applications of the first derivative applications of the second derivative curve sketching

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4 4 Applications of the First Applications of the First Derivative Derivative Applications of the Second Applications of the Second Derivative Derivative Curve Sketching Curve Sketching Optimization I Optimization I Optimization II Optimization II Applications of the Derivative Applications of the Derivative

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4. Applications of the First Derivative Applications of the Second Derivative Curve Sketching Optimization I Optimization II. Applications of the Derivative. 4.1. Applications of the First Derivative. Increasing and Decreasing Functions. - PowerPoint PPT Presentation

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Page 1: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

44

Applications of the First DerivativeApplications of the First Derivative

Applications of the Second DerivativeApplications of the Second Derivative

Curve SketchingCurve Sketching

Optimization IOptimization I

Optimization IIOptimization II

Applications of the DerivativeApplications of the Derivative

Page 2: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

4.14.1Applications of the First DerivativeApplications of the First Derivative

–7 –5 –3 –1 1 3 5 7

60

40

20

–20

–40

–60

x

yff((xx))

Relative Minimum

Relative Maximum

Page 3: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Increasing and Decreasing FunctionsIncreasing and Decreasing Functions

A function A function f f is is increasingincreasing on an on an intervalinterval ((aa, , bb)) if for any two numbers if for any two numbers xx11 and and xx22 in in ((aa, , bb)), ,

ff((xx11) < ) < ff((xx22)) wherever wherever xx11 << xx22..

xx

yy

aa

ff((xx22))

ff((xx11))

bbxx22xx11

Page 4: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Increasing and Decreasing FunctionsIncreasing and Decreasing Functions

A function A function f f is is decreasingdecreasing on an on an intervalinterval ((aa, , bb)) if for any two numbers if for any two numbers xx11 and and xx22 in in ((aa, , bb)), ,

ff((xx11) > ) > ff((xx22)) wherever wherever xx11 << xx22..

xx

yy

aa

ff((xx11))

ff((xx22))

bbxx22xx11

Page 5: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Theorem 1Theorem 1

If If f f ′′((xx) > 0) > 0 for each value of for each value of xx in an interval in an interval ((aa, , bb)), , then then ff is is increasingincreasing on on ((aa, , bb))..

If If f f ′′((xx) < 0) < 0 for each value of for each value of xx in an interval in an interval ((aa, , bb)), , then then ff is is decreasingdecreasing on on ((aa, , bb))..

If If ff ′′((xx) = 0) = 0 for each value of for each value of xx in an interval in an interval ((aa, , bb)), , then then ff is is constantconstant on on ((aa, , bb))..

Page 6: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExampleExample

Find the Find the intervalinterval where the function where the function ff((xx) = ) = xx22 is is increasingincreasing and the and the intervalinterval where it is where it is decreasingdecreasing..

SolutionSolution The derivative of The derivative of ff((xx) = ) = xx22

is is ff ′′((xx) = 2) = 2xx.. f f ′′((xx) = 2) = 2xx > 0 > 0 if if xx > 0 > 0

and and f f ′′((xx) = 2) = 2xx < 0 < 0 if if xx < 0 < 0.. Thus, Thus, f f is is increasingincreasing on the on the

intervalinterval (0, (0, )) and and decreasingdecreasing

on the on the intervalinterval (–(– , 0), 0). .

–2 –1 0 1 2x

5

4

3

2

1

y

ff((xx) = ) = xx22

Example 1, page 245

Page 7: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Determining the Intervals Where a Function Determining the Intervals Where a Function is Increasing or Decreasing is Increasing or Decreasing

1.1. Find Find all the values ofall the values of xx for which for which f f ′′((xx) = 0) = 0 or or f f ′′ is is discontinuousdiscontinuous and identify and identify the open intervalsthe open intervals determined by these numbers.determined by these numbers.

2.2. Select a Select a test numbertest number cc in each interval found in in each interval found in step 1step 1 and determine the sign of and determine the sign of f f ′′((cc) ) in that interval. in that interval.

a.a. If If f f ′′((cc) > 0) > 0, , ff is is increasingincreasing on that interval. on that interval.

b.b. If If f f ′′((cc) < 0) < 0, , ff is is decreasingdecreasing on that interval. on that interval.

Page 8: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the Determine the intervalsintervals where the function where the function

ff((xx) = ) = xx33 – 3 – 3xx22 – 24 – 24xx + 32 + 32

is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution

1.1. Find Find f f ′′ and solve for and solve for f f ′′((xx) = 0) = 0::

f f ′′((xx) = 3) = 3xx22 – 6 – 6xx – 24 = 3( – 24 = 3(xx + 2)( + 2)(xx – 4) = 0 – 4) = 0

✦ Thus, the Thus, the zeroszeros of of f f ′′ are are xx = –2 = –2 and and xx = 4 = 4..✦ These numbers divide the real line into the These numbers divide the real line into the intervalsintervals

(–(– , –2), –2), , (–(– 2, 4)2, 4), and , and (4, (4, ))..

Example 2, page 246

Page 9: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the Determine the intervalsintervals where the function where the function

ff((xx) = ) = xx33 – 3 – 3xx22 – 24 – 24xx + 32 + 32

is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution

2.2. To determine the sign of To determine the sign of f f ′′((xx)) in the intervals we found in the intervals we found (–(– , –2), –2), , (–(– 2, 4)2, 4), and , and (4, (4, )), we , we computecompute f f ′′((cc)) at a at a convenient convenient test pointtest point in each interval. in each interval.✦ Lets consider the values Lets consider the values –3–3, , 00,, and and 55::

f f ′′(–3)(–3) = 3(–3)= 3(–3)22 – 6(–3) – 24 = 27 +18 – 24 = 21 > 0 – 6(–3) – 24 = 27 +18 – 24 = 21 > 0

f f ′′(0)(0) = 3(0)= 3(0)22 – 6(0) – 24 = 0 +0 – 24 = –24 < 0 – 6(0) – 24 = 0 +0 – 24 = –24 < 0

f f ′′(5)(5) = 3(5)= 3(5)22 – 6(5) – 24 = 75 – 30 – 24 = 21 > 0 – 6(5) – 24 = 75 – 30 – 24 = 21 > 0

✦ Thus, we conclude that Thus, we conclude that f f is is increasingincreasing on the on the intervalsintervals (–(– , –2), –2), , (4, (4, )), and is , and is decreasingdecreasing on the on the intervalinterval (–(– 2, 2, 4)4)..Example 2, page 246

