Graphing and Optimization -...

274

I N T R O D U C T I O N

Since the derivative is associated with the slope of the graph of a function at a point,

we might expect that it is also associated with other properties of a graph. As we will

see in this and the next section, the derivative can tell us a great deal about the shape

of the graph of a function. In addition, our investigation will lead to methods for find-

ing absolute maximum and minimum values for functions that do not require graph-

ing. Manufacturing companies can use these methods to find production levels that will

minimize cost or maximize profit, pharmacologists can use them to find levels of drug

dosages that will produce maximum sensitivity, and so on.

5-1 First Derivative and Graphs

5-2 Second Derivative and Graphs

5-3 L’ Hôpital’s Rule

5-4 Curve-Sketching Techniques

5-5 Absolute Maxima and Minima

5-6 Optimization

Chapter 5 Review

Review Exercise

Graphing and Optimization

5CHAPTER

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Section 5-1 FIRST DERIVATIVE AND GRAPHS Increasing and Decreasing Functions Local Extrema First-Derivative Test Applications to Economics

Increasing and Decreasing FunctionsSign charts (Section 3-2) will be used throughout this chapter. You will find it help-ful to review the terminology and techniques for constructing sign charts now.

Explore & Discuss 1 Figure 1 shows the graph of and a sign chart for where

andf¿1x2 = 3x2

- 3 = 31x + 121x - 12f1x2 = x3

- 3x

f¿1x2,y = f1x2

S e c t i o n 5 - 1 First Derivative and Graphs 275

(1, )(, 1) (1, 1)

11

00

x

x2112

f (x)

2

1

1

2

f (x)

FIGURE 1

Discuss the relationship between the graph of f and the sign of over eachinterval on which has a constant sign. Also, describe the behavior of thegraph of f at each partition number for

As they are scanned from left to right, graphs of functions generally have rising andfalling sections. If you scan the graph of in Figure 1 from left to right,you will see that

• On the interval the graph of f is rising, f(x) is increasing,* and theslope of the graph is positive

• On the interval the graph of f is falling, is decreasing, and theslope of the graph is negative

• On the interval the graph of f is rising, is increasing, and the slopeof the graph is positive

• At and the slope of the graph is 0

In general, if (is positive) on the interval (a, b) (Fig. 2), then f(x) in-creases and the graph of f rises as we move from left to right over the interval;if (is negative) on an interval (a, b), then f(x) decreases and the graphof f falls as we move from left to right over the interval. We summarize these impor-tant results in Theorem 1.

1Ω2f¿1x2 6 01˚2 f¿1x2 7 0

[f¿1x2 = 0].x = 1,x = -1

[f¿1x2 7 0].f1x211, q2,

[f¿1x2 6 0].f1x21-1, 12,

[f¿1x2 7 0].1- q , -12,

f1x2 = x3- 3x

f¿.f¿1x2 f¿1x2

* Formally, we say that the function f is increasing on an interval (a, b) if wheneverand f is decreasing on (a, b) if whenever a 6 x1 6 x2 6 b.f(x2) 6 f(x1)a 6 x1 6 x2 6 b,

f(x2) 7 f(x1)

x

y

Slope 0Slope

positive Slopenegative

a b c

f

FIGURE 2

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276 C H A P T E R 5 Graphing and Optimization

x

f (x)

5

5

5

(A)

g(x)

5

5

5

(B)

x

FIGURE 3

DecreasingIncreasing

(4, )(, 4)

f (x)

f (x)

4x

0

Test Numbers

x

3

5 -2 122 12f ¿1x2

THEOREM 1 INCREASING AND DECREASING FUNCTIONSFor the interval (a, b),

f (x) Graph of f Examples

Increases Rises

Decreases Falls ΩΩ

˚˚

f ¿1x2

Explore & Discuss 2 The graphs of and are shown in Figure 3. Both functionschange from decreasing to increasing at Discuss the relationship betweenthe graph of each function at and the derivative of the function at x = 0.x = 0

x = 0.g1x2 = ƒ x ƒf1x2 = x2

E X A M P L E 1 Finding Intervals on Which a Function Is Increasing or Decreasing Giventhe function

(A) Which values of x correspond to horizontal tangent lines?

(B) For which values of x is f(x) increasing? Decreasing?

(C) Sketch a graph of f. Add any horizontal tangent lines.

SOLUTION (A)

Thus, a horizontal tangent line exists at only.

(B) We will construct a sign chart for to determine which values of x makeand which values make Recall from Section 3-2 that the

partition numbers for a function are the points where the function is 0 or dis-continuous. Thus, when constructing a sign chart for we must locate allpoints where or is discontinuous. From part (A), we know that

at Since is a polynomial, it is con-tinuous for all x. Thus, 4 is the only partition number. We construct a sign chartfor the intervals and using test numbers 3 and 5:14, q2,1- q , 42

f¿1x2 = 8 - 2xx = 4.f¿1x2 = 8 - 2x = 0f¿1x2f¿1x2 = 0

f¿1x2,f¿1x2 6 0.f¿1x2 7 0f¿1x2

x = 4

x = 4

f¿1x2 = 8 - 2x = 0

f1x2 = 8x - x2,

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Hence, f(x) is increasing on and decreasing on

(C)

14, q2.1- q , 42S e c t i o n 5 - 1 First Derivative and Graphs 277

x f(x)

0 0

2 12

4 16

6 12

8 0

f (x)

x1050

15

10

5 f (x)increasing

f (x)decreasing

Horizontaltangent line

MATCHED PROBLEM 1 Repeat Example 1 for f1x2 = x2- 6x + 10.

As Example 1 illustrates, the construction of a sign chart will play an important rolein using the derivative to analyze and sketch the graph of a function f. The partitionnumbers for are central to the construction of these sign charts and also to theanalysis of the graph of We already know that if then the graphof will have a horizontal tangent line at But the partition numbersfor also include the numbers c such that does not exist.* There are two pos-sibilities at this type of number: f(c) does not exist; or f(c) exists, but the slope of thetangent line at is undefined.x = c

f¿1c2f¿

x = c.y = f1x2 f¿1c2 = 0,y = f1x2.f¿

DEFINITION Critical Values

The values of x in the domain of f where or where does not exist arecalled the critical values of f.

f¿1x2f¿1x2 = 0

The critical values of f are always in the domain of f and are also partition numbers forbut may have partition numbers that are not critical values.If f is a polynomial, then both the partition numbers for and the critical values of f

are the solutions of f¿1x2 = 0.f¿

f¿f¿,

I N S I G H T

We will illustrate the process for locating critical values with examples.

E X A M P L E 2 Partition Numbers and Critical Values Find the critical values of f, the inter-vals on which f is increasing, and those on which f is decreasing, for

SOLUTION Begin by finding the partition number for

The partition number 0 is in the domain of f, so 0 is the only critical value of f.

f¿1x2 = 3x2= 0, only at x = 0

f¿1x2:f1x2 = 1 + x3.

* We are assuming that does not exist at any point of discontinuity of There do exist functions f such that is discontinuous at yet exists. However, we do not consider such functions in this book.

f ¿(c)x = c,f ¿

f ¿.f ¿(c)

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The sign chart for (partition number is 0) isf¿1x2 = 3x2


(0, )(, 0)

0

0

x

f (x)

f (x)

Increasing Increasing

Test Numbers

x

1 3 123 12-1

f œ1x2

Test Numbers

x

0

2 -13 12

-13 12f¿1x2

x

f (x)

11

2

FIGURE 4

(1, )(, 1)

1

ND

x

f (x)

f (x)

DecreasingDecreasing

x

f (x)

210

1

1

FIGURE 5

The sign chart indicates that f(x) is increasing on and Since f is con-tinuous at it follows that f(x) is increasing for all x. The graph of f is shown inFigure 4.

x = 0,10, q2.1- q , 02

MATCHED PROBLEM 2 Find the critical values of f, the intervals on which f is increasing, and those on whichf is decreasing, for f1x2 = 1 - x3.


SOLUTION

To find partition numbers for we note that is continuous for all x, except forvalues of x for which the denominator is 0; that is, does not exist and is dis-continuous at Since the numerator is the constant for any valueof x. Thus, is the only partition number for Since 1 is in the domain of f,

is also the only critical value of f. When constructing the sign chart for weuse the abbreviation ND to note the fact that is not defined at

The sign chart for (partition number is 1) isf¿1x2 = -1>3311 - x22>34x = 1.f¿1x2 f¿x = 1

f¿.x = 1-1, f¿1x2 Z 0x = 1.

f¿f¿112f¿f¿,

f¿1x2 = -

13

11 - x2-2>3=

-1

311 - x22>3

f1x2 = 11 - x21>3.

The sign chart indicates that f is decreasing on and Since f is con-tinuous at it follows that f(x) is decreasing for all x. Thus, a continuous func-tion can be decreasing (or increasing) on an interval containing values of x where

does not exist. The graph of f is shown in Figure 5. Notice that the undefinedderivative at results in a vertical tangent line at In general, a verticaltangent will occur at if f is continuous at and if becomes largerand larger as x approaches c.

f ¿(x) x cx cx = 1.x = 1

f ¿(x)

x = 1,11, q2.1- q , 12

MATCHED PROBLEM 3 Find the critical values of f, the intervals on which f is increasing, and those on whichf is decreasing, for f1x2 = 11 + x21>3.

E X A M P L E 4 Partition Numbers and Critical Values Find the critical values of f, the inter-

vals on which f is increasing, and those on which f is decreasing, for f1x2 =

1x - 2

.

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 278

SOLUTION

To find the partition numbers for note that for any x and is not de-fined at Thus, is the only partition number for However, is notin the domain of f. Consequently, is not a critical value of f. This function hasno critical values.

The sign chart for (partition number is 2) isf¿1x2 = -1>1x - 222x = 2

x = 2f¿.x = 2x = 2.f¿f¿1x2 Z 0f¿,

f¿1x2 = -1x - 22-2=

-1

1x - 222

f1x2 =

1x - 2

= 1x - 22-1


Test Numbers

x

1

3 -1 12-1 12

f ¿1x2

(2, )(, 2)

2

ND

x

f (x)

f (x)


x

f (x)

5

55

5

FIGURE 6

x

f (x)

4321

2

1

1

2

Test Numbers

x

1

4 -6 126 12f ¿1x2


(2, )(0, 2)

f (x)

f (x)

20x

ND

Thus, f is decreasing on and See the graph of f in Figure 6.12, q2.1- q , 22

MATCHED PROBLEM 4 Find the critical values for f, the intervals on which f is increasing, and those on which

f is decreasing, for f1x2 =

1x

.


SOLUTION The natural logarithm function ln x is defined on is definedonly for We have

Find a common denominator.

Subtract numerators.

Factor numerator.

Note that at and at 2 and is discontinuous at 0. These are thepartition numbers for Since the domain of f is are not criti-cal values. The remaining partition number, 2, is the only critical value for

The sign chart for (partition number is 2), isf¿1x2 =

212 - x212 + x2x

, x 7 0

f1x2.10, q2, 0 and -2f¿1x2. f¿1x2-2f¿1x2 = 0

=

212 - x212 + x2x

, x 7 0

=

8 - 2x2

x

=

8x

-

2x2

x

f¿1x2 =

8x

- 2x

f1x2 = 8 ln x - x2, x 7 0

x 7 0.10, q2, or x 7 0, so f1x2

f1x2 = 8 ln x - x2.

Thus, f is increasing on and decreasing on See the graph of f inFigure 7.

12, q2.10, 22FIGURE 7

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Explore & Discuss 3 A student examined the sign chart in Example 4 and concluded thatis decreasing for all x except However,

which seems to indicate that f is increasing. Discuss the differ-ence between the correct answer in Example 4 and the student’s answer. Explainwhy the student’s description of where f is decreasing is unacceptable.

-1 6 f132 = 1,f112 =x = 2.f1x2 = 1>1x - 22


x

y f (x)

c1 c2 c3 c4 c5 c6 c7

FIGURE 8

MATCHED PROBLEM 5 Find the critical values of f, the intervals on which f is increasing, and those on whichf is decreasing, for f1x2 = 5 ln x - x.

Example 4 illustrates two important ideas:

1. Do not assume that all partition numbers for the derivative are critical values of thefunction f. To be a critical value, a partition number must also be in the domain of f.

2. The values for which a function is increasing or decreasing must always be expressedin terms of open intervals that are subsets of the domain of the function.

f¿

I N S I G H T

Local ExtremaWhen the graph of a continuous function changes from rising to falling, a high point,or local maximum, occurs; when the graph changes from falling to rising, a low point, or local minimum, occurs. In Figure 8, high points occur at and and lowpoints occur at and In general, we call f(c) a local maximum if there exists aninterval (m, n) containing c such that

Note that this inequality need hold only for values of x near c—hence the use of theterm local.

f1x2 … f1c2 for all x in 1m, n2

c4.c2

c6,c3

The quantity f(c) is called a local minimum if there exists an interval (m, n) con-taining c such that

The quantity f (c) is called a local extremum if it is either a local maximum or a localminimum. A point on a graph where a local extremum occurs is also called a turningpoint. Thus, in Figure 8 we see that local maxima occur at and local minimaoccur at and and all four values produce local extrema. Also, the local maximum

is not the highest point on the graph in Figure 8. Later in this chapter, we con-sider the problem of finding the highest and lowest points on a graph, or absoluteextrema. For now, we are concerned only with locating local extrema.

f1c32c4,c2

c6,c3

f1x2 Ú f1c2 for all x in 1m, n2

E X A M P L E 6 Analyzing a graph Use the graph of f in Figure 9 to find the intervals on which fis increasing, those on which f is decreasing, any local maxima, and any local minima.

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SOLUTION The function f is increasing (the graph is rising) on and on andis decreasing (the graph is falling) on Because the graph changes from risingto falling at is a local maximum. Because the graph changes fromfalling to rising at is a local minimum.x = 3, f132 = -5

x = -1, f1-12 = 31-1, 32. 13, q21- q , -12


f (x)

x

5

55

5

FIGURE 9

g(x)

x

5

55

5

FIGURE 10

MATCHED PROBLEM 6 Use the graph of g in Figure 10 to find the intervals on which g is increasing, thoseon which g is decreasing, any local maxima, and any local minima.

How can we locate local maxima and minima if we are given the equation of afunction and not its graph? The key is to examine the critical values of the function.The local extrema of the function f in Figure 8 occur either at points where the de-rivative is 0 ( and ) or at points where the derivative does not exist ( and ). Inother words, local extrema occur only at critical values of f. Theorem 2 shows that thisis true in general.

c6c4c3c2

THEOREM 2 EXISTENCE OF LOCAL EXTREMAIf f is continuous on the interval (a, b), c is a number in (a, b), and f(c) is a local ex-tremum, then either or does not exist (is not defined).f¿1c2f¿1c2 = 0

Theorem 2 states that a local extremum can occur only at a critical value, but it doesnot imply that every critical value produces a local extremum. In Figure 8, and are critical values (the slope is 0), but the function does not have a local maximumor local minimum at either of these values.

Our strategy for finding local extrema is now clear: We find all critical values of fand test each one to see if it produces a local maximum, a local minimum, or neither.

First-Derivative TestIf exists on both sides of a critical value c, the sign of can be used todetermine whether the point (c, f(c)) is a local maximum, a local minimum, or

f¿1x2f¿1x2

c5c1

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neither. The various possibilities are summarized in the following box and areillustrated in Figure 11:


f (c) 0: Horizontal tangent

f (c)

f (x)

(A) f (c) is alocal minimum

(B) f (c) is alocal maximum

(C) f (c) is neithera local maximumnor a local minimum

(D) f (c) is neithera local maximumnor a local minimum

xc

f (c)

f (x)

xc

f (x)

f (c)

f (x)

xc

f (c)

f (x)

xc

0 f (x) 0 f (x) 0 f (x) 0

f (c) is not defined but f (c) is defined

(E) f (c) is a localminimum

(F) f (c) is a localmaximum

(G) f (c) is neithera local maximumnor a local minimum

f (c)

f (x)

(H) f (c) is neithera local maximumnor a local minimum

xc

f (c)

f (x)

xc

f (c)

f (x)

xc

f (c)

f (x)

xc

f (x) ND f (x) ND f (x) ND f (x) ND

FIGURE 11 Local extrema

Sign Chart f(c)

f(c) is local minimum.

If changes from negative to positive at c, thenf(c) is a local minimum.

f(c) is local maximum.

If changes from positive to negative at c, thenf(c) is a local maximum.

f ¿(x)

f ¿(x)

Increasing

x

DecreasingDecreasingf (x)

f (x)

m c n

x

f (x)

f (x)

m c n

Increasing Increasing

( (

( (

PROCEDURE First-Derivative Test for Local ExtremaLet c be a critical value of f [ f(c) is defined and either or is notdefined]. Construct a sign chart for close to and on either side of c.f¿1x2

f¿1c2f¿1c2 = 0

f(c) is not a local extremum.

If does not change sign at c, then f(c) is neithera local maximum nor a local minimum.

f(c) is not a local extremum.

If does not change sign at c, then f(c) is neithera local maximum nor a local minimum.

f ¿(x)

f ¿(x)

Decreasing Increasingf (x)

f (x)

x

x

m n

DecreasingIncreasingf (x)

f (x)

m c n

( (

( (

c

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The sign chart indicates that f increases on has a local maximum atdecreases on (1, 3), has a local minimum at and increases onThese facts are summarized in the following table:13, q2. x = 3,x = 1,

1- q , 12,


(3, )(, 1) (1, 3)

31

00

x

f (x)

f (x)


Localminimum

Localmaximum

Increasing

Test Numbers

x

0

2

4 9 12-3 12

9 12f ¿1x2

x f(x) Graph of f

Increasing Rising

0 Local maximum Horizontal tangent

(1, 3) Decreasing Falling

0 Local minimum Horizontal tangent

Increasing Rising+(3, q)

x = 3

-

x = 1

+(- q, 1)

f ¿1x2

E X A M P L E 7 Locating Local Extrema Given

(A) Find the critical values of f.

(B) Find the local maxima and minima.

(C) Sketch the graph of f.

SOLUTION (A) Find all numbers x in the domain of f where or does not exist.

exists for all x; the critical values are and

(B) The easiest way to apply the first-derivative test for local maxima and minima isto construct a sign chart for for all x. Partition numbers for areand (which also happen to be critical values of f).

Sign chart for f¿1x2 = 31x - 121x - 32:x = 3

x = 1f¿1x2f¿1x2x = 3.x = 1f¿1x2

x = 1 or x = 3

31x - 121x - 32 = 0

31x2- 4x + 32 = 0

f¿1x2 = 3x2- 12x + 9 = 0

f¿1x2f¿1x2 = 0

f1x2 = x3- 6x2

+ 9x + 1,

(C) We sketch a graph of f, using the information from part (B) and point-by-pointplotting.

x f(x)

0 1

1 5

2 3

3 1

4 5

x

f (x)

4321

6

5

4

3

2

1

Localmaximum

Localminimum

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How can you tell if you have found all the local extrema of a function? In gener-al, this can be a difficult question to answer. However, in the case of a polynomial func-tion, there is an easily determined upper limit on the number of local extrema. Sincethe local extrema are the x intercepts of the derivative, this limit is a consequence of the number of x intercepts of a polynomial. The relevant information is summarizedin the following theorem, which is stated without proof:


t10

Years

Bill

ions

per

yea

r

15

B(t)

0

5

5

10

FIGURE 12 Rate of change of the balance of trade

MATCHED PROBLEM 7 Given

(A) Find the critical values of f.(B) Find the local maxima and minima.(C) Sketch a graph of f.

f1x2 = x3- 9x2

+ 24x - 10,

THEOREM 3 INTERCEPTS AND LOCAL EXTREMA OF POLYNOMIAL FUNCTIONSIf is an nth-degree polynomial,then f has at most n x intercepts and at most local extrema.n - 1

f1x2 = anxn+ an - 1x

n - 1+

Á+ a1x + a0, an Z 0,

Theorem 3 does not guarantee that every nth-degree polynomial has exactly local extrema; it says only that there can never be more than local extrema. Forexample, the third-degree polynomial in Example 7 has two local extrema, while thethird-degree polynomial in Example 2 does not have any.

Applications to EconomicsIn addition to providing information for hand-sketching graphs, the derivative is animportant tool for analyzing graphs and discussing the interplay between a functionand its rate of change. The next two examples illustrate this process in the context ofsome applications to economics.

n - 1n - 1

E X A M P L E 8 Agricultural Exports and Imports Over the past several decades, the UnitedStates has exported more agricultural products than it has imported, maintaining apositive balance of trade in this area. However, the trade balance fluctuated consid-erably during that period.The graph in Figure 12 approximates the rate of change ofthe balance of trade over a 15-year period, where B(t) is the balance of trade (inbillions of dollars) and t is time (in years).

(A) Write a brief verbal description of the graph of including a discussionof any local extrema.

(B) Sketch a possible graph of y = B1t2.y = B1t2,

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SOLUTION (A) The graph of the derivative contains the same essential informationas a sign chart. That is, we see that is positive on (0, 4), 0 at nega-tive on (4, 12), 0 at and positive on (12, 15). Hence, the trade balanceincreases for the first 4 years to a local maximum, decreases for the next 8 yearsto a local minimum, and then increases for the final 3 years.

(B) Without additional information concerning the actual values of wecannot produce an accurate graph. However, we can sketch a possible graphthat illustrates the important features, as shown in Figure 13. The absence of ascale on the vertical axis is a consequence of the lack of information about thevalues of B(t).

y = B1t2,

t = 12,t = 4,B¿1t2y = B¿1t2


t50 10

Years

15

B(t)

FIGURE 13 Balance of trade

(A) Write a brief verbal description of the graph of including a discussionof any local extrema.

(B) Sketch a possible graph of y = S1t2.y = S1t2,

MATCHED PROBLEM 8 The graph in Figure 14 approximates the rate of change of the U.S. share of the totalworld production of motor vehicles over a 20-year period, where S(t) is the U.S.share (as a percentage) and t is time (in years).

t10

YearsPerc

ent p

er y

ear

2015

S(t)

0

5

3

FIGURE 14

E X A M P L E 9 Revenue Analysis The graph of the total revenue R(x) (in dollars) from the saleof x bookcases is shown in Figure 15.

(A) Write a brief verbal description of the graph of the marginal revenue functionincluding a discussion of any x intercepts.

(B) Sketch a possible graph of y = R¿1x2.y = R¿1x2,

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SOLUTION (A) The graph of indicates that R(x) increases on (0, 550), has a local max-imum at and decreases on (550, 1,000). Consequently, the marginalrevenue function must be positive on (0, 550), 0 at and negativeon (550, 1,000).

(B) A possible graph of illustrating the information summarized in part(A) is shown in Figure 16.

y = R¿1x2x = 550,R¿1x2x = 550,

y = R1x2


5000 1,000

R(x)

$40,000

$20,000

x

FIGURE 15 Revenue

x1,000500

R(x)

FIGURE 16 Marginal revenue

x

R(x)

200 400 600 800 1,000

$20,000

$40,000

$60,000

FIGURE 17

MATCHED PROBLEM 9 The graph of the total revenue R(x) (in dollars) from the sale of x desks is shown inFigure 17.

