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CALCULUSFor Business, Economics, and the Social and life Sciences Hoffmann, L.D. & Bradley, G.L.

Chapter 3Additional Applications of the Derivative

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Additional Applications of the Derivative

In this Chapter, we will encounter some important concepts.

Increasing and Decreasing Functions; Relative Extrema （相对极值）

Concavity （凹凸性） and Points of Inflection （拐点） .

Optimization （最优化）

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Section 3.1 Increasing and Decreasing FunctionsIncreasing and Decreasing FunctionIncreasing and Decreasing Function Let Let f(x)f(x) be a function be a functiondefined on the interval defined on the interval a<x<ba<x<b, and let , and let xx11 and and xx22 be two numbers be two numbers

in the interval, Then in the interval, Then ff((xx) is ) is increasingincreasing on the interval if on the interval if f(xf(x22)>f(x)>f(x11)) whenever whenever xx22>x>x11

ff((xx) is ) is decreasingdecreasing on the interval if on the interval if f(xf(x22)<f(x)<f(x11)) whenever whenever xx22 >x >x11

Monotonic Monotonic IncreasingIncreasing

MonotonicMonotonic

DecreasingDecreasing

4

( ) 0f x

Tangent line with positive slopeTangent line with positive slope f(x)f(x) will be increasing will be increasing

( ) 0f x

Tangent line with negative slopeTangent line with negative slope f(x)f(x) will be decreasing will be decreasing

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If for every x on some interval I, then f(x) is increasing on the interval.If for every x on some interval I, then f(x) is decreasing on the interval.If for every x on some interval I, then f(x) is constant on the interval.

( ) 0f x

0)( xf

0)( xf

How to determine all intervals of increase and decrease How to determine all intervals of increase and decrease for a function ? How to find all intervals on which the sign for a function ? How to find all intervals on which the sign of the derivative does not change.of the derivative does not change.

Intermediate value propertyIntermediate value property A continuous function cannot change sign without first becoming A continuous function cannot change sign without first becoming 0.0.

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Procedure for using the derivative to determine Procedure for using the derivative to determine intervals of increase and decrease for a function of intervals of increase and decrease for a function of ff..

Step 2. Choose a test number Step 2. Choose a test number cc from each interval from each interval a<x<b a<x<b determined in the step determined in the step 11 and evaluate . Then and evaluate . Then If the function If the function f(x)f(x) is increasing on is increasing on a<x<ba<x<b.. If the function If the function f(x)f(x) is decreasing on is decreasing on a<x<b.a<x<b.( ) 0f c

( ) 0f c )(cf

Step 1. Find all values of Step 1. Find all values of x x for which or isfor which or isnot continuous, and mark these numbers on a number line. not continuous, and mark these numbers on a number line. This divides the line into a number of open intervals. This divides the line into a number of open intervals.

)(xf ( ) 0f x

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Example 1

Find the intervals of increase and decrease for the functionFind the intervals of increase and decrease for the function3 2( ) 2 3 12 7f x x x x

Solution:Solution:2( ) 6 6 12 6( 2)( 1)f x x x x x The derivative of The derivative of f(x)f(x) is is

which is continuous everywhere, with where which is continuous everywhere, with where x=1x=1 and and x=-2. x=-2. The number The number -2-2 and and 11 divide divide xx axis into three open axis into three open intervals.intervals.x<-2x<-2, , -2<x<1-2<x<1 and and x>1.x>1.