Page 10: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

––77 –5 –5 –3 –3 –1 –1 11 33 55 77

ExamplesExamples

Determine the Determine the intervalsintervals where the function where the function

ff((xx) = ) = xx33 – 3 – 3xx22 – 24 – 24xx + 32 + 32

is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution So,So, f f increasesincreases on on (–(– , –2), –2), , (4, (4, )), and , and decreasesdecreases on on (–(– 2, 4)2, 4)::

6060

4040

2020

––2020

––4040

–– 6060

xx

yy

yy = = xx3 – 33 – 3xx22 – 24 – 24xx + 32 + 32

Example 2, page 246

Page 11: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples Determine the Determine the intervalsintervals where where is is

increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution1.1. Find Find f f ′′ and solve for and solve for f f ′′((xx) = 0) = 0::

✦ f f ′′((xx) = 0) = 0 when the when the numeratornumerator is equal to zero, so: is equal to zero, so:

✦ Thus, the Thus, the zeroszeros of of f f ′′ are are xx = –1 = –1 and and xx = 1 = 1..✦ Also note that Also note that f f ′′ is is not definednot defined at at xx = 0 = 0, so we have , so we have four four

intervalsintervals to consider: to consider: (–(– , –1), –1), , (–(– 1, 0)1, 0), , (0, 1)(0, 1), and , and (1, (1, ))..

1( )f x x

x

2

2 2

1 1( ) 1 0

xf x

x x

2

2

1 0

1

1

x

x

x

Example 4, page 247

Page 12: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the Determine the intervalsintervals where where is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution

2.2. To determine the sign of To determine the sign of f f ′′((xx)) in the intervals we found in the intervals we found (–(– , –1), –1), , (–(– 1, 0)1, 0), , (0, 1)(0, 1), and , and (1, (1, )), we , we computecompute f f ′′((cc)) at a at a convenient convenient test pointtest point in each interval. in each interval.✦ Lets consider the values Lets consider the values –2–2, , –1/2–1/2, , 1/21/2,, and and 22::

✦ So So ff is is increasingincreasing in the interval in the interval (–(– , –1), –1)..

1( )f x x

x

2

1 1 3( 2) 1 1 0

4 42f

Example 4, page 247

Page 13: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the Determine the intervalsintervals where where is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution

2.2. To determine the sign of To determine the sign of f f ′′((xx)) in the intervals we found in the intervals we found (–(– , –1), –1), , (–(– 1, 0)1, 0), , (0, 1)(0, 1), and , and (1, (1, )), we , we computecompute f f ′′((cc)) at a at a convenient convenient test pointtest point in each interval. in each interval.✦ Lets consider the values Lets consider the values –2–2, , –1/2–1/2, , 1/21/2,, and and 22::

✦ So So ff is is decreasingdecreasing in the interval in the interval (–(– 1, 0)1, 0). .

1( )f x x

x

12 2 11

42

1 1( ) 1 1 1 4 3 0f

Example 4, page 247

Page 14: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the Determine the intervalsintervals where where is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution

2.2. To determine the sign of To determine the sign of f f ′′((xx)) in the intervals we found in the intervals we found (–(– , –1), –1), , (–(– 1, 0)1, 0), , (0, 1)(0, 1), and , and (1, (1, )), we , we computecompute f f ′′((cc)) at a at a convenient convenient test pointtest point in each interval. in each interval.✦ Lets consider the values Lets consider the values –2–2, , –1/2–1/2, , 1/21/2,, and and 22::

✦ So So ff is is decreasingdecreasing in the interval in the interval (0, 1)(0, 1). .

1( )f x x

x

12 2 11

42

1 1( ) 1 1 1 4 3 0f

Example 4, page 247

Page 15: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the Determine the intervalsintervals where where is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution

2.2. To determine the sign of To determine the sign of f f ′′((xx)) in the intervals we found in the intervals we found (–(– , –1), –1), , (–(– 1, 0)1, 0), , (0, 1)(0, 1), and , and (1, (1, )), we , we computecompute f f ′′((cc)) at a at a convenient convenient test pointtest point in each interval. in each interval.✦ Lets consider the values Lets consider the values –2–2, , –1/2–1/2, , 1/21/2,, and and 22::

✦ So So ff is is increasingincreasing in the interval in the interval (1, (1, )). .

1( )f x x

x

2

1 1 3(2) 1 1 0

4 42f

Example 4, page 247

Page 16: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the Determine the intervalsintervals where where is is increasingincreasing and where it is and where it is decreasingdecreasing..

SolutionSolution Thus,Thus, f f isis increasingincreasing on on (–(– , –1), –1) and and (1, (1, )), and, and decreasingdecreasing

on on (–(– 1, 0)1, 0) and and(0, 1)(0, 1): :

1( )f x x

x

––44 –2 –2 22 44

44

22

––22

––44

xx

yy

Example 4, page 247

Page 17: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Relative ExtremaRelative Extrema

The first derivative may be used to help us The first derivative may be used to help us locatelocate high high pointspoints and and low pointslow points on the graph of on the graph of ff::✦ High pointsHigh points are called are called relative maximarelative maxima✦ Low pointsLow points are called are called relative minimarelative minima. .

Both high and low pointsBoth high and low points are called are called relative extremarelative extrema..

xx

yy

yy = = ff((xx))

Relative Relative MaximaMaxima

Relative Relative MinimaMinima

Page 18: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Relative ExtremaRelative Extrema

Relative MaximumRelative Maximum A function A function f f has a has a relative maximumrelative maximum at at x x == c c if if

there exists an open interval there exists an open interval ((aa, , bb)) containing containing cc such that such that ff((xx) ) ff((cc)) for all for all xx in in ((aa, , bb))..

xx11 xx22

xx

yy

yy = = ff((xx))

Relative Relative MaximaMaxima

Page 19: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Relative ExtremaRelative Extrema

Relative MinimumRelative Minimum A function A function f f has a has a relative minimumrelative minimum at at x x == c c if if

there exists an open interval there exists an open interval ((aa, , bb)) containing containing cc such that such that ff((xx) ) ff((cc)) for all for all xx in in ((aa, , bb))..

xx33 xx44

xx

yy

yy = = ff((xx))

Relative Relative MinimaMinima

Page 20: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Finding Relative ExtremaFinding Relative Extrema

Suppose that Suppose that ff has a has a relative maximumrelative maximum at at cc.. The The slope of the tangentslope of the tangent line to the graph line to the graph must must

change change fromfrom positivepositive toto negativenegative as as x x increases. increases. Therefore, theTherefore, the tangenttangent line to the graph of line to the graph of ff at pointat point