(A) Write a brief verbal description of the graph of the marginal revenue functionincluding a discussion of any x intercepts.

(B) Sketch a possible graph of y = R¿1x2.y = R¿1x2,

Comparing Examples 8 and 9, we see that we were able to obtain more informa-tion about the function from the graph of its derivative (Example 8) than we werewhen the process was reversed (Example 9). In the next section, we introduce someideas that will enable us to extract additional information about the derivative fromthe graph of the function.

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 286

Answers to Matched Problems 1. (A) Horizontal tangent line at (C)(B) Decreasing on

increasing on

2. Partition number: critical value: decreasing for all x3. Partition number: critical value: increasing for all x4. Partition number: no critical values; decreasing on and

5. Partition number: critical value: increasing on decreasing on

6. Increasing on decreasing on and local maximum at local minimum at

7. (A) Critical values: (B) Local maximum at

local minimum at (C)

x = 4x = 2;

x = 2, x = 4

x = -3x = 1;11, q2;1- q , -321-3, 12;15, q210, 52;x = 5;x = 5;

10, q21- q , 02x = 0;x = -1;x = -1;

x = 0;x = 0;

13, q21- q , 32;

x = 3.


50

10

5

x

f (x)

x

f (x)

50

5

10

8. (A) The U.S. share of the world marketdecreases for 6 years to a local min-imum, increases for the next 10 yearsto a local maximum, and then de-creases for the final 4 years.

(B)

9. (A) The marginal revenue is positive on (0, 450), 0 at and negative on (450, 1,000).

(B)

x = 450,

x1,000

R(x)

t

S(t)

1050 15 20

A Problems 1–8 refer to the following graph of :y = f1x2

x

f (x)

a b c

d

e f g h

1. Identify the intervals on which f(x) is increasing.

2. Identify the intervals on which f(x) is decreasing.

3. Identify the intervals on which

4. Identify the intervals on which

5. Identify the x coordinates of the points where

6. Identify the x coordinates of the points where doesnot exist.

7. Identify the x coordinates of the points where f(x) has alocal maximum.

8. Identify the x coordinates of the points where f(x) has alocal minimum.

f¿1x2f¿1x2 = 0.

f¿1x2 7 0.

f¿1x2 6 0.

Exercise 5-1

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 287


15.

In Problems 9 and 10, f(x) is continuous on and hascritical values at and d. Use the sign chart for to determine whether f has a local maximum, a local minimum,or neither at each critical value.

9.

f¿1x2x = a, b, c,1- q , q2

NDND 0f (x)

a b c dx

0

ND0 NDf (x)

a b c dx

0

In Problems 11–18, match the graph of f with one of the signcharts a–h in the figure.

11.

10.

x

f (x)

60

6

x

f (x)

60

6

x

f (x)

60

6

x

f (x)

60

6

x

f (x)

60

6

x

f (x)

60

6

x

f (x)

60

6

x

f (x)

60

6

12.

13.

17.

18.

14.

16.

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 288

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

Figure for 11–18

B In Problems 19–36, find the intervals on which f(x) isincreasing, the intervals on which f(x) is decreasing, and the local extrema.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31.

32.

33.

34. f1x2 = 1x + 22ex

f1x2 = 2x - x ln x

f1x2 = 1x2- 922>3

f1x2 = 4x1>3- x2>3

f1x2 = x ln x - x

f1x2 = 1x - 12e-x

f1x2 = x4+ 2x3

+ 5

f1x2 = 3x4- 4x3

+ 5

f1x2 = -2x3+ 3x2

+ 120x

f1x2 = 2x3- 3x2

- 36x

f1x2 = -x3- 4x + 8

f1x2 = x3+ 4x - 5

f1x2 = -3x2+ 12x - 5

f1x2 = -2x2- 16x - 25

f1x2 = -3x2- 12x

f1x2 = 2x2- 4x


35.

36.

In Problems 37–46, use a graphing calculator to approximatethe critical values of f(x) to two decimal places. Find the inter-vals on which f(x) is increasing, the intervals on which f(x) isdecreasing, and the local extrema.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

In Problems 47–54, find the intervals on which f(x) isincreasing and the intervals on which f(x) is decreasing. Thensketch the graph. Add horizontal tangent lines.

47.

48.

49.

50.

51.

52.

53.

54.

In Problems 55–62, f(x) is continuous on Use thegiven information to sketch the graph of f.

55.

56.

1- q , q2.

f1x2 = -x4+ 50x2

f1x2 = x4- 18x2

f1x2 = x3+ 3x2

+ 3x

f1x2 = 10 - 12x + 6x2- x3

f1x2 = x3- 12x + 2

f1x2 = x3- 3x + 1

f1x2 = 2x2- 8x + 9

f1x2 = 4 + 8x - x2

f1x2 =

ln xx

- 5x + x2

f1x2 = x1>3+ x4>3

- 2x

f1x2 = e-x- 3x2

f1x2 = ex- 2x2

f1x2 = 3x - x1>3- x4>3

f1x2 = x ln x - 1x - 223f1x2 = x4

+ 5x3- 15x

f1x2 = x4- 4x3

+ 9x

f1x2 = x4+ x2

- 9x

f1x2 = x4+ x2

+ x

f1x2 = x4>3- 7x1>3

f1x2 = 1x2- 3x - 424>3

f (x)

3x

0

f (x)

3x

ND

f (x)

3x

0

3

f (x)

x

ND

f (x)

3x

0

f (x)

3x

ND

3

f (x)

x

ND

f (x)

x

3

0

x 0 1 2

f(x) 1 2 3 1-1

-1-2

x 0 1 2

f(x) 1 3 2 1 -1

-1-2

0f (x)

1 1x

0

0f (x)

1 1x

0

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 289

57.

58.

59.

on and on and (0, 2)

60.

on and

61.is not defined;

on and on and (0, 1)

62.is not defined;

on and (0, 1);on and

Problems 63–68 involve functions and their derivatives,Use the graphs shown in figures (A) and (B) to match

each function with its derivative gj.fi

g1–g6.f1–f6

11, q21-1, 02f¿1x2 6 01- q , -12f¿1x2 7 0

f¿1-12 = 0, f¿112 = 0, f¿102f1-12 = 2, f102 = 0, f112 = 2;

1-1, 02f¿1x2 6 011, q2;1- q , -12f¿1x2 7 0

f¿1-12 = 0, f¿112 = 0, f¿102f1-12 = 2, f102 = 0, f112 = -2;

12, q21- q , -22, 1-2, 22,f¿1x2 7 0f¿1-22 = 0, f¿122 = 0;f1-22 = -1, f102 = 0, f122 = 1;

1-2, 02f¿1x2 6 012, q2;1- q , -22f¿1x2 7 0

f¿1-22 = 0, f¿102 = 0, f¿122 = 0;f1-22 = 4, f102 = 0, f122 = -4;


x 0 2 4

f(x) 2 1 2 1 0

-1-2

x 0 2 3

f(x) 0 2 0-1-3

-1-2

0NDf (x)

1 0 2x

0

00f (x)

1 0 2x

ND

x

f1(x)

5

55

5

x

f2(x)

5

55

5

x

f3(x)

5

55

5

x

f4(x)

5

55

5

x

f5(x)

5

55

5

x

f6(x)

5

55

5

Figure (A) for 63–68

x

g1(x)

5

55

5

x

g2(x)

5

55

5

x

g3(x)

5

55

5

x

g4(x)

5

55

5

x

g5(x)

55

5

x

g6(x)

5

55

5

Figure (B) for 63–68

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 290


63.

64.

65.

66.

67.

68.

In Problems 69–74, use the given graph of to find the intervals on which f is increasing, the intervals on which f is decreasing, and the local extrema. Sketch a possible graph of

69. 70.

y = f1x2.

y = f¿1x2f6

f5

f4

f3

f2

f1

71. 72.

73. 74.

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

x

f (x)

5

55

5

75. 76.

77. 78.

C In Problems 79–90, find the critical values, the intervals onwhich f(x) is increasing, the intervals on which f(x) isdecreasing, and the local extrema. Do not graph.

79.

80.

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

91. Let where k is a constant. Discuss thenumber of local extrema and the shape of the graph of f if

(A) (B) (C)

92. Let where k is a constant. Discuss thenumber of local extrema and the shape of the graph of f if

(A) (B) (C) k = 0k 6 0k 7 0

f1x2 = x4+ kx2,

k = 0k 6 0k 7 0

f1x2 = x3+ kx,

f1x2 =

-3x

x2+ 4

f1x2 =

2x2

x2+ 1

f1x2 = 614 - x22>3 + 4

f1x2 = 31x - 222>3 + 4

f1x2 = x31x - 522f1x2 = x41x - 622f1x2 =

x2

x + 1

f1x2 =

x2

x - 2

f1x2 = 3 -

4x

-

2

x2

f1x2 = 1 +

1x

+

1

x2

f1x2 =

9x

+ x

f1x2 = x +

4x

In Problems 75–78, use the given graph of to find the intervals on which the intervals on which

and the values of x for which Sketch apossible graph of y = f¿1x2.

f¿1x2 = 0.f¿1x2 6 0,f¿1x2 7 0,

y = f1x2

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 291


93. Profit analysis. The graph of the total profit P(x) (in dollars) from the sale of x cordless electricscrewdrivers is shown in the figure.

x1,000500

R(x)

40,000

20,000

20,000

40,000

Figure for 94

0x

1,000

500

P(x)

40,000

20,000

20,000

40,000

Figure for 93

t30 50 70

B(t)

0.02

0.01

0.01

0.02

0.03

0.04

Figure for 95

t30 50 7010

E(t)

0.01

0.01

0.02

0.03

Figure for 96

(A) Write a brief verbal description of the graph of the marginal profit function including a discussion of any x intercepts.

(B) Sketch a possible graph of

94. Revenue analysis. The graph of the total revenue R(x) (indollars) from the sale of x cordless electric screwdrivers isshown in the figure.

y = P¿1x2.y = P¿1x2,

(A) Write a brief verbal description of the graph of including a discussion of any localextrema.


96. Price analysis. The graph in the figure approximates therate of change of the price of eggs over a 70-month peri-od, where E(t) is the price of a dozen eggs (in dollars) andt is time (in months).

y = B1t2.y = B1t2,

(A) What is the average cost per toaster if xtoasters are produced in one day?

(B) Find the critical values of the intervals onwhich the average cost per toaster is decreasing, theintervals on which the average cost per toaster isincreasing, and the local extrema. Do not graph.

98. Average cost. A manufacturer incurs the following costs in producing x blenders in one day for fixed costs, $450; unit production cost, $30 per blender;equipment maintenance and repairs, dollars.

(A) What is the average cost per blender if xblenders are produced in one day?

(B) Find the critical values of the intervals onwhich the average cost per blender is decreasing, theintervals on which the average cost per blender isincreasing, and the local extrema. Do not graph.

C1x2,C1x2

0.08x2

0 6 x 6 200:

C1x2,C1x2

(A) Write a brief verbal description of the graph of themarginal revenue function including adiscussion of any x intercepts.


95. Price analysis. The graph in the figure approximates therate of change of the price of bacon over a 70-month peri-od, where B(t) is the price of a pound of sliced bacon (indollars) and t is time (in months).

y = R¿1x2.y = R¿1x2,

(A) Write a brief verbal description of the graph ofincluding a discussion of any local

extrema.


97. Average cost. A manufacturer incurs the following costsin producing x toasters in one day, for fixedcosts, $320; unit production cost, $20 per toaster; equip-ment maintenance and repairs, dollars. Thus, thecost of manufacturing x toasters in one day is given by

C1x2 = 0.05x2+ 20x + 320 0 6 x 6 150

0.05x2

0 6 x 6 150:

y = E1t2.y = E1t2,

Applications

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 292

99. Marginal analysis. Show that profit will be increasing overproduction intervals (a, b) for which marginal revenue isgreater than marginal cost. [Hint: ]

100. Marginal analysis. Show that profit will be decreasingover production intervals (a, b) for which marginalrevenue is less than marginal cost.

101. Medicine. A drug is injected into the bloodstream of apatient through the right arm. The concentration of thedrug in the bloodstream of the left arm t hours after theinjection is approximated by

Find the critical values of C(t), the intervals on which theconcentration of the drug is increasing, the intervals onwhich the concentration of the drug is decreasing, andthe local extrema. Do not graph.

102. Medicine. The concentration C(t), in milligrams per cubiccentimeter, of a particular drug in a patient’s bloodstreamis given by

C1t2 =

0.3t

t2+ 6t + 9

0 6 t 6 12

C1t2 =

0.28t

t2+ 4

0 6 t 6 24

P1x2 = R1x2 - C1x2

S e c t i o n 5 - 2 Second Derivative and Graphs 293

where t is the number of hours after the drug is takenorally. Find the critical values of C(t), the intervals onwhich the concentration of the drug is increasing, theintervals on which the concentration of the drug isdecreasing, and the local extrema. Do not graph.

103. Politics. Public awareness of a congressional candidatebefore and after a successful campaign was approxi-mated by

where t is time (in months) after the campaign startedand P(t) is the fraction of people in the congressionaldistrict who could recall the candidate’s (and later,congressman’s) name. Find the critical values of P(t), the time intervals on which the fraction is increasing, the time intervals on which the fraction is decreasing, and the local extrema. Do not graph.

P1t2 =

8.4t

t2+ 49

+ 0.1 0 6 t 6 24

Section 5-2 SECOND DERIVATIVE AND GRAPHS Using Concavity as a Graphing Tool Finding Inflection Points Analyzing Graphs Curve Sketching Point of Diminishing Returns

In Section 5-1, we saw that the derivative can be used when a graph is rising andfalling. Now we want to see what the second derivative (the derivative of the deriva-tive) can tell us about the shape of a graph.

Using Concavity as a Graphing ToolConsider the functions

for x in the interval Since

and

both functions are increasing on 10, q2.

g¿1x2 =

1

22x7 0 for 0 6 x 6 q

f¿1x2 = 2x 7 0 for 0 6 x 6 q

10, q2.f1x2 = x2 and g1x2 = 2x

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 293

Explore & Discuss 1 (A) Discuss the difference in the shapes of the graphs of f and g shown inFigure 1.


(1, 1) (1, 1)

(A) f (x) x2 (B) g(x) x

xx

g(x)f (x)

11

11

FIGURE 1

x 0.25 0.5 0.75 1

g¿1x2f ¿1x2

f (1) 2

f (.75) 1.5

g(1) .5

g(.75) .6

g(.5) .7

g(.25) 1f (.50) 1

f (.25) .5

(A) f (x) x2

f (x)

.25 .50 .75 1 .25 .50 .75 1x x

1 1

g(x)

(B) g(x) x

FIGURE 2

(B) Complete the following table, and discuss the relationship between the val-ues of the derivatives of f and g and the shapes of their graphs:

We use the term concave upward to describe a graph that opens upward and concavedownward to describe a graph that opens downward. Thus, the graph of f in Figure 1Ais concave upward, and the graph of g in Figure 1B is concave downward. Finding amathematical formulation of concavity will help us sketch and analyze graphs.

It will be instructive to examine the slopes of f and g at various points on theirgraphs (see Fig. 2). We can make two observations about each graph:

1. Looking at the graph of f in Figure 2A, we see that (the slope of thetangent line) is increasing and that the graph lies above each tangent line;

2. Looking at Figure 2B, we see that is decreasing and that the graph liesbelow each tangent line.

g¿1x2f¿1x2

With these ideas in mind, we state the general definition of concavity.

DEFINITION Concavity

The graph of a function f is concave upward on the interval (a, b) if is increasingon (a, b) and is concave downward on the interval (a, b) if is decreasing on (a, b).f¿1x2 f¿1x2

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 294

Geometrically, the graph is concave upward on (a, b) if it lies above its tangent linesin (a, b) and is concave downward on (a, b) if it lies below its tangent lines in (a, b).

How can we determine when is increasing or decreasing? In Section 5-1, weused the derivative of a function to determine when that function is increasing ordecreasing. Thus, to determine when the function is increasing or decreasing,we use the derivative of The derivative of the derivative of a function is calledthe second derivative of the function. Various notations for the second derivative aregiven in the following box:

f¿1x2. f¿1x2f¿1x2


NOTATION Second Derivative

For the second derivative of f, provided that it exists, is

Other notations for are

d2y

dx2 y–

f–1x2f–1x2 =

d

dx f¿1x2

y = f1x2,

Returning to the functions f and g discussed at the beginning of this section, wehave

For we see that thus, is increasing and the graph of f is con-cave upward (see Fig. 2A). For we also see that thus, is de-creasing and the graph of g is concave downward (see Fig. 2B). These ideas aresummarized in the following box:

g¿1x2g–1x2 6 0;x 7 0,f¿1x2f–1x2 7 0;x 7 0,

f–1x2 =

d

dx 2x = 2 g–1x2 =

d

dx 12

x-1>2= -

14

x-3>2= -

1

42x3

f¿1x2 = 2x g¿1x2 =

12

x-1>2=

1

22x

f1x2 = x2 g1x2 = 2x = x1>2

SUMMARY CONCAVITYFor the interval (a, b),

Graph of Examples

Increasing Concave upward

Decreasing Concave downward-

+

y = f(x)f ¿(x)f –(x)

Be careful not to confuse concavity with falling and rising. A graph that is concave upwardon an interval may be falling, rising, or both falling and rising on that interval. A similarstatement holds for a graph that is concave downward. See Figure 3 on the next page.

I N S I G H T

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 295


FIGURE 3 Concavity

E X A M P L E 1 Concavity of Graphs Determine the intervals on which the graph of each func-tion is concave upward and the intervals on which it is concave downward. Sketch agraph of each function.

(A) (B) (C)

SOLUTION (A) (B) (C)

h–1x2 = 6xg–1x2 = -

1

x2f–1x2 = ex

h¿1x2 = 3x2g¿1x2 =

1x

f¿1x2 = ex

h1x2 = x3g1x2 = ln xf1x2 = ex

h1x2 = x3g1x2 = ln xf1x2 = ex

Since onthe graph of[Fig. 4(A)]

is concave upward on1- q , q2.f1x2 = ex1- q , q2,f–1x2 7 0 The domain of

is and onthis interval, so the graphof [Fig. 4(B)] is concave downward on10, q2.

g1x2 = ln x

g–1x2 6 010, q2 g1x2 = ln x Since when and when the graph of [Fig.4(C)] is concave downwardon and concaveupward on 10, q2.1- q , 02

h1x2 = x3x 7 0,h–1x2 7 0

x 6 0h–1x2 6 0

x

f (x)

2

4

2

f (x) exConcaveupward

(A) Concave upwardfor all x.

g(x) ln x

Concavedownward

4

2

2

g(x)

x

(B) Concave downwardfor x 0.

Concavedownward

Concaveupwardh(x) x3

x

h(x)

1

11

1

(C) Concavity changesat the origin.

FIGURE 4

f (x) > 0 on (a, b)Concave upward

f (x) < 0 on (a, b)Concave downward

(A) f (x) is negativeand increasing.Graph of f is falling.

(B) f (x) increases fromnegative to positive.Graph of f falls, then rises.

(C) f (x) is positiveand increasing.Graph of f is rising.

(D) f (x) is positiveand decreasing.Graph of f is rising.

(E) f (x) decreases frompositive to negative.Graph of f rises, then falls.

(F) f (x) is negativeand decreasing.Graph of f is falling.

f f f

ff

f

a b a b a b

a b a b a b

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 296

Finding Inflection Points

Explore & Discuss 2 Discuss the relationship between the change in concavity of each of the follow-ing functions at and the second derivative at and near 0:

(A) (B) (C)

In general, an inflection point is a point on the graph of the function where the con-cavity changes (from upward to downward or from downward to upward). For theconcavity to change at a point, must change sign at that point. But in Section3-2, we saw that the partition numbers* identify the points where a function canchange sign. Thus, we have the following theorem:

f–1x2

h1x2 = x4g1x2 = x4>3f1x2 = x3

x = 0


*As we did with the first derivative, we assume that if is discontinuous at c, then does not exist.f–1c2f–

MATCHED PROBLEM 1 Determine the intervals on which the graph of each function is concave upward andthe intervals on which it is cancave downward. Sketch a graph of each function.

(A) (B) (C)

Refer to Example 1. The graphs of and never change con-cavity. But the graph of changes concavity at This point is called aninflection point.

10, 02.h1x2 = x3g1x2 = ln xf1x2 = ex

h1x2 = x1>3g1x2 = ln 1x

f1x2 = -e-x

THEOREM 1 INFLECTION POINTSIf is continuous on (a, b) and has an inflection point at then either

or does not exist.f–1c2f–1c2 = 0x = c,y = f1x2

Note that inflection points can occur only at partition numbers of but not everypartition number of produces an inflection point. Two additional requirementsmust be satisfied for an inflection point to occur:

A partition number c for produces an inflection point for the graph of f only if

1. changes sign at c and

2. c is in the domain of f.

Figure 5 illustrates several typical cases.

ffl1x2ffl

f–

f–,

(A) f (c) 0 (B) f (c) 0 (C) f (c) 0 (D) f (c) is not defined

c c c c

0 ND 0 0f (x) f (x) f (x) f (x)

FIGURE 5 Inflection points

If exists and changes sign at the tangent line at an inflection point(c, f(c)) will always lie below the graph on the side that is concave upward and abovethe graph on the side that is concave downward (see Fig. 5A, B, and C).

x = c,f–1x2f¿1c2

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 297

From the sign chart, we see that the graph of f has an inflection point at Thatis, the point

is an inflection point on the graph of f.

12, f1222 = 12, 32 f122 = 23- 6 # 22

+ 9 # 2 + 1 = 3

x = 2.


Test Numbers

x

1

3 6 12-6 12

f –1x2

Graph of f2

x

0

Concavedownward

Concaveupward

f (x)

Inflectionpoint

(, 2) (2, )

E X A M P L E 2 Locating Inflection Points Find the inflection point(s) of

SOLUTION Since inflection point(s) occur at values of x where changes sign, we con-struct a sign chart for We have

The sign chart for (partition number is 2) isf–1x2 = 61x - 22 f–1x2 = 6x - 12 = 61x - 22 f¿1x2 = 3x2

- 12x + 9

f1x2 = x3- 6x2

+ 9x + 1

f–1x2. f–1x2f1x2 = x3

- 6x2+ 9x + 1

MATCHED PROBLEM 2 Find the inflection point(s) of

f1x2 = x3- 9x2

+ 24x - 10

E X A M P L E 3 Locating Inflection Points Find the inflection point(s) of

SOLUTION First we find the domain of f (a good first step for most calculus problems involvingproperties of the graph of a function). Since ln x is defined only for is definedonly for

Use completing the square (Section 2-3).True for all x.

So the domain of f is Now we find and construct a sign chart for it.We have

=

2x2- 8x + 10 - 4x2

+ 16x - 16

1x2- 4x + 522

=

1x2- 4x + 522 - 12x - 4212x - 42

1x2- 4x + 522

f–1x2 =

1x2- 4x + 5212x - 42¿ - (2x - 421x2

- 4x + 52¿1x2

- 4x + 522

f¿1x2 =

2x - 4

x2- 4x + 5

f1x2 = ln1x2- 4x + 52

f–1x21- q , q2.1x - 222 + 1 7 0x2

- 4x + 5 7 0

x 7 0, f

f1x2 = ln 1x2- 4x + 52

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 298

The partition numbers for f–1x2 are x = 1 and x = 3.