( ) 0f x

Rising f is increasing 2x>1

Falling f is deceasing 0-2<x<1

Rising f is increasing - 3x<-2

Direction Direction

of graphof graph

Conclusion Conclusion TestTest

NumberNumber

Interval )(cf

0)3( f

0)0( f0)2( f

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Example 2

Find the interval of increase and decrease for the function Find the interval of increase and decrease for the function

2)(

2

x

xxf

Solution:Solution:

To be continued

The function is defined for The function is defined for xx≠2≠2, and its derivative is , and its derivative is

22

2

)2(

)4(

)2(

)1()2)(2()(

x

xx

x

xxxxf

which is discontinuous at which is discontinuous at x=2x=2 and has and has f’(x)=0f’(x)=0 at at x=0x=0 and and x=4x=4. . Thus, there are four intervals on which the sign of Thus, there are four intervals on which the sign of f’(x)f’(x) does not does not change: namely, change: namely, x<0x<0, , 0<x<20<x<2, , 2<x<42<x<4 and and x>4x>4. Choosing test . Choosing test numbers in these intervals (say, numbers in these intervals (say, -2,1,3-2,1,3, and , and 55, respectively), we , respectively), we find that find that

9

09

5)5( 03)3( 03)1( 0

4

3)2( ffff

We conclude that We conclude that f(x)f(x) is increasing for is increasing for x<0x<0 and for and for x>4x>4 and that and that it is decreasing for it is decreasing for 0<x<20<x<2 and for and for 2<x<42<x<4. .

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Relative (Local) ExtremaRelative (Local) Extrema The Graph of the function The Graph of the function f(x)f(x) i i

s said to be have a s said to be have a relative maximumrelative maximum at at x=cx=c if if f(c)f(c)≥≥f(x)f(x) f for all or all xx in interval in interval a<x<ba<x<b containing containing cc. Similarly the graph . Similarly the graph has a has a relative Minimumrelative Minimum at at x=cx=c if if f(c)f(c)≤≤ f(x) f(x) on such an inte on such an interval. Collectively, the relative maxima and minima of rval. Collectively, the relative maxima and minima of ff are are called its relative extremacalled its relative extrema （相对极值或局部极值）（相对极值或局部极值） . .

Peaks: C,E,Peaks: C,E,(Relative (Relative maxima)maxima)Valleys: B, D, Valleys: B, D, G (Relative G (Relative minima)minima)

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Critical Numbers and Critical PointsCritical Numbers and Critical Points （关键点）（关键点） : A nu: A number mber cc in the domain of in the domain of f(x)f(x) is called a critical number i is called a critical number if either or is not continuous. The correspof either or is not continuous. The corresponding point nding point (c,f(c))(c,f(c)) on the graph of on the graph of f(x)f(x) is called a critica is called a critical point for l point for f(x).f(x).

)(cf ( ) 0f c

Relative extrema can only occur at critical points!Relative extrema can only occur at critical points!

Since a function Since a function f(x)f(x) is increasing when is increasing when f’(x)>0f’(x)>0 and and decreasing when decreasing when f’(x)<0f’(x)<0, the only points where , the only points where f(x)f(x) can can have a relative extremum are where have a relative extremum are where f’(x)=0f’(x)=0 or or f’(x)f’(x) is is not continuous. Such points are so important that we not continuous. Such points are so important that we give them a special name. give them a special name.

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Not all critical points correspond to relative extrema!Not all critical points correspond to relative extrema!

Figure. Three critical points where Figure. Three critical points where f’f’((xx) = 0: (a) relative maximum,) = 0: (a) relative maximum, (b) relative minimum (c) not a relative extremum. (b) relative minimum (c) not a relative extremum.

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Not all critical points correspond to relative extrema!Not all critical points correspond to relative extrema!

FigureFigure Three critical points where Three critical points where f’f’((xx) is undefined:(a) relative m) is undefined:(a) relative maximum, (b) relative minimum (c) not a relative extremum.aximum, (b) relative minimum (c) not a relative extremum.