((cc, , ff((cc)))) must be horizontalmust be horizontal, so that , so that f f ′′((xx)) = 0 = 0 oror f f ′′((xx)) is is undefinedundefined..

xx

yy

ccaa bb

f f ′(′(xx)) < 0< 0

f f ′′((xx)) > 0> 0f f ′(′(xx)) = 0= 0

Page 21: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Finding Relative ExtremaFinding Relative Extrema

Suppose that Suppose that ff has a has a relative minimumrelative minimum at at cc.. The The slope of the tangentslope of the tangent line to the graph line to the graph must must

change change fromfrom negativenegative toto positivepositive as as x x increases. increases. Therefore, theTherefore, the tangenttangent line to the graph of line to the graph of ff at pointat point

((cc, , ff((cc)))) must be horizontalmust be horizontal, so that , so that f f ′′((xx)) = 0 = 0 oror f f ′′((xx)) is is undefined.undefined.

xx

yy

ccaa bb

f f ′(′(xx)) > 0> 0

f f ′(′(xx)) < 0< 0

f f ′(′(xx)) = 0= 0

Page 22: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Finding Relative ExtremaFinding Relative Extrema

In some cases In some cases a derivative does not exista derivative does not exist for for particular values of particular values of xx..

Extrema may existExtrema may exist at such points, as the graph at such points, as the graph below shows:below shows:

xx

yy

aa bb

Relative Relative MaximumMaximum

Relative Relative MinimumMinimum

Page 23: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Critical NumbersCritical Numbers

We refer to a number in the domain of We refer to a number in the domain of ff that that maymay give give rise to a rise to a relative extremumrelative extremum as a as a critical numbercritical number..

Critical number of Critical number of ff

✦ A critical number of a function A critical number of a function ff is any number is any number xx in the domain of in the domain of ff such that such that f f ′′((xx)) = 0= 0 or or f f ′′((xx)) does not exist.does not exist.

Page 24: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Critical NumbersCritical Numbers

The The graph belowgraph below shows us shows us several critical numbersseveral critical numbers.. At points At points aa, , bb, and , and cc, , f f ′′((xx)) = 0= 0.. There is a There is a cornercorner at point at point dd, so , so f f ′′((xx)) does not existdoes not exist there. there. The tangent to the curve at point The tangent to the curve at point ee is is verticalvertical, so , so f f ′′((xx))

does not existdoes not exist there either. there either. Note that points Note that points aa, , bb, and , and dd are are relative extremarelative extrema, while , while

points points cc and and ee are notare not..

xx

yy

aa bb

Relative Relative ExtremaExtrema

NotNot Relative Relative ExtremaExtrema

cc dd ee

Page 25: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

The First Derivative TestThe First Derivative Test

1.1. Determine the Determine the critical numberscritical numbers of of ff..2.2. Determine the Determine the signsign of of f f ′′((xx)) to the to the leftleft and and rightright of of

each critical point.each critical point.a.a. If If f f ′′((xx)) changes sign from changes sign from positive positive toto negative negative as as

we move across a critical number we move across a critical number cc, then , then ff((cc)) is a is a relative maximumrelative maximum..

b.b. If If f f ′′((xx)) changes sign from changes sign from negative negative to to positive positive as as we move across a critical number we move across a critical number cc, then , then ff((cc)) is a is a relative minimumrelative minimum..

c.c. If If f f ′′((xx)) does does notnot change sign change sign as we move across a as we move across a critical number critical number cc, then , then ff((cc)) is is notnot a relative a relative extremumextremum..

Procedure for Finding Relative Extrema of a Continuous Procedure for Finding Relative Extrema of a Continuous Function Function ff

Page 26: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Find the relative maxima and minima ofFind the relative maxima and minima of 2( )f x x

–2 –1 0 1 2x

5

4

3

2

1

yy

ff((xx) = ) = xx22

SolutionSolution The derivative of The derivative of ff is is f f ′′((xx) = 2) = 2xx. . Setting Setting f f ′′((xx) = 0) = 0 yields yields x x = 0= 0 as as

the only the only critical numbercritical number of of ff. . SinceSince f f ′′((xx) < 0) < 0 if if xx < 0 < 0

andand f f ′′((xx) > 0) > 0 if if xx > 0 > 0

we see that we see that f f ′′((xx) ) changes sign changes sign from from negativenegative to to positivepositive as as we move across we move across 00. .

Thus, Thus, ff(0) = 0(0) = 0 is a is a relative relative minimumminimum of of ff..

f f ′(′(xx) < 0) < 0 f f ′(′(xx) > 0) > 0

Relative Relative MinimumMinimum

Example 5, page 252

Page 27: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Find the relative maxima and minima ofFind the relative maxima and minima ofSolutionSolution The derivative of The derivative of ff is is f f ′′((xx) = 2/3) = 2/3xx––1/31/3. . f f ′′((xx) ) is is notnot defined defined at at x x = 0= 0, is , is continuouscontinuous everywhere else, everywhere else,

and is and is never equal to zeronever equal to zero in its domain. in its domain. Thus Thus x x = 0= 0 is the only is the only critical number critical number of of ff..

2/3( )f x x

SinceSince f f ′′((xx) < 0) < 0 if if xx < < 00 andand f f ′′((xx) > 0) > 0 if if xx > > 00 we see that we see that f f ′′((xx) ) changes changes sign from sign from negativenegative to to positivepositive as we move as we move across across 00. .

Thus, Thus, ff(0) = 0(0) = 0 is a is a relative minimumrelative minimum of of ff..

f´f´((xx) < 0) < 0 f´f´((xx) > 0) > 0

– – 44 – – 2 2 22 44

44

22

xx

yy

ff((xx) = ) = xx2/32/3

Relative Relative MinimumMinimum

Example 6, page 252

Page 28: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples Find the relative maxima and minima ofFind the relative maxima and minima of

SolutionSolution The derivative of The derivative of ff and equate to zero: and equate to zero:

The zeros ofThe zeros of f f ′′((xx) ) are are x x = –2= –2 and and x x = 4= 4.. f f ′′((xx) ) is is defined everywheredefined everywhere, so, so x x = –2= –2 and and x x = 4 = 4 are the onlyare the only

critical numberscritical numbers of of ff. .