=

-21x - 121x - 321x2

- 4x + 522

=

-2x2+ 8x - 6

1x2- 4x + 522


x

g(x)f (x)

22x

22

22

2 2

(A) f (x) x4 1x

(B) g(x)

FIGURE 6

Test Numbers

x

0

2

4 -

625

122 12

-

625

12f –1x2

(3, )

Concavedownward

1x

0

(, 1) (1, 3)

Concavedownward

Concaveupward

f (x)

3

0

Inflectionpoint

Inflectionpoint

The sign chart shows that the graph of f has inflection points at x = 1 and x = 3.

MATCHED PROBLEM 3 Find the inflection point(s) of

f1x2 = ln 1x2- 2x + 52

It is important to remember that the partition numbers for are only candidates forinflection points. The function f must be defined at and the second derivative mustchange sign at in order for the graph to have an inflection point at Forexample, consider

In each case, is a partition number for the second derivative, but neither the graphof nor the graph of has an inflection point at Function f does not have aninflection point at because does not change sign at (see Fig. 6A). Func-tion g does not have an inflection point at because g(0) is not defined (see Fig. 6B).x = 0

x = 0f–1x2x = 0x = 0.g1x2f1x2

x = 0

f–1x2 = 12x2 g–1x2 =

2

x3

f¿1x2 = 4x3 g¿1x2 = -

1

x2

f1x2 = x4 g1x2 =

1x

x = c.x = cx = c,

f–

I N S I G H T

Sign chart for f–1x2

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 299

Analyzing GraphsIn the next example, we combine increasing/decreasing properties with concavityproperties to analyze the graph of a function.


x (x) (Fig. 7) f(x) (Fig. 8)

Negative and increasing Decreasing and concave upward

Local maximum Inflection point

Negative and decreasing Decreasing and concave downward

Local minimum Inflection point

Negative and increasing Decreasing and concave upward

x intercept Local minimum

Positive and increasing Increasing and concave upward1 6 x 6 q

x = 1

0 6 x 6 1

x = 0

-2 6 x 6 0

x = -2

- q 6 x 6 -2

f œ

TABLE 1f (x)

x55

5

FIGURE 8

f (x)

x55

5

FIGURE 9

E X A M P L E 4 Analyzing a Graph Figure 7 shows the graph of the derivative of a function f.Use this graph to discuss the graph of f. Include a sketch of a possible graph of f.

x

f (x)

5

55

5

FIGURE 7

SOLUTION The sign of the derivative determines where the original function is increasing anddecreasing, and the increasing/decreasing properties of the derivative determine theconcavity of the original function. The relevant information obtained from the graphof is summarized in Table 1, and a possible graph of f is shown in Figure 8.f¿

MATCHED PROBLEM 4 Figure 9 shows the graph of the derivative of a function f. Use this graph to discussthe graph of f. Include a sketch of a possible graph of f.

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 300

Curve SketchingGraphing calculators and computers produce the graph of a function by plottingmany points. Although the technology is quite accurate, important points on a plotmany be difficult to identify. Using information gained from the function and itsderivatives, and plotting the important points—intercepts, local extrema, and inflec-tion points—we can sketch by hand a very good representation of the graph of .This graphing process is called curve sketching and is summarized next.

f1x2f1x2


Test Numbers

x

1

2 8 12-2 12

-10 12- 1

f ¿1x2x

Decreasing Increasing

(0, w) (w, )

f (x)

f (x)

0x

Decreasing

(, 0)

0 0

Localminimum

w

PROCEDURE Graphing Strategy (First Version)*Step 1. Analyze f(x). Find the domain and the intercepts. The x intercepts are the

solutions of and the y intercept is

Step 2. Analyze Find the partition numbers for, and critical values of, Construct a sign chart for determine the intervals on which f isincreasing and decreasing, and find local maxima and minima.

Step 3. Analyze Find the partition numbers for Construct a sign chartfor , determine the intervals on which the graph of f is concave upwardand concave downward, and find inflection points.

Step 4. Sketch the graph of f. Locate intercepts, local maxima and minima, andinflection points. Sketch in what you know from steps 1–3. Plot additionalpoints as needed and complete the sketch.

f–1x2 f–1x2.f–1x2.f¿1x2, f¿1x2.f¿1x2.

f102.f1x2 = 0

Example will illustrate the use of this strategy.

E X A M P L E 5 Using the Graphing Strategy Follow the graphing strategy and analyze thefunction

State all the pertinent information and sketch the graph of f.

SOLUTION Step 1. Analyze f(x). Since f is a polynomial, its domain is

Step 2. Analyze

Critical values of f(x): 0 and

Partition numbers for

Sign chart for f ¿1x2:f¿1x2: 0 and 32

32

f¿1x2 = 4x3- 6x2

= 4x21x -322f¿1x2.

y intercept: f102 = 0

x = 0, 2

x31x - 22 = 0

x4- 2x3

= 0

x intercept: f1x2 = 0

1- q . q2.

f1x2 = x4- 2x3

*We will modify this summary in Section 5-4 to include some additional information about the graph of f.

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 301

Thus, f(x) is decreasing on is increasing on and has a local minimumat

Step 3. Analyze

Partition numbers for 0 and 1

Sign chart for f–1x2:f–1x2:

f–1x2 = 12x2- 12x = 12x1x - 12f–1x2.

x =32.

132, q2,1- q , 322,302 C H A P T E R 5 Graphing and Optimization

x f(x)

0 0

1

2 0

-2716

32

- 1

f(x)

x

2

1 1 2

1

1

f (x)

Graph of f (x)

Basic shape

0

0

1

0

1.5

f (x)

0 0

x

(, 0) (1.5, )(0, 1)

Decreasing,concave upward

Decreasing,concave downward

Decreasing,concave upward

Increasing,concave upward

(1, 1.5)

FIGURE 10

Test Numbers

x

2 24 12-3 121

2

24 12- 1

f –1x2x

Concavedownward

(0, 1) (1, )

Graph of f

f (x)

0x

Concaveupward

Concaveupward

(, 0)

0 0

Inflectionpoint

Inflectionpoint

1

Thus, the graph of f is concave upward on and is concave down-ward on (0, 1), and has inflection points at and

Step 4. Sketch the graph of f.

x = 1.x = 011, q2,1- q , 02

MATCHED PROBLEM 5 Follow the graphing strategy and analyze the function . State all thepertinent information and sketch the graph of f.

f1x2 = x4+ 4x3

Refer to the solution to Example 5. Combining the sign charts for and (Fig. 10) partitions the real-number line into intervals on which neither nor changes sign. On each of these intervals, the graph of f(x) must have one of four basicshapes (see also Fig. 3, parts A, C, D, and F). This reduces sketching the graph of a func-tion to plotting the points identified in the graphing strategy and connecting them with oneof the basic shapes.

f–1x2f¿1x2f–1x2f¿1x2

I N S I G H T

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 302


E X A M P L E 6 Using the Graphing Strategy Follow the graphing strategy and analyze thefunction

State all the pertinent information and sketch the graph of f. Round any decimalvalues to two decimal places.

SOLUTION Step 1. Analyze f(x).

Since is defined for any x and any positive p, the domain of f is

The x intercepts of f are

y intercept:

Step 2. Analyze

Again,

Critical values of f :


Sign chart for f¿1x2:f: -8, 8

x = 23= 8 and x = 1-223 = -8.

= 51x1>3- 221x1>3

+ 22a2

- b2= 1a - b21a + b2 = 51x2>3

- 42 f¿1x2 = 5x2>3

- 20

f¿1x2.f102 = 0.

x = 0, x = aA203

b3

L 17.21, x = a -A203

b3

L -17.21

3xax1>3- A20

3 b ax1>3

+ A203

b = 0

1a2- b22 = 1a - b21a + b2 3xax2>3

-

203b = 0

3x5>3- 20x = 0

x intercepts: Solve f1x2 = 0

1- q , q2.xp

f1x2 = 3x5>3- 20x

f1x2 = 3x5>3- 20x

Test Numbers

x

0 20

12 6.21 1212

6.21 12-12

f ¿1x2f (x)

(8, )

8x

0

(, 8) (8, 8)

8

0

Increasing Decreasing Increasing

Localminimum

Localmaximum

So f is increasing on and decreasing on is alocal maximum, and f(8) is a local minimum.

Step 3. Analyze

Partition number for f–: 0

f–1x2 =

103

x-1>3=

10

3x1>3

f¿1x2 = 5x2>3- 20

f–1x2.1-8, 82, f1-821- q , -82 and 18, q2

BARNMC05_0132328186.QXD 02/21/2007 20:57 3

So f is concave downward on is concave upward on and has an in-flection point at

Step 4. Sketch the graph of f.

x = 0.10, q2,1- q , 02,


Test Numbers

x

8 1.67 12-1.67 12-8

f fl1x20

x

ND

(, 0) (0, )

Concavedownward

Concaveupward

f (x)

Inflectionpoint

x

0

8

17.21 0

-64

0

64-8

0-17.21

f1x2

x

f (x)

20101020

60

60

f (x) 3x5/3 20x

Point of Diminishing ReturnsIf a company decides to increase spending on advertising, it would expect sales toincrease. At first, sales will increase at an increasing rate and then increase at a de-creasing rate. The value of x where the rate of change of sales changes from increas-ing to decreasing is called the point of diminishing returns. This is also the pointwhere the rate of change has a maximum value. Money spent after this point may in-crease sales, but at a lower rate. The next example illustrates this concept.

MATCHED PROBLEM 6 Follow the graphing strategy and analyze the function State allthe pertinent information and sketch the graph of f. Round any decimal values totwo decimal places.

f1x2 = 3x2>3- x.

E X A M P L E 7 Maximum Rate of Change Currently, a discount appliance store is selling 200large-screen television sets monthly. If the store invests $x thousand in an advertis-ing campaign, the ad company estimates that sales will increase to

When is rate of change of sales increasing and when is it decreasing? What is thepoint of diminishing returns and the maximum rate of change of sales? Graph N and

on the same coordinate system.

SOLUTION The rate of change of sales with respect to advertising expenditures is

N¿1x2 = 9x2- x3

= x219 - x2

N¿

N1x2 = 3x3- 0.25x4

+ 200 0 … x … 9

Sign chart for f–1x2:

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 304

To determine when is increasing and decreasing, we find the derivativeof

The information obtained by analyzing the signs of and is summarizedin Table 2 (sign charts are omitted).

N–1x2N¿1x2N–1x2 = 18x - 3x2

= 3x16 - x2N¿1x2: N–1x2,N¿1x2


f (x)

x22

3

f (x) ex

g(x)

x3

2

11x

g(x) ln2

x61

y

N(x) 0 N(x) 0

y N(x)

y N(x)N(x) N(x)

Point of diminishing returns

200

400

600

800

2 3 4 5 7 8 9

x N(x)

Increasing Increasing, concave upward

0 Local maximum Inflection point

Decreasing Increasing, concave downward+-6 6 x 6 9

+x = 6

++0 6 x 6 6

N¿(x)N¿(x)N–(x)

TABLE 2

Examining Table 2, we see that is increasing on (0, 6) and decreasing on (6, 9). The point of diminishing returns is and the maximum rate of change is

Note that has a local maximum and N(x) has an inflection pointat x = 6.

N¿1x2N¿162 = 108.x = 6

N¿1x2

MATCHED PROBLEM 7 Repeat Example 7 for

N1x2 = 4x3- 0.25x4

+ 500 0 … x … 12

Answers to Matched Problems 1. (A) Concave downward on 1- q , q2

(B) Concave upward on 10, q2

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 305

(C) Concave upward on and concave downward on 10, q21- q , 02306 C H A P T E R 5 Graphing and Optimization

h(x) x1/3

22

h(x)

x

2

2

2. The only inflection point is

3. The inflection points are

4.

1-1, f1-122 = 1-1, ln 82 and 13, f1322 = 13, ln 82.13, f1322 = 13, 82.

x f(x)

Positive and decreasing Increasing and concave downward

Local minimum Inflection point

Positive and increasing Increasing and concave upward

Local maximum Inflection point

Positive and decreasing Increasing and concave downward

x intercept Local maximum

Negative and decreasing Decreasing and concave downward2 6 x 6 q

x = 2

1 6 x 6 2

x = 1

-1 6 x 6 1

x = -1

- q 6 x 6 -1

f ¿1x2

x

f(x)

21235

10

2030

2030

x f(x)

0

0 0

-16-2

-27-3

-4

f (x)

10

x20

30

x f(x)

0 0

8 4

27 0

5. x intercepts: y intercept:

Decreasing on increasing on local minimum at

Concave upward on and concave downward on

Inflection points at and x = 0x = -2

1-2, 0210, q2;1- q , -22

x = -31-3, q2;1- q , -32;

f102 = 0-4, 0;

6. x intercepts: 0, 27; y intercept: f(0) 0

Decreasing on increasing on local minimum: f(0) 0; local maximum: f(8) 4

Concave downward on no inflection points

1- q , 02 and 10, q2;10, 82;1- q , 02 and 18, q2;

f (x)

5

x55

5

BARNMC05_0132328186.QXD 02/21/2007 20:57 Page 306

7. is increasing on (0, 8) and decreasing on (8, 12). The point of diminishing returns isand the maximum rate of change is N¿182 = 256.x = 8

N¿1x2S e c t i o n 5 - 2 Second Derivative and Graphs 307

x8 12

y

N(x) 0 N(x) 0

y N(x)

y N(x)

Point of diminishing returns

400

800

1200

1600

2000

2400

N(x)N(x)

e f ha

b dc g

g(x)

x

f (x)

xa b

(A)

f (x)

xa b

(B)

f (x)

xa b

(C)

f (x)

xa b

(D)

A1. Use the graph of to identify

(A) Intervals on which the graph of f is concave upward(B) Intervals on which the graph of f is concave downward(C) Intervals on which (D) Intervals on which (E) Intervals on which is increasing(F) Intervals on which is decreasing(G) The x coordinates of inflection points(H) The x coordinates of local extrema for f ¿1x2

f¿1x2f¿1x2f–1x2 7 0f–1x2 6 0

y = f1x2In Problems 3–6, match the indicated conditions with one of the graphs (A)–(D) shown in the figure.

a b c d e f g h

f (x)

x

2. Use the graph of to identify

(A) Intervals on which the graph of g is concave upward(B) Intervals on which the graph of g is concave downward(C) Intervals on which (D) Intervals on which (E) Intervals on which is increasing(F) Intervals on which is decreasing(G) The x coordinates of inflection points(H) The x coordinates of local extrema for g¿1x2

g¿1x2g¿1x2g–1x2 7 0g–1x2 6 0

y = g1x2

3. and on (a, b)

4. and on (a, b)

5. and on (a, b)

6. and on (a, b)

In Problems 7–18, find the indicated derivative for eachfunction.

7. for

8. for

9. for

10. for

11. for

12. for

13. for

14. for

15. for

16. for

17. for

18. for

In Problems 19–28, find the intervals on which the graph of f is concave upward, the intervals on which the graph of f isconcave downward, and the inflection points.

19.

20. f1x2 = x4+ 6x

f1x2 = x4+ 6x2

y = x2 ln xy–

y =

ln x

x2y–

f1x2 = xe-xf–1x2f1x2 = e-x2

f–1x2y = 1x2

- 1625y–

y = 1x2+ 924y–

y = x3- 24x1>3d2y>dx2

y = x2- 18x1>2d2y>dx2

k1x2 = -6x-2+ 12x-3k–1x2

h1x2 = 2x-1- 3x-2h–1x2

g1x2 = -x3+ 2x2

- 3x + 9g–1x2f1x2 = 2x3

- 4x2+ 5x - 6f–1x2

f–1x2 6 0f¿1x2 6 0

f–1x2 7 0f¿1x2 6 0

f–1x2 6 0f¿1x2 7 0

f–1x2 7 0f¿1x2 7 0

Exercise 5-2

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 307

21.

22.

23.

24.

25.

26.

27.

28.

In Problems 29–36, f(x) is continuous on Use thegiven information to sketch the graph of f.

29.

1- q , q2.f1x2 = e3x

- 9ex

f1x2 = 8ex- e2x

f1x2 = ln 1x2+ 6x + 132

f1x2 = ln 1x2- 2x + 102

f1x2 = x4- 2x3

- 36x + 12

f1x2 = -x4+ 12x3

- 12x + 24

f1x2 = -x3- 5x2

+ 4x - 3

f1x2 = x3- 4x2

+ 5x - 2 33.

on and

on (0, 2);

on

on

34.

on

on and

on

on

35.and are not defined;

on (0, 1) and on and

and are not defined;on on and

36.

is not defined;

on (0, 1) and (1, 2);

on and

is not defined;

on

on

B In Problems 37–58, summarize the pertinent informationobtained by applying the graphing strategy and sketch thegraph of

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50. f1x2 = 3x5- 5x4

f1x2 = 2x6- 3x5

f1x2 = 1x2- 121x2

- 52f1x2 = 1x2

- 422f1x2 = 1x2

+ 321x2- 12

f1x2 = 1x2+ 3219 - x22

f1x2 = -4x1x + 223f1x2 = 16x1x - 123f1x2 = 0.25x4

- 2x3

f1x2 = -0.25x4+ x3

f1x2 = 11 - x21x2+ x + 42

f1x2 = 1x + 121x2- x + 22

f1x2 = 1x - 321x2- 6x - 32

f1x2 = 1x - 221x2- 4x - 82

y = f1x2.

11, q2f–1x2 6 0

1- q , 12;f–1x2 7 0

f–11212, q2;1- q , 02f¿1x2 6 0

f¿1x2 7 0

f¿102 = 0, f¿122 = 0, f¿112f102 = -2, f112 = 0, f122 = 4;

11, q21- q , -12f –1x2 6 01-1, 12;f –1x2 7 0

f–112f–1-121-1, 02;1- q , -12f¿1x2 6 0

11, q2;f¿1x2 7 0f¿112f¿102 = 0, f ¿1-12

f1-12 = 0, f102 = -2, f112 = 0;

10, q2f–1x2 6 0

1- q , 02;f–1x2 7 0

f–102 = 0;

12, q2;1- q , -22f¿1x2 6 0

1-2, 22;f¿1x2 7 0

f¿1-22 = 0, f¿122 = 0;

f1-22 = -2, f102 = 1, f122 = 4;

1- q , 12f–1x2 6 0

11, q2;f–1x2 7 0

f–112 = 0;

f¿1x2 6 0

12, q2;1- q , 02f¿1x2 7 0

f¿102 = 0, f¿122 = 0;

f102 = 2, f112 = 0, f122 = -2;


f (x)

2x

0

0

ND

f (x)

0x

ND

4

0

2

0

f (x)

2x

0

2

0

f (x)

1x

0

2

0

x 0 2 4

f(x) 0 3 1.5 0 -3-1

-1-2-4

x 0 2 4

f(x) 0 0 1 3- 1- 2

- 1- 2- 430.

f (x)

2x

0

2

0

f (x)

1x

0

2

0

x 0 1 2 4 5

f(x) 0 2 1 0- 1- 4

- 3

x 0 2 4 6

f(x) 0 3 0 0 3-2

-2-4

31.

f (x)

0x

ND

1

0

f (x)

0x

ND

2

0

4

0

32.

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 308


x

f(x)

5

55

5

x

f(x)

5

55

5

x

f(x)

5

55

5

x

f(x)

5

55

5

C(x)

x

$100,000

$50,000

500 1,000

Figure for 77 Production costs at plant A

75. Inflation. One commonly used measure of inflation is theannual rate of change of the Consumer Price Index (CPI).A newspaper headline proclaims that the rate of changeof inflation for consumer prices is increasing. What doesthis say about the shape of the graph of the CPI?

76. Inflation. Another commonly used measure of inflation is the annual rate of change of the Producer Price Index(PPI). A government report states that the rate of change of inflation for producer prices is decreasing.What does this say about the shape of the graph of the PPI?

77. Cost analysis. A company manufactures a variety oflighting fixtures at different locations. The total costC(x) (in dollars) of producing x desk lamps per week at plant A is shown in the figure. Discuss the graph of themarginal cost function and interpret the graph of

in terms of the efficiency of the production processat this plant.C¿(x)

C¿1x2

In Problems 63–70, apply steps 1–3 of the graphing strategyto f (x). Use a graphing calculator to approximate (to two deci-mal places) x intercepts, critical values, and x coordinates of in-flection points. Summarize all the pertinent information.

63.

64.

65.

66.

67.

68.

69.

70.

C In Problems 71–74, assume that f is a polynomial.

71. Explain how you can locate inflection points for the graphof by examining the graph of

72. Explain how you can determine where is increasingor decreasing by examining the graph of

73. Explain how you can locate local maxima and minima forthe graph of by examining the graph of

74. Explain how you can locate local maxima and minima forthe graph of by examining the graph ofy = f¿1x2.

y = f1x2

y = f1x2.y = f¿1x2

y = f1x2.f¿1x2

y = f¿1x2.y = f1x2

f1x2 = x5+ 4x4

- 7x3- 20x2

+ 20x - 20

f1x2 = 0.1x5+ 0.3x4

- 4x3- 5x2

+ 40x + 30

f1x2 = -x4+ x3

+ x2+ 6

f1x2 = -x4- x3

+ 2x2- 2x + 3

f1x2 = x4- 12x3

+ 28x2+ 76x - 50

f1x2 = x4- 21x3

+ 100x2+ 20x + 100

f1x2 = x4+ 2x3

- 5x2- 4x + 4

f1x2 = x4- 5x3

+ 3x2+ 8x - 5

51.

52.

53.

54.

55.

56.

57.

58.

In Problems 59–62, use the graph of to discuss thegraph of Organize your conclusions in a table(see Example 4), and sketch a possible graph of

59. 60.

61. 62.

y = f1x2.y = f1x2.

y = f ¿1x2f1x2 = 1 - ln1x - 32f1x2 = ln1x + 42 - 2

f1x2 = 5 - 3 ln x

f1x2 = -4 + 2 ln x

f1x2 = 2e0.5x+ e-0.5x

f1x2 = e0.5x+ 4e-0.5x

f1x2 = 2 - 3e-2x

f1x2 = 1 - e-x

78. Cost analysis. The company in Problem 77 produces the same lamp at another plant. The total cost C(x)(in dollars) of producing x desk lamps per week at plantB is shown in the figure. Discuss the graph of the marginalcost function and interpret the graph of in terms of the efficiency of the production process at

C¿(x)C¿1x2

Applications

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 309

plant B. Compare the production processes at the two plants.

It was found that the demand for the new hot dog is givenapproximately by

where x is the number of hot dogs (in thousands) that can be sold during one game at a price of $p.

