush with the top of the dish?

7. The parabolic dish is 32 in. in diameter and 8 in. deep, andthe diameter of the cardboard cylinder is 16 in.

8. The parabolic dish is 36 in. in diameter and 9 in. deep, andthe diameter of the cardboard cylinder is 18 in.

9. The parabolic dish is 30 in. in diameter and 7.5 in. deep, andthe diameter of the cardboard cylinder is 15 in.

10. The parabolic dish is 40 in. in diameter and 10 in. deep,and the diameter of the cardboard cylinder is 20 in.

11. The parabolic dish is 20 in. in diameter and 5 in. deep,and the diameter of the cardboard cylinder is 10 in.

12. The parabolic dish is 50 in. in diameter and 12.5 in. deep,and the diameter of the cardboard cylinder is 25 in. �


Alternate Example 4 Finding the Model and Solving

Grain is leaking through a hole in a storage bin at a constant rate of 5 cubic inches perminute. The grain forms a cone-shaped pile on the ground below. As it grows, the heightof the cone always remains equal to its radius. If the cone is one foot tall now, how tallwill it be in two hours?

Figure 1.23 A cone with equal height and radius.

SOLUTION

Reading the problem carefully, we realize that the formula for the volume of the cone is

needed. (Figure 1.23). From memory or by looking it up, we get the formula V r h .= 13

2π

A careful reading also reveals that the height and the radius are always equal, so we can

get volume directly as a function of height: V h .= 13

3π

When h = 12 in., the volume is V ( ) . .= =13

12 5763 3π π in

Two hours later, the volume will have grown by 120 5 6003

3min min

.i in. in.= The total

volume of the pile at that point will be (576π + 600) in.3. Finally, we use the volumeformula once again to solve for h:

V h

V

h

( )

=

= +

= +

13576 6003 576 600

3

3

π

πππ

hh

h

( )

.

= +

≈

3 576 600

12 43

3 ππ

inches


In Exercises 13–18, grain is leaking through a hole in a storage bin at the given constant rate. The grain forms a cone-shaped pile on theground below. As it grows, the height of the cone always remains equal to its radius. Given the height of the cone at the present time, how tallwill it be after the given elapsed time?

13. rate of leakage: 10 cubic inches per minute height at thepresent time: 15 inches elapsed time: 45 minutes

15. rate of leakage: 20 cubic inches per minute height at thepresent time: 10 inches elapsed time: 1 hour

17. rate of leakage: 15 cubic inches per minute height at thepresent time: 2 feet elapsed time: 45 minutes

14. rate of leakage: 12 cubic inches per minute height at thepresent time: 12 inches elapsed time: 12 minutes

16. rate of leakage: 6 cubic inches per minute height at thepresent time: 4 inches elapsed time: 2 hours

18. rate of leakage: 10 cubic inches per minute height at thepresent time: 2 feet elapsed time: 2 hours �

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