3 2( ) 3 24 32f x x x x 3 2( ) 3 24 32f x x x x

2

2

( ) 3 6 24 03( 2 8) 0

3( 4)( 2) 0

f x x xx x

x x

2

2

( ) 3 6 24 03( 2 8) 0

3( 4)( 2) 0

f x x xx x

x x

Example 7, page 253

Page 29: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples Find the relative maxima and minima ofFind the relative maxima and minima of

SolutionSolution Since Since f f ′′((xx) > 0) > 0 if if xx < –2 < –2 and and f f ′′((xx) < 0) < 0 if if 0 <0 < xx < 4 < 4,,

we see that we see that f f ′′((xx) ) changes sign from changes sign from positivepositive to to negativenegative as as we move across we move across –2–2. .

Thus, Thus, ff(–2) = 60(–2) = 60 is a is a relative maximumrelative maximum..

––77 ––5 5 ––33 ––11 11 33 55 77

6060

4040

2020

––2020

––4040

–– 6060

xx

yy

3 2( ) 3 24 32f x x x x 3 2( ) 3 24 32f x x x x

ff((xx))

Relative Relative MaximumMaximum

Example 7, page 253

Page 30: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples Find the relative maxima and minima ofFind the relative maxima and minima of

SolutionSolution Since Since f f ′′((xx) < 0) < 0 if if 0 <0 < xx < 4 < 4 and and f f ′′((xx) > 0) > 0 if, if, xx > 4 > 4

we see that we see that f f ′′((xx) ) changes sign from changes sign from negativenegative to to positivepositive as as we move across we move across 44. .

Thus, Thus, ff(4) = –(4) = – 4848 is a is a relative minimumrelative minimum..

3 2( ) 3 24 32f x x x x 3 2( ) 3 24 32f x x x x

––77 –5 –5 –3 –3 –1 –1 11 33 55 77

6060

4040

2020

––2020

––4040

–– 6060

xx

yyff((xx))

Relative Relative MinimumMinimum

Relative Relative MaximumMaximum

Example 7, page 253

Page 31: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

4.24.2Applications of the Second DerivativeApplications of the Second Derivative

x

y

y = f(x)

Relative Maxima

Relative Minima

Page 32: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ConcavityConcavity

A curve is said to be A curve is said to be concave upwardsconcave upwards when the when the slope of slope of tangent linetangent line to the curve is to the curve is increasingincreasing::

Thus, if Thus, if ff is differentiable on an interval is differentiable on an interval ((aa, , bb)), then, then ff is is concave upwardsconcave upwards on on ((aa, , bb)) if if f f ′′ is is increasingincreasing on on ((aa, , bb))..

xx

yy

Page 33: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ConcavityConcavity

A curve is said to be A curve is said to be concave downwardsconcave downwards when the when the slope of slope of tangent linetangent line to the curve is to the curve is decreasingdecreasing::

Thus, if Thus, if ff is differentiable on an interval is differentiable on an interval ((aa, , bb)), then, then ff is is concave downwardsconcave downwards on on ((aa, , bb)) if if ff ′′ is is decreasingdecreasing on on ((aa, , bb))..

xx

yy

Page 34: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Theorem 2Theorem 2

Recall that Recall that f f ″″((xx) ) measures the measures the rate of change of the sloperate of change of the slope f f ′′((xx)) of the tangent line to the graph of of the tangent line to the graph of ff at the point at the point ((xx, , ff((xx))))..

Thus, we can use Thus, we can use f f ″″((xx) ) to determine the to determine the concavityconcavity of of ff..

a.a. If If f f ″″((xx) > 0 ) > 0 for each value of for each value of xx in in ((aa, , bb)), then , then ff is is concave upwardconcave upward on on ((aa, , bb))..

b.b. If If f f ″″((xx) < 0 ) < 0 for each value of for each value of xx in in ((aa, , bb)), then , then ff is is concave downwardconcave downward on on ((aa, , bb))..

Page 35: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Steps in Determining the Concavity of Steps in Determining the Concavity of ff

1.1. Determine the values of Determine the values of xx for which for which f f ″″is is zerozero or where or where f f ″ ″ is is not definednot defined, and identify the open intervals , and identify the open intervals determined by these numbers.determined by these numbers.

2.2. Determine the sign of Determine the sign of f f ″ ″ in each interval found in in each interval found in step 1step 1. .

To do this compute To do this compute f f ″″((cc)), where , where cc is any conveniently is any conveniently chosen test number in the interval.chosen test number in the interval. If If f f ″″((cc) > 0) > 0, , ff is is concave upwardconcave upward on that interval. on that interval. If If f f ″″((cc) < 0) < 0, , ff is is concave downwardconcave downward on that interval. on that interval.

Page 36: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine Determine wherewhere the function the function

is is concave upwardconcave upward and where it is and where it is concave downwardconcave downward..

SolutionSolution Here,Here, andand Setting Setting f f ″″((cc) = 0) = 0 we find we find

which gives which gives xx = 1 = 1. . So we consider the intervals So we consider the intervals (–(– , 1), 1) and and (1, (1, ))::

✦ f f ″″((xx) < 0) < 0 when when xx < 1 < 1, so , so ff is is concave downwardconcave downward on on (–(– , 1), 1)..

✦ f f ″″((xx) > 0) > 0 when when xx > 1 > 1, so , so ff is is concave upwardconcave upward on on (1, (1, ))..

3 2( ) 3 24 32f x x x x 3 2( ) 3 24 32f x x x x

2( ) 3 6 24f x x x 2( ) 3 6 24f x x x ( ) 6 6f x x ( ) 6 6f x x

( ) 6 6 0

6( 1) 0

f x x

x

( ) 6 6 0

6( 1) 0

f x x

x

Example 1, page 265

Page 37: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine Determine wherewhere the function the function

is is concave upwardconcave upward and where it is and where it is concave downwardconcave downward..

SolutionSolution The graph confirms thatThe graph confirms that f f is is concave downwardconcave downward on on (–(– , 1), 1)

andand concave upwardconcave upward on on (1, (1, ))::

3 2( ) 3 24 32f x x x x 3 2( ) 3 24 32f x x x x

––77 –5 –5 –3 –3 –1 –1 11 33 55 77

6060

4040

2020

––2020

––4040

–– 6060

xx

yyff((xx))

Example 1, page 265

Page 38: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the intervals Determine the intervals wherewhere the function the function is is concave upwardconcave upward and and concave downwardconcave downward..

SolutionSolution Here,Here, andand

So,So, f f ″″cannot be zerocannot be zero and is and is not definednot defined at at xx = 0 = 0. . So we consider the intervals So we consider the intervals (–(– , 0), 0) and and (0, (0, ))::

✦ f f ″″((xx) < 0) < 0 when when xx < 0 < 0, so , so ff is is concave downwardconcave downward on on (–(– , 0), 0)..