P = 8 - 2 ln x 5 … x … 50


$100,000

$50,000

500 1,000

C(x)

x

Figure for 78 Production costs at plant B

79. Revenue. The marketing research department of a com-puter company used a large city to test market the firm’snew product. The department found that the relationshipbetween price p (dollars per unit) and the demand x(units per week) was given approximately by

Thus, weekly revenue can be approximated by

(A) Find the local extrema for the revenue function.(B) On which intervals is the graph of the revenue func-

tion concave upward? Concave downward?

80. Profit. Suppose that the cost equation for the company inProblem 79 is

(A) Find the local extrema for the profit function.(B) On which intervals is the graph of the profit function

concave upward? Concave downward?

81. Revenue. A cosmetics company is planning to introduceand promote a new lipstick line. After test marketing thenew line in a carefully selected large city, the marketingresearch department found that the demand in that city is given approximately by

where x thousand lipsticks were sold per week at a priceof $p each.

p = 10e-x 0 … x … 5

C1x2 = 830 + 396x

R1x2 = xp = 1,296x - 0.12x3 0 6 x 6 80

p = 1,296 - 0.12x2 0 6 x 6 80

(A) Find the local extrema for the revenue function.

(B) On which intervals is the graph of the revenue func-tion concave upward? Concave downward?

82. Revenue. A national food service runs food concessionsfor sporting events throughout the country. The company’smarketing research department chose a particularfootball stadium to test market a new jumbo hot dog.

(A) Find the local extrema for the revenue function.(B) On which intervals is the graph of the revenue func-

tion concave upward? Concave downward?

83. Production: point of diminishing returns. A T-shirt manu-facturer is planning to expand its workforce. It estimatesthat the number of T-shirts produced by hiring x newworkers is given by

When is the rate of change of T-shirt productionincreasing and when is it decreasing? What is the point of diminishing returns and the maximum rate of changeof T-shirt production? Graph T and on the samecoordinate system.

84. Production: point of diminishing returns. A baseball capmanufacturer is planning to expand its workforce. Itestimates that the number of baseball caps produced byhiring x new workers is given by

When is the rate of change of baseball cap productionincreasing and when is it decreasing? What is the point ofdiminishing returns and the maximum rate of change ofbaseball cap production? Graph T and on the samecoordinate system.

85. Advertising: point of diminishing returns. A companyestimates that it will sell N(x) units of a product afterspending $x thousand on advertising, as given by

When is the rate of change of sales increasing and when isit decreasing? What is the point of diminishing returnsand the maximum rate of change of sales? Graph N and


86. Advertising: point of diminishing returns. A companyestimates that it will sell N(x) units of a product afterspending $x thousand on advertising, as given by


on the same coordinate system.N¿

N1x2 = -0.25x4+ 13x3

- 180x2+ 10,000 15 … x … 24

N¿

N1x2 = -0.25x4+ 23x3

- 540x2+ 80,000 24 … x … 45

T ¿

T1x2 = -0.25x4+ 6x3 0 … x … 18

T ¿

T1x2 = -0.25x4+ 5x3 0 … x … 15

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 310

87. Advertising. An automobile dealer uses television adver-tising to promote car sales. On the basis of past records,the dealer arrived at the following data, where x is thenumber of ads placed monthly and y is the number of cars sold that month:

(B) How many ads should the store manager place each month to maximize the rate of change of sales with respect to the number of ads, and howmany CDs can the manager expect to sell with this number of ads? Round answers to the nearest integer.

89. Population growth: bacteria. A drug that stimulatesreproduction is introduced into a colony of bacteria.After t minutes, the number of bacteria is givenapproximately by

(A) When is the rate of growth, increasing?Decreasing?

(B) Find the inflection points for the graph of N.

(C) Sketch the graphs of N and on the samecoordinate system.

(D) What is the maximum rate of growth?

90. Drug sensitivity. One hour after x milligrams of a particu-lar drug are given to a person, the change in body temper-ature T(x), in degrees Fahrenheit, is given by

The rate at which T(x) changes with respect to thesize of the dosage x is called the sensitivity of the body tothe dosage.

(A) When is increasing? Decreasing?(B) Where does the graph of T have inflection points?(C) Sketch the graphs of T and on the same

coordinate system.(D) What is the maximum value of

91. Learning. The time T (in minutes) it takes a person tolearn a list of length n is

(A) When is the rate of change of T with respect to thelength of the list increasing? Decreasing?

(B) Where does the graph of T have inflection points?Graph T and on the same coordinate system.

(C) What is the minimum value of T ¿1n2?T ¿

T1n2 = 0.08n3- 1.2n2

+ 6n n Ú 0

T¿1x2?T ¿

T¿1x2

T¿1x2

T1x2 = x2 a1 -

x

9b 0 … x … 6

N¿

N¿1t2,N1t2 = 1,000 + 30t2

- t3 0 … t … 20

S e c t i o n 5 - 3 L’Hôpital’s Rule 311

Number of Ads Number of Carsx y

10 325

12 339

20 417

30 546

35 615

40 682

50 795

(A) Enter the data in a graphing calculator and find acubic regression equation for the number of cars soldmonthly as a function of the number of ads.

(B) How many ads should the dealer place each month tomaximize the rate of change of sales with respect tothe number of ads, and how many cars can the dealerexpect to sell with this number of ads? Round answersto the nearest integer.

88. Advertising. A music store advertises on the radio topromote sales of CDs. The store manager used pastrecords to determine the following data, where x is thenumber of ads placed monthly and y is the number ofCDs sold that month.

Number of Ads Number of CDs x y

10 345

14 488

20 746

30 1,228

40 1,671

50 1,955

(A) Enter the data in a graphing calculator and find acubic regression equation for the number of CDssold monthly as a function of the number of ads.

Section 5-3 L’HÔPITAL’S RULE Introduction L’Hôpital’s Rule and the Indeterminate Form 00 One-Sided Limits and Limits at L’Hôpital’s Rule and the Indeterminate Form

IntroductionThe ability to evaluate a wide variety of different types of limits is one of the skillsthat are necessary to apply the techniques of calculus successfully. Limits play afundamental role in the development of the derivative and are an important

q>qq

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 311


x

y

3

3

3

3

y x

xBlim x 0xB0

lim x

lim x xB

(A) y x

x

y

3

5

3

y x2

xBlim x2 0xB0

lim x2

lim x2 xB

(B) y x2

x

y

3

3

3

y 1x

1x

xBlim

xB0

limxB0

limxB0

xB

(C) y

1x lim 0

1x

lim 01x

1x

Does not exist1x

x

y

33

y 1x2

1

1x2

xBlim

xB0

limxB0

limxB0

xB

(D) y

lim 0

lim 0

1x2

1x2

1x2

1x2

1x2

FIGURE 1 Limits involving powers of x

graphing tool. In order to deal effectively with graphs, we need to develop someadditional methods for evaluating limits.

In this section, we discuss a powerful technique for evaluating limits of quotientscalled L’Hôpital’s rule. The rule is named after the French mathematician Marquis deL’Hôpital (1661–1704), who is generally credited with writing the first calculus text-book. To use L’Hôpital’s rule, it is necessary to be familiar with the limit propertiesof some basic functions. Figure 1 reviews some limits involving powers of x that wehave discussed earlier.

The limits in Figure 1 are easily extended to functions of the form and In general, if n is an odd integer, limits involving or

as x approaches behave, respectively, like the limits of x andas x approaches If n is an even integer, limits involving these

expressions behave, respectively, like the limits of and 1x2 as x approaches0 1or ; q2. x2

0 1or ; q2.1>x c 1or ; q21>1x - c2n 1x - c2ng1x2 = 1>1x - c2n.1x - c2nf1x2 =

E X A M P L E 1 Limits Involving Powers of

(A) Compare with in Figure 1.

(B) Compare with in Figure 1.

(C) Compare with in Figure 1.

(D) Compare with in Figure 1.limx: - q

xlimx: - q

3x3= - q

limx: q

1

x 2limx: q

4

1x - 926 = 0

limx:0-

1x

limx: -1-

4

1x + 123 = - q

limx:0

1x 2lim

x:2

5

1x - 224 = q

x c

MATCHED PROBLEM 1 Evaluate each limit.

(A) (B)

(C) (D) limx: q

5x4limx:- q

3

1x + 223

limx: -4

6

1x + 426limx:3 +

7

1x - 325

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 312

Figure 2 reviews some limits of exponential and logarithmic functions.The limits in Figure 2 also generalize to other simple exponential and logarithmic

forms.


x

y

3

5

3

y ex

xBlim ex 1xB0

lim ex

lim ex 0xB

(A) y ex

x

y

3

5

3

y ex

xBlim ex 1xB0

lim ex 0

lim ex xB

(B) y ex

x

y

5

3

0

3

y ln x

xBlim ln x lim ln x

(C) y ln x

xB0

FIGURE 2 Limits involving exponential and logarithmic functions

E X A M P L E 2 Limits Involving Exponential and Logarithmic Forms

(A) Compare with in Figure 2.

(B) Compare with in Figure 2.

(C) Compare with in Figure 2.

(D) Compare with in Figure 2.limx:0+ ln xlim

x:2 +

ln1x - 22 = - q

limx: q

ln xlimx: q

ln1x + 42 = q

limx: q

e-xlimx: q

4e-5x= 0

limx : q

e xlimx: q

2e3x= q

MATCHED PROBLEM 2 Evaluate each limit.

(A) (B)

(C) (D) limx: q

ln1x - 102limx: - 4+

ln1x + 42lim

x:- q

3e2xlimx: - q

2e- 6x

Now that we have reviewed the limit properties of some basic functions, we areready to consider the main topic of this section: L’Hôpital’s rule—a powerful tool forevaluating certain types of limits.

Explore & Discuss 1 It follows from Figure 2 that

Does

exist? If so, what is its value? Use graphical methods to support your answers.

L’Hôpital’s Rule and the Indeterminate Form Recall that the limit

is a indeterminate form if

limx:c

f1x2 = 0 and limx:c

g1x2 = 0

0>0limx:c

f1x2g1x2

0/0

limx: q

ln11 + 3e-x2

e-x

limx: q

e-x= 0 and lim

x: q

ln11 + 3e-x2 = 0

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 313

The quotient property for limits in Section 3-1 does not apply, since If we are dealing with a indeterminate form, the limit may or may not exist,

and we cannot tell which is true without further investigation.Each of the following is a 00 indeterminate form:

The first limit can be evaluated by performing some algebraic simplifications, such as

The second cannot. Instead, we turn to the powerful L’Hôpital’s rule, which we nowstate without proof. This rule can be used whenever a limit is a 00 indeterminateform.

limx:2

x2

- 4x - 2

= limx:2

1x - 221x + 22

x - 2= lim

x:21x + 22 = 4

limx:2

x2

- 4x - 2

and limx:1

ex

- e

x - 1

0>0limx:c

g1x2 = 0.


THEOREM 1 L’HÔPITAL’S RULE FOR INDETERMINATE FORMS: VERSION 1For c a real number,if and then

provided that the second limit exists or is or - q .+ q

lim

x:c

f 1x2g1x2 = lim

x:c

f ¿1x2g¿1x2

limx:c g1x2 = 0,lim

x:c f1x2 = 0

0/0

The use of L’Hôpital’s rule is best illustrated through examples.

E X A M P L E 3 L’Hôpital’s Rule Evaluate .

SOLUTION Step 1. Check to see if L’Hôpital’s rule applies:

L’Hôpital’s rule does apply.

Step 2. Apply L’Hôpital’s rule:

Apply L’Hôpital’s rule.

e x is continuous at x 1.

=

e1

1= e

= limx:1

ex

1

limx:1

ex

- e

x - 1 = lim

x:1

d

dx1e x

- e2d

dx1x - 12

limx:11e x

- e2 = e1- e = 0 and lim

x:11x - 12 = 1 - 1 = 0

limx:1

ex

- e

x - 1

MATCHED PROBLEM 3 Evaluate limx:4

ex

- e4

x - 4.

00 form

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 314

The functions

of Example 3 are different functions (see Fig. 3), but both functions have the samelimit e as x approaches 1. Although is undefined at the graph of providesa check of the answer to Example 3.

y1x = 1,y1

y1 =

e x- e

x - 1and y2 =

e x

1


In L’Hôpital’s rule, the symbol represents the derivative of divided bythe derivative of not the derivative of the quotient

When applying L’Hôpital’s rule to a indeterminate form, be certain that youdifferentiate the numerator and denominator separately.

0>0f1x2>g1x2.g1x2,

f1x2f¿1x2>g¿1x2I N S I G H T

FIGURE 3

2

4

2

y

y2y1

x

E X A M P L E 4 L’Hôpital’s Rule Evaluate .





Simplify.

= q

= limx:0

1

2x211 + x22

= limx:0

2x

1 + x2

4x3

1

4x3

1

4x3

limx:0

ln11 + x22

x4 = limx:0

2x

1 + x2

4x3

limx:0

ln11 + x22x4 = lim

x:0

d

dx ln11 + x22

d

dx x4

limx:0

ln11 + x22 = ln 1 = 0 and limx:0

x4= 0

limx:0

ln11 + x22

x4

00 form

MATCHED PROBLEM 4 Evaluate .limx:1

ln x

1x - 123

E X A M P L E 5 L’Hôpital’s Rule May Not Be Applicable Evaluate .


L’Hôpital’s rule does not apply.Step 2. Evaluate by another method. The quotient property for limits from Sec-

tion 3-1 does apply, and we have

limx:1

ln x = ln 1 = 0, but limx:1

x = 1 Z 0

limx:1

ln x

x

Multiply numerator and denominator by 1/4x3.

Apply Theorem 1 in Section 3-3 andcompare with Fig. 1(D).

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 315

Note that applying L’Hôpital’s rule would give us an incorrect result:

limx:1

ln xx

Z lim x:1

d

dx ln x

d

dx x

= limx:1

1>x1

= 1

limx:1

ln xx

=

limx:1

ln x

limx:1

x=

ln 11

=

01

= 0


MATCHED PROBLEM 5 Evaluate .limx:0

x

e x

As Example 5 illustrates, all limits involving quotients are not indeterminate forms.

You must always check to see if L’Hôpital’s rule applies before you use it.

0>0I N S I G H T

E X A M P L E 6 Repeated Application of L’Hôpital’s Rule Evaluate


L’Hôpital’s rule does apply.Step 2. Apply L’Hôpital’s rule:

Since and , the new limit obtained is also a indeterminate form, and L’Hôpital’s rule can be applied again.Step 3. Apply L’Hôpital’s rule again:

Thus,

limx:0

x2

ex- 1 - x

= limx:0

2x

ex- 1

= limx:0

2ex = 2

limx:0

2x

ex- 1

= limx:0

d

dx 2x

d

dx1ex

- 12 = lim

x:0 2ex =

2

e0 = 2

0>0limx:01e x- 12 = 0limx:02x = 0

limx:0

x2

ex- 1 - x

= limx:0

d

dx x2

d

dx1ex

- 1 - x2 = lim

x:0

2x

ex- 1

limx:0

x2= 0 and lim

x:01ex

- 1 - x2 = 0

limx:0

x2

ex- 1 - x

form00

form00

MATCHED PROBLEM 6 Evaluate: limx:0

e2x

- 1 - 2x

x2

One-Sided Limits and Limits at qqIn addition to examining the limit as x approaches c, we have discussed one-sidedlimits and limits at q in Chapter 3. L’Hôpital’s rule is valid in these cases also.

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 316

For example, if and then

provided that the second limit exists or is or Similar rules can be writtenfor and x : - q .x : c+, x : c-,

- q .+ q

limx: q

f1x2g1x2 = lim

x: q

f¿1x2g¿1x2

limx: q

g1x2 = 0,limx: q

f1x2 = 0


THEOREM 2 L’HÔPITAL’S RULE FOR INDETERMINATE FORMS: VERSION 2(FOR ONE-SIDED LIMITS AND LIMITS AT INFINITY)The first version of L’Hôpital’s rule (Theorem 1) remains valid if the symbol is replaced everywhere it occurs with one of the following symbols:

x : c+ x : c- x : q x : - q

x : c

0/0

E X A M P L E 7 L’Hôpital’s Rule for One-Sided Limits Evaluate .





Simplify.

The limit as is because has a vertical asymptote at (Theorem 1, Section 3-3) and x1x - 12 7 0 for x 7 1.

x = 11>2x1x - 12qx : 1+

= q

= limx:1+

1

2x1x - 12

= limx:1+

1>x

21x - 12

limx:1+

ln x

1x - 122 = limx:1+

d

dx1ln x2

d

dx1x - 122

limx:1 +

ln x = 0 and limx:1 +

1x - 122 = 0

limx:1+

ln x

1x - 122

form00

MATCHED PROBLEM 7 Evaluate .limx:1-

ln x

1x - 122

E X A M P L E 8 L’Hôpital’s Rule for Limits at Infinity Evaluate .



limx: q

ln11 + e-x2 = ln11 + 02 = ln 1 = 0 and limx: q

e-x= 0

limx: q

ln11 + e-x2

e-x

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 317



Multiply numerator anddenominator by .

=

11 + 0

= 1

limx: q

e- x= 0lim

x: q

ln11 + e-x2

e-x

= limx: q

1

1 + e-x

-ex = limx: q

-e-x>11 + e-x2

-e-x

limx: q

ln 11 + e-x2

e-x = limx: q

d

dx 3ln 11 + e-x24

d

dx e-x


form00

MATCHED PROBLEM 8 Evaluate .limx: - q

ln11 + 2e x2ex

L’Hôpital’s Rule and the Indeterminate Form In Section 3-3, we discussed techniques for evaluating limits of rational functionssuch as

(1)

Each of these limits is an indeterminate form. In general, if and then

is called an indeterminate form. Furthermore, can be replaced in allthree limits above with or It can be shown thatL’Hôpital’s rule also applies to these indeterminate forms.q>q x : - q .x : c+, x : c-, x : q ,

x : c

limx:c

f 1x2g1x2

limx:cg1x2 = ; q ,limx:c f1x2 = ; qq>q

limx: q

2x2

x3+ 3

limx: q

4x3

2x2+ 5

limx: q

3x3

5x3+ 6

THEOREM 3 L’HÔPITAL’S RULE FOR THE INDETERMINATE FORM : VERSION 3Versions 1 and 2 of L’Hôpital’s rule for the indeterminate form are also valid if thelimit of f and the limit of g are both infinite; that is, both and are permissiblefor either limit.

- q+ q

0>0

For example, if and , then L’Hôpital’s rulecan be applied to

Explore & Discuss 2 Evaluate each of the limits in (1) in two ways:

1. Use Theorem 4 in Section 3-3.

2. Use L’Hôpital’s rule.

Given a choice, which method would you choose? Why?

limx:c+ 3f1x2>g1x24. limx:c+ g1x2 = - qlimx:c+ f1x2 = q

E X A M P L E 9 L’Hôpital’s Rule for the Indeterminate Form Evaluate .



limx: q

ln x = q and limx: q

x2= q

limx: q

ln x

x2

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 318



Simplify.

See Figure 1(D).

= 0

limx: q

ln x

x2 = limx: q

1

2x2

= limx: q

1>x2x

limx: q

ln x

x2 = limx: q

d

dx1ln x2d

dx x2


formqq

MATCHED PROBLEM 9 Evaluate .limx: q

ln x

x

E X A M P L E 1 0 L’Hôpital’s Rule for the Indeterminate Form Evaluate .




Since and this limit is an indeterminate formand L’Hôpital’s rule can be applied again.

Step 3. Apply L’Hôpital’s rule again:

Thus,

limx: q

ex

x2 = limx: q

ex

2x= lim

x: q

ex

2= q

limx: q

ex

2x = lim

x: q

d

dx ex

d

dx 2x

= limx: q

ex

2= q

q>qlimx: q 2x = q ,limx: q e x= q

limx: q

ex

x2 = limx: q

d

dx ex

d

dx x2

= limx: q

ex

2x

limx: q

e x= q and lim

x: q

x2= q

limx: q

ex

x2

formqq

formqq

MATCHED PROBLEM 10 Evaluate .limx: q

e2x

x2

Explore & Discuss 3 Let n be a positive integer. Explain how L’Hôpital’s rule can be used to show that

Does this imply that increases more rapidly than any power of x?ex

limx: q

ex

xn = q

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 319


The three versions of L’Hôpital’s rule cover a multitude of limits—far too many to re-member case by case. Instead, we suggest you use the following pattern, common to allversions, as a memory aid:

1. All versions involve three limits: lim f(x)g(x), lim f(x), and lim g(x).2. The independent variable x must behave the same in all three limits. The acceptable

behaviors are

3. The form of lim f(x)g(x) must be or and both lim f(x) and lim g(x) must

approach 0 or both must approach ; q .

; q

; q

00

x : c, x : c+, x : c-, x : q , or x : - q .

I N S I G H T

Answers to Matched Problems 1. (A) q (B) q (C) 0 (D) q

A Use L’Hôpital’s rule to find each limit in Problems 1–14.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

In Problems 15–18, explain why L’Hôpital’s rule does notapply. If the limit exists, find it by other means.

15. 16.

17. 18.

B Find each limit in Problems 19–42. Note that L’Hôpital’s ruledoes not apply to every problem, and some problems willrequire more than one application of L’Hôpital’s rule.

19. 20.

21. 22.

23. 24.

25. 26. limx:0+

ln11 + x21xlim

x:0+

ln11 + 1x2x

limx:0-

ln11 + 2x2x2

limx:0+

ln11 + x22x3

limx: -1

ln1x + 22

x + 2limx:2

ln1x - 12

x - 1

limx:0

3x + 1 - e3x

x2limx:0

e4x

- 1 - 4x

x2

limx: -3

x2

1x + 325limx:2

x + 2

1x - 224

limx: q

e-x

ln xlimx:1

x2

+ 5x + 4

x3+ 1

limx: q

ln x

x4limx: q

x2

ln x

limx: q

x

e4xlim

x: q

e3x

x

limx: q

3x4

+ 6

2x2+ 5

limx: q

2x2

+ 7

5x3+ 9

limx:0

e2x

- 1x

limx:0

ln11 + 4x2

x

limx:1

ln x

x - 1limx:0

ex

- 1x

limx:4

x2

- 8x + 16

x2- 5x + 4

limx:2

x2

+ x - 6

x2+ 6x - 16

limx:1

x6

- 1

x5- 1

limx:2

x4

- 16

x3- 8

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 42.

C 43. Find .

[Hint: Write

44. Find .

[Hint: Write

In Problems 45–48, n is a positive integer. Find each limit.

45. 46.

47. 48.

In Problems 49–52, show that the repeated application of L’Hôpital’s rule does not lead to a solution. Then use algebraic manipulation to evaluate each limit. [Hint: For and

]

49. 50.