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The First Derivative Test for Relative ExtremaThe First Derivative Test for Relative Extrema Let Let cc be a critical number for be a critical number for f(x)f(x) [that is, [that is, f(c)f(c) is defined is defined and either or is not continuous]. Then the criticaand either or is not continuous]. Then the critical point l point (c,f(c))(c,f(c)) is is

( ) 0f x )(cf

A A relative maximumrelative maximum if to the if to the left of left of cc and to the right of and to the right of cc

( ) 0f x

( ) 0f x c 0f 0f

A A relative minimumrelative minimum if to the if to the left of left of cc and to the right of and to the right of cc

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

Not a relative extremumNot a relative extremum if if has the same sign on both sides of has the same sign on both sides of cc

)(xf c 0f 0f

c 0f 0f

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Find all critical numbers of the function Find all critical numbers of the function and classify each critical point as a relative maximum, a and classify each critical point as a relative maximum, a relative minimum, or neither. relative minimum, or neither.

Example 3

342)( 24 xxxf

Solution:Solution:3( ) 8 8 8 ( 1)( 1)f x x x x x x

The derivative exists for all The derivative exists for all xx, the only critical numbers , the only critical numbers are where that is, are where that is, x=0,x=-1,x=1x=0,x=-1,x=1. These . These numbers divide that numbers divide that xx axis into four intervals, axis into four intervals, x<-1x<-1, , -1<x<0-1<x<0, , 0<x<10<x<1, , x>1.x>1.

( ) 0f x

1 1 15( 5) 960 0 ( ) 3 0 ( ) 0 (2) 48 0

2 4 8f f f f

Choose a test number in each of these intervalsChoose a test number in each of these intervals

To be continued

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Thus the graph of Thus the graph of ff falls for falls for x<-1x<-1 and for and for 0<x<10<x<1, and rises for , and rises for -1<x<0-1<x<0 and for and for x>1x>1, so , so x=0x=0 is relative maximum, is relative maximum, x=1x=1 and and x=-1x=-1 are are relative minimum.relative minimum.

-1 min-1 min

- - - - - - - - + + + + + + - - - - - - - - + + + + + +

0 max0 max 1 min1 min

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Step 4Step 4 Sketch the graph of f as a smooth curve joining the critical Sketch the graph of f as a smooth curve joining the criticalpoints in such way that it rise where , falls where points in such way that it rise where , falls where and has a horizontal tangent where . and has a horizontal tangent where .

Step 3Step 3.. Plot the critical point Plot the critical point P(c,f(c))P(c,f(c)) on a coordinate plane, on a coordinate plane, with a “cap” at with a “cap” at PP if it is a relative maximum or a “cup” if it is a relative maximum or a “cup” if if PP is a relative minimum. Plot intercepts and other key points that is a relative minimum. Plot intercepts and other key points that can be easily found.can be easily found.

Step 2Step 2. Find and each critical number, analyze the sign of . Find and each critical number, analyze the sign of derivative to determine intervals of increase and decrease for derivative to determine intervals of increase and decrease for f(x).f(x).

A Procedure for Sketching the Graph of a A Procedure for Sketching the Graph of a Continuous Continuous Function Function f(x)f(x) Using the Derivative Using the Derivative

Step 1Step 1.. Determine the domain of Determine the domain of f(x).f(x).

( )f x

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

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The derivative exists for all The derivative exists for all xx, the only critical numbers , the only critical numbers are where that is, are where that is, x=0x=0, , x=-3x=-3. These numbers . These numbers divide that divide that xx axis into three intervals, axis into three intervals, x<-3x<-3, , -3<x<0-3<x<0, , x>0x>0. . Choose test number in each interval (say, Choose test number in each interval (say, -5-5, , -1-1 and and 11 respectively) respectively)

Example 4

Sketch the graph of the functionSketch the graph of the function 4 3 2( ) 8 18 8f x x x x

Solution:Solution:3 2 2( ) 4 24 36 4 ( 3)f x x x x x x

( ) 0f x

Thus the graph of Thus the graph of ff has a horizontal tangents where has a horizontal tangents where xx is is -3-3 and and 00, and it is falling in the interval , and it is falling in the interval x<-3x<-3 and and -3<x<0-3<x<0 and is rising for and is rising for x>0.x>0.