✦ f f ″″((xx) > 0) > 0 when when xx > 0 > 0, so , so ff is is concave upwardconcave upward on on (0, (0, ))..

1( )f x x

x

1( )f x x

x

2

1( ) 1f x

x 2

1( ) 1f x

x 3

2( )f x

x 3

2( )f x

x

Example 2, page 266

Page 39: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Determine the intervals Determine the intervals wherewhere the function the functionis is concave upwardconcave upward and and concave downwardconcave downward..

SolutionSolution The graph confirms thatThe graph confirms that f f is is concave downwardconcave downward on on (–(– , 0), 0)

andand concave upwardconcave upward on on (0, (0, ))::

–4 –2 2 4

4

2

–2

–4

x

y

1( )f x x

x

1( )f x x

x

Example 2, page 266

Page 40: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Inflection PointInflection Point

A A pointpoint on the graph of a continuous function where the on the graph of a continuous function where the tangent line exists and where the tangent line exists and where the concavity changesconcavity changes is is called an called an inflection pointinflection point..

Examples:Examples:

xx

yy

Inflection Inflection PointPoint

Page 41: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Inflection PointInflection Point

A A pointpoint on the graph of a continuous function where the on the graph of a continuous function where the tangent line exists and where the tangent line exists and where the concavity changesconcavity changes is is called an called an inflection pointinflection point..

Examples:Examples:

xx

yy

Inflection Inflection PointPoint

Page 42: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Inflection PointInflection Point

A A pointpoint on the graph of a continuous function where the on the graph of a continuous function where the tangent line exists and where the tangent line exists and where the concavity changesconcavity changes is is called an called an inflection pointinflection point..

Examples:Examples:

xx

yy

Inflection Inflection PointPoint

Page 43: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Finding Inflection PointsFinding Inflection Points

To find inflection points:To find inflection points:

1.1. Compute Compute f f ″″((xx))..

2.2. Determine the numbers in the domain of Determine the numbers in the domain of ff for for which which f f ″″((xx) = 0 ) = 0 or or f f ″″((xx)) does not exist. does not exist.

3.3. Determine the sign of Determine the sign of f f ″″((xx) ) to the left and right of to the left and right of each number each number cc found in found in step 2step 2..

If there is a change in the sign of If there is a change in the sign of f f ″″((xx) ) as we move as we move across across xx = = cc, then , then ((cc, , ff((cc)))) is an inflection point of is an inflection point of ff..

Page 44: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Find the points of inflection of the functionFind the points of inflection of the functionSolutionSolution We haveWe have andand So, So, f f ″″ is is continuous everywherecontinuous everywhere and is and is zerozero for for xx = 0 = 0.. We see that We see that f f ″″((xx) < 0 ) < 0 whenwhen xx < 0 < 0, and , and f f ″″((xx) > 0 ) > 0 whenwhen xx > 0 > 0.. Thus, we find that the graph of Thus, we find that the graph of ff ::

3( )f x x

2( ) 3f x x ( ) 6f x x

✦ Has one and only Has one and only inflection inflection pointpoint at at ff(0) = 0(0) = 0..

✦ IsIs concave downwardconcave downward on on the interval the interval (–(– , 0), 0)..

✦ IsIs concave upwardconcave upward on the on the interval interval (0, (0, )).. – –22 22

44

22

––22

––44

xx

yyff((xx))

Inflection Inflection PointPoint

Example 3, page 268

Page 45: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples Find the points of inflection of the functionFind the points of inflection of the function

SolutionSolution We haveWe have andand

So, So, f f ″″ is is not definednot defined at at xx = 1 = 1, and , and f f ″″ is is never equal to zeronever equal to zero.. We see that We see that f f ″″((xx) < 0 ) < 0 whenwhen xx < 1 < 1, and , and f f ″″((xx) > 0 ) > 0 whenwhen xx > 1 > 1.. Thus, we find that the graph of Thus, we find that the graph of ff : :

5/3( ) 1f x x

2/35( ) 1

3f x x

1/3

10( )

9 1f x

x

✦ Has one and only Has one and only inflection pointinflection point at at ff(1) = 0(1) = 0..

✦ IsIs concave downwardconcave downward on on the interval the interval (–(– , 1), 1)..

✦ IsIs concave upwardconcave upward on the on the interval interval (1, (1, ))..

– –22 –1 –1 1 1 22

33

22

11

––11

––22

––33

xx

yy

ff((xx))Inflection Inflection

PointPoint

Example 4, page 268

Page 46: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Applied Example:Applied Example: Effect of Advertising on Sales Effect of Advertising on Sales

The The total salestotal sales SS (in thousands of dollars)(in thousands of dollars) of Arctic Air Co., of Arctic Air Co., which makes automobile air conditioners, is related to the which makes automobile air conditioners, is related to the amount of money amount of money xx (in thousands of dollars)(in thousands of dollars) the company the company spends on spends on advertisingadvertising its product by the formula its product by the formula

Find the Find the inflection pointinflection point of the function of the function SS.. Discuss the Discuss the meaningmeaning of this point. of this point.

3 2( ) 0.01 1.5 200 (0 100) S x x x x 3 2( ) 0.01 1.5 200 (0 100) S x x x x

Applied Example 7, page 270

Page 47: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Applied Example:Applied Example: Effect of Advertising on Sales Effect of Advertising on Sales

SolutionSolution The first two derivatives of The first two derivatives of SS are given by are given by

Setting Setting SS ″″((xx) = 0) = 0 gives gives xx = 50 = 50. . So So (50, (50, SS(50))(50)) is the is the only candidateonly candidate for an for an inflection pointinflection point.. Since Since SS ″″((xx) > 0) > 0 for for xx < 50 < 50, and , and SS ″″((xx) < 0) < 0 for for xx > 50 > 50, the point , the point

(50, 2700)(50, 2700) is, in fact, an is, in fact, an inflection pointinflection point of of SS. . This means that the firm experiences This means that the firm experiences diminishing returns diminishing returns

on advertisingon advertising beyond beyond $50,000$50,000::✦ Every additional dollar spent on advertising increases Every additional dollar spent on advertising increases

sales by less than previously spent dollars.sales by less than previously spent dollars.