51. 52. limx: q

x2

321x3+ 122lim

x: - q

32x3

+ 1x

limx: - q

x24 + x2

limx: q

21 + x2

x

n 7 0, n1xn= x.

x 7 0

limx: q

xn

exlimx: q

ex

xn

limx: q

xn

ln xlim

x: q

ln xxn

1x ln x = 1ln x2>x- 1>2.4lim

x:0+

11x ln x2x ln x = 1ln x2>x- 1.4

limx:0+

1x ln x2limx:0

e2x

- 1 - 2x - 2x2

x3limx:0

ex

- e-x- 2x

x3

limx: q

ln11 + 2e-x2ln11 + e-x2lim

x: q

e-x

ln11 + 4e-x2

limx: - q

1 + e-x

1 + x2limx: q

1 + e-x

1 + x2

limx: q

e3x

x3limx: q

x2

e2x

limx: q

4x2

+ 9x

5x2+ 8

limx: q

3x2

+ 5x

4x3+ 7

limx:1+

x3- x2

- x + 1

x3- 3x2

+ 3x - 1lim

x:2-

x3- 12x + 16

x3- 6x2

+ 12x - 8

limx:3

x3

+ 3x2- x - 3

x2+ 6x + 9

limx: -1

x3

+ x2- x - 1

x3+ 4x2

+ 5x + 2

limx:1

2x3

- 3x2+ 1

x3- 3x + 2

limx: -2

x2

+ 2x + 1

x2+ x + 1

EXERCISE 5-3

2. (A) q (B) 0 (C) q (D) q

3. 4. q 5. 0 6. 2 7. - qe4 8. 2 9. 0 10. q

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 320

Section 5-4 CURVE-SKETCHING TECHNIQUES Modifying the Graphing Strategy Using the Graphing Strategy Modeling Average Cost

When we summarized the graphing strategy in Section 5-2, we omitted one impor-tant topic: asymptotes. Polynomial functions do not have any asymptotes. Asymp-totes of rational functions were discussed in Section 3-3, but what about all the otherfunctions, such as logarithmic and exponential functions? Since investigating asymp-totes always involves limits, we can now use L’Hôpital’s rule (Section 5-3) as a tool forfinding asymptotes of many different types of functions.

Modifying the Graphing StrategyThe first version of the graphing strategy in Section 5-2 made no mention of asymp-totes. Including information about asymptotes produces the following (and final)version of the graphing strategy:

S e c t i o n 5 - 4 Curve-Sketching Techniques 321

PROCEDURE Graphing Strategy (Final Version)Step 1. Analyze

(A) Find the domain of f.

(B) Find the intercepts.

(C) Find asympotes.

Step 2. Analyze Find the partition numbers for, and critical values of, Construct a sign chart for determine the intervals on which f isincreasing and decreasing, and find local maxima and minima.

Step 3. Analyze Find the partition numbers of Construct a sign chartfor determine the intervals on which the graph of f is concave upwardand concave downward, and find inflection points.

Step 4. Sketch the graph of f. Draw asymptotes and locate intercepts, local maximaand minima, and inflection points. Sketch in what you know from steps 1–3.Plot additional points as needed and complete the sketch.

f–(x),f–(x).f–(x).

f¿(x),f¿(x).f¿(x).

f(x).

Using the Graphing Strategy

From now on, you should always use the final version of the graphing strategy. If a functiondoes not have any asymptotes, simply state this fact.

I N S I G H T

We will illustrate the graphing strategy with several examples:

E X A M P L E 1 Using the Graphing Strategy Use the graphing strategy to analyze the func-tion State all the pertinent information and sketch thegraph of f.

SOLUTION Step 1. Analyze f(x).

(A) Domain: All real x, except

(B) y intercept:

x intercepts: Since a fraction is 0 when its numerator is 0 and its denomi-nator is not 0, the x intercept is x = 1.

f102 =

0 - 10 - 2

=

12

x = 2

f1x2 =

x - 1x - 2

f1x2 = 1x - 12>1x - 22.

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 321

(C) Horizontal asymptote:

Thus, the line is a horizontal asymptote.

Vertical asymptote: The denominator is 0 for and the numerator isnot 0 for this value. Therefore, the line is a vertical asymptote.

Step 2. Analyze

Critical values of f(x): None

Partition number for

Sign chart for f¿1x2:f¿1x2: x = 2

f ¿1x2 =

1x - 22112 - 1x - 121121x - 222 =

-1

1x - 222f¿1x2.x = 2

x = 2,

y = 1

amxm

bnxn =

xx

= 1


Thus, f(x) is decreasing on and There are no local extrema.

Step 3. Analyze

Partition number for

Sign chart for f–1x2:f–1x2: x = 2

f–1x2 =

2

1x - 223f–1x2.12, q2.1- q , 22

Thus, the graph of f is concave downward on and concave upward on Since f(2) is not defined, there is no inflection point at even though changes sign at

Step 4. Sketch the graph of f. Insert intercepts and asymptotes, and plot a few addi-tional points (for functions with asymptotes, plotting additional points is oftenhelpful). Then sketch the graph:

x = 2.f–1x2x = 2,12, q2.1- q , 22

2

ND


(2, )(, 2)

f (x)

f (x)

x

2

ND

Concaveupward

Concavedownward

(2, )(, 2)

Graph of f

f (x)

x

x

f (x)

0 2 5

5

Test Numbers

x

1

3 -1 12-1 12

f œ1x2

Test Numbers

x

1

3 2 12-2 12

f fl1x2

x f(x)

0

1 0

3

3 2

4 32

52

-132

12

34-2

MATCHED PROBLEM 1 Follow the graphing strategy and analyze the function . State allthe pertinent information and sketch the graph of f.

f1x2 = 2x>11 - x2

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 322


E X A M P L E 2 Using the Graphing Strategy Use the graphing strategy to analyze the function

State all pertinent information and sketch the graph of g.

SOLUTION Step 1. Analyze g(x).

(A) Domain: All real x, except

(B) x intercept:

y intercept: Since 0 is not in the domain of g, there is no y intercept.

(C) Horizontal asymptote: (the x axis)

Vertical asymptote: The denominator of g(x) is 0 at and thenumerator is not. So the line (the y axis) is a vertical asymptote.

Step 2. Analyze

Critical values of


Sign chart for g¿1x2:g¿1x2: x = 0, x = 1

g1x2: x = 1

=

211 - x2x3

g¿1x2 = -2x-2+ 2x-3

= -

2

x2 +

2

x3 =

-2x + 2

x3

g1x2 =

2x - 1

x2 =

2x

-

1

x2 = 2x-1- x-2

g¿1x2.x = 0

x = 0

y = 0

x =

12

= 0.5

x = 0

g1x2 =

2x - 1

x2

x

(0, 1) (1, )

g(x)

0x

(, 0)

ND 0

1

Function f(x) is decreasing on and is increasing on (0, 1), and has alocal maximum at

Step 3. Analyze


Sign chart for g–1x2:g–1x2: x = 0, x =

32

= 1.5

g–1x2 = 4x-3- 6x-4

=

4

x3 -

6

x4 =

4x - 6

x4 =

212x - 32x4

g¿1x2 = -2x-2+ 2x-3

g–1x2.x = 1.

11, q2,1- q , 02

x

(0, 1.5) (1.5, )

0x

(, 0)

ND 0

1.5

g(x)

Function g(x) is concave downward on and (0, 1.5), is concave upward onand has an inflection point at x = 1.5.11.5, q2, 1- q , 02

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 323

Step 4. Sketch the graph of g. Plot key points, note that the coordinate axes areasymptotes, and sketch the graph.


x

g(x)

1

510 5 10

Horizontalasymptote

Verticalasymptote

1

2

3

x g(x)

0.5 0

1 1

1.5 0.89

10 0.19

- 3- 1

- 0.21- 10

MATCHED PROBLEM 2 Use the graphing strategy to analyze the function

State all pertinent information and sketch the graph of h.

h1x2 =

4x + 3

x2

E X A M P L E 3 Graphing Strategy Follow the steps of the graphing strategy and analyze thefunction . State all the pertinent information and sketch the graph of f.

SOLUTION Step 1. Analyze f(x):

(A) Domain: All real numbers

(B) y intercept:

x intercept: for only, since for all x.

(C) Vertical asymptotes: None

Horizontal asymptotes: We use tables to determine the nature of the graphof f as and x : - q :x : q

ex7 0x = 0xex

= 0

f102 = 0

f1x2 = xex.

f1x2 = xex

Step 2. Analyze

Critical value of f(x):

Partition number for Sign chart for f¿1x2:

f¿1x2: -1

-1

= xex+ ex

= ex1x + 12 f¿1x2 = x

d

dx ex

+ ex d

dx x

f¿1x2:

x 1 5 10

f(x) 2.72 742.07 220,264.66 : q

: q

x

f(x) : 0-0.000 45-0.03-0.37

: - q-10-5-1

Test Numbers

x

0 1 1212- e-2

- 2

f œ1x2

f (x)1

x

0


f (x)

(, 1) (1, )

Thus, f(x) decreases on has a local minimum at and increaseson (Since for all x, we do not have to evaluate to conclude that

when using the test number )-2.-e-26 0

e-2ex7 01-1, q2. x = -1,1- q , -12,

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 324

Step 3. Analyze

Sign chart for (partition number is ):-2f–1x2 = ex

+ 1x + 12ex= ex1x + 22

f–1x2 = ex d

dx 1x + 12 + 1x + 12

d

dx ex

f–1x2:S e c t i o n 5 - 4 Curve-Sketching Techniques 325

Test Numbers

x

12e-1-1

12-e-3-3

f–1x2Graph of f

2x

0

Concavedownward

Concaveupward

f (x)

Inflectionpoint

(, 2) (2, )

Thus, the graph of f is concave downward on has an inflection point atand is concave upward on

Step 4. Sketch the graph of f, using the information from steps 1 to 3:

1-2, q2.x = -2,1- q , -22,

x f(x)

2 0.27

1 0.37

0 0

f (x)

x14

3

2

1

MATCHED PROBLEM 3 Analyze the function State all the pertinent information and sketchthe graph of f.

f1x2 = xe-0.5x.

Remember, if p is any real number, then The sign of the exponent p does notmatter. This is a useful fact to remember when you work with sign charts involvingexponential forms.

ep7 0.

I N S I G H T

Explore & Discuss 1 Refer to the discussion of vertical asymptotes in the solution of Example 3. Weused tables of values to estimate limits at infinity and determine horizontalasymptotes. In some cases, the functions involved in these limits can be writtenin a form that allows us to use L’Hôpital’s rule.

formRewrite as a fraction.

form


Simplify.

Property of ex

= 0

= limx: -q

(-ex)

= limx: -q

1

-e-x

= limx: -q

x

e-x

- q>q limx: - q

f1x2 = limx: -q

xex

- q # 0

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 325

Use algebraic manipulation and L’Hôpital’s rule to verify the value of each of thefollowing limits:

(A)

(B)

(C) limx:0+

x ln x = 0

limx:0+

x2(ln x - 0.5) = 0

limx: q

xe-0.5x= 0


E X A M P L E 4 Graphing Strategy Let Follow the steps in the graphingstrategy and analyze this function. State all the pertinent information and sketch thegraph of f.

SOLUTION Step 1. Analyze

(A) Domain:

(B) y intercept: None [ f(0) is not defined.]

x intercept: Solve

Discard, since 0 is not in the domain of f.

if and only if x intercept

(C) Asymptotes: None.The following tables suggest the nature of the graphas and as x : q :x : 0+

x = e0.5

x = ea.ln x = a ln x = 0.5

ln x - 0.5 = 0 or x2= 0

x21ln x - 0.52 = 0

10, q2f1x2: f1x2 = x2 ln x - 0.5x2

= x21ln x - 0.52.

f1x2 = x2 ln x - 0.5x2.

Step 2. Analyze

Critical value of f(x): 1Partition number for Sign chart for f¿1x2:

f¿1x2: 1

= 2x ln x

= x + 2x ln x - x

= x2 1x

+ 1ln x2 2x - 0.512x2 f¿1x2 = x2

d

dx ln x + 1ln x2

d

dx x2

- 0.5 d

dx x2

f¿1x2:

x 0.1 0.01 0.001

f(x) : 0-0.00007-0.00051-0.0280

: 0+

x 10 100 1,000

f(x) 180 41,000 6,400,000 : q

: q

Test Numbers

x

0.5

2 0.7726 1212-0.2983

f œ1x2

1x

0


(0, 1) (1, )

The function f(x) decreases on (0, 1), has a local minimum at and increases on11, q2. x = 1,

See Explore &Discuss 1(B).

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 326

Step 3. Analyze

Sign chart for (partition number is ):e-1f¿1x2 x = e-1

L 0.3679 ln x = -1

2 ln x = -2 = 2 + 2 ln x = 0

= 2x 1x

+ 1ln x2 2 f–1x2 = 2x

d

dx 1ln x2 + 1ln x2

d

dx 12x2

f–1x2:S e c t i o n 5 - 4 Curve-Sketching Techniques 327

Test Numbers

x

0.2

1 2 1212-1.2189

f œ1x2

x f(x)

1

0e0.5

-0.5

-1.5e-2e-1

e1x

0

Concavedownward

Concaveupward

(0, e1) (e1, )

f (x)

x1 2

0.5

0.5

The graph of f(x) is concave downward on has an inflection point at and is concave upward on

Step 4. Sketch the graph of f, using the information from steps 1 to 3:

1e-1, q2. x = e-1,10, e-12,

MATCHED PROBLEM 4 Analyze the function State all pertinent information and sketch thegraph of f.

f1x2 = x ln x.

Modeling Average Cost

E X A M P L E 5 Average Cost Given the cost function where x is the num-ber of items produced, use the graphing strategy to analyze the graph of the average costfunction. State all the pertinent information and sketch the graph of the average costfunction.Find the marginal cost function and graph it on the same set of coordinate axes.

SOLUTION The average cost function is

Step 1. Analyze

(A) Domain: Since negative values of x do not make sense and is notdefined, the domain is the set of positive real numbers.

(B) Intercepts: None

(C) Horizontal asymptote:amxm

bnxn =

0.5x2

x= 0.5x

C102C1x2.

C1x2 =

5,000 + 0.5x2

x=

5,000x

+ 0.5x

C1x2 = 5,000 + 0.5x2,

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 327

Thus, there is no horizontal asymptote.

Vertical asymptote: The line is a vertical asymptote, since thedenominator is 0 and the numerator is not 0 for

Oblique asymptotes: If a graph approaches a line that is neither horizon-tal nor vertical as x approaches or , that line is called an obliqueasymptote. If x is a large positive number, then 5,000x is very small and

That is,

This implies that the graph of approaches the line as xapproaches That line is an oblique asymptote for the graph of[More generally, whenever is a rational function forwhich the degree of is 1 more than the degree of , we can usepolynomial long division to write , where thedegree of is less than the degree of The line is thenan oblique asymptote for the graph of .]

Step 2. Analyze

Critical value for

Partition numbers for 0 and 100

Sign chart for C¿1x2:C¿1x2:

C1x2: 100

=

0.51x - 10021x + 1002x2

=

0.5x2- 5,000

x2

C¿1x2 = -

5,000

x2 + 0.5

C¿1x2.y = f1x2 y = mx + bd1x2.r1x2 f1x2 = mx + b + r1x2>d1x2d1x2n1x2 f1x2 = n1x2>d1x2 y = C1x2.q .

y = 0.5xy = C1x2lim

x: q

3C1x2 - 0.5x4 = limx: q

5,000

x= 0

C1x2 =

5,000x

+ 0.5x L 0.5x

- qq

x = 0.x = 0


IncreasingDecreasing

(100, )(0, 100)

C(x)

C(x)

1000x

0

Localminimum

Test Numbers

x

50

125 0.18 12-1.5 12C

œ1x2

Thus, is decreasing on (0, 100), is increasing on and has a local mini-mum at

Step 3. Analyze

is positive for all positive x; therefore, the graph of is concave upwardon

Step 4. Sketch the graph of The graph of is shown in Figure 1.CC.

10, q2. y = C1x2C–1x2C–1x2: C–1x2 =

10,000

x3 .

x = 100.1100, q2,C1x2

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 328

The marginal cost function is The graph of this linear function is alsoshown in Figure 1.

The graph in Figure 1 illustrates an important principle in economics:

The minimum average cost occurs when the average cost is equal to the marginalcost.

C¿1x2 = x.


x

y

100 200 300 400

100

200

C(x) x

Minimum average cost

5,000x

C(x) 0.5x

y 0.5x(oblique asymptote)

FIGURE 1

MATCHED PROBLEM 5 Given the cost function where x is the number of itemsproduced,

(A) Use the graphing strategy to analyze the graph of the average cost function.State all the pertinent information and sketch the graph of the average costfunction. Find the marginal cost function and graph it on the same set of coor-dinate axes. Include any oblique asymptotes.

(B) Find the minimum average cost.

C1x2 = 1,600 + 0.25x2,

Answers to Matched Problems 1. Domain: All real x, except y intercept: x intercept: 0Horizontal asymptote: vertical asymptote:Increasing on and Concave upward on concave downward on 11, q21- q , 12;

11, q21- q , 12x = 1y = -2;

f102 = 0;x = 1

x

f(x)

55

5

5

x f(x)

0 0

2

2

5 -52

-4

-632

12

-1-1

2. Domain: All real x, except

x intercept:

h(0) is not definedVertical asymptote: (the y axis)Horizontal asymptote: (the x axis)Increasing on Decreasing on and Local minimum at Concave upward on and 10, q21-2.25, 02

x = 1.510, q21- q , -1.52

1-1.5, 02y = 0

x = 0

= -

34

= -0.75

x = 0

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 329

3. Domain: y intercept: x intercept: Horizontal asymptote: (the x axis)Increasing on Decreasing on Local maximum at Concave downward on Concave upward on Inflection point at

4. Domain: y intercept: None [f(0) is not defined]x intercept: Increasing on Decreasing on Local minimum at Concave upward on 10, q2

x = e-1L 0.368

10, e-121e-1, q2

x = 1

10, q2x = 414, q21- q , 42

x = 212, q21- q , 22

y = 0x = 0f102 = 0

1- q , q2


x 5 10 100

f(x) 8.05 23.03 460.52 : q

: q

x 0.1 0.01 0.001 0.000 1

f(x) : 0-0.000 92-0.006 9-0.046-0.23

: 0

x211

f (x)

1

2

1

x822

f (x)

1

1

4 6

Concave downward on Inflection point at x = -2.25

1- q , -2.252

5. (A) Domain:Intercepts: NoneVertical asymptote: oblique asymptote:Decreasing on (0, 80); increasing on local minimum at Concave upward on 10, q2

x = 80180, q2;y = 0.25xx = 0;

10, q2

h(x)

x

5

10 10

x h(x)

0

2 2.75

10 0.43

-0.75

-1.33-1.5

-1.19-2.25

-0.37-10

x

y

80 160 240

40

80

120C(x) 0.5x

y 0.25x(oblique asymptote)

1,600x

C(x) 0.25x

(B) Minimum average cost is 40 at x = 80.

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 330


A1. Use the graph of f to identify

4. Domain: All real x; limx: - q

f1x2 = -3; limx: q

f1x2 = 3

Exercise 5-4

f (x)

xc d eba

L

x

f(x)

b ca

d e

f g h

M

L

(A) the intervals on which (B) the intervals on which (C) the intervals on which f(x) is increasing(D) the intervals on which f(x) is decreasing(E) the x coordinate(s) of the point(s) where f(x) has a

local maximum(F) the x coordinate(s) of the point(s) where f(x) has a

local minimum(G) the intervals on which (H) the intervals on which (I) the intervals on which the graph of f is concave upward(J) the intervals on which the graph of f is concave

downward(K) the x coordinate(s) of the inflection point(s)(L) the horizontal asymptote(s)(M) the vertical asymptote(s)

2. Repeat Problem 1 for the following graph of f:

f–1x2 7 0f–1x2 6 0

f¿1x2 7 0f¿1x2 6 0

In Problems 3–10, use the given information to sketch thegraph of f. Assume that f is continuous on its domain and thatall intercepts are included in the table of values.

3. Domain: All real x; limx: ; q

f1x2 = 2

x 0 2 4

f(x) 0 0 0-2-2

-2-4

f (x)

f (x)

2x

0

2

0

0

ND

4x

0

0

4

0ND

x 0 1 2

f (x) 0 2 0 0-2

-1-2

x 0 4 6

f(x) 0 0 3 2

-4

x 0 2

f(x) 0 0 0-2

-2-4

f (x)

f (x)

1x

0

1

0

0

ND

2x

0

0

ND

2

0

f (x)

2x

ND

4

0

2x

ND

6

0f (x)

f (x)

2x

0

1

ND

1x

ND f (x)

5. Domain: All real x, except lim

x: - 2+

f1x2 = - q ; limx: q

f1x2 = 1lim

x: - 2-

f1x2 = q ;x = -2;

6. Domain: All real x, except lim

x:1+

f1x2 = q ; limx: q

f1x2 = -2x = 1; lim

x:1-

f1x2 = q ;

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 331



on and on on

vertical asymptote: horizontal asymptote:


on and on on

vertical asymptote: horizontal asymptote:

9. Domain: All real x, except and

on and on on and on (0, 2) and

vertical asymptotes: and horizontal asymptote:

10. Domain: All real x, except and

on and (0, 1);on and on and

vertical asymptotes: and horizontal asymptote:

B In Problems 11–56, summarize the pertinent informationobtained by applying the graphing strategy and sketch thegraph of

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34. f1x2 =

x2- 5x - 6

x2f1x2 =

x2+ x - 2

x2

f1x2 =

3 - 2x

x2f1x2 =

x2+ 2x

f1x2 =

x

1x - 222f1x2 =

-5x

1x - 122

f1x2 =

2x

x2- 9

f1x2 =

2x

1 - x2

f1x2 =

x2

1 + x2f1x2 =

1

1 + x2

f1x2 =

1

x2- 4

f1x2 =

x

x2- 4

f1x2 = ln1x2+ 42f1x2 = x - ln x

f1x2 = ln12x + 42f1x2 = ln11 - x2f1x2 = 10xe-0.1xf1x2 = 5xe-0.2x

f1x2 = 3 + 7e-0.2xf1x2 = 5 + 5e-0.1x

f1x2 =

2 + x

3 - xf1x2 =

x

x - 2

f1x2 =

2x - 4x + 2

f1x2 =

x + 3x - 3

y = f1x2.

y = 0x = 1;x = -1

11, q2;1- q , -12, 1-1, 12,f–1x2 7 011, q2;1-1, 02f¿1x2 6 0

1- q , -12f¿1x2 7 0f1-22 = 1, f102 = 0, f122 = 1;

x = 1;x = -1

y = 0x = 2;x = -2

12, q2;f–1x2 7 01-2, 02;1- q , -22f–1x2 6 0

1-2, 22;f¿1x2 7 012, q2;1- q , -22f¿1x2 6 0

f1-32 = -1, f102 = 0, f132 = 1;x = 2;x = -2

y = -1x = 1;

11, q2;f–1x2 7 01- q , 12;f–1x2 6 0

11, q2;1- q , 12f¿1x2 6 0f102 = -2, f122 = 0;

x = 1;

y = 1x = -1;

1-1, q2;1- q , -12; f–1x2 6 0f–1x2 7 01-1, q2;1- q , -12f¿1x2 7 0

f1-32 = 2, f1-22 = 3, f102 = -1, f112 = 0;x = -1;

35. 36.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

C In Problems 57–64, show that the line is an obliqueasymptote for the graph of summarize all pertinentinformation obtained by applying the graphing strategy, andsketch the graph of

57. 58.

59. 60.

61. 62.

63. 64.