( 5) 80 0 ( 1) 16 0 (1) 64 0f f f

To be continued

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-3-3 neitherneither

- - - - - - - - + + + + + +- - - - - - - -

0 0 minmin

f(-3)=19 f(0)=-8f(-3)=19 f(0)=-8 Plot a “cup” at the critical point Plot a “cup” at the critical point (0,-8)(0,-8)

Plot a “twist” at Plot a “twist” at (-3,19)(-3,19) to indicate a galling graph with to indicate a galling graph with a horizontal tangent at this pointa horizontal tangent at this point .

Complete the sketch by passing a smooth curve through the Complete the sketch by passing a smooth curve through the Critical point in the directions indicated by arrowCritical point in the directions indicated by arrow

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Example 5

The revenue derived from the sale of a new kind of The revenue derived from the sale of a new kind of motorized skateboard motorized skateboard tt weeks after its introduction is weeks after its introduction is given by given by

2

2

63( )

63

t tR t

t

million dollars. When does maximum revenue occur? million dollars. When does maximum revenue occur? What is the maximum revenue? What is the maximum revenue?

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Solution:Solution:

Differentiating Differentiating R(t)R(t) by the quotient rule, we get by the quotient rule, we get

77 MaxMax

++++++ - - - - - t00 6363

By setting the numerator in this expression for By setting the numerator in this expression for R’(t)R’(t) equal to equal to 0, 0, we find that t=7 is the only solution in the interval we find that t=7 is the only solution in the interval 0≤t≤63, 0≤t≤63, and and hence is the only critical number of hence is the only critical number of R(t)R(t) in its domain. The in its domain. The critical number divides the domain into two intervals: critical number divides the domain into two intervals: 0≤t<7 0≤t<7 andand 7<t≤63. 7<t≤63.

2

2

63(7) (7)(7) 3.5

(7) 63R

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Increase and decrease of the slopes are our concern!Increase and decrease of the slopes are our concern!

Section 3.2 Concavity （凹凸性） and Points of Inflection （拐点）

Figure: The output Figure: The output QQ((tt) of a factory worker ) of a factory worker tt hours after coming hours after coming to work.to work.

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ConcavityConcavity: If the function : If the function f(x)f(x) is differentiable on the int is differentiable on the interval erval a<x<ba<x<b then the graph of then the graph of ff is is Concave upwardConcave upward （凹的）（凹的） on on a<x<ba<x<b if is incr if is increasing on the intervaleasing on the interval Concave downwardConcave downward （凸的）（凸的） on on a<x<ba<x<b if is d if is decreasing on the intervalecreasing on the interval

)(xf

)(xf

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A graph is concave upward on the interval if it lies A graph is concave upward on the interval if it lies above all its tangent lines on the interval and concave above all its tangent lines on the interval and concave downward on an Interval where it lies below all its downward on an Interval where it lies below all its tangent lines.tangent lines.

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Step 2Step 2. Choose a test number c from each interval . Choose a test number c from each interval a<x<ba<x<b determined in the step determined in the step 1 1 and evaluate . Then and evaluate . Then If , the graph of If , the graph of f(x)f(x) is concave upward on is concave upward on a<x<ba<x<b.. If , the graph of If , the graph of f(x)f(x) is concave downward on is concave downward on a<x<b.a<x<b.

Determining Intervals of Concavity Using the Sign of f’’

Step 1Step 1. Find all values of . Find all values of xx for which or for which or is not continuous, and mark these numbers on a number is not continuous, and mark these numbers on a number line. This divides the line into a number of open line. This divides the line into a number of open intervals.intervals.