2( ) 0.03 3 ( ) 0.06 3S x x x S x x and 2( ) 0.03 3 ( ) 0.06 3S x x x S x x and

Applied Example 7, page 270

Page 48: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

The Second Derivative TestThe Second Derivative Test

MaximaMaxima occur when a curve is occur when a curve is concave downwardsconcave downwards, while , while minimaminima occur when a curve is occur when a curve is concave upwardsconcave upwards. .

This is the basis of the This is the basis of the second derivativesecond derivative testtest::

1.1. Compute Compute f f ′′((xx)) and and f f ″″((xx))..

2.2. Find all the Find all the critical numberscritical numbers of of ff at which at which f f ′′((xx)) = 0= 0..

3.3. Compute Compute f f ″″((cc)), if it exists, for each critical number , if it exists, for each critical number cc..

1.1. If If f f ″″((cc)) < 0< 0, then , then ff has a has a relative maximumrelative maximum at at cc..

2.2. If If f f ″″((cc)) > 0> 0, then , then ff has a has a relative minimumrelative minimum at at cc..

3.3. If If f f ″″((cc)) = 0= 0, then the test fails (it is inconclusive)., then the test fails (it is inconclusive).

Page 49: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExampleExample

Determine the relative extrema of the functionDetermine the relative extrema of the function

SolutionSolution We haveWe have

so so f f ′′((xx) = 0) = 0 gives the gives the critical numberscritical numbers xx = –2 = –2 and and xx = 4 = 4.. Next, we haveNext, we have

which we use to which we use to test the critical numberstest the critical numbers::

so, so, ff(–2) = 60(–2) = 60 is a is a relative maximumrelative maximum of of ff..

so, so, ff(4) = –48(4) = –48 is a is a relative minimumrelative minimum of of ff..

3 2( ) 3 24 32f x x x x 3 2( ) 3 24 32f x x x x

2( ) 3 6 24 3( 4)( 2)f x x x x x 2( ) 3 6 24 3( 4)( 2)f x x x x x

( ) 6 6f x x

( 2) 6( 2) 6 12 6 18 0f ( 2) 6( 2) 6 12 6 18 0f

(4) 6(4) 6 24 6 18 0f (4) 6(4) 6 24 6 18 0f

Example 9, page 272

Page 50: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

4.34.3Curve SketchingCurve Sketching

x

y

y = –1

x = –2 x = 2

Page 51: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Vertical AsymptotesVertical Asymptotes

The line The line xx = = aa is a is a vertical asymptotevertical asymptote of the of the graph of a function graph of a function ff if either if either

oror

lim ( ) or x a

f x

lim ( ) or x a

f x

lim ( ) or x a

f x

lim ( ) or x a

f x

Page 52: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Finding Vertical Asymptotes of Rational FunctionsFinding Vertical Asymptotes of Rational Functions

Suppose Suppose f f is a is a rational functionrational function

where where PP and and QQ are are polynomial functionspolynomial functions.. Then, the line Then, the line x = ax = a is a is a vertical asymptotevertical asymptote of of

the graph of the graph of ff if if QQ((aa) = 0) = 0 but but PP((aa) ) ≠ 0≠ 0..

( )( )

( )

P xf x

Q x

( )( )

( )

P xf x

Q x

Page 53: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExampleExample

Find the Find the vertical asymptotesvertical asymptotes of the graph of the function of the graph of the function

SolutionSolution ff is a is a rational functionrational function with with PP((xx) = ) = xx22 and and QQ((xx) = 4 – ) = 4 – xx22.. The zeros of The zeros of QQ are found by solving are found by solving

giving giving xx = –2 = –2 and and xx = 2 = 2..

Examine Examine xx = –2 = –2:: PP(–2) = (–2)(–2) = (–2)22 = 4 = 4 ≠ 0≠ 0, so , so xx = –2 = –2 is a is a vertical asymptotevertical asymptote..

Examine Examine xx = 2 = 2:: PP(2) = (2)(2) = (2)22 = 4 = 4 ≠ 0≠ 0, so , so xx = 2 = 2 is also a is also a vertical asymptotevertical asymptote..

2

2( )

4

xf x

x

2

2( )

4

xf x

x

24 0

(2 )(2 ) 0

x

x x

24 0

(2 )(2 ) 0

x

x x

Example 1, page 285

Page 54: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Horizontal AsymptotesHorizontal Asymptotes

The line The line y = by = b is a is a horizontal asymptotehorizontal asymptote of the of the graph of a function graph of a function ff if either if either

lim ( ) lim ( )or x x

f x b f x b

lim ( ) lim ( )or x x

f x b f x b

Page 55: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExampleExample

Find the Find the horizontal asymptoteshorizontal asymptotes of the graph of the function of the graph of the function

SolutionSolution We computeWe compute

and so and so yy = –1 = –1 is a is a horizontal asymptotehorizontal asymptote.. We computeWe compute

also yielding also yielding yy = –1 = –1 as a as a horizontal asymptotehorizontal asymptote..

2

2( )

4

xf x

x

2

2( )

4

xf x

x

2

2

2

1 1lim lim 1

44 0 11x x

x

xx

2

2

2

1 1lim lim 1

44 0 11x x

x

xx

2

2

2

1 1lim lim 1

44 0 11x x

x

xx

2

2

2

1 1lim lim 1

44 0 11x x

x

xx

Example 2, page 287

Page 56: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExampleExample

Find the Find the horizontal asymptoteshorizontal asymptotes of the graph of the function of the graph of the function

SolutionSolution So, the graph of So, the graph of ff has has twotwo vertical asymptotesvertical asymptotes xx = –2 = –2

and and xx = 2 = 2, and , and oneone horizontal asymptotehorizontal asymptote yy = –1 = –1::

2

2( )

4

xf x

x

2

2( )

4

xf x

x

xx

yy

y y = –1 = –1

x x = –2 = –2 x x = 2 = 2

ff((xx) )

Example 2, page 287

Page 57: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Asymptotes and PolynomialsAsymptotes and Polynomials

A polynomial function has A polynomial function has no vertical asymptotesno vertical asymptotes.. To see this, note that a polynomial function To see this, note that a polynomial function PP((xx)) can can

be written as a be written as a rational functionrational function with a denominator with a denominator equal to equal to 11. .

Thus, Thus,

Since the Since the denominator isdenominator is never zeronever zero, , PP has has no vertical no vertical asymptotesasymptotes..

( )( )

1

P xP x

( )( )

1

P xP x

Page 58: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Asymptotes and PolynomialsAsymptotes and Polynomials

A polynomial function has A polynomial function has no horizontal asymptotesno horizontal asymptotes.. If If PP((xx)) is a polynomial of degree greater or equal to is a polynomial of degree greater or equal to 11, ,

thenthen

are either are either infinityinfinity or or minus infinityminus infinity; that is, ; that is, they do they do not existnot exist..

