SECTION 2.5 Transformations of Functions

Section 2.5 • Transformations of Functions • 235

SECTION 2.5 Transformations of Functions

1 Recognize graphs ofcommon functions.

Have you seen Terminator 2, The Mask, or The Matrix? These were among thefirst films to use spectacular effects in which a character or object having oneshape was transformed in a fluid fashion into a quite different shape. The namefor such a transformation is morphing. The effect allows a real actor to beseamlessly transformed into a computer-generated animation. The animationcan be made to perform impossible feats before it is morphed back to theconventionally filmed image.

Like transformed movie images, the graph of one function can be turned intothe graph of a different function.To do this,we need to rely on a function’s equation.Knowing that a graph is a transformation of a familiar graph makes graphing easier.

Graphs of Common FunctionsTable 2.4 below and on page 236 gives names to six frequently encounteredfunctions in algebra.The table shows each function’s graph and lists characteristicsof the function. Study the shape of each graph and take a few minutes to verify thefunction’s characteristics from its graph. Knowing these graphs is essential foranalyzing their transformations into more complicated graphs.

Constant Function Identity Function Standard Quadratic Function

• Domain: • Domain: • Domain:

• Range: the single number c • Range: • Range:

• Constant on • Increasing on • Decreasing on and increasing on

• Even function • Odd function • Even function(0, q)(-q, 0)(-q, q)(-q, q)

[0, q)(-q, q)

(-q, q)(-q, q)(-q, q)

1

2

−1

−2

1 2−1−2

y

xf(x) = x2

1

2

−1

−2

1 2−1−2

y

x

f(x) = x

1

2

−1

−2

1 2−1−2

y

x

f(x) = c

Table 2.4 Algebra’s Common Graphs

Objectives1. Recognize graphs of

common functions.

2. Use vertical shifts tograph functions.

3. Use horizontal shifts tograph functions.

4. Use reflections to graphfunctions.

5. Use vertical stretchingand shrinking to graphfunctions.

6. Graph functions involvinga sequence oftransformations.

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 235

236 • Chapter 2 • Functions and Graphs

2 Use vertical shifts to graphfunctions.

Standard Cubic Function Square Root Function Absolute Value Function

• Domain: • Domain: • Domain:

• Range: • Range: • Range:

• Increasing on • Increasing on • Decreasing on and increasing on

• Odd function • Neither even nor odd • Even function(0, q)(-q, 0)(0, q)(-q, q)

[0, q)[0, q)(-q, q)

(-q, q)[0, q)(-q, q)

1

2

−1

−2

1 2−1−2

y

x

f(x) = |x|

1

2

−1

−2

1 2−1−2

y

x

f(x) = �x

1

2

−1

−2

1 2−1−2

y

x

f(x) = x3

Table 2.4 Algebra’s Common Graphs (continued)

The graph of can begradually morphed into the graphof by usinganimation to graph

for Byselecting many values for c, wecan create an animated sequencein which change appears to occurcontinuously.

0 � c � 2.f(x) = x2+ c

g(x) = x2+ 2

f(x) = x2

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

g(x) = x2 + 2

h(x) = x2 − 3

f(x) = x2

Figure 2.40 Vertical shifts

Vertical ShiftsLet’s begin by looking at three graphs whose shapes are the same. Figure 2.40shows the graphs. The black graph in the middle is the standard quadraticfunction, Now, look at the blue graph on the top. The equation ofthis graph, adds 2 to the right side of What effectdoes this have on the graph of f? It shifts the graph vertically up by 2 units.

Finally, look at the red graph on the bottom of Figure 2.40. The equation of thisgraph, subtracts 3 from the right side of What effectdoes this have on the graph of f? It shifts the graph vertically down by 3 units.

In general, if c is positive, shifts the graph of f upward cunits and shifts the graph of f downward c units. These are calledvertical shifts of the graph of f.

y = f(x) - cy = f(x) + c

The graph of h shifts the graph of f down 3 units.

h(x) = x2- 3 = f(x) - 3

f(x) = x2.h(x) = x2- 3,

The graph of g shifts the graph of f up 2 units.

g(x) = x2+ 2 = f(x) + 2

f(x) = x2.g(x) = x2+ 2,

f(x) = x2.

The study of how changing a function’s equation can affect its graph can beexplored with a graphing utility. Use your graphing utility to verify the hand-drawngraphs as you read this section.

Discovery

BLITMC02B.131013599_235-278 12/30/02 11:12 AM 36


−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

g(x) = |x| − 4

f(x) = |x|

Figure 2.41

x

0

1

2 (2, -2)@2 @ - 4 = -2(2, 2)@2 @ = 2

(1, -3)@1 @ - 4 = -3(1, 1)@1 @ = 1

(0, -3)@0 @ - 4 = -4(0, 0)@0 @ = 0

(-1, -3)@-1 @ - 4 = -3(-1, 1)@-1 @ = 1- 1

(-2, -2)@-2 @ - 4 = -2(-2, 2)@-2 @ = 2- 2

(x, g(x))= �x� - 4 = f(x) - 4(x, f(x))y = f(x) = �x�y = g(x)

Vertical ShiftsLet f be a function and c a positive real number.

• The graph of is the graph of shifted c unitsvertically upward.

• The graph of is the graph of shifted c unitsvertically downward.

EXAMPLE 1 Vertical Shift Down

Use the graph of to obtain the graph of

Solution The graph of has the same shape as the graph ofHowever, it is shifted down vertically 4 units. We have constructed a

table showing some of the coordinates for f and g. The graphs of f and g areshown in Figure 2.41.

f(x) = �x�.g(x) = �x� - 4

g(x) = �x� - 4.f(x) = �x�

x

y y

x

c

c

y = f(x)

y = f(x) + c

y = f(x) − c

y = f(x)

y = f(x)y = f(x) - c

y = f(x)y = f(x) + c


Horizontal ShiftsWe return to the graph of the standard quadratic function. InFigure 2.42 on the next page, the graph of function f is in the middle of the threegraphs.Turn the page and verify this observation.

f(x) = x2,

g(x) = �x� + 3.f(x) = �x�Check

Point1

3 Use horizontal shifts tograph functions.

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 237


We know that positivenumbers are to the right ofzero on a number line andnegative numbers are to theleft of zero. This positive-negative orientation doesnot apply to horizontalshifts. A positive numbercauses a shift to the left anda negative number causes ashift to the right.

Study Tip

−1

1234

8765

−2

2 3 4 5−1−2−3−4−5

y

x

f(x) = x2

g(x) = (x − 3)2h(x) = (x + 2)2

Figure 2.42 Horizontal shifts

By contrast to the vertical shift situation, this time there are graphs to the left andto the right of the graph of f. Look at the blue graph on the right.The equation ofthis graph, subtracts 3 from each value of x in the domain of

What effect does this have on the graph of f? It shifts the graphhorizontally to the right by 3 units.

Now, look at the red graph on the left in Figure 2.42.The equation of this graph,adds 2 to each value of x in the domain of What

effect does this have on the graph of f? It shifts the graph horizontally to theleft by 2 units.

In general, if c is positive, shifts the graph of f to the left c unitsand shifts the graph of f to the right c units. These are calledhorizontal shifts of the graph of f.

Horizontal ShiftsLet f be a function and c a positive real number.

• The graph of is the graph of shifted to the leftc units.

• The graph of is the graph of shifted to theright c units.

EXAMPLE 2 Horizontal Shift to the Left


Solution Compare the equations for and Theequation for g adds 5 to each value of x in the domain of f.

The graph of has the same shape as the graph of However, it is shifted horizontally to the left 5 units. We have created tables onthe next page showing some of the coordinates for f and g. As shown in Figure2.43, every point in the graph of g is exactly 5 units to the left of a correspondingpoint on the graph of f.

f(x) = 1x .g(x) = 1x + 5

The graph of g shifts the graph of f 5 units to the left.

y = g(x) = 1x + 5 = f(x + 5)

g(x) = 1x + 5 .f(x) = 1x

g(x) = 1x + 5 .f(x) = 1x

y y

x x

c c

y = f(x + c)

y = f(x − c)

y = f(x)

y = f(x)

y = f(x)y = f(x - c)

y = f(x)y = f(x + c)

y = f(x - c)y = f(x + c)

The graph of h shifts the graph of f 2 units to the left.

h(x) = (x + 2)2= f(x + 2)

f(x) = x2.h(x) = (x + 2)2,

The graph of g shifts the graph of f 3 units to the right.

g(x) = (x - 3)2= f(x - 3)

f(x) = x2.g(x) = (x - 3)2,

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

f(x) = �xg(x) = �x + 5

Figure 2.43 Shifting fiveunits left

f(x) = 1x

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 238


x

0

1

4 (4, 2)14 = 2

(1, 1)11 = 1

(0, 0)10 = 0

(x, f(x))y = f(x) = 1x x

(-1, 2)1-1 + 5 = 14 = 2- 1

(-4, 1)1-4 + 5 = 11 = 1- 4

(-5, 0)1-5 + 5 = 10 = 0- 5

(x, g(x))y = g(x) = 1x + 5


Some functions can be graphed by combining horizontal and verticalshifts.These functions will be variations of a function whose equation you knowhow to graph, such as the standard quadratic function, the standard cubicfunction, the square root function, or the absolute value function.

In our next example, we will use the graph of the standard quadraticfunction, to obtain the graph of We willgraph three functions:

EXAMPLE 3 Combining Horizontal and Vertical Shifts


SolutionStep 1 Graph The graph of the standard quadratic function isshown in Figure 2.44(a). We’ve identified three points on the graph.

Step 2 Graph Because we add 1 to each value of x in thedomain of the standard quadratic function, we shift the graph of fhorizontally one unit to the left.This is shown in Figure 2.44(b). Notice that everypoint in the graph in Figure 2.44(b) has an x-coordinate that is one less than thex-coordinate for the corresponding point in the graph in Figure 2.44(a).

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

f(x) = x 2.

h(x) = (x + 1)2- 3.f(x) = x2

Start by graphingthe standard

quadratic function.

Shift the graphof f horizontallyone unit to the

left.

Shift the graph ofg vertically down 3

units.

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

h(x) = (x + 1)2- 3.f(x) = x2,

g(x) = 1x - 4 .f(x) = 1x

CheckPoint

2

−1

12345

−2−3

1 2 3 4−1−2−3−4

y

x

f(x) = x2

(2, 4)

(0, 0)

(−2, 4)

Figure 2.44

−1

12345

−2−3

1 2 3 4−2−3−4

y

x

g(x) = (x + 1)2

(−3, 4) (1, 4)

(−1, 0)

(b) The graph of g(x) = (x + 1)2(a) The graph of f(x) = x2

−1

12345

−2

1 2 3 4−1−2−3−4

y

x

h(x) = (x + 1)2 − 3

(1, 1)(−3, 1)

(−1, −3)

(c) The graph of h(x) = (x + 1)2- 3

Work Example 3 by firstshifting the graph of

three units down,graphing Now, shift this graph one unitleft to graph

Didyou obtain the graph inFigure 2.44(c)? What can youconclude?

h(x) = (x + 1)2- 3.

g(x) = x2- 3.

f(x) = x2

Discovery

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 239

5

−5

5−5

y

x

f(x) = x2

(1, 1)

(1, −1)

(2, −4)(−2, −4)

(2, 4)

(−1, −1)

(−1, 1)

(−2, 4)

g(x) = −x2

1234

−2−3−4

2 3 4−2−3−4

Figure 2.45 Reflections about thex-axis


4 Use reflections to graphfunctions.

Step 3 Graph Because we subtract 3, we shift the graphin Figure 2.44(b) vertically down 3 units. The graph is shown in Figure 2.44(c).Notice that every point in the graph in Figure 2.44(c) has a y-coordinate that isthree less than the y-coordinate of the corresponding point in the graph inFigure 2.44(b).


Reflections of Graphs

This photograph shows a reflection of an old bridge in a Maryland river. Thisperfect reflection occurs because the surface of the water is absolutely still. Amild breeze rippling the water’s surface would distort the reflection.

Is it possible for graphs to have mirror-like qualities? Yes. Figure 2.45 showsthe graphs of and The graph of g is a reflection about thex-axis of the graph of f. In general, the graph of reflects the graph of fabout the x-axis. Thus, the graph of g is a reflection of the graph of f about the x-axis because

Reflection about the x-AxisThe graph of is the graph of reflected about the x-axis.y = f(x)y = - f(x)

g(x) = - x2= - f(x).

y = - f(x)g(x) = - x2.f(x) = x2

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

Point3

h(x) = (x + 1)2- 3.