0)( xf )(xf

)(cf

0)( cf

0)( cf

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Example 6

Determine intervals of concavity for the function Determine intervals of concavity for the function 3752)( 46 xxxxf

Solution:Solution:

We find that and We find that and 72012)( 35 xxxf

)1)(1(60)1(606060)( 22224 xxxxxxxxf

The second derivative The second derivative f’’(x)f’’(x) is continuous for all is continuous for all xx and and f’’(x)=0f’’(x)=0 f for or x=0x=0, , x=1x=1, and , and x=-1x=-1. These numbers divide the . These numbers divide the xx axis into four axis into four intervals on which intervals on which f’’(x)f’’(x) does not change sign; namely, does not change sign; namely, x<1x<1, -, -1<1<x<0x<0, , 0<x<10<x<1, and , and x>1x>1. Evaluating . Evaluating f’’(x)f’’(x) at test numbers in each o at test numbers in each of these intervals (say, at f these intervals (say, at x=-2x=-2, , x=-1/2, x=1/2x=-1/2, x=1/2, and , and x=5x=5, respectiv, respectively), we find that ely), we find that

to be continued

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-1-1

- - - - - - -+ + + + - - - - - + + + + +

00 11

Type of concavityType of concavity

Sign of Sign of xx

Thus, the graph of Thus, the graph of f(x)f(x) is concave up for is concave up for x<-1x<-1 and for and for x>1x>1 and and concave down for -1<x<0 and for 0<x<1, as indicated in this coconcave down for -1<x<0 and for 0<x<1, as indicated in this concavity diagram. ncavity diagram.

0000,36)5( 04

45

2

1

04

45

2

1 0720)2(

ff

ff

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NoteNote Don’t confuse the concavity of a graph with its Don’t confuse the concavity of a graph with its “direction” (rising or falling). A function may be increasing or “direction” (rising or falling). A function may be increasing or decreasing on an interval regardless of whether its graph is decreasing on an interval regardless of whether its graph is concave upward or concave downward on the interval.concave upward or concave downward on the interval.

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Inflection Point Inflection Point （拐点） （拐点） An inflection point is a point An inflection point is a point (c,f(c))(c,f(c)) on the graph of a function on the graph of a function ff where the concavity where the concavity changes. At such a point, either changes. At such a point, either f’’(c)=0f’’(c)=0 or or f’’(c)f’’(c) is not is not continuous. continuous.

is not continuous.

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Example 7

In each case, find all inflection points of the given In each case, find all inflection points of the given function.function.

3

145 )( b. 153)( a. xxgxxxf

Solution:Solution:

to be continued

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- - - - - - - - - - - + + + + +

00No inflectionNo inflection

11inflection inflection

Type of concavityType of concavitySign of

to be continued

32

- - - - - - + + + +

00inflectioninflection

Type of concavityType of concavity

Sign ofSign of

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For example, if For example, if f(x)=1/xf(x)=1/x, then , then

, so , so f’’(x)<0f’’(x)<0 if if x<0 x<0 and and f’’(x)>0f’’(x)>0 if if x>0. x>0. The The concavity changes from concavity changes from downward to upward at downward to upward at x=0x=0 but there is no inflection point but there is no inflection point at at x=0x=0 since since f(0)f(0) is not is not defined, defined,

NoteNote: A function can have an inflection point only : A function can have an inflection point only where it is continuous!!where it is continuous!!

3/2)( xxf

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For instance, if , For instance, if , then then f(0)=0f(0)=0 and and so so f’’(0)=0f’’(0)=0. . However, However, f’’(x)>0f’’(x)>0 for any for any number number x≠0x≠0 , so the graph , so the graph of of ff is always concave is always concave upward, and there is no upward, and there is no inflection point at inflection point at (0,0).(0,0).