Therefore,Therefore, P P has has no horizontal asymptotesno horizontal asymptotes..

lim ( ) lim ( ) and x x

P x P x

lim ( ) lim ( ) and x x

P x P x

Page 59: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

A Guide to Sketching a CurveA Guide to Sketching a Curve

1.1. Determine the Determine the domaindomain of of ff..

2.2. Find the Find the xx-- and and yy-intercepts-intercepts of of ff..

3.3. Determine the behavior of Determine the behavior of ff for large for large absolute absolute valuesvalues of of xx..

4.4. Find all Find all horizontalhorizontal and and vertical asymptotesvertical asymptotes of of ff..

5.5. Determine the Determine the intervalsintervals where where f f is is increasingincreasing and and where where ff is is decreasingdecreasing..

6.6. Find the Find the relative extremarelative extrema of of ff..

7.7. Determine the Determine the concavityconcavity of of ff..

8.8. Find the Find the inflection pointsinflection points of of ff..

9.9. Plot a few Plot a few additional pointsadditional points to help further identify to help further identify the shape of the graph of the shape of the graph of ff and and sketch the graphsketch the graph..

Page 60: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

1.1. The The domaindomain of of f f is the interval is the interval (–(– , , ))..

2.2. By setting By setting xx = 0 = 0, we find that the , we find that the yy-intercept -intercept is is 22. . (The (The xx-intercept -intercept is is not readily obtainablenot readily obtainable))

3.3. Since Since

we see that we see that ff decreases without bounddecreases without bound as as xx decreases decreases without boundwithout bound and that and that ff increases without boundincreases without bound when when xx increases without boundincreases without bound..

4.4. Since Since ff is a is a polynomial functionpolynomial function, there are , there are no asymptotesno asymptotes..

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

3 2

3 2

lim 6 9 2

lim 6 9 2

x

x

x x x

x x x

3 2

3 2

lim 6 9 2

lim 6 9 2

x

x

x x x

x x x

Example 3, page 288

Page 61: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

5.5.

Setting Setting f f ′′((xx) = 0) = 0 gives gives xx = 1 = 1 and and xx = 3 = 3 as as critical pointscritical points. .

TestingTesting with different values of with different values of xx we find that we find that f f ′′((xx) > 0) > 0 when when xx < 1 < 1, , f f ′′((xx) < 0) < 0 when when 1 < 1 < xx < 3 < 3, and , and f f ′′((xx) > 0) > 0 when when xx > 3 > 3..

Thus, Thus, ff is is increasingincreasing in the in the intervalsintervals (–(– , 1), 1) and and (3, (3, )),,

and and ff is is decreasingdecreasing in the in the intervalinterval (1, 3)(1, 3). .

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

2( ) 3 12 9 3( 3)( 1)f x x x x x 2( ) 3 12 9 3( 3)( 1)f x x x x x

Example 3, page 288

Page 62: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

6.6. f f ′′ changes from changes from positive to negativepositive to negative as we go across as we go across xx = 1 = 1, , so a so a relative maximumrelative maximum of of ff occurs at occurs at (1, (1, ff(1))(1)) == (1, 6)(1, 6)..

f f ′′ changes from changes from negativenegative to positive to positive as we go across as we go across xx = 3 = 3, , so a so a relative minimumrelative minimum of of ff occurs at occurs at (3, (3, ff(3))(3)) == (1, 2)(1, 2)..

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

Example 3, page 288

Page 63: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

7.7.

which is equal to zero when which is equal to zero when xx = 2 = 2..

TestingTesting with different values of with different values of xx we find that we find that f f ″″((xx) < 0) < 0 when when xx < 2 < 2 and and f f ″″((xx) > 0) > 0 when when 2 < 2 < xx..

Thus, Thus, ff is is concave downwardconcave downward in the in the intervalinterval (–(– , 2), 2) and and

concave upwardconcave upward in the in the intervalinterval (2, (2, ))..

8.8. Since Since f f ″″(2) = 0(2) = 0, we have an , we have an inflection pointinflection point at at (2, (2, ff(2))(2)) == (2, 4)(2, 4)..

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

( ) 6 12 6( 2)f x x x ( ) 6 12 6( 2)f x x x

Example 3, page 288

Page 64: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution SummarizingSummarizing, we’ve found the following:, we’ve found the following:

✦ DomainDomain: : (–(– , , ))..✦ InterceptIntercept: : (0, 2)(0, 2)..✦

✦ AsymptotesAsymptotes: : NoneNone..✦ ff is is increasingincreasing in the in the intervalsintervals (–(– , 1), 1) and and (3, (3, )), and , and

ff is is decreasingdecreasing in the in the intervalinterval (1, 3)(1, 3)..✦ A A relative maximumrelative maximum of of ff occurs at occurs at (1, 6)(1, 6)..✦ A A relative minimumrelative minimum of of ff occurs at occurs at (1, 2)(1, 2)..✦ ff is is concave downwardconcave downward in the in the intervalinterval (–(– , 2), 2) and and

ff is is concave upwardconcave upward in the in the intervalinterval (2, (2, ))..✦ ff has an has an inflection pointinflection point at at (2, 4)(2, 4)..

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

lim ( ) lim ( ) a d nx x

f x f x

lim ( ) lim ( ) a d nx x

f x f x

Example 3, page 288

Page 65: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

77

66

55

44

33

22

11

xx

yy

11 22 33 44 55 66

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution Sketch Sketch the graph:the graph:

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

(1, 6)(1, 6) Relative maximumRelative maximum

(2, 4)(2, 4) Inflection pointInflection point

(3, 2)(3, 2) Relative minimumRelative minimum

InterceptIntercept

Example 3, page 288

Page 66: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

77

66

55

44

33

22

11

xx

yy

11 22 33 44 55 66

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution Sketch Sketch the graph:the graph:

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

Example 3, page 288

Page 67: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

1.1. f f is undefined when is undefined when xx = 1 = 1, so the domain of, so the domain of f f is the set of is the set of all real numbers other than all real numbers other than xx = 1 = 1..

2.2. Setting Setting yy = 0 = 0, gives an , gives an xx-intercept -intercept of of –1–1. . Setting Setting xx = 0 = 0, gives an , gives an yy-intercept -intercept of of –1–1. .