−1

12345

−2−3

1 2 3 4−1−2−3−4

y

x

f(x) = x2

(2, 4)

(0, 0)

(−2, 4)

Figure 2.44, repeated

−1

12345

−2−3

1 2 3 4−2−3−4

y

x

g(x) = (x + 1)2

(−3, 4) (1, 4)

(−1, 0)

(b) The graph of g(x) = (x + 1)2(a) The graph of f(x) = x2

−1

12345

−2

1 2 3 4−1−2−3−4

y

x

h(x) = (x + 1)2 − 3

(1, 1)(−3, 1)

(−1, −3)

(c) The graph of h(x) = (x + 1)2- 3

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 240


−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

g(x) = −�x

f(x) = �x

x

0

1

4 (4, -2)- 14 = -2(4, 2)14 = 2

(1, -1)- 11 = -1(1, 1)11 = 1

(0, 0)- 10 = 0(0, 0)10 = 0

(x, g(x))g(x) = - 1x(x, f(x))f(x) = 1x

EXAMPLE 4 Reflection about the x-Axis


Solution Compare the equations for and Thegraph of g is a reflection about the x-axis of the graph of f because

We have created a table showing some of the coordinates for f and g. Thegraphs of f and g are shown in Figure 2.46.

g(x) = - 1x = - f(x).

g(x) = - 1x .f(x) = 1x

g(x) = - 1x .f(x) = 1x

Figure 2.46 Reflecting about the x-axis

f(x) = 2x

−1

12345

−2−3−4−5

1 2 3 4 8765−1−2

y

x

f(x) = �x − 1

(1, 0)

(5, 2)(2, 1)

Figure 2.48

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

f(x) = �xh(x) = �−x

(−1, 1) (1, 1)(−4, 2) (4, 2)

x

0

1

4 (4, 2)14 = 2

(1, 1)11 = 1

(0, 0)10 = 0

(x, f(x))f(x) = 1x x

0

(-4, 2)1-(-4) = 14 = 2- 4

(-1, 1)1-(-1) = 11 = 1- 1

(0, 0)1-0 = 10 = 0

(x, h(x))h(x) = 1-x


It is also possible to reflect graphs about the y-axis.

Reflection about the y-AxisThe graph of is the graph of reflected about the y-axis.

EXAMPLE 5 Reflection about the y-Axis


Solution Compare the equations for and The graphof h is a reflection about the y-axis of the graph of f because

We have created tables showing some of the coordinates for f and h.The graphsof f and h are shown in Figure 2.47.

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

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

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

y = f(x)y = f(- x)

g(x) = - �x�.f(x) = �x�CheckPoint

4

Use the graph of in Figure 2.48 to obtain thegraph of h(x) = 1- x - 1 .

f(x) = 1x - 1CheckPoint

5

Figure 2.47 Reflecting about the y-axis

f(x) = 2x

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2345

−3−2−1

−4−5

21 3 4 5−2 −1−3−4−5

y

x

f(x) = |x|

g(x) = 2 |x|

Figure 2.50 Stretching f(x) = ∑x∑

−1

1234

765

−2−3

1 2 3 4 5−1−2−3−4−5

y

x

g(x) = 2x2

f(x) = x2

12h(x) = x2

Figure 2.49 Stretching and shrinkingf(x) = x2

x

0

1

2 (2, 4)2 @2 @ = 4(2, 2)@2 @ = 2

(1, 2)2 @1 @ = 2(1, 1)@1 @ = 1

(0, 0)2 @0 @ = 0(0, 0)@0 @ = 0

(-1, 2)2 @-1 @ = 2(-1, 1)@-1 @ = 1-1

(-2, 4)2 @-2 @ = 4(-2, 2)@-2 @ = 2-2

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

5 Use vertical stretching andshrinking to graphfunctions.

Use the graph of to obtain the graph of g(x) = 3�x�.

f(x) = �x�CheckPoint

6

Vertical Stretching and ShrinkingMorphing does much more than move an image horizontally, vertically, or aboutan axis. An object having one shape is transformed into a different shape.Horizontal shifts, vertical shifts, and reflections do not change the basic shape of agraph. How can we shrink and stretch graphs, thereby altering their basic shapes?

Look at the three graphs in Figure 2.49. The black graph in the middle isthe graph of the standard quadratic function, Now, look at the bluegraph on the top. The equation of this graph is Thus, for each x, the y-coordinate of g is 2 times as large as the corresponding y-coordinate on thegraph of f.The result is a narrower graph.We say that the graph of g is obtainedby vertically stretching the graph of f. Now, look at the red graph on the bottom.The equation of this graph is or Thus, for each x, they-coordinate of h is one-half as large as the corresponding y-coordinate on thegraph of f. The result is a wider graph. We say that the graph of h is obtained byvertically shrinking the graph of f.

These observations can be summarized as follows:

Stretching and Shrinking GraphsLet f be a function and c a positive real number.

• If the graph of is the graph of verticallystretched by multiplying each of its y-coordinates by c.

• If the graph of is the graph of vertically shrunk by multiplying each of its y-coordinates by c.

EXAMPLE 6 Vertically Stretching a Graph


Solution The graph of is obtained by vertically stretching the graphof We have constructed a table showing some of the coordinates forf and g. Observe that the y-coordinate on the graph of g is twice as large as thecorresponding y-coordinate on the graph of f. The graphs of f and g are shownin Figure 2.50.

f(x) = �x�.g(x) = 2�x�

g(x) = 2�x�.f(x) = �x�

y = f(x)y = cf(x)0 6 c 6 1,

y = f(x)y = cf(x)c 7 1,

h(x) =12 f(x).h(x) =

12 x2,

g(x) = 2x2.f(x) = x2.

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12345

−2−1

−3−4−5

1 2 4 5−1−2−3−4−5

y

x

f(x) = |x|

12h(x) = |x|

Figure 2.51 Shrinking f(x) = ∑x∑

x

0

1

2 (2, 1)12 ∑2∑ = 1(2, 2)∑2∑ = 2

(1, 12)12 ∑1∑ =

12(1, 1)∑1∑ = 1

(0, 0)12 ∑0∑ = 0(0, 0)∑0∑ = 0

(-1, 12)12 ∑-1∑ =

12(-1, 1)∑-1∑ = 1- 1

(-2, 1)12 ∑-2∑ = 1(-2, 2)∑-2∑ = 2- 2

(x, h(x))h(x) =12 ∑x∑ =

12 f (x)(x, f(x))f(x) = ∑x∑

To Graph: Draw the Graph of f and: Changes in the Equation of

Vertical shifts

Raise the graph of f by c units. c is added to

Lower the graph of f by c units. c is subtracted from f(x).

Horizontal shifts

Shift the graph of f to the left c units. x is replaced with

Shift the graph of f to the right c units. x is replaced with

Reflection about the x-axis Reflect the graph of f about f(x) is multiplied by the x-axis.

Reflection about the y-axis Reflect the graph of f about x is replaced with the y-axis.

Vertical stretching or shrinking

Multiply each y-coordinate f(x) is multiplied by c, of by c, verticallystretching the graph of f.

Multiply each y-coordinate f(x) is multiplied by c, of by c, verticallyshrinking the graph of f.

y = f(x)0 6 c 6 1.y = cf(x), 0 6 c 6 1

y = f(x)c 7 1.y = cf(x), c 7 1

y = f(- x)- x.

y = - f(x)- 1.

x - c.y = f(x - c)

x + c.y = f(x + c)

y = f(x) - c

f(x).y = f(x) + c

y = f(x)

Table 2.5 Summary of TransformationsIn each case, c represents a positive real number.

EXAMPLE 7 Vertically Shrinking a Graph


Solution The graph of is obtained by vertically shrinking the graphof We have constructed a table showing some of the coordinates forf and h. Observe that the y-coordinate on the graph of h is one-half thecorresponding y-coordinate on the graph of f. The graphs of f and h are shownin Figure 2.51.

f(x) = �x�.h(x) =

12 ∑x∑

h(x) =12 ∑x∑.f(x) = ∑x∑


Sequences of TransformationsTable 2.5 summarizes the procedures for transforming the graph of y = f(x).

h(x) =14 ∑x∑.f(x) = �x�

CheckPoint

7

6 Graph functions involving a sequence oftransformations.

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 243

3. 4.

5. 6.

7. 8.

9. 10. g(x) =12 (x - 1)2g(x) = 2(x - 2)2

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

+ 1

h(x) = -(x - 1)2h(x) = –(x - 2)2

g(x) = (x - 1)2g(x) = (x - 2)2


A function involving more than one transformation can be graphed byperforming transformations in the following order:

1. Horizontal shifting2. Vertical stretching or shrinking3. Reflecting4. Vertical shifting

EXAMPLE 8 Graphing Using a Sequence of Transformations

Use the graph of to graph

Solution The following sequence of steps is illustrated in Figure 2.52.We beginwith the graph of

Step 1 Horizontal Shifting Graph Because x is replaced withthe graph of is shifted 1 unit to the left.

Step 2 Vertical Stretching or Shrinking Because the equation isnot multiplied by a constant in no stretching or shrinkingis involved.

Step 3 Reflecting We are interested in graphing orWe have now graphed We can graph

by noting that x is replaced with Thus, we graphby reflecting the graph of about the y-axis.

Step 4 Vertical Shifting We can use the graph of to get thegraph of Because 3 is added, shift the graph of

up 3 units.y = 11 - xg(x) = 11 - x + 3.

y = 11 - x

y = 1x + 1y = 1- x + 1- x.y = 1- x + 1

y = 1x + 1 .y = 1-x + 1 + 3.y = 11 - x + 3,

g(x) = 11 - x + 3,y = 1x + 1

f(x) = 1xx + 1,y = 1x + 1 .

f(x) = 1x .

g(x) = 11 - x + 3.f(x) = 1x

−1

1234

−2−3−4

1 2 3 54−1−2−3

y

x

f(x) = �x

(0, 0)

(4, 2)(1, 1)

Figure 2.52 Using to graph g(x) = 11 - x + 3f(x) = 1x

−1

123

5

−2−3

1 2 3−1−2−3−5 −4

y

x

(0, 4)(−3, 5)

(1, 3)g(x) = �1 − x + 3

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

(1, 0)

(0, 1)

(−3, 2)y = �−x + 1 = �1 − x

−1

234

−2−3−4

1 2 3 4−2−3−4

y

x

(−1, 0)

(0, 1) (3, 2)

y = �x + 1

Practice ExercisesIn Exercises 1–10, begin by graphing the standard quadratic function, Then use

transformations of this graph to graph the given function.

1. 2. g(x) = x2- 1g(x) = x2

- 2

f(x) = x2.

EXERCISE SET 2.5

Use the graph of to graph g(x) = - (x - 2)2+ 3.f(x) = x2

CheckPoint

8

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 244

Exercise Set 2.5 • 245

In Exercises 11–22, begin by graphing the square rootfunction, Then use transformations of thisgraph to graph the given function.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

In Exercises 23–34, begin by graphing the absolute valuefunction, f(x)=|x|. Then use transformations of this graphto graph the given function.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

In Exercises 35–44, begin by graphing the standard cubicfunction, Then use transformations of this graphto graph the given function.

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

In Exercises 45–52, use the graph of the function f to sketchthe graph of the given function g.

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = f(x)

r(x) = (x - 2)3+ 1r(x) = (x - 3)3

+ 2

h(x) =14 x3h(x) =

12 x3

h(x) = -(x - 2)3h(x) = -x3

g(x) = (x - 2)3g(x) = (x - 3)3

g(x) = x3- 2g(x) = x3

- 3

f(x) = x3.

h(x) = 2�x + 3�h(x) = 2�x + 4�

g(x) = - �x + 4� + 2g(x) = - �x + 4� + 1

h(x) = - �x + 3�h(x) = - �x + 4�

h(x) = �x + 3� - 2h(x) = �x + 4� - 2

g(x) = �x + 3�g(x) = �x + 4�

g(x) = �x� + 3g(x) = �x� + 4

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

g(x) = 21x + 1g(x) =12 1x + 2

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

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

g(x) = 1x + 1g(x) = 1x + 2

g(x) = 1x + 1g(x) = 1x + 2

f(x) = 1x .45. 46. g(x)=f(x)+2

47. 48. g(x)=f(x+2)

49. 50. g(x)=

51. 52. g(x)=–f(x+2)

In Exercises 53–56, write a possible equation for the functionwhose graph is shown. Each graph shows a transformationof a common function.

53.

54.

55.

56.

[−1, 9, 1] by [−1, 5, 1]

[−5, 3, 1] by [−5, 10, 1]

[−3, 3, 1] by [−6, 6, 1]

[−2, 8, 1] by [−1, 4, 1]

g(x) =12 f(x + 1)

12 f(x)g(x) = -f(x)

g(x) = f(x + 1)

g(x) = f(x) + 1

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 245


a. Describe how the graph of f can be obtained usingtransformations of the square root function

Then sketch the graph of f over theinterval If applicable, use a graphing utilityto verify your hand-drawn graph.

b. According to the model, how much was loaned in2000? Round to the nearest tenth of a billion. Howwell does the model describe the actual data?

c. Use the model to find the average rate of change, inbillions of dollars per year, between 1993 and 1995Round to the nearest tenth.

d. Use the model to find the average rate of change, inbillions of dollars per year, between 1998 and 2000.Round to the nearest tenth. How does this comparewith you answer in part (c)? How is this differenceshown by your graph?

e. Rewrite the function so that it represents theamount, , in billions of dollars, of new studentloans x years after 1995.