In Problems 65–72, summarize all pertinent informationobtained by applying the graphing strategy, and sketch thegraph of [Note: These rational functions are notreduced to lowest terms.]

65.

66.

67.

68.

69.

70.

71.

72. f1x2 =

x2+ x - 2

x2+ 4x + 4

f1x2 =

x2+ x - 2

x2- 2x + 1

f1x2 =

x3- 5x2

- 6x

x2+ 3x + 2

f1x2 =

x3- 5x2

+ 6x

x2- x - 2

f1x2 =

2x2+ 11x + 14

x2- 4

f1x2 =

2x2+ x - 15

x2- 9

f1x2 =

x2+ x - 6

x2- x - 12

f1x2 =

x2+ x - 6

x2- 6x + 8

y = f1x2.

f1x2 = x -

16

x3f1x2 = x +

1x

+

4

x3

f1x2 = x +

27

x3f1x2 = x -

9

x3

f1x2 = x +

32

x2f1x2 = x -

4

x2

f1x2 = x -

9x

f1x2 = x +

4x

y = f1x2.y = f1x2,

y = x

f1x2 =

x3

x2- 12

f1x2 =

x3

3 - x2

f1x2 =

x3

1x + 222f1x2 =

x3

1x - 422

f1x2 =

x

x2- 36

f1x2 =

x

x2- 4

f1x2 =

1

3 - 2x - x2f1x2 =

1

x2+ 2x - 8

f1x2 =

x

ln xf1x2 = 1ln x22

f1x2 =

ln xx

f1x2 = x2 ln x

f1x2 = e-2x2f1x2 = e-0.5x2

f1x2 = 1x - 22exf1x2 = 13 - x2ex

f1x2 =

x3

4 - xf1x2 =

x3

x - 2

f1x2 =

2x2+ 5

4 - x2f1x2 =

3x2+ 2

x2- 9

f1x2 =

x2

2 + xf1x2 =

x2

x - 1

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 332


73. Revenue. The marketing research department for acomputer company used a large city to test market thefirm’s new product. The department found that the rela-tionship between price p (dollars per unit) and demand x(units sold per week) was given approximately by

Thus, weekly revenue can be approximated by

Graph the revenue function R.

74. Profit. Suppose that the cost function C(x) (in dollars)for the company in Problem 73 is

(A) Write an equation for the profit P(x).(B) Graph the profit function P.

75. Pollution. In Silicon Valley (California), a number ofcomputer-related manufacturing firms were found to becontaminating underground water supplies with toxicchemicals stored in leaking underground containers.A water quality control agency ordered the companiesto take immediate corrective action and to contribute toa monetary pool for testing and cleanup of the under-ground contamination. Suppose that the required mone-tary pool (in millions of dollars) for the testing andcleanup is estimated to be given by

where x is the percentage (expressed as a decimal frac-tion) of the total contaminant removed.

(A) Where is P(x) increasing? Decreasing?(B) Where is the graph of P concave upward? Downward?(C) Find any horizontal and vertical asymptotes.(D) Find the x and y intercepts.(E) Sketch a graph of P.

76. Employee training. A company producing computercomponents has established that, on the average, a newemployee can assemble N(t) components per day after tdays of on-the-job training, as given by

(A) Where is N(t) increasing? Decreasing?(B) Where is the graph of N concave upward?

Downward?(C) Find any horizontal and vertical asymptotes.(D) Find the intercepts.(E) Sketch a graph of N.

77. Replacement time. An office copier has an initial price of$3,200. A maintenance/service contract costs $300 for thefirst year and increases $100 per year thereafter. It can beshown that the total cost of the copier (in dollars) after nyears is given by

C1n2 = 3,200 + 250n + 50n2 n Ú 1

N1t2 =

100t

t + 9 t Ú 0

P1x2 =

2x

1 - x 0 … x 6 1

C1x2 = 830 + 396x

R1x2 = xp = 1,296x - 0.12x3 0 … x … 80

p = 1,296 - 0.12x2 0 … x … 80

(A) Write an expression for the average cost per year,for n years.

(B) Graph the average cost function found in part (A).(C) When is the average cost per year minimum? (This time

is frequently referred to as the replacement time forthis piece of equipment.)

78. Construction costs. The management of a manufacturingplant wishes to add a fenced-in rectangular storage yardof 20,000 square feet, using the plant building as one sideof the yard (see the figure). If x is the distance (in feet)from the building to the fence parallel to the building,show that the length of the fence required for the yard isgiven by

L1x2 = 2x +

20,000x

x 7 0

C1n2,

Applications

x

Storage yard

Figure for 78

(A) Graph L.

(B) What are the dimensions of the rectangle requiringthe least amount of fencing?

79. Average and marginal costs. The total daily cost (indollars) of producing x park benches is given by

(A) Sketch the graphs of the average cost function andthe marginal cost function on the same set of coordi-nate axes. Include any oblique asymptotes.


80. Average and marginal costs. The total daily cost (indollars) of producing x picnic tables is given by

(A) Sketch the graphs of the average cost function andthe marginal cost function on the same set of coordi-nate axes. Include any oblique asymptotes.


C1x2 = 500 + 2x + 0.2x2

C1x2 = 1,000 + 5x + 0.1x2

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 333


Number of Pizzas Total Costs x y

50 395

100 475

150 640

200 910

250 1,140

300 1,450

Number of Pizzas Total Costsx y

50 595

100 755

150 1,110

200 1,380

250 1,875

300 2,410

81. Minimizing average costs. The data in the table give thetotal daily costs y (in dollars) of producing x pepperonipizzas at various production levels.

(B) Use the regression equation from part (A) to findthe minimum average cost (to the nearest cent) andthe corresponding production level (to the nearestinteger).

83. Medicine. A drug is injected into the bloodstream of apatient through her right arm. The concentration of thedrug in the bloodstream of the left arm t hours after theinjection is given by

Graph C.

84. Physiology. In a study on the speed of muscle contrac-tion in frogs under various loads, researchers W. O.Fems and J. Marsh found that the speed of contractiondecreases with increasing loads. More precisely, theyfound that the relationship between speed of contraction,S (in centimeters per second), and load w (in grams) isgiven approximately by

Graph S.

85. Psychology: retention. An experiment on retention isconducted in a psychology class. Each student in the classis given one day to memorize the same list of 30 specialcharacters. The lists are turned in at the end of the day,and for each succeeding day for 30 days, each student isasked to turn in a list of as many of the symbols as can berecalled. Averages are taken, and it is found that

provides a good approximation of the average numberN(t) of symbols retained after t days. Graph N.

N1t2 =

5t + 20t

t Ú 1

S1w2 =

26 + 0.06w

w w Ú 5

C1t2 =

0.14t

t2+ 1

(A) Enter the data into a graphing calculator and find aquadratic regression equation for the total costs.


82. Minimizing average costs. The data in the table give thetotal daily costs y (in dollars) of producing x deluxepizzas at various production levels.

(A) Enter the data into a graphing calculator and find aquadratic regression equation for the total costs.

Section 5-5 ABSOLUTE MAXIMA AND MINIMA Absolute Maxima and Minima Second Derivative and Extrema

We are now ready to consider one of the most important applications of the deriva-tive: the use of derivatives to find the absolute maximum or minimum value of afunction. An economist may be interested in the price or production level of a com-modity that will bring a maximum profit; a doctor may be interested in the time ittakes for a drug to reach its maximum concentration in the bloodstream after aninjection; and a city planner might be interested in the location of heavy industry ina city in order to produce minimum pollution in residential and business areas. In thissection, we develop the procedures needed to find the absolute maximum andabsolute minimum values of a function.

Absolute Maxima and MinimaRecall that f(c) is a local maximum value if for x near c and a local min-imum value if for x near c. Now we are interested in finding the largestand the smallest values of f(x) throughout its domain.

f1x2 Ú f1c2 f1x2 … f1c2

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 334

Figure 1 illustrates some typical examples.

S e c t i o n 5 - 5 Absolute Maxima and Minima 335

DEFINITION Absolute Maxima and MinimaIf for all x in the domain of f, then f(c) is called the absolute maximumvalue of f.

If for all x in the domain of f, then f(x) is called the absolute minimumvalue of f.

f1c2 … f1x2f1c2 Ú f1x2

x

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

xx2 2

x3

3f (x) 4x

(A) No absolute maximum or minimum One local maximum at x 2 One local minimum at x 2

f (x) 4 x2

(B) Absolute maximum at x 0 No absolute minimum

f (x) x2/3

(C) Absolute minimum at x 0 No absolute maximum

FIGURE 1

Explore & Discuss 1 Functions f, g, and h, along with their graphs, are shown in Figure 2.

If f(c) is the absolute maximum value of a function f, then f(c) is obviously a “value” of f.It is common practice to omit “value” and to refer to f(c) as the absolute maximumof f. In either usage, note that c is a value of x in the domain of f where the absolutemaximum occurs. It is incorrect to refer to c as the absolute maximum. Collectively, theabsolute maximum and the absolute minimum are referred to as absolute extrema.

I N S I G H T

x x

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

1

1

1

1

1 1 1 1 1 1

5

5

x

(A) f (x) 1 x (B) g(x) x 1x

xx

(C) h(x) x

FIGURE 2

(A) Which of these functions are continuous on

(B) Find the absolute maximum and the absolute minimum of each function onif they exist, and the corresponding values of x that produce these

absolute extrema.

(C) Suppose that a function p is continuous on and satisfies and Sketch a possible graph for p. Does the function yougraphed have an absolute maximum? An absolute minimum? Can youmodify your sketch so that p does not have an absolute maximum or anabsolute minimum on 3-1, 14?

p112 = 0.p1-12 = 03-1, 14

3-1, 14,3-1, 14?

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 335


f(x)

50

100

150

][5 10

Absolutemaximum

f (12) 154

Absoluteminimumf (2) 24

(A) [a, b] [2, 12]

x

f(x)

50

100

150

][5 10

Absolutemaximumf (5) 105

f (9) 73Absoluteminimum

(B) [a, b] [4, 10]

x

f(x)

50

100

150

][5 10

Absolutemaximumf (5) 105

f (8) 78Absoluteminimum

(C) [a, b] [4, 8]

x

f(x)

50

100

150

][5 10

f (3) 73 f (9)Absoluteminimum

(D) [a, b] [3, 11]

xa 2 a 4a 4b 12 b 8b 10 b 11a 3

Absolutemaximum

f (5) 105 f (11)

FIGURE 3 Absolute extrema for for various closed intervalsf1x2 = x3- 21x2

+ 135x - 170

In many practical problems, the domain of a function is restricted because of prac-tical or physical considerations. If the domain is restricted to some closed interval, asis often the case, then Theorem 1 applies.

THEOREM 1 EXTREME VALUE THEOREMA function f that is continuous on a closed interval [a, b] has both an absolute max-imum value and an absolute minimum value on that interval.

It is important to understand that the absolute maximum and minimum values de-pend on both the function f and the interval [a, b]. Figure 3 illustrates four cases.

In all four cases illustrated in Figure 3, the absolute maximum value and absoluteminimum value occur at a critical value or an endpoint. This property is general-ized in Theorem 2. Note that both the absolute maximum value and the absoluteminimum value are unique, but each can occur at more than one point in the interval(Fig. 3D).

THEOREM 2 LOCATING ABSOLUTE EXTREMAAbsolute extrema (if they exist) must always occur at critical values or at endpoints.

Thus, to find the absolute maximum or minimum value of a continuous function ona closed interval, we simply identify the endpoints and the critical values in the in-terval, evaluate the function at each, and then choose the largest and smallest valuesout of this group.

PROCEDURE Finding Absolute Extrema on a Closed IntervalStep 1. Check to make certain that f is continuous over [a, b].

Step 2. Find the critical values in the interval (a, b).

Step 3. Evaluate f at the endpoints a and b and at the critical values found in step 2.

Step 4. The absolute maximum f(x) on [a, b] is the largest of the values found instep 3.

Step 5. The absolute minimum f(x) on [a, b] is the smallest of the values found instep 3.

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 336


E X A M P L E 1 Finding Absolute Extrema Find the absolute maximum and absolute minimumvalues of

on each of the following intervals:

(A) (B) (C)

SOLUTION (A) The function is continuous for all values of x.

Thus, and are critical values in the interval Evaluate f atthe endpoints and critical values ( and 4), and choose the maximumand minimum from these:

Absolute minimum

Absolute maximum

(B) Interval: 3-4, 24 f142 = 69

f112 = -12

f1-32 = 20

f1-62 = -61

-6, -3, 1,1-6, 42.x = 1x = -3

f¿1x2 = 3x2+ 6x - 9 = 31x - 121x + 32

3-2, 243-4, 243-6, 44

f1x2 = x3+ 3x2

- 9x - 7

x f(x)

13

20 Absolute maximum1 Absolute minimum2 -5

-12

-3

-4

The critical value is not included in this table, because it is not in the interval 3-2, 24.

x = -3

MATCHED PROBLEM 1 Find the absolute maximum and absolute minimum values of

on each of the following intervals:

(A) (B) (C) 3-3, 143-3, 343-5, 54

f1x2 = x3- 12x

Now, suppose that we want to find the absolute maximum or minimum value ofa function that is continuous on an interval that is not closed. Since Theorem 1 nolonger applies, we cannot be certain that the absolute maximum or minimumvalue exists. Figure 4 illustrates several ways that functions can fail to have absoluteextrema.

x f (x)

15 Absolute maximum1 Absolute minimum2 - 5

- 12

- 2

(C) Interval: 3-2, 24

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 337

In general, the best procedure to follow in searching for absolute extrema on aninterval that is not of the form [a, b] is to sketch a graph of the function. However,many applications can be solved with a new tool that does not require any graphing.

Second Derivative and ExtremaThe second derivative can be used to classify the local extrema of a function. Supposethat f is a function satisfying and First, note that if it follows from the properties of limits* that in some interval (m, n) con-taining c. Thus, the graph of f must be concave upward in this interval. But this im-plies that is increasing in the interval. Since must change fromnegative to positive at and f(c) is a local minimum (see Fig. 5). Reasoning inthe same fashion, we conclude that if and then f(c) is a localmaximum. Of course, it is possible that both and In this case,the second derivative cannot be used to determine the shape of the graph around

may be a local minimum, a local maximum, or neither.x = c; f1c2f–1c2 = 0.f¿1c2 = 0

f–1c2 6 0,f¿1c2 = 0x = c,

f¿1c2 = 0, f¿1x2f¿1x2f–1x2 7 0

f–1c2 7 0,f–1c2 7 0.f¿1c2 = 0


* Actually, we are assuming that is continuous in an interval containing c. It is very unlikely that wewill encounter a function for which exists but is not continuous in an interval containing c.f–1x2f–1c2

f–1x2

x

x

x

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

1

1

1

0

2

1

2

3

4

5

1

1 2

2123 3

xf (x)

1 x2

f (x) 8x 2x2 31 x 2

x1 x2

f (x)

1 x 1

(A) No absolute extrema on (, ): 1 f (x) 1 for all x f (x) 1 or 1 for any x

(B) No absolute extrema on (1, 2): 3 f (x) 5 for x (1, 2) f (x) 3 or 5 for any x (1, 2)

(C) No absolute extrema on (1, 1): Graph has vertical asymptotes at x 1 and x 1

1

2

11

FIGURE 4 Functions with no absolute extrema

f (c) 0

f (c) 0

f (x) 0

f (c) 0

f (c) 0

f (x) 0f (x) 0f (x) 0

(A) f (c) 0 and f (c) 0 implies f (c) is a local minimum

(B) f (c) 0 and f (c) 0 implies f (c) is a local maximum

m c n m c n

FIGURE 5 Second derivative and local extrema

The sign of the second derivative thus provides a simple test for identifying localmaxima and minima. This test is most useful when we do not want to draw the graphof the function. If we are interested in drawing the graph and have already con-structed the sign chart for the first-derivative test can be used to identify thelocal extrema.

f¿1x2,

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 338


Graph of f is: f(c) Example

0 Concave upward Local minimum

0 Concave downward Local maximum

0 0 ? Test does not apply

-

+

f –1c2f ¿1c2

E X A M P L E 2 Testing Local Extrema Find the local maxima and minima for each function. Usethe second-derivative test when it applies.

(A)

(B)

(C)

SOLUTION (A) Take first and second derivatives and find critical values:

Critical values are and

f has a local maximum atf has a local minimum at

(B)

Critical value:

f has a local maximum at

(C)

Critical values are and

The second-derivative test fails at both critical values, sothe first-derivative test must be used.

Sign chart for (partition numbers are 0 and 10):f¿1x2 = x31x - 1022 f –1102 = 0

f–102 = 0

x = 10.x = 0

f–1x2 = 5x4- 80x3

+ 300x2

f¿1x2 = x5- 20x4

+ 100x3= x31x - 1022

f1x2 =16 x6

- 4x5+ 25x4

x = 5.

f –152 = e- 11-0.22 6 0

x = 1>0.2 = 5

= e- 0.2x(0.04x - 0.4)

f–1x2 = e- 0.2x(-0.2)11 - 0.2x2 + e-0.2x(-0.2)

= e- 0.2x11 - 0.2x2 f¿1x2 = e- 0.2x

+ xe- 0.2x1-0.22 f1x2 = xe- 0.2x

x = 3. f–132 = 6 7 0

x = 1. f–112 = -6 6 0

x = 3.x = 1

f–1x2 = 6x - 12 = 61x - 22 f¿1x2 = 3x2

- 12x + 9 = 31x - 121x - 32 f1x2 = x3

- 6x2+ 9x + 1

f1x2 =16 x6

- 4x5+ 25x4

f1x2 = xe-0.2x

f1x2 = x3- 6x2

+ 9x + 1

Test Numbers

x

1

11 1,331 1281 12

-121 12-1

f œ1x2

IncreasingDecreasing

(10, )(, 0)

f (x)

f (x)

0x

0

Increasing10

0

(0, 10)

RESULT Second-Derivative TestLet c be a critical value of f(x).

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 339

From the chart, we see that f(x) has a local minimum at and does nothave a local extremum at x = 10.

x = 0


The solution of many optimization problems involves searching for an absolute ex-tremum. If the function in question has only one critical value, then the second-derivative test not only classifies the local extremum, but also guarantees that thelocal extremum is, in fact, the absolute extremum.

MATCHED PROBLEM 2 Find the local maxima and minima for each function. Use the second-derivative testwhen it applies.

(A)

(B)

(C) f1x2 = 10x6- 24x5

+ 15x4

f1x2 = ex- 5x

f1x2 = x3- 9x2

+ 24x - 10

The second-derivative test does not apply if or if is not defined. As Ex-ample 2C illustrates, if then f(c) may or may not be a local extremum. Someother method, such as the first-derivative test, must be used when or doesnot exist.

f–1c2f–1c2 = 0f–1c2 = 0,

f–1c2f–1c2 = 0

I N S I G H T

THEOREM 3 SECOND-DERIVATIVE TEST FOR ABSOLUTE EXTREMUMLet f be continuous on an interval I with only one critical value c in I.

If and then f(c) is the absolute minimum of f on I.

If and then f(c) is the absolute maximum of f on I.

f–1c2 6 0,f¿1c2 = 0

f–1c2 7 0,f¿1c2 = 0

x)(I

x)(I

Since the second-derivative test cannot be applied when or doesnot exist, Theorem 3 makes no mention of these cases.

f–1c2f–1c2 = 0

E X A M P L E 3 Finding an Absolute Extremum on an Open Interval Find the absolute min-imum value of each function on

(A)

(B)

SOLUTION (A)

Critical values areand .

The only critical value in the interval is Since is the absolute minimum value of f on 10, q2.f122 = 4

f¿122 = 1 7 0,x = 2.10, q2 f–1x2 =

8

x3

x = 2x = -2 f¿1x2 = 1 -

4

x2 =

x2- 4

x2 =

1x - 221x + 22x2

f1x2 = x +

4x

f1x2 = (ln x)2- 3 ln x

f1x2 = x +

4x

(0, q).

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 340

(B)

Critical value is .

The only critical value in the interval is Since is the absolute minimum value of f on10, q2. 2/e3

7 0, f(e3/2) = -2.25f–1e3/22 =

x = e3/2.10, q2 f–1x2 =

x 2x

- (2 ln x - 3)

x2 =

5 - 2 ln x

x2

x = e3/2 f¿1x2 = (2 ln x) 1x

-

3x

=

2 ln x - 3x

f1x2 = (ln x)2- 3 ln x


MATCHED PROBLEM 3 Find the absolute maximum value of each function on

(A) (B)

Answers to Matched Problems 1. (A) Absolute maximum: absolute minimum:(B) Absolute maximum: absolute minimum: (C) Absolute maximum: absolute minimum:

2. (A) f(2) is a local maximum; f(4) is a local minimum.(B) is a local minimum.(C) f(0) is a local minimum; there is no local extremum at .

3. (A) (B) f152 = 5 ln 5 - 5f1152 = 12 - 215

x = 1f1ln 52 = 5 - 5 ln 5

f112 = -11f1-22 = 16;f122 = -16f1-22 = 16;

f1-52 = -65f152 = 65;

f1x2 = 5 ln x - xf1x2 = 12 - x -

5x

10, q2.

A Problems 1–10 refer to the graph of shown here.Find the absolute minimum and the absolute maximum overthe indicated interval.

y = f(x) 19. 20.

21. 22.

23. 24.

25. 26.

B In Problems 27–50, find the indicated extremum of eachfunction on the given interval.

27. Absolute minimum value on for

28. Absolute maximum value on for






f1x2 = 2x4- 8x3

10, q2f1x2 = 12 - x21x + 122

30, q2f1x2 = 1x + 421x - 222

30, q2f1x2 = x3

- 6x230, q2

f1x2 = 3x2- x3

30, q2f1x2 = 6x - x2

+ 4

30, q2f1x2 = 2x2

- 8x + 6

30, q2

f1x2 =

9 - x2

x2+ 4

f1x2 =

x2- 1

x2+ 1

f1x2 =

-8x

x2+ 4

f1x2 =

2x

x2+ 1

f1x2 =

1

x2+ 1

f1x2 =

x2

x2+ 1

f1x2 = x +

25x

f1x2 = x +

16x

Exercise 5-5

x5 10

5

10

15

f(x)

1. [0, 10] 2. [2, 8] 3. [0, 8] 4. [2, 10]

5. [1, 10] 6. [0, 9] 7. [1, 9] 8. [0, 2]

9. [2, 5] 10. [5, 8]

In Problems 11–26, find the absolute maximum and minimum,if either exists, for each function.

11. 12.

13. 14.

15. 16.