4)( xxf 212)( xxf

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Behavior of Graph Behavior of Graph ff((xx) at an inflection point ) at an inflection point PP((c,fc,f((cc))))

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A graph showing concavity and inflection pointsA graph showing concavity and inflection points

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Example 8

Determine where the function Determine where the function

15181223)( 234 xxxxxf

Is increasing and decreasing, and where its graph is conIs increasing and decreasing, and where its graph is concave up and concave down. Find all relative extrema ancave up and concave down. Find all relative extrema and points of inflection, and sketch the graph. d points of inflection, and sketch the graph. Solution:Solution:

First, note that since First, note that since f(x)f(x) is a polynomial, it is continuous for all is a polynomial, it is continuous for all xx, , as are the derivatives as are the derivatives f’(x)f’(x) and and f’’(x).f’’(x). The first derivative of The first derivative of f(x)f(x) is is

)32()1(61824612)( 223 xxxxxxf

and and f’(x)=0f’(x)=0 only when only when x=1x=1 and and x=-1.5x=-1.5. The sign of . The sign of f’(x) f’(x) does not does not change for change for x<-1.5x<-1.5, nor in the interval , nor in the interval -1.5<x<1-1.5<x<1, nor for , nor for x>1x>1. .

to be continued

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-1.5-1.5minmin

1 1 NeitherNeither

- - - - - - - - + + + + + + + + + + + +Sign of Sign of f’f’(x)(x)

Evaluating Evaluating f’(x)f’(x) at test numbers in each interval (say, at at test numbers in each interval (say, at -2,0-2,0, an, and d 33), you obtain the arrow diagram shown. Note that there is a re), you obtain the arrow diagram shown. Note that there is a relative minimum at lative minimum at x=-1.5x=-1.5 but no extremum at but no extremum at x=1x=1. .

xx

The second derivative is and The second derivative is and f’’(x)=0f’’(x)=0 only when only when x=1x=1 and and x=-2/3x=-2/3. The sign of . The sign of f’’(x)f’’(x) does not does not change on each of the intervals change on each of the intervals x<-2/3x<-2/3, , -2/3<x<1-2/3<x<1, and , and x>1x>1. Eva. Evaluating luating f’’(x)f’’(x) at test numbers in each interval, we obtain the conc at test numbers in each interval, we obtain the concavity diagram. avity diagram.

)23)(1(12241236)( 2 xxxxxf

to be continued

39

- - - - - + + + + +

-2/3-2/3inflectioninflection

11inflectioninflection

Type of concavityType of concavity

Sign ofSign of + + + + +

The patterns in these two diagrams suggest that there is The patterns in these two diagrams suggest that there is a relative minimum at a relative minimum at x=-1.5x=-1.5 and inflection points at and inflection points at x=-2/3x=-2/3 and and x=1 x=1 (since the concavity changes at both (since the concavity changes at both points). points).

to be continued

40

41

42

43

Example 9

Solution:Solution:

to be continued

44

45

Example 10

Solution:Solution:

to be continued

46

33 MaxMax

+ + + + - - - - - -tt

00 44

47

Section 3.3 Curve Sketching limits involving infinitylimits involving infinity: Limits at infinity and infinite limits: Limits at infinity and infinite limitsOur goal is to see how limits involving infinity may be interpreted Our goal is to see how limits involving infinity may be interpreted as graphical features!!as graphical features!!

48

垂直渐进线垂直渐进线

49

Example 11

Determine all vertical asymptotes of the graph of Determine all vertical asymptotes of the graph of

xx

xxf

3

9)(

2

2

Solution:Solution:

Let and be the numerator and Let and be the numerator and denominator of denominator of f(x),f(x), respectively. The respectively. The q(x)=0q(x)=0 when when x=-3x=-3 and and when when x=0x=0. However, for . However, for x=-3x=-3, we also have , we also have p(-3)=0p(-3)=0 and and

9)( 2 xxp xxxq 3)( 2

23

lim3

9lim

32

2

3

x

x

xx

xxx

This means that the graph of This means that the graph of f(x)f(x) has a “hole” at the point has a “hole” at the point (-3,2)(-3,2) and and x=-3x=-3 is not a vertical asymptote of the graph. is not a vertical asymptote of the graph.