Writing in Mathematics59. What must be done to a function’s equation so that its

graph is shifted vertically upward?60. What must be done to a function’s equation so that its

graph is shifted horizontally to the right?61. What must be done to a function’s equation so that its

graph is reflected about the x-axis?62. What must be done to a function’s equation so that its

graph is reflected about the y-axis?63. What must be done to a function’s equation so that its

graph is stretched?

Technology Exercises64. a. Use a graphing utility to graph

b. Graph and in the same viewing rectangle.

c. Describe the relationship among the graphs of f, g, h,and k, with emphasis on different values of x for pointson all four graphs that give the same y-coordinate.

d. Generalize by describing the relationship betweenthe graph of f and the graph of g, where

for e. Try out your generalization by sketching the graphs

of f(cx) for and for afunction of your choice.

65. a. Use a graphing utility to graph b. Graph and and

in the same viewing rectangle.c. Describe the relationship among the graphs of f, g, and

h, with emphasis on different values of x for points onall three graphs that give the same y-coordinate.

d. Generalize by describing the relationship between thegraph of f and the graph of g, where for

e. Try out your generalization by sketching the graphsof f(cx) for and and for afunction of your choice.

c =14c =

12 ,c = 1,

0 6 c 6 1.g(x) = f(cx)

fA14 xBh(x) =g(x) = fA12 xB,f(x) = x2

+ 1,f(x) = x2

+ 1.

c = 4c = 3,c = 2,c = 1,

c 7 1.g(x) = f(cx)

k(x) = f(4x)h(x) = f(3x),g(x) = f(2x),f(x) = x2

+ 1,f(x) = x2

+ 1.

f(x)

0 � x � 9.f(x) = 1x.Application Exercises

57. The function models the medianheight, in inches, of boys who are x months of age.The graph of f is shown.

a. Describe how the graph can be obtained using trans-formations of the square root function

b. According to the model, what is the median height ofboys who are 48 months, or four years, old? Use acalculator and round to the nearest tenth of an inch.Theactual median height for boys at 48 months is 40.8 inches.How well does the model describe the actual height?

c. Use the model to find the average rate of change, ininches per month, between birth and 10 months.Round to the nearest tenth.

d. Use the model to find the average rate of change, ininches per month, between 50 and 60 months. Roundto the nearest tenth. How does this compare withyour answer in part (c)? How is this difference shownby the graph?

58. The graph shows the amount of money, in billions ofdollars, of new student loans from 1993 through 2000.

Source: U.S. Department of Education

The data shown can be modeled by the function where is the amount, in

billion of dollars, of new student loans x years after 1993.f(x)f(x) = 6.751x + 12,

Loa

n A

mou

nt (

billi

ons)

Amount of New Student Loans

Year20001999199819971996199519941993

30.028.727.225.524.0

22.0

18.0

12.0

10

20

30

$40

f(x) = 1x.

100 20

Age (months)

30 40 50 60

10

30

40

50

Med

ian

Hei

ght (

inch

es) f(x) = 2.9�x + 20.1

y

x

20

f(x),f(x) = 2.91x + 20.1

BLITMC02B.131013599_235-278 12/30/02 11:12 AM Page 246


Critical Thinking Exercises66. Which one of the following is true?

a. If and then the graphof g is a translation of three units to the right andthree units upward of the graph of f.

b. If and then f and g haveidentical graphs.

c. If and then the graphof g can be obtained from the graph of f bystretching f five units followed by a downward shiftof two units.

d. If and then thegraph of g can be obtained from the graph of f bymoving f three units to the right, reflecting in the x-axis,and then moving the resulting graph down four units.

In Exercises 67–70, functions f and g are graphed in the samerectangular coordinate system. If g is obtained from f througha sequence of transformations, find an equation for g.

67.

68.

−1

12345

−2−3−4−5

1 2 3 4 765−1−2−3

y

x

g(x) = ?

f(x) = |x|

−1

12345

−2−3−4−5

1 2 3−1−2−3−4−7 −6 −5

y

x

f(x) = x2

g(x) = ?

g(x) = -(x - 3)3- 4,f(x) = x3

g(x) = 5(x2- 2),f(x) = x2

g(x) = 1-x ,f(x) = - 1x

g(x) = �x + 3� + 3,f(x) = �x�

69.

70.

For Exercises 71–74, assume that (a, b) is a point on thegraph of f. What is the corresponding point on the graph ofeach of the following functions?

71. 72.73. 74.

Group Exercise75. This activity is a group research project on morphing and

should result in a presentation made by group membersto the entire class. Be sure to include morphing images thatwill intrigue class members. You should have no problemfinding an array of fascinating images online.Also includea discussion of films using spectacular morphing effects.Rent videos of these films and show appropriate excerpts.

y = f(x) - 3y = f(x - 3)

y = 2f(x)y = f(-x)

−1

123

5

−2−3−4−5

1 2 3 5−1−2−3−5

y

x

f(x) = �16 − x2

g(x) = ?

−1

12345

−2−3−4−5

1 2 3 4 86 75−1−2

y

x

g(x) = ?

f(x) = �x

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 247

Objectives1. Combine functions

arithmetically, specifyingdomains.

2. Form compositefunctions.

3. Determine domains forcomposite functions.

4. Write functions ascompositions.


1 Combine functionsarithmetically, specifyingdomains.

SECTION 2.6 Combinations of Functions; Composite Functions

They say a fool and his money are soon parted and the rest of us just wait to betaxed. It’s hard to believe that the United States was a low-tax country in theearly part of the twentieth century. Figure 2.53 shows how the tax burden hasgrown since then. We can use the information shown to illustrate how twofunctions can be combined to form a new function. In this section, you willlearn how to combine functions to obtain new functions.

Per

Cap

ita

Tax

U.S. Per Capita Tax Burden in 2000 Dollars

Year1970 1980 1990 20001960195019401930192019101900

10,000

$11,000

9000

8000

7000

6000

5000

4000

3000

2000

1000

Federal tax

State and local tax

Figure 2.53 Source: Tax Foundation

Combinations of Functions

To begin our discussion, take a look at the information shown for the year 2000.The total per capita tax burden is approximately $10,500. The per capita stateand local tax is approximately $3400.The per capita federal tax is the differencebetween these amounts.

Per capita federal tax = $10,500 - $3400 = $7100

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 248

We can think of this subtraction as the subtraction of function values.We do thisby introducing the following functions:

Using Figure 2.53, we see that

We can subtract these function values by introducing a new function, ,defined by the subtraction of T(x) and S(x). Thus,

total per capita tax in year x minus state and local per capita tax in year x.

For example,

Figure 2.53 illustrates that information involving differences of functionsoften appears in graphs seen in newspapers and magazines. Like numbers andalgebraic expressions, two functions can be added, subtracted multiplied, ordivided as long as there are numbers common to the domains of both functions.The common domain for functions T and S in Figure 2.53 is

Because functions are usually given as equations, we perform operationsby carrying out these operations with the algebraic expressions that appear onthe right side of the equations. For example, we can combine the following twofunctions using addition:

and

To do so, we add the terms to the right of the equal sign for f(x) to the terms tothe right of the equal sign for g(x). Here is how it’s done:

Add terms for f(x) and g(x).

Combine like terms.

Arrange terms in descending powers of x.

The name of this new function is Thus, the sum is the functiondefined by The domain of consists of thenumbers x that are in the domain of f and in the domain of g. Because neither fnor g contains division or even roots, the domain of each function is the set of allreal numbers. Thus, the domain of is also the set of all real numbers.f + g

f + g(f + g) (x) = x2+ 2x - 3.

f + gf + g.

= x2+ 2x - 3

= 2x - 3 + x2

= (2x + 1) + (x2- 4)

(f + g) (x) = f(x) + g(x)

g(x) = x2- 4.f(x) = 2x + 1

{1900, 1901, 1902, 1903, p , 2000}.

In 2000, the difference between total tax and state and localtax was $7100. This is the per capita federal tax.

(T - S) (2000) = T(2000) - S(2000) = $10, 500 - $3400 = $7100.

(T - S) (x) = T(x) - S(x) =

T - S

T(2000) = $10,500 and S(2000) = $3400.

Let S(x) = per capita state and local tax in year x.

Let T(x) = total per capita tax in year x.

Section 2.6 • Combinations of Functions; Composite Functions • 249

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 249

EXAMPLE 1 Finding the Sum of Two Functions

Let and Find:

a. b.

Solutiona. Thus,

b. We find by substituting 3 for x in the equation for

This is the equation for

Let and Find:

a. b.

Here is a general definition for function addition:

The Sum of FunctionsLet f and g be two functions. The sum is the function defined by

The domain of is the set of all real numbers that are common to thedomain of f and the domain of g.

EXAMPLE 2 Adding Functions and Determining the Domain

Let and Find:

a. b. the domain of

Solutiona.b. The domain of is the set of all real numbers that are common to the

domain of f and the domain of g.Thus, we must find the domains of f and g.We will do so for f first.

Note that is a function involving the square root ofBecause the square root of a negative quantity is not a real number,

the value of must be nonnegative. Thus, the domain of f is all x suchthat Equivalently, the domain is or

Likewise, is also a square root function. Because thesquare root of a negative quantity is not a real number, the value of must be nonnegative. Thus, the domain of g is all x such that Equivalently, the domain is or

Now, we can use a number line to determine the domain of Figure 2.54 shows the domain of f in blue and the domain of g in red.Can yousee that all real numbers greater than or equal to 2 are common to bothdomains? This is shown in purple on the number line. Thus, the domain of

is [2, q).f + g

f + g.[2, q).{x�x � 2},

x - 2 � 0.x - 2

g(x) = 1x - 2[–3, q).{x�x � - 3},x + 3 � 0.

x + 3x + 3.

f(x) = 1x + 3

f + g

(f + g)(x) = f(x) + g(x) = 1x + 3 + 1x - 2

f + g.(f + g)(x)

g(x) = 1x - 2 .f(x) = 1x + 3

f + g

(f + g)(x) = f(x) + g(x).

f + g

(f + g)(4).(f + g)(x)

g(x) = 2x + 7.f(x) = 3x2+ 4x - 1Check

Point1

(f + g)(3) = 32+ 4 � 3 + 2 = 9 + 12 + 2 = 23

Substitute 3 for x.

f + g.(f + g)(x) = x2+ 4x + 2

f + g.(f + g)(3)

(f + g)(x) = x2+ 4x + 2.

(f + g)(x) = f(x) + g(x) = (x2- 3) + (4x + 5) = x2

+ 4x + 2.

(f + g)(3).(f + g)(x)

g(x) = 4x + 5.f(x) = x2- 3


Domain of gDomain of g

Domain of f

Domain off + g2−3

Figure 2.54 Finding the domain ofthe sum f + g

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 250

Let and Find:

a. b. the domain of

We can also combine functions using subtraction, multiplication, and divisionby performing operations with the algebraic expressions that appear on the rightside of the equations. For example, the functions and

can be combined to form the difference, product, and quotient of fand g. Here’s how it’s done.

Just like the domain for f+g, the domain for each of these functions consists ofall real numbers that are common to the domains of f and g. In the case of the

quotient function we must remember not to divide by 0, so we add the

further restriction that g(x) Z 0.

The following definitions summarize our discussion:

Definitions: Sum, Difference, Product, and Quotient of FunctionsLet f and g be two functions. The sum the difference theproduct fg, and the quotient arefunctions whose domains are the set of all real numbers common to thedomains of f and g, defined as follows:

1. Sum:

2. Difference:

3. Product:

4. Quotient: provided g(x) Z 0a f

gb (x) =

f(x)

g(x) ,

(fg)(x) = f(x) �g(x)

(f - g)(x) = f(x) - g(x)

(f + g)(x) = f(x) + g(x)

fg

f - g,f + g,

f(x)

g(x) ,

a f

gb (x) =

f(x)

g(x)=

x + 3x - 1

, x Z 1

= (x + 3)(x - 1) = x2+ 2x - 3

(fg)(x) = f(x) � g(x)

= (x + 3) - (x - 1) = x + 3 - x + 1 = 4

(f - g)(x) = f(x) - g(x)

g(x) = x - 1f(x) = x + 3

f + g.(f + g)(x)

g(x) = 1x + 1 .f(x) = 1x - 3CheckPoint

2


The graph on the left is the graph of

in a by [0, 8, 1] viewing rectangle. Thegraph reveals what we discovered algebraically inExample 2(b). The domain of this function is[2, q).

[- 3, 10, 1]

y = 1x + 3 + 1x - 2

Technology

Domain [2, ∞)[

Difference: f − g

Product: fg

Quotient: fg−−

BLITMC02B.131013599_235-278_tg 1/13/03 12:09 PM Page 251

EXAMPLE 3 Combining Functions

If and find:

a. b. c.

Determine the domain for each function.

Solutiona. This is the definition of the

difference

Subtract g(x) from f(x).

Perform the subtraction.