17. 18. f1x2 = x4- 4x3f1x2 = 8x3

- 2x4

f1x2 = -x3- 2xf1x2 = x3

+ x

f1x2 = -x2+ 2x + 4f1x2 = -x2

- 6x + 9

f1x2 = x2+ 4x - 3f1x2 = x2

- 2x + 3

BARNMC05_0132328186.QXD 02/21/2007 20:58 Page 341
















f1x2 = x2(3 - ln x)

(0, q)

f1x2 = 4x ln x - 7x

(0, q)

f1x2 = 5x - 2x ln x

(0, q)

f1x2 =

ex

x

(0, q)

f1x2 =

x3

ex

(0, q)

f1x2 =

x4

ex

(0, q)

f1x2 =

ex

x2

(0, q)

f1x2 = 2x +

5x

+

4

x3

(0, q)

f1x2 = x +

1x

+

30

x3

(0, q)

f1x2 = 20 - 4x -

250

x2

(0, q)

f1x2 = 10 + 2x +

64

x2

(0, q)

f1x2 = 4 + x +

9x

(0, q)

f1x2 = 20 - 3x -

12x

(0, q)

f1x2 = 4x3- 8x4

(0, q) 48. Absolute minimum value on for



In Problems 51–56, find the absolute maximum and minimum,if either exists, for each function on the indicated intervals.

51.

(A) (B) (C) [2, 5]

52.

(A) (B) (C)

53.

(A) [0, 3] (B) [1, 7] (C) [3, 6]

54.

(A) (B) [0, 2] (C)

55.

(A) (B) [0, 4] (C)

56.

(A) (B) (C) [1, 3]

In Problems 57–64, describe the graph of f at the given pointrelative to the existence of a local maximum or minimum withone of the following phrases: “Local maximum,” “Local mini-mum,” “Neither,” or “Unable to determine from the giveninformation.” Assume that f(x) is continuous on

57. (2, f(2)) if and

58. (4, f(4)) if and

59. if and

60. if and

61. (6, f(6)) if and does not exist

62. (5, f(5)) if and does not exist

63. if and

64. (1, f(1)) if and f–112 7 0f¿112 = 0

f–1-22 6 0f¿1-22 = 01-2, f1-222f–152f¿152 = 0

f–162f¿162 = 1

f–1-12 6 0f¿1-12 = 01-1, f1-122f–1-32 = 0f¿1-32 = 01-3, f1-322

f–142 6 0f¿142 = 1

f–122 7 0f¿122 = 0

1- q , q2.

3-1, 143-4, 44f1x2 = x4

- 18x2+ 32

3-1, 143-1, 24f1x2 = x4

- 4x3+ 5

3-3, 443-1, 34f1x2 = x4

- 8x2+ 16

f1x2 = 1x - 121x - 523 + 1

3-2, 143-2, 343-3, 44f1x2 = 2x3

- 3x2- 12x + 24

3-1, 343-1, 54f1x2 = x3

- 6x2+ 9x - 6

f1x2 = ln(x2e-x)

(0, q)

f1x2 = ln(xe-x)

(0, q)

f1x2 = x3(ln x - 2)

(0, q)

Section 5-6 OPTIMIZATION Area and Perimeter Maximizing Revenue and Profit Inventory Control

Now we want to use the calculus tools we have developed to solve optimizationproblems—problems that involve finding the absolute maximum value or the ab-solute minimum value of a function. As you work through this section, note that thestatement of the problem does not usually include the function that is to be optimized.Often, it is your responsibility to find the function and then to find its absoluteextremum.

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 342

Area and PerimeterThe techniques used to solve optimization problems are best illustrated throughexamples.

S e c t i o n 5 - 6 Optimization 343

E X A M P L E 1 Maximizing Area A homeowner has $320 to spend on building a fence around arectangular garden. Three sides of the fence will be constructed with wire fencing ata cost of $2 per linear foot. The fourth side is to be constructed with wood fencing at a cost of $6 per linear foot. Find the dimensions and the area of the largest gardenthat can be enclosed with $320 worth of fencing.

SOLUTION To begin, we draw a figure (Fig. 1), introduce variables, and look for relationshipsamong the variables.

Since we don’t know the dimensions of the garden, the lengths of fencing are rep-resented by the variables x and y. The costs of the fencing materials are fixed and arethus represented by constants.

Now we look for relationships among the variables. The area of the garden is

while the cost of the fencing is

The problem states that the homeowner has $320 to spend on fencing. We make theassumption that enclosing the largest area will use all of the money available for fenc-ing. The problem has now been reduced to

Before we can use calculus techniques to find the maximum area A, we must expressA as a function of a single variable. We use the cost equation to eliminate one of thevariables in the expression for the area (we choose to eliminate y—either will work).

Now we consider the permissible values of x. Because x is one of the dimensions ofa rectangle, x must satisfy

Length is always nonnegative.

And because is also a dimension of a rectangle, y must satisfy

Width is always nonnegative.

We summarize the preceding discussion by stating the following model for this opti-mization problem:

Next, we find any critical values of A:

Critical value x =

804

= 20

80 = 4x

A¿1x2 = 80 - 4x = 0

Maximize A1x2 = 80x - 2x2 for 0 … x … 40

40 Ú x or x … 40

80 Ú 2x

y = 80 - 2x Ú 0

y = 80 - 2x

x Ú 0

A = xy = x180 - 2x2 = 80x - 2x2

y = 80 - 2x

4y = 320 - 8x

8x + 4y = 320

Maximize A = xy subject to 8x + 4y = 320

= 8x + 4y

C = 2y + 2x + 2y + 6x

A = xy

x

y$6

$2$2

$2

FIGURE 1

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 343

Since A(x) is continuous on [0, 40], the absolute maximum value of A, if it exists,must occur at a critical value or an endpoint. Evaluating A at these values (Table 1),we see that the maximum area is 800 when

Finally, we must answer the questions posed in the problem. The dimensions of the gar-den with the maximum area of 800 square feet are 20 feet by 40 feet, with one 20-footside with wood fencing.

x = 20 and y = 80 - 21202 = 40


x A(x)

0 0

20 800

40 0

TABLE 1

MATCHED PROBLEM 1 Repeat Example 1 if the wood fencing costs $8 per linear foot and all otherinformation remains the same.

We summarize the steps we followed in the solution to Example 1 in the followingbox:

PROCEDURE Strategy for Solving Optimization ProblemsStep 1. Introduce variables, look for relationships among the variables, and con-

struct a mathematical model of the form

Maximize (or minimize) f(x) on the interval I

Step 2. Find the critical values of f(x).

Step 3. Use the procedures developed in Section 5-5 to find the absolute maximum(or minimum) value of f(x) on the interval I and the value(s) of x where thisoccurs.

Step 4. Use the solution to the mathematical model to answer all the questions askedin the problem.

E X A M P L E 2 Minimizing Perimeter Refer to Example 1.The homeowner judges that an areaof 800 square feet for the garden is too small and decides to increase the area to 1,250square feet. What is the minimum cost of building a fence that will enclose a gardenwith area 1,250 square feet? What are the dimensions of this garden? Assume that thecost of fencing remains unchanged.

SOLUTION Refer to Figure 1 in the solution for Example 1. This time we want to minimize the costof the fencing that will enclose 1,250 square feet. The problem can be expressed as

Since x and y represent distances, we know that and But neither vari-able can equal 0, because their product must be 1,250.

Solve the area equation for y.

Substitute for y in the cost equation.

The model for this problem is

= 8x + 5,000x-1

Minimize C1x2 = 8x +

5,000x

for x 7 0

= 8x +

5,000x

x 7 0

C1x2 = 8x + 4 1,250

x

y =

1,250x

xy = 1,250

y Ú 0.x Ú 0

Minimize C = 8x + 4y subject to xy = 1,250

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 344

The negative square root is discarded,since

We use the second derivative to determine the behavior at

The second-derivative test shows that C(x) has a local minimum at and sinceis the only critical value of C(25) must be the absolute minimum value

of C(x) for When the cost is

and

The minimal cost for enclosing a 1,250-square-foot garden is $400, and the dimensionsare 25 feet by 50 feet, with one 25-foot side with wood fencing.

y =

1,25025

= 50

C1252 = 81252 +

5,00025

= 200 + 200 = $400

x = 25,x 7 0.x 7 0,x = 25

x = 25,

C–1252 =

10,000

253 = 0.64 7 0

C–1x2 = 0 + 10,000x-3=

10,000

x3

C¿1x2 = 8 - 5,000x-2

x = 25.

x 7 0. x = 2625 = 25

x2=

5,0008

= 625

8 =

5,000

x2

= 8 -

5,000

x2 = 0

C¿1x2 = 8 - 5,000x-2


MATCHED PROBLEM 2 Repeat Example 2 if the homeowner wants to enclose an 1,800-square-foot gardenand all other data remain unchanged.

The restrictions on the variables in the solutions of Examples 1 and 2 are typical of prob-lems involving areas or perimeters (or the cost of the perimeter):

Cost of fencing (Example 1)

Area of garden (Example 2)

The equation in Example 1 restricts the values of x to

The endpoints are included in the interval for our convenience (a closed interval is easi-er to work with than an open one). The area function is defined at each endpoint, so it doesno harm to include them.

The equation in Example 2 restricts the values of x to

Neither endpoint can be included in this interval. We cannot include 0 because the areais not defined when , and we can never include as an endpoint. Remember, isnot a number, but a symbol which denotes that the interval is unbounded.

qqx = 0

x 7 0 or 10, q2

0 … x … 40 or 30, 404

xy = 1,250

8x + 4y = 320

I N S I G H T

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 345

Maximizing Revenue and Profit


E X A M P L E 3 Maximizing Revenue An office supply company sells x mechanical pencils peryear at $p per pencil.The price–demand equation for these pencils is What price should the company charge for the pencils to maximize revenue? What isthe maximum revenue?

SOLUTION

Both price and demand must be nonnegative, so

The mathematical model for this problem is

Critical value

Use the second-derivative test for absolute extrema:

When the demand is the price is

The company will realize a maximum revenue of $25,000 when the price of a pencilis $5.

p = 10 - 0.001x10 - 0.00115,0002 = $5

x = 5,000,

Max R1x2 = R15,0002 = $25,000

R–1x2 = -0.002 6 0 for all x

x =

100.002

= 5,000

10 = 0.002x 10 - 0.002x = 0

R¿1x2 = 10 - 0.002x

Maximize R1x2 = 10x - 0.001x2 0 … x … 10,000

10,000 Ú x 10 Ú 0.001x

x Ú 0 and p = 10 - 0.001x Ú 0

= 10x - 0.001x2

R1x2 = 110 - 0.001x2x Revenue = price * demand

p = 10 - 0.001x.

MATCHED PROBLEM 3 An office supply company sells x heavy-duty paper shredders per year at $p per shred-der. The price–demand equation for these shredders is

What price should the company charge for the shredders to maximize revenue? Whatis the maximum revenue?

p = 300 -

x

30

E X A M P L E 4 Maximizing Profit The total annual cost of manufacturing x mechanical pencilsfor the office supply company in Example 3 is

What is the company’s maximum profit? What should the company charge for eachpencil, and how many pencils should be produced?

SOLUTION Using the revenue model in Example 3, we have

= 8x - 0.001x2- 5,000

= 10x - 0.001x2- 5,000 - 2x

P1x2 = R1x2 - C1x2 Profit = Revenue - Cost

C1x2 = 5,000 + 2x

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 346

The mathematical model for profit is

(The restrictions on x come from the revenue model in Example 3.)

Critical value

Since is the only critical value and

Using the price–demand equation from Example 3 with we find that

Thus, a maximum profit of $11,000 is realized when 4,000 pencils are manufacturedannually and sold for $6 each.

The results in Examples 3 and 4 are illustrated in Figure 2.

p = 10 - 0.001xp = 10 - 0.00114,0002 = $6

x = 4,000,

Max P1x2 = P14,0002 = $11,000

P–1x2 6 0,x = 4,000

P–1x2 = -0.002 6 0 for all x

x =

80.002

= 4,000

8 = 0.002x

P¿1x2 = 8 - 0.002x = 0

Maximize P1x2 = 8x - 0.001x2- 5,000 0 … x … 10,000


Revenue

Cost

x

R C

10,000

20,000

30,000

Cos

t and

rev

enue

(do

llars

)

Minimumcost

5,0000 10,000

Loss

Profit

Profit

Maximumprofit

Maximumrevenue

Loss

Production (number of pencils per year)

FIGURE 2

In Figure 2, notice that the maximum revenue and the maximum profit occur at differentproduction levels. The profit is maximum when

that is, when the marginal revenue is equal to the marginal cost. Notice that the slopes ofthe revenue function and the cost function are the same at this production level.

P¿1x2 = R¿1x2 - C¿1x2 = 0

I N S I G H T

MATCHED PROBLEM 4 The annual cost of manufacturing x paper shredders for the office supply companyin Matched Problem 3 is What is the company’s maximumprofit? What should it charge for each shredder, and how many shredders should itproduce?

C1x2 = 90,000 + 30x.

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 347


E X A M P L E 5 Maximizing Profit The government decides to tax the company in Example 4 $2 for each pencil produced.Taking into account this additional cost, how many pen-cils should the company manufacture each week to maximize its weekly profit? Whatis the maximum weekly profit? How much should the company charge for the pencilsto realize the maximum weekly profit?

SOLUTION The tax of $2 per unit changes the company’s cost equation:

The new profit function is

Thus, we must solve the following equation:

Critical value

Using the price–demand equation (Example 3) with we find that

Thus, the company’s maximum profit is $4,000 when 3,000 pencils are produced andsold weekly at a price of $7.

Even though the tax caused the company’s cost to increase by $2 per pencil,the price that the company should charge to maximize its profit increases by only $1.The company must absorb the other $1, with a resulting decrease of $7,000 in maxi-mum profit.

p = 10 - 0.001xp = 10 - 0.00113,0002 = $7

x = 3,000,

Max P1x2 = P13,0002 = $4,000

P–1x2 = -0.002 6 0 for all x

x = 3,000 6 - 0.002x = 0

P¿1x2 = 6 - 0.002x

Maximize P1x2 = 6x - 0.001x2- 5,000 0 … x … 10,000

= 6x - 0.001x2- 5,000

= 10x - 0.001x2- 5,000 - 4x

P1x2 = R1x2 - C1x2

= 5,000 + 4x

= 5,000 + 2x + 2x

C1x2 = original cost + tax

MATCHED PROBLEM 5 The government decides to tax the office supply company in Matched Problem 4 $20 for each shredder produced. Taking into account this additional cost, how manyshredders should the company manufacture each week to maximize its weekly profit?What is the maximum weekly profit? How much should the company charge for theshredders to realize the maximum weekly profit?

E X A M P L E 6 Maximizing Revenue When a management training company prices its seminaron management techniques at $400 per person, 1,000 people will attend the seminar.The company estimates that for each $5 reduction in the price, an additional 20 peoplewill attend the seminar. How much should the company charge for the seminar inorder to maximize its revenue? What is the maximum revenue?

SOLUTION Let x represent the number of $5 price reductions. Then

R1x2 = 1400 - 5x2 * 11,000 + 20x2 Revenue = 1price per customer21number of customers2

1,000 + 20x = number of customers

400 - 5x = price per customer

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 348

Since price cannot be negative, we have

A negative value of x would result in a price increase. Since the problem is stated interms of price reductions, we must restrict x so that Putting all this together,we have the following model:

Critical value

Since R(x) is continuous on the interval [0, 80], we can determine the behavior ofthe graph by constructing a table. Table 2 shows that is the absolute maximum revenue. The price of attending the seminar at is

The company should charge $325 for the seminar in order toreceive a maximum revenue of $422,500.400 - 51152 = $325.

x = 15R1152 = $422,500

x = 15

3,000 = 200x

R¿1x2 = 3,000 - 200x = 0

R1x2 = 400,000 + 3,000x - 100x2

Maximize R1x2 = 1400 - 5x211,000 + 20x2 for 0 … x … 80

x Ú 0.

80 Ú x or x … 80

400 Ú 5x

400 - 5x Ú 0


MATCHED PROBLEM 6 A walnut grower estimates from past records that if 20 trees are planted per acre,each tree will average 60 pounds of nuts per year. If, for each additional tree plantedper acre, the average yield per tree drops 2 pounds, how many trees should be plant-ed to maximize the yield per acre? What is the maximum yield?

TABLE 2

x R(x)

0 400,000

15 422,500

80 0

Explore & Discuss 1 In Example 6, letting x be the number of $5 price reductions produced a simpleand direct solution to the problem. However, this is not the most obviouschoice for a variable. Suppose that we proceed as follows:

Let x be the new price and let y be the attendance at that price level.Then thetotal revenue is given by xy.

(A) Find y when and when Find the equation of the linethrough these two points.

(B) Use the equation from part (A) to express the revenue in terms of either xor y, and use the expression you came up with to solve Example 6.

(C) Compare this method of solution to the one used in Example 6 with respectto ease of comprehension and ease of computation.

x = 395.x = 400

E X A M P L E 7 Maximizing Revenue After additional analysis, the management training compa-ny in Example 6 decides that its estimate of attendance was too high. Its new estimateis that only 10 additional people will attend the seminar for each $5 decrease in price.All other information remains the same.How much should the company charge for theseminar now in order to maximize revenue? What is the new maximum revenue?

SOLUTION Under the new assumption, the model becomes

Critical value

Note that is not in the interval [0, 80]. Since R(x) is continuous on [0, 80],we can use a table to find the absolute maximum revenue. Table 3 shows that themaximum revenue is The company should leave the price at $400.Any $5 decreases in price will lower the revenue.

R102 = $400,000.

x = -10

x = -10

-1,000 = 100x

R¿1x2 = -1,000 - 100x = 0

= 400,000 - 1,000x - 50x2

Maximize R1x2 = 1400 - 5x211,000 + 10x2 0 … x … 80

x R(x)

0 400,000

80 0

TABLE 3

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 349

Inventory Control


MATCHED PROBLEM 7 After further analysis, the walnut grower in Matched Problem 6 determines that eachadditional tree planted will reduce the average yield by 4 pounds. All other infor-mation remains the same. How many additional trees per acre should the growerplant now in order to maximize the yield? What is the new maximum yield?

The solution in Example 7 is called an endpoint solution, because the optimal value occursat the endpoint of an interval rather than at a critical value in the interior of the interval.It is always important to verify that the optimal value has been found.

I N S I G H T

E X A M P L E 8 Inventory Control A recording company anticipates that there will be a demandfor 20,000 copies of a certain compact disk (CD) during the next year. It costs the com-pany $0.50 to store a CD for one year. Each time it must make additional CDs, itcosts $200 to set up the equipment. How many CDs should the company make duringeach production run to minimize its total storage and setup costs?

SOLUTION This type of problem is called an inventory control problem. One of the basicassumptions made in such problems is that the demand is uniform. For example, if thereare 250 working days in a year, the daily demand would beThe company could decide to produce all 20,000 CDs at the beginning of the year. Thiswould certainly minimize the setup costs, but would result in very large storage costs.At the other extreme, the company could produce 80 CDs each day. This would mini-mize the storage costs, but would result in very large setup costs. Somewhere betweenthese two extremes is the optimal solution that will minimize the total storage and setupcosts. Let

It is easy to see that the total setup cost for the year is 200y, but what is the total stor-age cost? If the demand is uniform, the number of CDs in storage between produc-tion runs will decrease from x to 0, and the average number in storage each day is x2.This result is illustrated in Figure 3.

y = number of production runs

x = number of CDs manufactured during each production run

20,000 , 250 = 80 CDs.

Number in storage

Production run

x

0First Second Third Fourth

Averagenumber in

storagex2

FIGURE 3

Since it costs $0.50 to store a CD for one year, the total storage cost isand the total cost is

C = 200y + 0.25x

total cost = setup cost + storage cost

0.51x>22 = 0.25x

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 350

In order to write the total cost C as a function of one variable, we must find a rela-tionship between x and y. If the company produces x CDs in each of y production runs,the total number of CDs produced is xy. Thus,

Certainly, x must be at least 1 and cannot exceed 20,000. Therefore, we must solve thefollowing equation:

is not a critical value, since

Thus,

The company will minimize its total cost by making 4,000 CDs five times during theyear.

y =

20,0004,000

= 5

Min C1x2 = C14,0002 = 2,000

C–1x2 =

8,000,000

x3 7 0 for x H 11, 20,00021 … x … 20,000. x = 4,000

-4,000 x2= 16,000,000

x2=

4,000,0000.25

-4,000,000

x2 + 0.25 = 0

C¿1x2 = -

4,000,000

x2 + 0.25

C1x2 =

4,000,000x

+ 0.25x

Minimize C1x2 = 200a20,000xb + 0.25x 1 … x … 20,000

y =

20,000x

xy = 20,000


MATCHED PROBLEM 8 Repeat Example 8 if it costs $250 to set up a production run and $0.40 to store a CDfor one year.

Answers to Matched Problems 1. The dimensions of the garden with the maximum area of 640 square feet are 16 feet by40 feet, with one 16-foot side with wood fencing.

2. The minimal cost for enclosing a 1,800-square-foot garden is $480, and the dimensions are30 feet by 60 feet, with one 30-foot side with wood fencing.

3. The company will realize a maximum revenue of $675,000 when the price of a shredder is$150.

4. A maximum profit of $456,750 is realized when 4,050 shredders are manufactured annuallyand sold for $165 each.

5. A maximum profit of $378,750 is realized when 3,075 shredders are manufactured annuallyand sold for $175 each.

6. The maximum yield is 1,250 pounds per acre when 5 additional trees are planted on eachacre.

7. The maximum yield is 1,200 pounds when no additional trees are planted.

8. The company should produce 5,000 CDs four times a year.

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 351


Preliminary word problems:

1. How would you divide a 10-inch line so that the productof the two lengths is a maximum?

2. What quantity should be added to 5 and subtracted from5 to produce the maximum product of the results?

3. Find two numbers whose difference is 30 and whoseproduct is a minimum.

4. Find two positive numbers whose sum is 60 and whoseproduct is a maximum.

5. Find the dimensions of a rectangle with perimeter100 centimeters that has maximum area. Find themaximum area.

6. Find the dimensions of a rectangle of area 225 squarecentimeters that has the least perimeter. What is theperimeter?

Problems 7–10 refer to a rectangular area enclosed by a fencethat costs $B per foot. Discuss the existence of a solution andthe economical implications of each optimization problem.

7. Given a fixed area, minimize the cost of the fencing.

8. Given a fixed area, maximize the cost of the fencing.

9. Given a fixed amount to spend on fencing, maximize theenclosed area.

10. Given a fixed amount to spend on fencing, minimize theenclosed area.

11. Maximum revenue and profit. A company manufacturesand sells x videophones per week. The weekly price–demand and cost equations are, respectively,

(A) What price should the company charge for thephones, and how many phones should be producedto maximize the weekly revenue? What is themaximum weekly revenue?

(B) What is the maximum weekly profit? How muchshould the company charge for the phones, and howmany phones should be produced to realize the max-imum weekly profit?

12. Maximum revenue and profit. A company manufacturesand sells x digital cameras per week. The weekly price–demand and cost equations are, respectively,

(A) What price should the company charge for thecameras, and how many cameras should be producedto maximize the weekly revenue? What is themaximum revenue?

(B) What is the maximum weekly profit? How muchshould the company charge for the cameras, and howmany cameras should be produced to realize themaximum weekly profit?

13. Maximum revenue and profit. A company manufacturesand sells x television sets per month. The monthly costand price–demand equations are

p = 200 -

x

30 0 … x … 6,000

C1x2 = 72,000 + 60x

p = 400 - 0.4x and C1x2 = 2,000 + 160x

p = 500 - 0.5x and C1x2 = 20,000 + 135x

(A) Find the maximum revenue.(B) Find the maximum profit, the production level that

will realize the maximum profit, and the price thecompany should charge for each television set.