to be continued

50

On the other hand, for On the other hand, for x=0x=0 we have we have q(0)=0q(0)=0 but but p(0)p(0)≠0, ≠0, which swhich suggests that the uggests that the yy axis is a vertical asymptote for the graph of axis is a vertical asymptote for the graph of ff(x).(x). This asymptote behavior is verified by noting that This asymptote behavior is verified by noting that

xx

x

xx

xx 3

9lim and

3

9lim

2

2

0x2

2

0

51

水平渐进线水平渐进线

52

Example 12

Determine all horizontal asymptotes of the graph of Determine all horizontal asymptotes of the graph of

1)(

2

2

xx

xxf

Solution:Solution:

Dividing each term in the rational function Dividing each term in the rational function f(x)f(x) by (the by (the highest power of x in the denominator), we find that highest power of x in the denominator), we find that

2x

rulepowerreciprocalxx

xxxxx

xx

xx

xxf

x

xxx

1/1/11

1lim

/1//

/lim

1lim)(lim

2

2222

22

2

2

and similarly, and similarly, 1

1lim)(lim

2

2

xx

xxf

xx

to be continued

53

Thus, the graph of Thus, the graph of f(x)f(x) has has y=1y=1 as a horizontal asymptote. as a horizontal asymptote.

NOTENOTE: The graph of a function : The graph of a function f(x)f(x) can never cross a vertical can never cross a vertical asymptote asymptote x=cx=c because at least one of the one-sided limits because at least one of the one-sided limits must be infinite. However, it is possible for a must be infinite. However, it is possible for a graph to cross its horizontal asymptotes. graph to cross its horizontal asymptotes.

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to be continued

55

56

Example 13

Sketch the graph of the function Sketch the graph of the function

2)1()(

x

xxf

Solution:Solution:

to be continued

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-1 -1 AsymptoteAsymptote

- - - - - - + + + + - - - - - -

1 1 MaxMax

to be continued

58

- - - - - - - - - - - + + + + +

-1-1No inflectionNo inflection

22inflectioninflection

Type of concavityType of concavity

Sign of Sign of

to be continued

59

7. The vertical asymptote (dashed line) breaks the graph into two parts. join the features in each separate part by a smooth curve to obtain the completed graph.

Exercise

Sketch the graph of Sketch the graph of 152

3)(

2

2

xx

xxf

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Section 3.4 Optimization （最优化）Absolute Maxima and Minima Absolute Maxima and Minima （极大值和极小值）（极大值和极小值） of of a Functiona Function Let Let ff be a Function defined on an interval be a Function defined on an interval II t that contains the number c. Then hat contains the number c. Then ff(c) is the absolute maximum of (c) is the absolute maximum of ff on on I I if if ff(c) (c) ≥≥ ff(x) for (x) for all all xx in in II ff(c) is the absolute minimum of (c) is the absolute minimum of f f on on II if if ff(c) (c) ≤≤ ff(x) for (x) for all all xx in in II

Absolute extrema often Absolute extrema often coincide with relative coincide with relative extrema but not always!extrema but not always!

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The Extreme Value PropertyThe Extreme Value Property （极值定理）（极值定理） A function A function f(x)f(x) that that is continuous on the closed Interval is continuous on the closed Interval aa≤≤xx≤≤bb attains its absolute ex attains its absolute extrema on the interval either at an endpoint of the interval trema on the interval either at an endpoint of the interval (a or b)(a or b) or at a critical number or at a critical number cc such that such that a<c<ba<c<b. .