Combine like terms and arrangeterms in descending powers of x.

b. This is the definition of theproduct fg.

Multiply f(x) and g(x).

Multiply each term in thesecond factor by 2x and–1, respectively.

Use the distributiveproperty.

Rearrange terms so thatlike terms are adjacent.

Combine like terms.

c. This is the definition of the quotient

Divide the algebraic expressions for f(x) and g(x).

Because the equations for f and g do not involve division or contain evenroots, the domain of both f and g is the set of all real numbers.Thus, the domainof f-g and fg is the set of all real numbers. However, for the denominatorcannot equal zero. We can factor the denominator as follows:

We see that the domain for is the set of all real numbers except and,

If and find:

a. b. c.

Determine the domain for each function.

a f

gb (x).(fg)(x)(f - g)(x)

g(x) = x2- 1,f(x) = x - 5Check

Point3

x Z 1}.1: {x�x Z -2-2f

g

Because x + 2 ≠ 0,x ≠ −2.

Because x − 1 ≠ 0,x ≠ 1.

a f

gb (x) =

2x - 1x2

+ x - 2=

2x - 1(x + 2)(x - 1)

.

fg ,

=

2x - 1x2

+ x - 2

fg

. a f

gb (x) =

f(x)

g(x)

= 2x3+ x2

- 5x + 2

= 2x3+ (2x2

- x2) + (-4x - x) + 2

= 2x3+ 2x2

- 4x - x2- x + 2

= 2x(x2+ x - 2) - 1(x2

+ x - 2)

= (2x - 1)(x2+ x - 2)

(fg)(x) = f(x) � g(x)

= -x2+ x + 1

= 2x - 1 - x2- x + 2

= (2x - 1) - (x2+ x - 2)

f - g.(f - g)(x) = f(x) - g(x)

¢ f

g≤ (x).(fg)(x)(f - g)(x)

g(x) = x2+ x - 2,f(x) = 2x - 1


If the function can besimplified, determine thedomain before simplifying.

EXAMPLE:

1

=

(x + 2)(x - 2)

(x - 2)

1

= x + 2

fg−−Domain of is {x x ≠ 2}. |

¢ f

g≤ (x) =

x2- 4

x - 2

g(x) = x - 2

f(x) = x2- 4 and

fg

Study Tip

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 252

Replace x with 1400.


Composite FunctionsThere is another way of combining two functions. To help understand this newcombination, suppose that your computer store is having a sale.The models thatare on sale cost either $300 less than the regular price or 85% of the regularprice. If x represents the computer’s regular price, both discounts can bedescribed with the following functions:

At the store, you bargain with the salesperson. Eventually, she makes an offeryou can’t refuse: The sale price is 85% of the regular price followed by a $300reduction:

In terms of functions f and g, this offer can be obtained by taking the output ofg(x)=0.85x, namely 0.85x, and using it as the input of f:

Because 0.85x is g(x), we can write this last equation as

We read this equation as “f of g of x is equal to 0.85x-300.” We call f Ag(x) Bthe composition of the function f with g, or a composite function. This compositefunction is written Thus,

Like all functions, we can evaluate for a specified value of x in the function’sdomain. For example, here’s how to find the value of this function at 1400:

This composite function describes the offer you cannot refuse.

This means that a computer that regularly sells for $1400 is on sale for $890subject to both discounts.

Before you run out to buy a new computer, let’s generalize our discussionof the computer’s double discount and define the composition of any twofunctions.

(f � g)(1400) = 0.85(1400) - 300 = 1190 - 300 = 890.

(f � g)(x) = 0.85x - 300

f � g

(f � g)(x) = f(g(x)) = 0.85x - 300.

f � g.

f(g(x)) = 0.85x - 300.

f(0.85x) = 0.85x - 300.

f(x) = x - 300

85% ofthe regular

price

followed bya $300

reduction

0.85x - 300.

The computer is on salefor $300 less thanits regular price.

The computer ison sale for 85%

of its regular price.

g(x) = 0.85x.f(x) = x - 300

2 Form composite functions.

Replace x with 0.85x , the output of g(x) = 0.85x.

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 253


EXAMPLE 4 Forming Composite Functions

Given and find:

a. b.

Solutiona. We begin with the composition of f with g. Because

means f Ag(x) B , we must replace each occurrence of x in the equation for fwith g(x).

This is the given equation for f.

Because replace g(x)with

Use the distributive property.

Simplify.

Thus, (f � g)(x) = 3x2+ 14.

= 3x2+ 14

= 3x2+ 18 - 4

x2+ 6

g(x) = x2+ 6,= 3(x2

+ 6) - 4

(f � g)(x) = f(g(x)) = 3g(x) - 4

f(x) = 3x - 4

(f � g)(x)(f � g)(x),

(g � f)(x).(f � g)(x)

g(x) = x2+ 6,f(x) = 3x - 4

The Composition of FunctionsThe composition of the function f with g is denoted by and is definedby the equation

The domain of the composite function is the set of all x such that1. x is in the domain of g and2. g(x) is in the domain of f.

The composition of f with g, is pictured as a machine with inputsand outputs in Figure 2.55. The diagram indicates that the output of g, or g(x),becomes the input for “machine” f. If g(x) is not in the domain of f, it cannotbe input into machine f, and so g(x) must be discarded.

f � g,

f � g

(f � g)(x) = f(g(x)).

f � g

Inputs, x

g(x)

Outputsf(g(x)) or (f �g)(x)

g

f

Figure 2.55 Inputting one functioninto a second function

Replace x with g(x).

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 254

b. Next, we find the composition of g with f. Because means g Af(x) B , we must replace each occurrence of x in the equation forg with f(x).

This is the given equation for g.

Because replacef(x) with

Use to square

Simplify.

Thus, Notice that is not the samefunction as

Given and find:

a. b.

We need to be careful in determining the domain for the compositefunction

The following values must be excluded from the input x:

• If x is not in the domain of g, it must not be in the domain of • Any x for which g(x) is not in the domain of f must not be in the domain of

EXAMPLE 5 Forming a Composite Function and Finding Its Domain

Given and , find:

a. b. the domain of

Solutiona. Because means we must replace x in

with g(x).

Thus, .(f � g) (x) =

2x

3 - x

Simplify the complexfraction by multiplying

by , or 1.xx−−

g(x) = 3x−−

(f � g) (x) = f(g(x)) =

2g(x) - 1

=

23x - 1

=

23x - 1

�xx

=

2x

3 - x

f(x) =

2x - 1

f(g(x)), (f � g) (x)

f � g.(f � g) (x)

g(x) =

3x

f(x) =

2x - 1

f � g.

f � g.

(f � g) (x) = f(g(x)).

(g � f)(x).(f � g)(x)

g(x) = x2- 1,f(x) = 5x + 6Check

Point4

(g � f)(x).(f � g)(x)(g � f)(x) = 9x2

- 24x + 22.

= 9x2- 24x + 22

3x - 4.(A - B)2

= A2- 2AB + B2

= 9x2- 24x + 16 + 6

3x - 4.f(x) = 3x - 4,= (3x - 4)2

+ 6

(g � f)(x) = g(f(x)) = (f(x))2+ 6

Replace x with f(x).

g(x) = x2+ 6

(g � f)(x)(g � f)(x),


3 Determine domains forcomposite functions.

The procedure for simplifyingcomplex fractions can be foundin Section P.6, pages 66–67.

Study Tip

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 255

b. We determine the domain of in two steps.(f � g) (x)


Rules for Excluding Numbers from Applying the Rules to

the Domain of and

If x is not in the domain of g, it must The domain of g is . Thus, 0 not be in the domain of must be excluded from the domain of

Any x for which g(x) is not in The domain of f is This meansthe domain of f must not be in we must exclude from the domainthe domain of of any x for which

Set g(x) equal to 1.

Multiply both sides by x.3 must be excluded from the domain of f � g.3 = x

3x

= 1

g(x) = 1.f � gf � g.

Ex∑x Z 1F .f � g.f � g.

Ex∑x Z 0Fg(x) =

3x

f(x) =

2x - 1

(f � g) (x) = f(g(x))

The domain of is and

Given and , find:

a. b. the domain of

Decomposing FunctionsWhen you form a composite function, you “compose” two functions to form a newfunction. It is also possible to reverse this process. That is, you can “decompose” agiven function and express it as a composition of two functions. Although there ismore than one way to do this, there is often a “natural” selection that comes to mindfirst. For example, consider the function h defined by

The function h takes and raises it to the power 5. A natural wayto write h as a composition of two functions is to raise the function

to the power 5. Thus, if we let

EXAMPLE 6 Writing a Function as a Composition

Express as a composition of two functions:

Solution The function h takes and takes its cube root.A natural way towrite h as a composition of two functions is to take the cube root of the function

. Thus, we let

and .g(x) = x2+ 1f(x) = 23 x

g(x) = x2+ 1

x2+ 1

h(x) = 23 x2+ 1.

(f � g)(x) = f(g(x)) = f(3x2- 4x + 1) = (3x2

- 4x + 1)5.

f(x) = x5 and g(x) = 3x2- 4x + 1, then

g(x) = 3x2- 4x + 1

3x2- 4x + 1

h(x) = (3x2- 4x + 1)5.

f � g.(f � g) (x)

g(x) =

1x

f(x) =

4x + 2Check

Point5

x Z 3F .Ex�x Z 0f � g

4 Write functions ascompositions

Suppose the form of function h is

Function h can be expressed as acomposition, using

g(x) = algebraic expression.f(x) = xpower

f � g,

h(x) = (algebraic expression)power.

Study Tip

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 256


We can check this composition by finding This should give the originalfunction, namely

Express as a composition of two functions:

h(x) = 2x2+ 5.

CheckPoint

6

(f � g) (x) = f(g(x)) = f(x2+ 1) = 23 x2

+ 1 = h(x)

h(x) = 23 x2+ 1.

(f � g) (x).

EXERCISE SET 2.6

21.

22.23.

24.

25.

26.

27.

28.

In Exercises 29–38, find:a. b. the domain of .

29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

In Exercises 39–46, express the given function h as acomposition of two functions f and g so that

.

39. 40.

41. 42.

43. 44.

45. 46. h(x) =

14x + 5

h(x) =

12x - 3

h(x) = ∑3x - 4∑h(x) = ∑2x - 5∑

h(x) = 25x2+ 3h(x) = 23 x2

- 9

h(x) = (2x - 5)3h(x) = (3x - 1)4

h(x) = (f � g) (x)

f(x) = 9 – x2, g(x) = 2x2- 9

f(x) = 4 - x2, g(x) = 2x2- 4

f(x) = x2+ 1, g(x) = 22 - x

f(x) = x2+ 4, g(x) = 21 - x

f(x) = 2x, g(x) = x - 3

f(x) = 2x, g(x) = x + 3

f(x) =

x

x + 5, g(x) =

6x

f(x) =

x

x + 1, g(x) =

4x

f(x) =

5x + 4

, g(x) =

1x

f(x) =

2x + 3

, g(x) =

1x

f � g(f � g) (x)

f(x) = 6x – 3, g(x) =

x + 36

f(x) = 2x – 3, g(x) =

x + 32

f(x) = 2x, g(x) = x + 2

f(x) = 2x, g(x) = x - 1

f(x) = x2+ 1, g(x) = x2

- 3

f(x) = x2+ 2, g(x) = x2

- 2f(x) = 7x + 1, g(x) = 2x2

- 9

f(x) = 4x - 3, g(x) = 5x2- 2Practice Exercises

1. If and find:a. b.

2. If and find:a. b.

3. Let and find:a. b. the domain of .

4. Let and find:a. b. the domain of

In Exercises 5–16, find , and Determinethe domain for each function.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

In Exercises 17–28, find:

a.b.c.

17.

18.

19.

20. f(x) = 5x + 2, g(x) = 3x - 4

f(x) = x + 4, g(x) = 2x + 1

f(x) = 3x, g(x) = x - 5

f(x) = 2x, g(x) = x + 7

(f � g) (2).(g � f)(x)(f � g) (x)

f(x) = 2x + 6, g(x) = 2x - 3

f(x) = 2x + 4, g(x) = 2x - 1

f(x) = 6 -

1x

, g(x) =

1x

f(x) = 2 +

1x

, g(x) =

1x

f(x) =

1x

, g(x) = x - 5

f(x) = 2x, g(x) = x - 4

f(x) = 6x2- x - 1, g(x) = x - 1

f(x) = 2x2- x - 3, g(x) = x + 1

f(x) = x - 6, g(x) = 5x2

f(x) = x - 5, g(x) = 3x2

f(x) = 3x - 4, g(x) = x + 2

f(x) = 2x + 3, g(x) = x - 1

fg.f + g, f - g, fg

f + g.(f + g) (x)g(x) = 2x + 5,f(x) = 2x – 8

f + g(f + g) (x)g(x) = 2x + 2,f(x) = 2x – 6

(f + g) (5).(f + g) (x)g(x) = 4x - 1,f(x) = 3x2 – 2x + 1

(f + g) (4).(f + g) (x)g(x) = 3x + 7,f(x) = 2x2

- 5

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 257


60. Find an estimate of What does thismean in terms of the variables in this situation?

61. Using the information shown in the graph, what is thedomain of

62. Using the information shown in the graph, what is thedomain of

Consider the following functions:

population of the world’s more developedregions in year x

population of the world’s less developed regionsin year x

total world population in year x.