(C) If the government decides to tax the company $5 foreach set it produces, how many sets should the com-pany manufacture each month to maximize its profit?What is the maximum profit? What should thecompany charge for each set?

14. Maximum revenue and profit. Repeat Problem 13 for

15. Maximum profit. The following table contains price–demand and total cost data for the production of radialarm saws, where p is the wholesale price (in dollars) of asaw for an annual demand of x saws and C is the totalcost (in dollars) of producing x saws:

p = 200 -

x

50 0 … x … 10,000

C1x2 = 60,000 + 60x

Exercise 5-6

x p C

2,300 98 145,000

3,300 84 170,000

4,500 67 190,000

5,200 51 210,000

(A) Find a quadratic regression equation for the price–demand data, using x as the independent variable.

(B) Find a linear regression equation for the cost data,using x as the independent variable.

(C) What is the maximum profit? What is the wholesaleprice per saw that should be charged to realize themaximum profit? Round answers to the nearestdollar.

16. Maximum profit. The following table contains price–demand and total cost data for the production of air-brushes, where p is the wholesale price (in dollars) of anairbrush for an annual demand of x airbrushes and C isthe total cost (in dollars) of producing x airbrushes:

x p C

950 240 130,000

1,200 210 150,000

1,800 160 180,000

2,050 120 190,000

(A) Find a quadratic regression equation for the price–demand data, using x as the independent variable.

(B) Find a linear regression equation for the cost data,using x as the independent variable.

(C) What is the maximum profit? What is the wholesaleprice per airbrush that should be charged to realizethe maximum profit? Round answers to the nearestdollar.

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 352


17. Maximum revenue. A deli sells 640 sandwiches per dayat a price of $8 each.

(A) A market survey shows that for every $0.10 reduc-tion in price, 40 more sandwiches will be sold. Howmuch should the deli charge for a sandwich in orderto maximize revenue?

(B) A different market survey shows that for every$0.20 reduction in the original $8 price, 15 more sand-wiches will be sold. Now how much should the delicharge for a sandwich in order to maximize revenue?

18. Maximum revenue. A university student center sells1,600 cups of coffee per day at a price of $2.40.

(A) A market survey shows that for every $0.05 reduc-tion in price, 50 more cups of coffee will be sold.How much should the student center charge for acup of coffee in order to maximize revenue?

(B) A different market survey shows that for every $0.10reduction in the original $2.40 price, 60 more cups ofcoffee will be sold. Now how much should the stu-dent center charge for a cup of coffee in order tomaximize revenue?

19. Car rental. A car rental agency rents 200 cars per day at arate of $30 per day. For each $1 increase in rate, 5 fewercars are rented. At what rate should the cars be rentedto produce the maximum income? What is the maximumincome?

20. Rental income. A 300-room hotel in Las Vegas is filled tocapacity every night at $80 a room. For each $1 increase inrent, 3 fewer rooms are rented. If each rented room costs$10 to service per day, how much should the managementcharge for each room to maximize gross profit? What isthe maximum gross profit?

21. Agriculture. A commercial cherry grower estimates frompast records that if 30 trees are planted per acre, eachtree will yield an average of 50 pounds of cherries perseason. If, for each additional tree planted per acre (up to20), the average yield per tree is reduced by 1 pound, howmany trees should be planted per acre to obtain the max-imum yield per acre? What is the maximum yield?

22. Agriculture. A commercial pear grower must decide onthe optimum time to have fruit picked and sold. If the pearsare picked now, they will bring 30¢ per pound, with eachtree yielding an average of 60 pounds of salable pears. Ifthe average yield per tree increases 6 pounds per tree perweek for the next 4 weeks, but the price drops 2¢ per poundper week, when should the pears be picked to realize themaximum return per tree? What is the maximum return?

23. Manufacturing. A candy box is to be made out of a pieceof cardboard that measures 8 by 12 inches. Squares ofequal size will be cut out of each corner, and then theends and sides will be folded up to form a rectangularbox. What size square should be cut from each corner toobtain a maximum volume?

24. Packaging. A parcel delivery service will deliver a pack-age only if the length plus girth (distance around) doesnot exceed 108 inches.

(A) Find the dimensions of a rectangular box with squareends that satisfies the delivery service’s restriction andhas maximum volume. What is the maximum volume?

(B) Find the dimensions (radius and height) of a cylin-drical container that meets the delivery service’s

Length

Girth

Figure for 24

requirement and has maximum volume. What is themaximum volume?

25. Construction costs. A fence is to be built to enclose arectangular area of 800 square feet. The fence along threesides is to be made of material that costs $6 per foot. Thematerial for the fourth side costs $18 per foot. Find thedimensions of the rectangle that will allow the mosteconomical fence to be built.

26. Construction costs. The owner of a retail lumber storewants to construct a fence to enclose an outdoor storagearea adjacent to the store, using all of the store as part ofone side of the area (see the figure). Find the dimensionsthat will enclose the largest area if(A) 240 feet of fencing material are used.(B) 400 feet of fencing material are used.

100 ft

Figure for 26

27. Inventory control. A paint manufacturer has a uniformannual demand for 16,000 cans of automobile primer. Itcosts $4 to store one can of paint for one year and $500 toset up the plant for production of the primer. How manytimes a year should the company produce this primer inorder to minimize the total storage and setup costs?

28. Inventory control. A pharmacy has a uniform annualdemand for 200 bottles of a certain antibiotic. It costs $10to store one bottle for one year and $40 to place an order.How many times during the year should the pharmacyorder the antibiotic in order to minimize the total storageand reorder costs?

29. Inventory control. A publishing company sells 50,000copies of a certain book each year. It costs the company$1 to store a book for one year. Each time it must printadditional copies, it costs the company $1,000 to set upthe presses. How many books should the company pro-duce during each printing in order to minimize its totalstorage and setup costs?

30. Operational costs. The cost per hour for fuel to run atrain is dollars, where v is the speed of the train inmiles per hour. (Note that the cost goes up as the square

v2>4

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 353


of the speed.) Other costs, including labor, are $300 perhour. How fast should the train travel on a 360-mile tripto minimize the total cost for the trip?

31. Construction costs. A freshwater pipeline is to be run froma source on the edge of a lake to a small resort communityon an island 5 miles offshore, as indicated in the figure.

of a large lake to an island 5 miles offshore and is thenreleased (see the figure).

Island

5 mi

10 miles10 x

x

Figure for 31

(A) If it costs 1.4 times as much to lay the pipe in the lakeas it does on land, what should x be (in miles) to min-imize the total cost of the project?

(B) If it costs only 1.1 times as much to lay the pipe inthe lake as it does on land, what should x be to mini-mize the total cost of the project? [Note: Comparewith Problem 36.]

32. Manufacturing costs. A manufacturer wants to producecans that will hold 12 ounces (approximately 22 cubicinches) in the form of a right circular cylinder. Find thedimensions (radius of an end and height) of the can thatwill use the smallest amount of material. Assume that thecircular ends are cut out of squares, with the corner por-tions wasted, and the sides are made from rectangles, withno waste.

33. Bacteria control. A lake used for recreational swimmingis treated periodically to control harmful bacteria growth.Suppose that t days after a treatment, the concentrationof bacteria per cubic centimeter is given by

How many days after a treatment will the concentrationbe minimal? What is the minimum concentration?

34. Drug concentration. The concentration C(t), in milli-grams per cubic centimeter, of a particular drug in apatient’s bloodstream is given by

where t is the number of hours after the drug is taken. Howmany hours after the drug is taken will the concentrationbe maximum? What is the maximum concentration?

35. Laboratory management. A laboratory uses 500 whitemice each year for experimental purposes. It costs $4 tofeed a mouse for one year. Each time mice are orderedfrom a supplier, there is a service charge of $10 for pro-cessing the order. How many mice should be orderedeach time to minimize the total cost of feeding the miceand of placing orders for the mice?

36. Bird flights. Some birds tend to avoid flights over largebodies of water during daylight hours. (Scientists specu-late that more energy is required to fly over water thanland, because air generally rises over land and falls overwater during the day.) Suppose that an adult bird withthis tendency is taken from its nesting area on the edge

C1t2 =

0.16t

t2+ 4t + 4

C1t2 = 30t2- 240t + 500 0 … t … 8

Flight pathNesting

area

Lake

Island

5 mi

10 miles10 x

x

Figure for 36

(A) If it takes 1.4 times as much energy to fly over wateras land, how far up the shore (x, in miles) should thebird head to minimize the total energy expended inreturning to the nesting area?

(B) If it takes only 1.1 times as much energy to fly overwater as land, how far up the shore should the birdhead to minimize the total energy expended in re-turning to the nesting area? [Note: Compare withProblem 31.]

37. Botany. If it is known from past experiments that theheight (in feet) of a certain plant after t months is givenapproximately by

how long, on the average, will it take a plant to reach itsmaximum height? What is the maximum height?

38. Pollution. Two heavily industrial areas are located10 miles apart, as indicated in the figure. If the concentra-tion of particulate matter (in parts per million) decreasesas the reciprocal of the square of the distance from thesource, and if area emits eight times the particulatematter as the concentration of particulate matter atany point between the two areas is given by

How far from will the concentration of particulatematter between the two areas be at a minimum?

A1

C1x2 =

8k

x2 +

k

110 - x22 0.5 … x … 9.5, k 7 0

A2,A1

H1t2 = 4t1>2- 2t 0 … t … 2

x

A1 A2

10 x

Figure for 38

39. Politics. In a newly incorporated city, it is estimated thatthe voting population (in thousands) will increase accord-ing to

where t is time in years. When will the rate of increase bemost rapid?

40. Learning. A large grocery chain found that, on the aver-age, a checker can recall P% of a given price list x hoursafter starting work, as given approximately by

At what time x does the checker recall a maximumpercentage? What is the maximum?

P1x2 = 96x - 24x2 0 … x … 3

N1t2 = 30 + 12t2- t3 0 … t … 8

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 354

Review Exercise 355

CHAPTER 5 REVIEW

5-1 First Derivative and Graphs• Increasing and decreasing properties of a function can be determined by examining a sign chart for the

derivative.

• A number c is a partition number for if or is discontinuous at c. If c is also in thedomain of , then c is a critical value.

• Increasing and decreasing properties and local extrema for can be determined by examining thegraph of .

• The first-derivative test is used to locate extrema of a function.

5-2 Second Derivative and Graphs• The second derivative of a function f can be used to determine the concavity of the graph of f.

• Inflection points on a graph are points where the concavity changes.

• The concavity of the graph of can also be determined by an examination of the graph of .

• A graphing strategy is used to organize the information obtained from the first and second derivatives.

5-3 L’Hôpital’s Rule• Limits at infinity and infinite limits involving powers of , , and ln x are reviewed.

• The first version of L’Hôpital’s rule applies to limits involving the indeterminate form as .

• You must always check that L’Hôpital’s rule applies.

• L’Hôpital’s rule can be applied more than once.

• L’Hôpital’s rule applies to one-sided limits.

• L’Hôpital’s rule applies to limits at infinity.

• L’Hôpital’s rule applies to limits involving the indeterminate form .

5-4 Curve-Sketching Techniques• The graphing strategy first used in Section 5-2 is expanded to include horizontal and vertical asymptotes.

• If is a rational function with the degree of 1 more than the degree of , thenthe graph of has an oblique asymptote of the form

5-5 Absolute Maxima and Minima• The steps involved in finding the absolute maximum and absolute minimum values of a continuous

function on a closed interval are listed in a procedure.

• The second-derivative test for local extrema can be used to test critical values, but it does not work in allcases.

• If a function is continuous on an interval I and has only one critical value in I, the second-derivative testfor absolute extrema can be used to find the absolute extrema, but it not does work in all cases.

5-6 Optimization• The methods used to solve optimization problem are summarized and illustrated by examples.

y = mx + b.f(x)d(x)n(x)f(x) = n(x)>d(x)

q>q

x : c0>0exx - c

f¿(x)f(x)

f¿(x)f(x)

f(x)f¿(x)f¿(c) = 0f¿(x)

Ex. 1, p. 276

Ex. 2, p. 277Ex. 3, p. 278Ex. 4, p. 278Ex. 5, p. 279Ex. 6, p. 280Ex. 7, p. 283

Ex. 1, p. 296Ex. 2, p. 298Ex. 3, p. 298Ex. 4, p. 300

Ex. 5, p. 301Ex. 6, p. 303

Ex. 1, p. 312Ex. 2, p. 313Ex. 3, p. 314Ex. 4, p. 315Ex. 5, p. 315

Ex. 6, p. 316

Ex. 7, p. 317

Ex. 8, p. 317

Ex. 9, p. 318Ex. 10, p. 319

Ex. 1, p. 321Ex. 2, p. 323Ex. 3, p. 324Ex. 4, p. 326Ex. 5, p. 327

Ex. 1, .p. 337

Ex. 2, p. 339

Ex. 3, p. 340

Ex. 1, p. 343Ex. 2, p. 344Ex. 3, p. 346Ex. 4, p. 346Ex. 5, p. 348Ex. 6, p. 348Ex. 7, p. 349Ex. 8, p. 350

Examples

Important Terms, Symbols, and Concepts

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 355


10. Domain: All real x;

11. Find for

12. Find for

In Problems 13–22, summarize all the pertinent informationobtained by applying the final version of the graphing strategy(Section 5–4) to f, and sketch the graph of f.

13. f1x2 = x3- 18x2

+ 81x

y = 3x +

4x

.y–

f1x2 = x4+ 5x3.f–1x2

limx: -q

f1x2 = 2; limx: q

f1x2 = 2

f–1x2 7 0 on 1-2, 22;f–1x2 6 0 on 1- q , -22 and 12, q2;f–1-22 = 0, f–122 = 0;

f¿1x2 7 0 on 10, q2;f¿102 = 0; f¿1x2 6 0 on 1- q , 02;f1-22 = 1, f102 = 0, f122 = 1;

f (x)

2x

0

0

0

f (x)

1x

0

0

0

2

ND

2

ND

(C)

x

f (x)

5

55

5

5

55

5

f (x)

x

x

f (x)

5

55

5

In Problems 9 and 10, use the given information to sketch thegraph of f. Assume that f is continuous on its domain and thatall intercepts are included in the information given.

9. Domain: All real x

x 0 2 3

f(x) 0 3 2 0 0-3

-1-2-3

14.

15. 16.

17. 18.

19. 20.

21. 22.

Find each limit in Problems 23–32.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. Use the graph of shown here to discuss thegraph of Organizeyour conclusions in a table(see Example 4, Section 5-2).Sketch a possible graph ofy = f1x2.

y = f1x2.y = f¿1x2

limx:0

ln(1 + 6x)

ln(1 + 3x)lim

x: q

ln(1 + 6x)

ln(1 + 3x)

limx: q

ln x

x5lim

x:0 +

11 + x - 11x

limx:0

ex

+ e- x- 2

x2limx: q

e4x

x2

limx:0

ln(1 + x)

1 + xlim

x:0 -

ln(1 + x)

x2

limx:2

x2

- 5x + 6

x2+ x - 6

limx:0

e3x

- 1x

f1x2 = x3 ln xf1x2 = 5 - 5e- x

f1x2 =

x3

x2+ 3

f1x2 =

x

1x + 222

f1x2 =

x2

x2+ 27

f1x2 =

3x

x + 2

f1x2 = 1x - 1231x + 32f1x2 = 8x3- 2x4

f1x2 = 1x + 421x - 222

34. Refer to the proceding graph of . Which of thefollowing could be the graph of

(A) (B)

y = f–1x2?y = f¿1x2

Work through all the problems in this chapter review, and checkyour answers in the back of the book. Answers to all review prob-lems are there, along with section numbers in italics to indicatewhere each type of problem is discussed. Where weaknesses showup, review appropriate sections in the text.

A Problems 1–8 refer to the graph of that follows.Identify the points or intervals on the x axis that produce theindicated behavior.

1. f(x) is increasing. 2.

3. The graph of f is concave downward.

f¿1x2 6 0

y = f1x2

4. Local minima 5. Absolute maxima

6. appears to be 0. 7. does not exist.

8. Inflection points

f¿1x2f¿1x2

x

f(x)

a c1 c2

c3

c4 c5 c6 c7

b

REVIEW EXERCISE

x

f (x)

5

55

5

Figure for 33 and 34

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 356

Applications 357

C43. Find the absolute maximum for if

Graph f and on the same coordinate system for

44. Find two positive numbers whose product is 400 and whosesum is a minimum. What is the minimum sum?

In Problems 45 and 46, apply the graphing strategy andsummarize the pertinent information. Round anyapproximate values to two decimal places.

45.

46.

47. Find the absolute maximum value, if it exists, for


f(x) =

ln xex x 7 0

f(x) = 3x - x2+ e- x x 7 0

f1x2 = 0.25x4- 5x3

+ 31x2- 70x

f1x2 = x4+ x3

- 4x2- 3x + 4

0 … x … 4.f¿

f1x2 = 6x2- x3

+ 8

f¿1x235. Use the second-derivative test to find any local extrema for

36. Find the absolute maximum and absolute minimum, ifeither exists, for

37. Find the absolute minimum, if it exists, for



40. Let be a polynomial function with local minimaat and Must f have at least one localmaximum between a and b? Justify your answer.

41. The derivative of is Sincefor is it correct to say that f(x) is

decreasing for all x except Explain.

42. Discuss the difference between a partition number forand a critical value of f(x), and illustrate with

examples.f¿1x2

x = 0?x Z 0,f¿1x2 6 0

f¿1x2 = -x-2.f1x2 = x-1

x = b, a 6 b.x = ay = f1x2

f1x2 = 10xe- 2x x 7 0

f1x2 = 11x - 2x ln x x 7 0

y = f1x2 = x2+

16

x2 x 7 0

y = f1x2 = x3- 12x + 12 -3 … x … 5

f1x2 = x3- 6x2

- 15x + 12

Figure for 49

49. Price analysis. The graph in the figure approximates therate of change of the price of tomatoes over a 60-monthperiod, where p(t) is the price of a pound of tomatoesand t is time (in months).

(A) Write a brief verbal description of the graph ofincluding a discussion of local extrema and

inflection points.(B) Sketch a possible graph of y = p1t2.

y = p1t2,

company manufacture each month to maximize itsprofit? What is the maximum monthly profit? Howmuch should the company charge for each stove?

51. Construction. A fence is to be built to enclose arectangular area. The fence along three sides is to bemade of material that costs $5 per foot. The material forthe fourth side costs $15 per foot.(A) If the area is 5,000 square feet, find the dimensions

of the rectangle that will allow the most economicalfence to be built.

(B) If $3,000 is available for the fencing, find the dimen-sions of the rectangle that will enclose the most area.

52. Rental income. A 200-room hotel in Fresno is filled tocapacity every night at a rate of $40 per room. For each $1increase in the nightly rate, 4 fewer rooms are rented. Ifeach rented room costs $8 a day to service, how muchshould the management charge per room in order to max-imize gross profit? What is the maximum gross profit?

53. Inventory control. A computer store sells 7,200 boxes offloppy disks annually. It costs the store $0.20 to store a boxof disks for one year. Each time it reorders disks, the storemust pay a $5.00 service charge for processing the order.How many times during the year should the store orderdisks to minimize the total storage and reorder costs?

APPLICATIONS

x

0.02

0.02

0.04

0.04

0.06

0.08

15 30 45 60

p(t)

50. Maximum revenue and profit. A company manufacturesand sells x electric stoves per month. The monthly costand price–demand equations are, respectively,

(A) Find the maximum revenue.(B) How many stoves should the company manufacture

each month to maximize its profit? What is themaximum monthly profit? How much should thecompany charge for each stove?

(C) If the government decides to tax the company $20 foreach stove it produces, how many stoves should the

p = 500 - 0.025x 0 … x … 20,000

C1x2 = 350x + 50,000

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 357

54. Average cost. The total cost of producing x garbage dis-posals per day is given by

Find the minimum average cost. Graph the average costand the marginal cost functions on the same coordinatesystem. Include any oblique asymptotes.

55. Average cost. The cost of producing x window fans isgiven by

Find the minimum average cost.

56. Marginal analysis. The price–demand equation for washing machines at an appliance store is

where x is the monthly demand and p is the price in dollars.Find the production level and price per unit that producesthe maximum revenue.What is the maximum revenue?

57. Maximum revenue. Graph the revenue function fromProblem 56 for

58. Maximum profit. Refer to Problem 56. If the washingmachines cost the store $220 each, find the price (to thenearest cent) that maximizes the profit. What is the maxi-mum profit (to the nearest dollar)?

59. Maximum profit. The data in the table show the dailydemand x for cream puffs at a state fair at various pricelevels p. If it costs $1 to make a cream puff, use logarith-mic regression to find the price (to thenearest cent) that maximizes profit.

1p = a + b ln x2

0 … x … 100.

p1x2 = 1,000e-0.02x

C1x2 = 200 + 50x - 50 ln x x Ú 1

C1x2 = 4,000 + 10x + 0.1x2

Demand Price per Cream Puff ($)x p

3,125 1.99

3,879 1.89

5,263 1.79

5,792 1.69

6,748 1.59

8,120 1.49

12 feet

10 feet

(A) Enter the data into a graphing calculator and find aquadratic regression equation for the total cost.


62. Advertising—point of diminishing returns. A companyestimates that it will sell units of a product afterspending $x thousand on advertising, as given by



63. Advertising. A chain of appliance stores uses televisionads to promote the sales of refrigerators. Analyzing pastrecords produced the data in the following table, where xis the number of ads placed monthly and y is the numberof refrigerators sold that month:

N¿

9 … x … 24

N1x2 = -0.25x4+ 11x3

- 108x2+ 3,000

N(x)

Dozens of Cookies Total Cost x y

50 119

100 187

150 248

200 382

250 505

300 695


Number of Ads Number of Refrigerators x y

10 271

20 427

25 526

30 629

45 887

48 917

(A) Enter the data into a graphing calculator, set the cal-culator to display two decimal places, and find a cubicregression equation for the number of refrigeratorssold monthly as a function of the number of ads.

(B) How many ads should be placed each month tomaximize the rate of change of sales with respect tothe number of ads, and how many refrigerators canbe expected to be sold with that number of ads?Round answers to the nearest integer.

64. Bacteria control. If t days after a treatment the bacteriacount per cubic centimeter in a body of water is given by

in how many days will the count be a minimum?

65. Politics. In a new suburb, it is estimated that the numberof registered voters will grow according to

where t is time in years and N is in thousands. When willthe rate of increase be maximum?

N = 10 + 6t2- t3 0 … t … 5

C1t2 = 20t2- 120t + 800 0 … t … 9

61. Average cost. The data in the table give the total dailycost y (in dollars) of producing x dozen chocolate chipcookies at various production levels.

60. Construction costs. The ceiling supports in a newdiscount department store are 12 feet apart. Lights are tobe hung from these supports by chains in the shape of a“Y.” If the lights are 10 feet below the ceiling, what is theshortest length of chain that can be used to support theselights?

BARNMC05_0132328186.QXD 02/21/2007 20:59 Page 358

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