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Example 14

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Example 14

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Example 15

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Two General Principle of Marginal AnalysisTwo General Principle of Marginal Analysis

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Explanation in EconomicsExplanation in Economics The marginal cost ( The marginal cost (MCMC) ) is approximately the same as the cost of producing is approximately the same as the cost of producing one additional unit. If the additional unit costs less to one additional unit. If the additional unit costs less to produce than the average cost (produce than the average cost (ACAC) of the existing ) of the existing units (If units (If MC<ACMC<AC), then this less-expensive unit will ), then this less-expensive unit will cause the average cost per unit to decrease. On the cause the average cost per unit to decrease. On the other hand, if the additional unit costs more than the other hand, if the additional unit costs more than the average cost of the existing units (if average cost of the existing units (if MC>ACMC>AC), then ), then this more-expensive unit will cause the average cost this more-expensive unit will cause the average cost per unit to increase. However (if per unit to increase. However (if MC=ACMC=AC), then the ), then the average cost will neither increase nor decrease, which average cost will neither increase nor decrease, which means means (AC)’=0(AC)’=0

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Price Elasticity of DemandPrice Elasticity of Demand （需求的价格弹（需求的价格弹性）性）

In general, an increase in the unit price of a commodity In general, an increase in the unit price of a commodity will result in decreased demand, but the sensitivity or will result in decreased demand, but the sensitivity or responsiveness of demand to a change in price varies responsiveness of demand to a change in price varies from one product to another. For some products, such as from one product to another. For some products, such as soap, flashlight batteries, and salt, small percentage soap, flashlight batteries, and salt, small percentage changes in price have little effect on demand. For other changes in price have little effect on demand. For other products, such as airline tickets, designer furniture, and products, such as airline tickets, designer furniture, and home loans, small percentage changes in price can affect home loans, small percentage changes in price can affect demand dramatically. demand dramatically.

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Example 16

Suppose the demand Suppose the demand qq and price and price p p for a certain commodity are for a certain commodity are related by the linear equation related by the linear equation q=240-2pq=240-2p (for (for 0≤p≤1200≤p≤120).).

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Example 16

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b. b. The total revenue, The total revenue, R=pqR=pq, increases when demand is inelasti, increases when demand is inelastic; that is, when c; that is, when 0≤ p<10. 0≤ p<10. For this range of prices, a specified For this range of prices, a specified percentage increase in price results in a smaller percentage depercentage increase in price results in a smaller percentage decrease in demand, so the bookstore will take in more money fcrease in demand, so the bookstore will take in more money for each increase in price up to or each increase in price up to $10$10 per copy. per copy.

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However, for the price range However, for the price range 10<p10<p≤√300≤√300 , the demand is , the demand is elastic, so the revenue is decreasing. If the book is priced in this elastic, so the revenue is decreasing. If the book is priced in this range, a specified percentage increase in price results in a larger range, a specified percentage increase in price results in a larger percentage decrease in demand. Thus, if the bookstore increases percentage decrease in demand. Thus, if the bookstore increases the price beyond the price beyond $10$10 per copy, it will lose revenue. per copy, it will lose revenue.

This means that the optimal price is This means that the optimal price is $10$10 per copy, which per copy, which corresponds to unit elasticity. corresponds to unit elasticity.

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SummaryFind the intervals of increase and decrease for the fFind the intervals of increase and decrease for the function unction f(x); f(x); Critical numbers and critical points; FiCritical numbers and critical points; First derivative test for relative extrema; Sketch the grarst derivative test for relative extrema; Sketch the graph of a continuous function.ph of a continuous function.

Concavity, concave upward and concave downwarConcavity, concave upward and concave downward; Using the second derivative to find intervals of cod; Using the second derivative to find intervals of concavity; The points of inflection; The second derivatincavity; The points of inflection; The second derivative test for relative extrema. Sketch the graph of a cove test for relative extrema. Sketch the graph of a continuous function. ntinuous function.

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Limits involving infinity: Vertical Asymptotes and Limits involving infinity: Vertical Asymptotes and Horizontal Asymptotes (rational functions)Horizontal Asymptotes (rational functions) Optimization: Absolute maximum and absolute Optimization: Absolute maximum and absolute minimum, marginal analysis criterion for maximum, minimum, marginal analysis criterion for maximum, profit and minimal average cost, elasticity of demandprofit and minimal average cost, elasticity of demand