Use these functions and the graph shown to answer Exercises63–66.

Source: Population Reference Bureau

63. What does the function represent?

64. What does the function represent?

65. Use the graph to estimate

66. Use the graph to estimate

67. A company that sells radios has a yearly fixed cost of$600,000. It costs the company $45 to produce each radio.Each radio will sell for $65. The company’s costs andrevenue are modeled by the following functions:

This function models the company’s costs.This function models the company’s revenue.

Find and interpret and (R - C) (40,000).

(R - C) (30,000)(R - C) (20,000),

R(x) = 65x.

C(x) = 600, 000 + 45x

(h - g) (2000).

(f + g) (2000).

h - g

f + g

Year1750 1850 1950 2050

y

x0

1

2

3

4

5

6

7

8

9

10

Wor

ld P

opul

atio

n (b

illio

ns)

World Population

Lessdeveloped

regions

Moredeveloped

regions

h(x) =

g(x) =

f(x) =

D - C?

D + C?

(D - C) (2000).In Exercises 47–58, use the graphs of f and g to evaluate eachfunction.

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

57. 58.

Application ExercisesIt seems that Phideau’s medical bills are costing us

an arm and a paw. The graph shows veterinary costs, inbillions of dollars, for dogs and cats in five selected years. Let

D(x) = veterinary costs, in billions of dollars, for dogsin year x

C(x) = veterinary costs, in billions of dollars, for cats in year x.

Use the graph to solve Exercises 59–62.

Source: American Veterinary Medical Association

59. Find an estimate of What does thismean in terms of the variables in this situation?

(D + C) (2000).

Cos

t (bi

llion

s)

Veterinary Costs in the U.S.

Year

20001983 1987 1991 1996

$15

12

9

6

3

Dogs Cats

(g � f) (-1)(g � f) (0)

(f � g) (1)(f � g) (2)

(fg) (-2)(fg) (-4)

a f

gb (-5)a f

gb (-6)

(g - f) (2)(f - g) (2)

(f + g) (-4)(f + g) (-3)

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = g(x)

y = f(x)

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 258

Technology Exercises77. The function models the

U.S. population ages 65 and older, f (t), in millions, t yearsafter 1990. The function models the total yearly cost of Medicare, g(t), in billions

of dollars, t years after 1990. Graph the function in a

[0, 15, 1] by [0, 60, 1] viewing rectangle. What does theshape of the graph indicate about the per capita costs ofMedicare for the U.S. population ages 65 and over withincreasing time?

78. Graph and in thesame[ 10, 1] by [ , 10, 1] viewing rectangle.Then usethe feature to trace along What happens at

Explain why this occurs.

79. Graph , and inthe same [–5, 5, 1] by [–5, 5, 1] viewing rectangle. If represents f and represents g, use the graph of to findthe domain of Then verify your observationalgebraically.


a. If and then and .

b. There can never be two functions f and g, wherefor which

c. If and then .d. If and then

81. Prove that if f and g are even functions, then fg is also aneven function.

82. Define two functions f and g so that .

83. Use the graphs given in Exercises 63–66 to create a graphthat shows the population, in billions, of less developedregions from 1950 through 2050.

Group Exercise84. Consult an almanac, newspaper, magazine, or the Internet

to find data displayed in a graph in the style of Figure 2.53on page 248. Using the two graphs that group membersfind most interesting, introduce two functions that arerelated to the graphs. Then write and solve a probleminvolving function subtraction for each selected graph. Ifyou are not sure where to begin, reread page 248–249 orlook at Exercises 63–66 in this exercise set.

f � g = g � f

(5) = g(2).(f � g)g(x) = 2x - 1, f(x) = 1x

(f � g) (4) = 35g(4) = 7f(7) = 5(f � g) (x) = (g � f) (x)f Z g,

(f � g) (5) = -25(x) = -x2(f � g)g(x) = 1x2

- 4,f(x) = x2- 4

f � gy3y2

y1

y3 = y22

- 4y2 = 24 - x2y1 = x2- 4,

x = 0?y3.TRACE

-10-10,y3 = y1 , y2y1 = x2

- 2x, y2 = x,

g

f

g(t) = 0.54t2+ 12.64t + 107.1

f(t) = -0.14t2+ 0.51t + 31.6


68. A department store has two locations in a city. From 1998through 2002, the profits for each of the store’s two branchesare modeled by the functions and

In each model, x represents thenumber of years after 1998 and f and g represent the profit,in millions of dollars.a. What is the slope of f? Describe what this means.b. What is the slope of g? Describe what this means.c. Find What is the slope of this function? What

does this mean?

69. The regular price of a computer is x dollars. Let and

a. Describe what the functions f and g model in terms ofthe price of the computer.

b. Find and describe what this models interms of the price of the computer.

c. Repeat part (b) for d. Which composite function models the greater

discount on the computer, or Explain.

70. The regular price of a pair of jeans is x dollars. Letand

a. Describe what functions f and g model in terms of theprice of the jeans.

b. Find and describe what this models interms of the price of the jeans.

c. Repeat part (b) for d. Which composite function models the greater

discount on the jeans, or Explain.

Writing in Mathematics

71. If equations for functions f and g are given, explain howto find

72. If the equations of two functions are given, explain howto obtain the quotient function and its domain.

73. If equations for functions f and g are given, describe twoways to find

74. Explain how to use the graphs in Figure 2.53 on page 248to estimate the per capita federal tax for any one of theyears shown on the horizontal axis.

75. Describe a procedure for finding . What is thename of this function?

76. Describe the values of x that must be excluded from thedomain of .(f � g) (x)

(f � g) (x)

(f - g) (3).

f + g.

g � f?f � g

(g � f) (x).

(f � g) (x)

g(x) = 0.6x.f(x) = x - 5

g � f?f � g

(g � f) (x).

(f � g) (x)

g(x) = 0.75x.x - 400f(x) =

f + g.

g(x) = 0.51x + 11.14.f(x) = -0.44x + 13.62

BLITMC02B.131013599_235-278 12/30/02 11:13 AM 9

In most societies, women say they prefer to marry men who are older thanthemselves, whereas men say they prefer women who are younger. Evolutionarypsychologists attribute these preferences to female concern with a partner'smaterial resources and male concern with a partner's fertility (Source: David M.Buss, Psychological Inquiry, 6, 1–30). When the man is considerably older than thewoman, people rarely comment. However, when the woman is older, as in therelationship between actors Susan Sarandon and Tim Robbins, people take notice.

Figure 2.56 shows the preferred age in a mate in five selected countries.We can focus on the data for the women and define a function.


Objectives1. Verify inverse functions.

2. Find the inverse of afunction.

3. Use the horizontal linetest to determine if afunction has an inversefunction.

4. Use the graph of a one-to-one function to graphits inverse function.

SECTION 2.7 Inverse Functions

Preferred Age in a Mate

Zambia Colombia

Yea

rs O

lder

Tha

n Se

lf

Pre

ferr

ed A

geY

ears

You

nger

Tha

n Se

lf

Poland Italy UnitedStates

Country

0

�1

�2

3

4

6

5

1

�4

�5

�8

�6

�7

�3

2

Men Women

4.2 4.5

3.3 3.32.5

Figure 2.56Source: Carole Wade and Carol Tavris, Psychology Sixth Edition, Prentice Hall, 2000

Let the domain of the function be the set of the five countries shown in the graph.Let the range be the set of the average number of years women in each of therespective countries prefer men who are older than themselves. The function canbe written as follows:

f:E(Zambia, 4.2), (Colombia, 4.5), (Poland, 3.3), (Italy, 3.3), (U.S., 2.5)F .

BLITMC02B.131013599_235-278 12/30/02 11:13 AM 0

Different second components

Same first component

Now let's “undo” f by interchanging the first and second components ineach of its ordered pairs. Switching the inputs and outputs of f, we obtain thefollowing relation:

Section 2.7 • Inverse Functions • 261

Can you see that this relation is not a function? Two of its ordered pairs havethe same first component and different second components. This violates thedefinition of a function.

If a function f is a set of ordered pairs, (x, y), then the changes produced byf can be “undone” by reversing the components of all the ordered pairs. Theresulting relation, (y, x), may or may not be a function. In this section, we willdevelop these ideas by studying functions whose compositions have a special“undoing” relationship.

Inverse FunctionsHere are two functions that describe situations related to the price of a computer,x:

Function f subtracts $300 from the computer’s price and function g adds $300to the computer’s price. Let’s see what f(g(x))does. Put g(x) into f:


Because replace with

=x.

Using f(g(x)), the computer’s price, x, went through two changes: thefirst, an increase; the second, a decrease:

The final price of the computer, x, is identical to its starting price, x.In general, if the changes made to x by function g are undone by the

changes made by function f, then

Assume, also, that this “undoing” takes place in the other direction:

Under these conditions, we say that each function is the inverse function of theother. The fact that g is the inverse of f is expressed by renaming g as read“f-inverse.” For example, the inverse functions

are usually named as follows:

f(x) = x - 300 f-1(x) = x + 300.

f(x) = x - 300 g(x) = x + 300

f-1,

g(f(x)) = x.

f(g(x)) = x.

x + 300 - 300.

This is the computer’soriginal price.

x + 300.g(x)g(x) = x + 300,= x + 300 - 300

f(g(x)) = g(x) - 300


f(x) = x - 300

g(x) = x + 300.f(x) = x - 300

Undoing f:E(4.2, Zambia), (4.5, Colombia), (3.3, Poland), (3.3, Italy), (2.5, U.S.)F .

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 261


With these ideas in mind, we present the formal definition of the inverseof a function:

Definition of the Inverse of a Function

Let f and g be two functions such that

for every x in the domain of g

and

for every x in the domain of f.

The function g is the inverse of the function f, and is denoted by (read“f-inverse”). Thus, and The domain of f isequal to the range of and vice versa.

EXAMPLE 1 Verifying Inverse Functions

Show that each function is an inverse of the other:

Solution To show that f and g are inverses of each other, we must show thatand We begin with f Ag(x) B .


Next, we find g Af(x) B .This is the given equation for g.

Because g is the inverse of f (and vice versa), we can use inverse notation andwrite

Notice how undoes the change produced by f: f changes x by multiplying by5 and undoes this by dividing by 5.


f(x) = 7x and g(x) =

x

7.

CheckPoint

1

f-1f-1

f(x) = 5x and f-1(x) =

x

5.

g(f(x)) =

f(x)

5=

5x

5= x

g(x) =

x

5

f(g(x)) = 5g(x) = 5 a x

5b = x

f(x) = 5x

g(f(x)) = x.f(g(x)) = x

f(x) = 5x and g(x) =

x

5.

f-1,f-1(f(x)) = x.f(f-1(x)) = x

f-1

g(f(x)) = x

f(g(x)) = x


Replace x with f(x).

The following partial tables of coordinates numericallyillustrate that inverse functionsreverse each other'scoordinates.

Using the first two tables, thefollowing table shows howinverse functions undo oneanother.

Study Tip

x –5 –2 0 3–25 –10 0 15f(x) = 5x

x –5 –2 0 3

=3=0=-2=-5

=g(15)=g(0)=g(-10)=g(-25)

g(f(3))g(f(0))g(f(-2))g(f(-5))g(f(x))

x –25 –10 0 15

–5 –2 0 3g(x) =

x

5

1 Verify inverse functions.

The notation representsthe inverse function of f. The–1 is not an exponent. The

notation does not mean

f-1Z

1f

.

1f

:f - 1

f - 1

Study Tip

Each output of the compositefunction is identical to the input.

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 262

EXAMPLE 2 Verifying Inverse Functions


Solution To show that f and g are inverses of each other, we must show thatand We begin with f Ag(x) B .

This is the equation for f.

Next, we find g Af(x) B .This is the equation for g.

Because g is the inverse of f (and vice versa), we can use inverse notation and write

Notice how undoes the changes produced by f: f changes x by multiplyingby 3 and adding 2, and undoes this by subtracting 2 and dividing by 3. This“undoing” process is illustrated in Figure 2.57.


Finding the Inverse of a FunctionThe definition of the inverse of a function tells us that the domain of f is equal tothe range of and vice versa. This means that if the function f is the set ofordered pairs (x, y), then the inverse of f is the set of ordered pairs (y, x). If afunction is defined by an equation, we can obtain the equation for the inverseof f, by interchanging the role of x and y in the equation for the function f.

Finding the Inverse of a FunctionThe equation for the inverse of a function f can be found as follows:

1. Replace f(x) with y in the equation for f(x).2. Interchange x and y.3. Solve for y. If this equation does not define y as a function of x, the function

f does not have an inverse function and this procedure ends. If this equationdoes define y as a function of x, the function f has an inverse function.

4. If f has an inverse function, replace y in step 3 by We can verifyour result by showing that and f–1(f(x)) = x.f(f-1(x)) = x

f-1(x).

f-1,

f-1,

f(x) = 4x - 7 and g(x) =

x + 74

.

CheckPoint

2

f-1f-1

f(x) = 3x + 2 and f-1(x) =

x - 23

.

gAf(x)B =

f(x) - 2

3=

(3x + 2) - 2

3=

3x

3= x

g(x) =

x - 23

f(g(x)) = 3g(x) + 2 = 3 a x - 23

b + 2 = x - 2 + 2 = x

f(x) = 3x + 2

g(f(x)) = x.f(g(x)) = x

g(x) =

x - 23

.f(x) = 3x + 2 and



Replace x with f (x).

x − 23

x − 2 x

3x 3x + 2x

f −1

f

Multiply by 3.Add 2.

Subtract 2.Divide by 3.

Figure 2.57 undoes the changesproduced by f.

f-1

The procedure for finding afunction’s inverse uses a switch-and-solve strategy. Switch x andy, then solve for y.

Study Tip

2 Find the inverse of afunction.

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 263

EXAMPLE 3 Finding the Inverse of a Function

Find the inverse of

Solution

Step 1 Replace f(x) with y :

Step 2 Interchange x and y :

This is the inverse function.

Step 3 Solve for y :

Add 5 to both sides.

Divide both sides by 7.

Step 4 Replace y with

The equation is written with on the left.

Thus, the inverse of is

The inverse function, undoes the changes produced by f. f changes xby multiplying by 7 and subtracting 5. undoes this by adding 5 and dividingby 7.

Find the inverse of f(x)=2x+7.

EXAMPLE 4 Finding the Equation of the Inverse

Find the inverse of

SolutionStep 1 Replace f(x) with y :


Step 3 Solve for y :

Step 4 Replace y with

Thus, the inverse of

Find the inverse of f(x) = 4x3- 1.

CheckPoint

4

f(x) = x3+ 1 is f-1(x) = 13 x - 1 .

f-1(x) = 13 x - 1 .f -1(x):

13 x - 1 = y

13 x - 1 = 23 y3

x - 1 = y3

x = y3+ 1

y = x3+ 1

f(x) = x3+ 1.

CheckPoint

3

f-1f-1,

f-1(x) =

x + 57

.f(x) = 7x - 5

f –1f-1(x) =

x + 57

f -1(x):

x + 57

= y

x + 5 = 7y

x = 7y - 5

y = 7x - 5

f(x) = 7x - 5.


In Example 3, we found thatif then

Verify this result by showingthat

and

f-1(f(x)) = x.

f(f-1(x)) = x

f-1(x) =

x + 57

.

f(x) = 7x - 5,

Discovery

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 264

The Horizontal Line Test and One-to-One FunctionsLet’s see what happens if we try to find the inverse of the standard quadraticfunction,

Step 1 Replace f(x) with y :


Step 3 Solve for y : We apply the square root method to solve for y.

We obtain

The ; in this last equation shows that for certain values of x (all positive realnumbers), there are two values of y. Because this equation does not represent y asa function of x, the standard quadratic function does not have an inverse function.

Can we look at the graph of a function and tell if it represents a function withan inverse? Yes. The graph of the standard quadratic function is shown inFigure 2.58. Four units above the x-axis, a horizontal line is drawn. This lineintersects the graph at two of its points, and (2, 4). Because inversefunctions have ordered pairs with the coordinates reversed, let’s see what happensif we reverse these coordinates. We obtain and (4, 2). A function providesexactly one output for each input. However, the input 4 is associated with twooutputs, and 2.The points and (4, 2) do not define a function.

If any horizontal line, such as the one in Figure 2.58, intersects a graph at twoor more points, these points will not define a function when their coordinates arereversed.This suggests the horizontal line test for inverse functions:

The Horizontal Line Test For Inverse FunctionsA function f has an inverse that is a function, if there is no horizontalline that intersects the graph of the function f at more than one point.

EXAMPLE 5 Applying the Horizontal Line Test

Which of the following graphs represent functions that have inverse functions?

Solution Can you see that horizontal lines can be drawn in parts (b) and (c) thatintersect the graphs more than once? This is illustrated in the figure at the top ofthe next page.These graphs do not pass the horizontal line test.The graphs in parts(b) and (c) are not the graphs of functions with inverse functions. By contrast, nohorizontal line can be drawn in parts (a) and (d) that intersect the graphs morethan once. These graphs pass the horizontal line test. Thus, the graphs inparts (a) and (d) represent functions that have inverse functions.

y

x

y

x

y

x

y

x

(a) (b) (c) (d)

f-1,

(4, -2)-2

(4, -2)

(-2, 4)

y = ; 1x .

y2= x

x = y2.

y = x2.

f(x) = x2.


3 Use the horizontal line testto determine if a functionhas an inverse function.

−1

1234

8765

−2

1 2 3 4 5−1−2−3−4−5

y

x

f(x) = x2

(2, 4)(−2, 4)

Figure 2.58 The horizontal lineintersects the graph twice

How might you restrict thedomain of graphed in Figure 2.58, so thatthe remaining portion of thegraph passes the horizontalline test?

f(x) = x2,

Discovery

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 265

Has an inverse No inverse No inverse Has an inverse function function function function

Which of the following graphs represent functions that have inverse functions?

A function passes the horizontal line test when no two different orderedpairs have the same second component. This means that if then

Such a function is called a one-to-one function. Thus, a one-to-one function is a function in which no two different ordered pairs have the samesecond component. Only one-to-one functions have inverse functions. Anyfunction that passes the horizontal line test is a one-to-one function. Any one-to-one function has a graph that passes the horizontal line test.

Graphs of f and There is a relationship between the graph of a one-to-one function, f, and itsinverse, Because inverse functions have ordered pairs with the coordinatesreversed, if the point (a, b) is on the graph of f, then the point (b, a) is on thegraph of The points (a, b) and (b, a) are symmetric with respect to the line

Thus, the graph of is a reflection of the graph of f about the lineThis is illustrated in Figure 2.59.y = x.

f -1y = x.f-1.

f-1.

f -1

f(x1) Z f(x2).x1 Z x2,

y

x

y

x

y

x

(a) (b) (c)

CheckPoint

5

y

x

y

x

y

x

y

x

(a) (b) (c) (d)


Graph of f

Graph of f−1

y = x

y

x

(a, b)

(b, a)Figure 2.59 The graph of is areflection of the graph of f abouty = x.

f-1

4 Use the graph of a one-to-one function to graph itsinverse function.

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 266


−1

12345

−3

−5

1 2 3 4 5−1−2−4−5

y = x

(−3, −2)

(−1, 0)

(−2, −3)

(0, �1)

(4, 2)

(2, 4)

x

y

y = f(x)

y = f−1(x)

Figure 2.61 The graphs of f and f-1

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = f(x) (4, 2)

(−3, −2)

(−1, 0)

Figure 2.60

EXAMPLE 6 Graphing the Inverse Function

Use the graph of f in Figure 2.60 to draw the graph of its inverse function.

Solution We begin by noting that no horizontal line intersects the graph of fat more than one point, so f does have an inverse function. Because the points

and (4, 2) are on the graph of f, the graph of the inversefunction, has points with these ordered pairs reversed.Thus,and (2, 4) are on the graph of We can use these points to graph Thegraph of is shown in Figure 2.61. Note that the graph of is the reflectionof the graph of f about the line

Use the graph of f in the figure below to draw the graph of itsinverse function.

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5(1, −1)

(0, 0)(−2, 1)

y = x

x

y

CheckPoint

6

y = x.f-1f-1

f-1.f-1.(-2, -3), (0, -1),f-1,

(-3, -2), (-1, 0),

EXERCISE SET 2.7

Practice ExercisesIn Exercises 1–10, find and and determine whether each pair of functions f and g are

inverses of each other.

1. and

2. and

3. and

4. and

5. and

6. and

7. and

8. and

9. and

10. and g(x) = x3+ 4f(x) = 23 x - 4

g(x) = -xf(x) = -x

g(x) =

2x

+ 5f(x) =

2x - 5

g(x) =

3x

+ 4f(x) =

3x - 4

g(x) =

x + 37

f(x) = 3x - 7

g(x) =

x + 59

f(x) = 5x - 9

g(x) =

x - 94

f(x) = 4x + 9

g(x) =

x - 83

f(x) = 3x + 8

g(x) =

x

6f(x) = 6x

g(x) =

x

4f(x) = 4x

g(f(x))f(g(x))

The functions in Exercises 11–30 are all one-to-one. For eachfunction:

a. Find an equation for the inverse function.b. Verify that your equation is correct by showing that

and

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. for

26. for

27. 28.

29. 30. f(x) = x3�5f(x) = 23 x - 4 + 3

f(x) =

2x - 3x + 1

f(x) =

2x + 1x - 3

x � 0f(x) = x2- 1,

x � 0f(x) = x2+ 1,

f(x) = 23 xf(x) = 2x

f(x) =

2x

f(x) =

1x

f(x) = (x - 1)3f(x) = (x + 2)3

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

+ 2

f(x) = 3x - 1f(x) = 2x + 3

f(x) = 4xf(x) = 2x

f(x) = x + 5f(x) = x + 3

f-1(f(x)) = x.f(f-1(x)) = x

f-1(x),

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 267

Which graphs in Exercises 31–36 represent functions thathave inverse functions?31. 32.

33. 34.

35. 36.

In Exercises 37–40, use the graph of f to draw the graph of itsinverse function.

37.

38.

−1

1234

−2−3−4

1 2 3 54−1−2−3

y

x

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

y

x

y

x

y

x

y

x

y

x

y

x


39.

40.

Application Exercises

41. Refer of Figure 2.56 on page 260. Recall that the bargraphs in the figure show the preferred age in a mate infive selected countries.a. Consider a function, f,whose domain is the set of the five

countries shown in the graph. Let the range be the set ofthe average number of years men in each of therespective countries prefer women who are youngerthan themselves. (You will need to use the graph toestimate these values. Assume that the bars for Polandand Italy have the same length. Round to the nearesttenth of a year.) Write function f as a set of ordered pairs.

b. Write the relation that is the inverse of f as a set ofordered pairs. Is this relation a function? Explainyour answer.

42. The bar graph shows the percentage of land owned bythe federal government in western states in which thegovernment owns at least half of the land.

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

Percentage of Land Owned bythe Federal Government

Nevada

Utah

Idaho

Alaska

Oregon

Wyoming 50%

52%

62%

62%

65%

83%

Source: Bureau of Land Management

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 268

a. Consider a function, f, whose domain is the set of sixstates shown. Let the range be the percentage of landowned by the federal government in each of therespective states. Write the function f as a set ofordered pairs.

b. Write the relation that is the inverse of f as a set ofordered pairs. Is this relation a function? Explainyour answer.

43. The graph represents the probability of two people in thesame room sharing a birthday as a function of the numberof people in the room. Call the function f.

a. Explain why f has an inverse that is a function.b. Describe in practical terms the meaning of

and

44. The graph shows the average age at which women in theUnited States marry for the first time over a 110-year period.

Source: U.S. Census Bureau

a. Does this graph have an inverse that is a function?What does this mean about the average age at whichU.S. women marry during the period shown?

b. Identify two or more years in which U.S. womenmarried for the first time at the same average age.What is a reasonable estimate of this average age?

24

Age

at F

irst

Mar

riag

e

23

25

26

22

21

20

1900 19701890 19201910 19401930

Year

19601950 200019901980

Average Age at First Marriage for U.S. Women

f-1(0.7).f-1(0.5), f-1(0.25),

Number of Persons0 10 20 30 40 50 60

y

x0.10.20.30.40.50.60.70.80.91.0

Pro

babi

lity

of T

wo

Peo

ple

Shar

ing

the

Sam

e B

irth

day

y = f(x)

45. The formula

is used to convert from x degrees Celsius to y degreesFahrenheit. The formula

is used to convert from x degrees Fahrenheit to y degreesCelsius. Show that f and g are inverse functions.

46. One yardstick for measuring how steadily—if slowly—athletic performance improved is the mile run. In 1923,the record for the mile was a comparatively sleepy 4minutes, 10.4 seconds. In 1954, Roger Bannister of Britaincracked the 4-minute mark, coming in at 3 minutes, 59.4seconds. In the half-century since, about 0.3 second peryear has been shaved off Bannister’s record.

Source: U.S.A. Track and Field

a. Consider the following information:

• In 1954, the record was 3 minutes, 59.4 seconds, or239.4 seconds.

• The record has decreased by 0.3 second per yearsince then.

Use this information to write a function, f, that modelsthe mile record, f(x), in seconds, x years after 1954.

b. Find the inverse of the mile-record function. Describewhat each variable in the inverse function represents.

c. According to the inverse model, how many years after1954 will someone run a 3-minute, or 180-second, mile?In which year will this occur?

Writing in Mathematics

47. Explain how to determine if two functions are inversesof each other.

48. Describe how to find the inverse of a one-to-onefunction.

49. What is the horizontal line test and what does itindicate?

50. Describe how to use the graph of a one-to-one functionto draw the graph of its inverse function.

51. How can a graphing utility be used to visually determineif two functions are inverses of each other?

y = g(x) =

59

(x - 32)

y = f(x) =

95

x + 32


Mile Records

1886 4:12.3 1958 3:54.51923 4:10.4 1966 3:51.31933 4:07.6 1979 3:48.91945 4:01.3 1985 3:46.31954 3:59.4 1999 3:43.1

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 269

Technology ExercisesIn Exercises 52–60, use a graphing utility to graph the function.Use the graph to determine whether the function has an inversethat is a function (that is, whether the function is one-to-one).

52. 53.

54. 55.

56. 57.58. 59.60.

In Exercises 61–63, use a graphing utility to graph f and g inthe same viewing rectangle. In addition, graph the line and visually determine if f and g are inverses.

61.

62.

63.


a. The inverse of {(1, 4), (2, 7)} is {(2, 7), (1, 4)}.

b. The function is one-to-one.

c. If then

d. The domain of f is the same as the range of f-1.

f-1(x) =

13x

.f(x) = 3x,

f(x) = 5

f(x) = 23 x - 2, g(x) = (x + 2)3

f(x) =

1x

+ 2, g(x) =

1x - 2

g(x) = 0.25x - 1f(x) = 4x + 4,

y = x

f(x) = x3+ x + 1

f(x) = -216 - x2f(x) = (x - 1)3

f(x) = ∑x - 2∑f(x) = int(x - 2)

f(x) =

x4

4f(x) =

x3

2

f(x) = 23 2 - xf(x) = x2- 1

65. If and find and

66. Show that

is its own inverse.

67. Freedom 7 was the spacecraft that carried the firstAmerican into space in 1961. Total flight time was 15minutes, and the spacecraft reached a maximum height of116 miles. Consider a function, s, that expresses Freedom7’s height, s(t), in miles, after t minutes. Is s a one-to-onefunction? Explain your answer.

68. If find x satisfying

Group Exercise

69. In Tom Stoppard’s play Arcadia, the characters dreamand talk about mathematics, including ideas involvinggraphing, composite functions, symmetry, and lack ofsymmetry in things that are tangled, mysterious, andunpredictable. Group members should read the play.Present a report on the ideas discussed by the charactersthat are related to concepts that we studied in this chapter.Bring in a copy of the play and read appropriate excerpts.

8 + f-1(x - 1) = 10.f(2) = 6,

f(x) =

3x - 25x - 3

(g-1 � f-1) (x).(f � g)-1(x)g(x) = x + 5, f(x) = 3x


DEFINITIONS AND CONCEPTS EXAMPLES

2.1 Lines and Slopea. The slope, m, of the line through and is Ex. 1, p.177

b. Equations of lines include point-slope form, slope-intercept form, Ex. 2 & 3, p.179;and general form, The equation of a horizontal line is Ex. 5 & 6, p.182

a vertical line is c. Parallel lines have equal slopes. Perpendicular lines have slopes that are negative reciprocals. Ex. 8 & 9, p.184–185

2.2 Distance and Midpoint Formulas; Circlesa. The distance, d, between the points and is given by Ex. 1, p.194

b. The midpoint of the line segment whose endpoints are and is the point with Ex. 2, p.195

coordinates a x1 + x2

2,

y1 + y2

2b .

(x2,y2)(x1,y1)

d = 2(x2 - x1)2

+ (y2 - y1)2.

(x2, y2)(x1, y1)

x = a.y = b;Ax + By + C = 0.y = mx + b,

y - y1 = m(x - x1),

m =

y2 - y1

x2 - x1 .

(x2, y2)(x1, y1)

CHAPTER SUMMARY, REVIEW, AND TEST

Summary

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 270

c. The standard form of the equation of a circle with center (h, k) and radius r is Ex. 3, p. 196

Ex. 4 & 5, p. 197d. The general form of the equation of a circle is e. To convert from the general form to the standard form of a circle’s equation, complete the square Ex. 6, p. 198

on x and y.

2.3 Basics of Functionsa. A relation is any set of ordered pairs. The set of first components is the domain and the set of Ex. 1, p. 202

second components is the range.b. A function is a correspondence from a first set, called the domain, to a second set, called the range, Ex. 2, p. 203

such that each element in the domain corresponds to exactly one element in the range. If any element in a relation’s domain corresponds to more than one element in the range, the relation is not a function.

c. Functions are usually given in terms of equations involving x and y, in which x is the independent Ex. 3, p. 205variable and y is the dependent variable. If an equation is solved for y and more than one value of y can be obtained for a given x, then the equation does not define y as a function of x. If an equation defines a function the value of the function at x, f(x),often replaces y.

d. The difference quotient is

Ex. 5, p. 207

e. If a function f does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of f(x) is a real number. Exclude from the function’s domain real Ex. 7, p. 210numbers that cause division by zero and real numbers that result in an even root of a negative number.

2.4 Graphs of Functionsa. The graph of a function is the graph of its ordered pairs. Ex. 1, p. 215b. The vertical line test for functions: If any vertical line intersects a graph in more than one point, Ex. 3, p. 217

the graph does not define y as a function of x.c. A function is increasing on intervals where its graph rises, decreasing on intervals where it falls, and Ex. 5, p. 220

constant on intervals where it neither rises nor falls. Precise definitions are given in the box on page 220.d. If the graph of a function is given, we can often visually locate the number(s) at which the function Fig 2.32, p. 222

has a relative maximum or relative minimum. Precise definitions are given in the box on page 221.e. The average rate of change of f from to is Ex. 6, p. 222

f. The graph of an even function in which is symmetric with respect to the y-axis. Ex. 7, p. 224The graph of an odd function in which is symmetric with respect to the origin.

g. The graph of where int(x) is the greatest integer that is less than or equal to x,has function values that form discontinuous steps, shown in Figure 2.39 on page 227. If where n is an integer, then int

2.5 Transformations of Functionsa. Table 2.4 on pages 235–236 shows the graphs of the constant function, the identity function,

the standard quadratic function, the standard cubic function,the square root function, and the absolute value function, The table also lists characteristics of each function.

b. Table 2.5 on page 243 summarizes how to graph a function using vertical shifts, Ex. 1 & 2, p. 237–238;horizontal shifts, reflections about the x-axis, reflections about the Ex. 3, p. 239;y-axis, y=f(–x), vertical stretching, and vertical shrinking, Ex. 4–7, p. 241–243

c. A function involving more than one transformation can be graphed in the following order: Ex. 8, p. 244(1) horizontal shifting; (2) vertical stretching or shrinking; (3) reflecting; (4) vertical shifting.

y = cf(x), 0 6 c 6 1.c 7 1,y = cf(x),y = -f(x),y = f(x ; c),

y = f(x) ; c,

f(x) = �x�.f(x) = 1x ,f(x) = x3,f(x) = x2,f(x) = x,

f(x) = c,

(x) = n.n � x 6 n + 1,

f(x) = int (x),

f(-x) = -f(x)f(-x) = f(x)

f(x2) - f(x1)

x2 - x1.

x2x1

f(x + h) - f(x)

h, h Z 0.

x2+ y2

+ Dx + Ey + F = 0.(x - h)2

+ (y - k)2= r2.

Chapter Summary, Review, and Test • 271

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 271

2.6 Combinations of Functions; Composite and Inverse Functions

a. When functions are given as equations, they can be added, subtracted, multiplied, or divided by Ex. 1, p. 250;performing operations with the algebraic expressions that appear on the right side of the equations. Ex. 2, p. 250;

Definitions for the sum f + g, the difference f – g, the product f g, and the quotient functions are Ex. 3, p. 252given in the box on page 251.

b. The composition of functions f and g, is defined by The domain of the Ex. 4, p. 254;composite function f � g is given in the box on page 256.This composite function is obtained by Ex. 5, p. 255replacing each occurrence of x in the equation for f with g(x).

2.7 Inverse Functionsa. If and function g is the inverse of function f, denoted and read “f Ex. 1, p. 262;

inverse.”Thus, to show that f and g are inverses of each other, one must show and Ex. 2, p. 263

b. The procedure for finding a function’s inverse uses a switch-and-solve strategy. Switch x and y, then Ex. 3 & 4, p. 264solve for y.The procedure is given in the box on page 263.

c. The horizontal line test for inverse functions: A function f has an inverse that is a function, if Ex. 5, p. 265there is no horizontal line that intersects the graph of the function f at more than one point.

d. A one-to-one function is one in which no two different ordered pairs have the same second component. Only one-to-one functions have inverse functions.

e. If the point (a, b) is on the graph of f, then the point (b, a) is on the graph of The graph of is a Ex. 6, p. 267reflection of the graph of f about the line y = x.

f-1f-1.

f-1,

g(f(x)) = x.f(g(x)) = x

f-1g(f(x)) = x, f(g(x)) = x

(f � g)(x) = f(g(x)).f � g,

f

g


Review Exercises

2.1In Exercises 1–4, find the slope of the line passing througheach pair of points or state that the slope is undefined. Thenindicate whether the line through the points rises, falls, ishorizontal, or is vertical.

1. (3, 2) and (5, 1) 2. and3. and 4. and

In Exercises 5–6, use the given conditions to write an equationfor each line in point-slope form and slope-intercept form.

5. Passing through with slope 6. Passing through (1, 6) and

In Exercises 7–10, give the slope and y-intercept of each linewhose equation is given. Then graph the line.

7. 8.9. 10.

11. Corporations in the United States are doing quite well, thankyou. The scatter plot in the next column shows corporateprofits, in billions of dollars, from 1990 through 2000. Alsoshown is a line that passes through or near the points.

2y - 8 = 02x + 3y + 6 = 0y = -4x + 5y =

25 x - 1

(-1, 2)-6(-3, 2)

(-2, 10)(-2, 5)A6, 14BA-3, 14B(-3, -4)(-1, -2)

Source: U.S Department of Labor

a. Use the two points whose coordinates are shown bythe voice balloons to find the point-slope equation ofthe line that models corporate profits, y, in billions ofdollars, x years after 1990.

b. Write the equation in part (a) in slope-intercept form.c. Use the linear model to predict corporate profits in 2010.

12. The scatter plot on the next page shows the number ofminutes each that 16 people exercise per week and thenumber of headaches per month each person experiences.

y

x

Years after 1990121086420

$1000

800

600

400

200

(2, 480)

(8, 720)

Pro

fits

(bill

ions

)

Corporate Profits

BLITMC02B.131013599_235-278 12/30/02 11:13 AM Page 272

a. Draw a line that fits the data so that the spread of thedata points around the line is as small as possible.

b. Use the coordinates of two points along your line towrite its point-slope and slope-intercept equations.

c. Use the equation in part (b) to predict the numberof headaches per month for a person exercising130 minutes per week.


13. Passing through and parallel to the line whoseequation is

14. Passing through and perpendicular to the linewhose equation is

2.2In Exercises 15–16, find the distance between each pair ofpoints. If necessary, round answers to two decimal places.

15. 16.

In Exercises 17–18, find the midpoint of each line segmentwith the given endpoints.

17. (2, 6) and 18. and

In Exercises 19–20, write the standard form of the equationof the circle with the given center and radius.

19. 20. Center

In Exercises 21–23, give the center and radius of each circleand graph its equation.

21. 22.

23.

2.3In Exercises 24–26, determine whether each relation is afunction. Give the domain and range for each relation.

24. E(2, 7), (3, 7), (5, 7)F 25.

26. E(12, 13), (14, 15), (12, 19)FIn Exercises 27–29, determine whether each equation definesy as a function of x.

27. 28.

29. 2x + y2= 6

3x2+ y = 142x + y = 8

{(1, 10) (2, 500), (13, �)}

x2+ y2

- 4x + 2y - 4 = 0

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

= 9x2+ y2

= 1

(-2, 4), r = 6Center (0, 0), r = 3

(-15, 2)(4, -6)(-12, 4)

(-4, 3) and (-2, 5)(-2, -3) and (3, 9)

y =13 x + 4

(-3, 6)

3x + y - 9 = 0(4, -7)

9876543210

20 40 60

Minutes per Week Spent Exercising

Num

ber

of H

eada

ches

per

Mon

th

80 100 120

y

x

In Exercises 30–33, evaluate each function at the given valuesof the independent variable and simplify.

30.a. f(4) b. c.

31.a. g(0) b.c. d.

32.

a. g(13) b. g(0) c.

33.

a. f(–2) b. f(1) c. f(2)

In Exercises 34–35, find and simplify the difference quotient

for the given function.

34. 35.

In Exercises 36–40, find the domain of each function.

36. 37.

38. 39.

40.

2.4Graph the functions in Exercises 41–42. Use the integer valuesof x given to the right of the function to obtain the ordered pairs.Use the graph to specify the function’s domain and range.

41.

42.

In Exercises 43–45, use the graph to determine a. thefunction’s domain; b. the function’s range; c. the x-intercepts,if any; d. the y-intercept, if any; e. intervals on which thefunction is increasing, decreasing, or constant; and f. thefunction values indicated below the graphs.

43.

f(-2) = ? f(3) = ?

−1

123

−2−3−4−5

1 2 3 4 5−1−2−3

y

x

y = f(x)

x = -1, 0, 1, 2, 3, 4f(x) = �2 - x�

x = -1, 0, 1, 2, 3, 4f(x) = x2- 4x + 4

g(x) =

1x - 2x - 5

f(x) =

x

x2- 1

h(x) = 18 - 2x

g(x) =

4x - 7

f(x) = x2+ 6x - 3

f(x) = x2- 13x + 5f(x) = 8x - 11

f(x + h) - f(x)

h, h Z 0

f(x) = c x2- 1

x - 112

if x Z 1

if x = 1

g(-3)

g(x) = b1x - 44 - x

if x � 4if x 6 4

g(-x)g(x - 1)g(-2)

g(x) = 3x2- 5x + 2

f(-x)f(x + 3)f(x) = 5 - 7x


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44.

45.

In Exercises 46–47, find:

a. The numbers, if any, at which f has a relative maximum.What are these relative maxima?

b. The numbers, if any, at which f has a relative minimum.What are these relative minima?

46. Use the graph in Exercise 43.

47. Use the graph in Exercise 44.

In Exercises 48–51, use the vertical line test to identify graphsin which y is a function of x.

48. 49.

50. 51.

52. Find the average rate of change of fromx1 = 5 to x2 = 9.

f(x) = x2- 4x

y

x

y

x

y

x

y

x

f(-9) = ? f(14) = ?

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

y = f(x)

f(-2) = ? f(6) = ?

−1

123

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x−1

123

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = f(x)

53. The graph shows annual spending per uniformed memberof the U.S. military in inflation-adjusted dollars. Find theaverage rate of change of spending per year from 1955through 2000. Round to the nearest dollar per year.

Source: Center for Strategic and Budgetary Assessments

In Exercises 54–56, determine whether each function is even,odd, or neither. State each function’s symmetry. If you areusing a graphing utility, graph the function and verify itspossible symmetry.

54. 55.

56.

57. The graph shows the height, in meters, of a vulture interms of its time, in seconds, in flight.

a. Is the vulture’s height a function of time? Use thegraph to explain why or why not.

b. On which interval is the function decreasing?Describe what this means in practical terms.

c. On which intervals is the function constant? Whatdoes this mean for each of these intervals?

d. On which interval is the function increasing? Whatdoes this mean?

Time (seconds)5 10 15 20 3025

y

x

Hei

ght (

met

ers)

10

20

30

50

40

f(x) = 2x21 - x2

f(x) = x4- 2x2

+ 1f(x) = x3- 5x

Spending per Uniformed Memberof the U.S. Military

Year2000199519901985198019751970196519601955

10,000

20,000

30,000

40,000

50,000

60,000

70,000

$80,000

(2000, $74,300)

(1955, $21,800)

Cold War

VietnamWar

Persian GulfWar

Ann

ual S

pend

ing

per

Mem

ber

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58. A cargo service charges a flat fee of $5 plus $1.50 for eachpound or fraction of a pound. Graph shipping cost, C(x), indollars, as a function of weight, x, in pounds, for

2.5In Exercises 59–61, begin by graphing the standard quadraticfunction, Then use transformations of this graphto graph the given function.

59. 60.61.

In Exercises 62–64, begin by graphing the square rootfunction, Then use transformations of thisgraph to graph the given function.

62. 63.64.

In Exercises 65–67, begin by graphing the absolute valuefunction, Then use transformations of this graphto graph the given function.

65. 66.67.

In Exercises 68–70, begin by graphing the standard cubicfunction, Then use transformations of this graphto graph the given function.

68. 69.70.

In Exercises 71–73, use the graph of the function f to sketchthe graph of the given function g.

71. 72.73.

2.6In Exercises 74–76, find and Determinethe domain for each function.

74.75.76.

In Exercises 77–78, find a. b.c. (f � g)(3).

(g � f)(x);(f � g)(x);

f(x) = 1x + 7 , g(x) = 1x - 2g(x) = x2

- 1f(x) = x2+ x + 1,

g(x) = x - 5f(x) = 3x - 1,

fg .f + g, f - g, fg,

g(x) = -2 + 2f(x + 2)

g(x) =12 f(x - 1)g(x) = f(x + 2) + 3

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

y = f(x)

r(x) =14 x3

- 1h(x) = -(x + 1)3g(x) =

12 (x - 1)3

f(x) = x3.

r(x) =12 ∑x + 2∑

h(x) = -∑x - 1∑ + 1g(x) = ∑x + 2∑ - 3

f(x) = ∑x∑.

r(x) = 21x + 2h(x) = 13 - xg(x) = 1x + 3

f(x) = 1x .

r(x) = -(x + 1)2

h(x) = (x + 2)2g(x) = x2+ 2

f(x) = x2.

0 6 x � 5.

77.78.

In Exercises 79–80, find a. b. the domain of

79.

80.

In Exercises 81–82, express the given function h as acomposition of two functions f and g so that

81. 82.

2.7In Exercises 83–84, find f Ag(x)B and g Af(x)B anddetermine whether each pair of functions f and g areinverses of each other.

83. and

84. and

The functions in Exercises 85–87 are all one-to-one. For eachfunction:

a. Find an equation for the inverse function.b. Verify that your equation is correct by showing that

and

85. 86.

87.

Which graphs in Exercises 88–91 represent functions thathave inverse functions?

88.

89.

−1

1234

−2−3−4

1 2 3 4 5−1−2−3−4−5

y

x

−1

12345

−2−3

1 2 3 4−1−2−3−4

y

x

f(x) = 8x3+ 1

f(x) = 1x + 2f(x) = 4x - 3

f-1(f(x)) = x.f(f-1(x)) = x

f-1(x),

g(x) =

2 - x

5f(x) = 2 - 5x

g(x) =

53

x - 2f(x) =

35

x +

12

h(x) =327x + 4h(x) = (x2

+ 2x - 1)4

h(x) = (f � g)(x).

f(x) = 1x - 1 , g(x) = x + 3

f(x) =

x + 1x - 2

, g(x) =

1x

(f � g).(f � g)(x);

f(x) = 1x , g(x) = x + 1g(x) = 4x - 1f(x) = x2

+ 3,

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Chapter 2 Test


1. Passing through (2, 1) and

2. Passing through and perpendicular to the linewhose equation is

3. Strong demand plus higher fuel and labor costs aredriving up the price of flying. The graph shows thenational averages for one-way fares. Also shown is a linethat models that data.

Source: American Express

a. Use the two points whose coordinates are shown bythe voice balloons to write the slope-interceptequation of the line that models the average one-wayfare, y, in dollars, x years after 1995.

b. According to the model, what will the nationalaverage for one-way fares be in 2008?

Fare

(do

llars

)

National Averages forOne-Way Airline Fares:

Business Travel

Years after 1995543210

300

400

500

$600

(0, 320)

(5, 530)

y = - 14 x + 5

(-4, 6)

(-1, -8)

4. Give the center and radius of the circle whose equationis and graph the equation.

5. List by letter all relations that are not functions.a. E (7, 5), (8, 5), (9, 5) Fb. E (5, 7), (5, 8), (5, 9) Fc.

d.

e.

6. If find and simplify.f(x - 1)f(x) = x2- 2x + 5,

y

x�5 5

x2+ y2

= 100

y

x

x2+ y2

+ 4x - 6y - 3 = 0

90.

91.

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

−1

1234

−2−3−4

1 2 3 4 5−1−2−3−4−5

y

x

92. Use the graph of f in the figure shown to draw the graphof its inverse function.

−1

1234

−2−3−4

1 2 3 4−1−2−3−4

y

x

y = f(x)

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7. If find and g(7).

8. If find the domain of f.

9. If find and simplify the difference

quotient

10. Use the graph of function f to answer the following questions.

a. What is b. What is the domain of f?c. What is the range of f?d. On which interval or intervals is f increasing?e. On which interval or intervals is f decreasing?f. For what number does f have a relative maximum?

What is the relative maximum?g. For what number does f have a relative minimum?

What is the relative minimum?h. What are the x-intercepts?i. What is the y-intercept?

11. Find the average rate of change of from

12. Determine whether is even, odd, orneither. Use your answer to explain why the graph in thefigure shown cannot be the graph of f.

13. The figure at the top of the next column shows how thegraph of is obtained from the graphof Describe this process, using the graph of gin your description.

f(x) = x2.h(x) = -2(x - 3)2

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

f(x) = x4- x2

x1 = 6 to x2 = 10.f(x) = 3x2

- 5

f(4) - f(-3)?

−1

12345

−2−3−4−5

1 2 3 4 5−1−2−3−4−5

y

x

y = f(x)

f(x + h) - f(x)

h.

f(x) = x2+ 11x - 7,

f(x) = 112 - 3x ,

g(-1)g(x) = b1x - 33 - x

if x � 3if x 6 3

,

14. Begin by graphing the absolute value function,Then use transformations of this graph to graph g(x)=

If and find eachfunction or function value in Exercises 15–19.

15. 16. and its domain

17. 18.

19.

20. If and find and the

domain of

21. Express as a composition of twofunctions f and g so that

22. If find the equation for Thenverify that your equation is correct by showing that

and

23. A function f models the amount given to charity as afunction of income.The graph of f is shown in the figure.

a. Explain why f has an inverse that is a function.b. Find f(80).c. Describe in practical terms the meaning of f-1(2000).

Income (thousands of dollars)0 20 40 60 80 100

y

x

1000

2000

3000

4000

5000

Am

ount

Giv

en to

Cha

rity

f-1(f(x)) = x.f(f-1(x)) = x

f-1(x).f(x) = 1x - 2 ,

h(x) = (f � g)(x).h(x) = (2x + 13)7

f � g.

(f � g)(x)g(x) =

2x

,f(x) =

7x - 4

f(g(2))

(g � f)(x)(f � g)(x)

¢ f

g≤ (x)(f - g)(x)

g(x) = 5x - 2,f(x) = x2+ 3x - 4

12 ∑x + 1∑ + 3.

f(x) = �x�.

x

y

1 2 3 4 5−1

−1

1g(x) = (x − 3)2

f(x) = x2 h(x) = −2(x − 3)2

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Cumulative Review Exercises (Chapters P–2)Simplify each expression in Exercises 1 and 2.

1. 2.

3. Factor:

In Exercises 4 and 5, perform the operations and simplify.

4. 5.

Solve each equation in Exercises 6–9.

6.

7.

8. 9.

Solve each inequality in Exercises 10 and 11. Express theanswer in interval notation.

10. 11.

12. Write the point-slope form and the slope-intercept formof the line passing through and perpendicularto the line whose equation is y = - 14 x +

13 .

(-2, 5)

x + 3x - 2

� 2x

2- 3 �

x

4+ 2

x2�3- x1�3

- 6 = 01x + 2 = x

3(4x - 1) = 4 - 6(x - 3)

(x + 3)(x - 4) = 8

4 +

2x

4 -

2x

x - 3x + 4

+

x

x - 2

x3- 4x2

+ 2x - 8.

5412

4x2y

2x5y-3

13. Graph and then use transformations of thisgraph to graph in the samerectangular coordinate system.

14. If find the equation for

15. If find and simplify.

16. Solve for c :

17. You invested $6000 in two accounts paying 7% and 9%annual interest, respectively. At the end of the year, thetotal interest from these investments was $510. How muchwas invested at each rate?

18. For a summer sales job, you are choosing between twopay arrangements: a weekly salary of $200 plus 5%commission on sales, or a straight 15% commission. Forhow many dollars of sales will the earnings be the sameregardless of the pay arrangement?

19. The length of a rectangular garden is 2 feet more thantwice its width. If 22 feet of fencing is needed to enclosethe garden, what are its dimensions?

20. On the first five tests you have scores of 61, 95, 71, 83, and80.The last test, a final exam, counts as two grades.Whatscore do you need on the final in order to have an averagescore of 80?

A =

cd

c + d .

f(x + h) - f(x)

hf(x) = 3 - x2,

f-1(x).f(x) = 2 + 1x - 3 ,

g(x) = 1x - 3 + 4f(x) = 1x

24. Use a graphing utility to graph

in a by viewing rectangle. Usethe graph to answer the following questions.

a. Is f one-to-one? Explain.b. Is f even, odd, or neither? Explain.c. What is the range of f?

[-30, 70, 10][-10, 10, 1]

f(x) =

x3

3+ x2

- 15x + 3d. On which interval or intervals is f increasing?e. On which interval or intervals is f decreasing?f. For what number does f have a relative maximum?

What is the relative maximum?g. For what number does f have a relative minimum?

What is the relative minimum?

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SECTION 2.5 Transformations of Functions

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Transcript of SECTION 2.5 Transformations of Functions