1( )

1

xf x

x

1

( )1

xf x

x

Example 4, page 290

Page 68: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

3.3. Since Since

we see that we see that ff((xx)) approaches the line approaches the line yy = 1 = 1 as as ||xx|| becomes becomes arbitrarily largearbitrarily large..✦ For For xx > 1 > 1, , ff((xx)) > 1> 1, so , so ff approaches the line approaches the line yy = 1 = 1 from abovefrom above..✦ For For xx < 1 < 1, , ff((xx)) < 1< 1, so , so ff approaches the line approaches the line yy = 1 = 1 from belowfrom below..

4.4. From step three we conclude that From step three we conclude that yy = 1 = 1 is a is a horizontal horizontal asymptoteasymptote of of ff..

Also, Also, the straight linethe straight line xx = 1 = 1 is a is a vertical asymptotevertical asymptote of of ff. .

1( )

1

xf x

x

1

( )1

xf x

x

1 1

lim 1 lim 11 1

and x x

x x

x x

1 1

lim 1 lim 11 1

and x x

x x

x x

Example 4, page 290

Page 69: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

5.5.

So, So, f f ′′((xx) ) is is discontinuousdiscontinuous at at xx = 1 = 1 and is and is never equal to zeronever equal to zero..

Testing we find that Testing we find that f f ′′((xx) < 0 ) < 0 wherever it is defined.wherever it is defined. From From step 5step 5 we see that there are we see that there are no critical numbersno critical numbers of of ff,,

since since f f ′′((xx) ) is never equal to zerois never equal to zero..

2 2

( 1)(1) ( 1)(1) 2( )

( 1) ( 1)

x xf x

x x

2 2

( 1)(1) ( 1)(1) 2( )

( 1) ( 1)

x xf x

x x

1( )

1

xf x

x

1

( )1

xf x

x

Example 4, page 290

Page 70: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution

7.7.

TestingTesting with different values of with different values of xx we find that we find that f f ″″((xx) < 0) < 0 when when xx < 1 < 1 and and f f ″″((xx) > 0) > 0 when when 1 < 1 < xx..

Thus, Thus, ff is is concave downwardconcave downward in the in the intervalinterval (–(– , 1), 1) and and

concave upwardconcave upward in the in the intervalinterval (1, (1, ))..

8.8. From From point 7point 7 we see there are we see there are nono inflection pointsinflection points of of ff, since , since f f ″″((xx) ) is is never equal to zeronever equal to zero..

2 33

4( ) 2( 1) 4( 1)

( 1)

df x x x

dx x 2 3

3

4( ) 2( 1) 4( 1)

( 1)

df x x x

dx x

1( )

1

xf x

x

1

( )1

xf x

x

Example 4, page 290

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ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution SummarizingSummarizing, we’ve found the following:, we’ve found the following:

✦ DomainDomain: : (–(– , 1) , 1) (1, (1, ))..✦ InterceptIntercept: : (0, –1)(0, –1); ; (–1, 0)(–1, 0). . ✦

✦ AsymptotesAsymptotes:: xx = 1 = 1 is a is a vertical asymptotevertical asymptote..yy = 1 = 1 is a is a horizontal asymptotehorizontal asymptote..

✦ ff is is decreasingdecreasing everywhereeverywhere in the domain of in the domain of ff..✦ Relative extremaRelative extrema: : NoneNone..✦ ff is is concave downwardconcave downward in the in the intervalinterval (–(– , 1), 1) and and

ff is is concave upwardconcave upward in the in the intervalinterval (1, (1, ))..✦ ff has has nono inflection pointsinflection points..

lim ( ) 1 lim ( ) 1 a nd x x

f x f x

lim ( ) 1 lim ( ) 1 a nd x x

f x f x

1( )

1

xf x

x

1

( )1

xf x

x

Example 4, page 290

Page 72: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

Sketch the graph of the functionSketch the graph of the function

SolutionSolution Sketch Sketch the graph:the graph:

1( )

1

xf x

x

1

( )1

xf x

x

44

22

––22

xx

yy

– –22 22 44

1( )

1

xf x

x

1

( )1

xf x

x

yy = 1 = 1

xx = 1 = 1

Example 4, page 290

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4.44.4Optimization IOptimization I

x

y

Relative minimum

Relative maximum

Absolute minimum

Absolute maximum

a x1 x2 x3 b

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Absolute ExtremaAbsolute Extrema

The The absolute extremaabsolute extrema of a function of a function ff

✦ If If ff((xx)) ff((cc)) for all for all xx in the domain of in the domain of ff, then , then ff((cc)) is called the is called the absolute maximumabsolute maximum value of value of ff..

✦ If If ff((xx)) ff((cc)) for all for all xx in the domain of in the domain of ff, then , then ff((cc)) is called the is called the absolute minimumabsolute minimum value of value of ff..

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ExamplesExamples

f f has an has an absolute minimumabsolute minimum at at (0, 0)(0, 0)::

Absolute minimumAbsolute minimum

44

22

xx

yy

2( )f x x 2( )f x x

––22 –1 –1 11 22

Page 76: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

ExamplesExamples

f f has an has an absolute maximumabsolute maximum at at (0, 4)(0, 4)::

Absolute maximumAbsolute maximum

44

22

xx

yy

2( ) 4f x x 2( ) 4f x x

––22 –1 –1 11 22

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ExamplesExamples

f f has an has an absolute minimumabsolute minimum at at

and an and an absolute maximumabsolute maximum at at

11

1/21/2

––1/21/2

––11

xx

yy

––1 1 11

2( ) 1f x x x 2( ) 1f x x x

( 2 / 2, 1 / 2): ( 2 / 2, 1 / 2):

Absolute minimumAbsolute minimum

Absolute maximumAbsolute maximum

( 2 / 2,1 / 2):( 2 / 2,1 / 2):

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ExamplesExamples

f f has has no absolute extremano absolute extrema::

77

66

55

44

33

22

11

xx

yy

11 22 33 44 55 66

3 2( ) 6 9 2f x x x x 3 2( ) 6 9 2f x x x x

Page 79: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Theorem 3Theorem 3

Absolute Extrema in a Closed IntervalAbsolute Extrema in a Closed Interval If a function If a function f f is is continuouscontinuous on a on a closed interval closed interval

[[aa, , bb]], then , then ff has both an has both an absolute maximumabsolute maximum value and an value and an absolute minimumabsolute minimum value on value on [[aa, , bb]]..

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ExampleExample

The The relative minimumrelative minimum of of ff at at xx33 is also the is also the absolute absolute minimumminimum of of ff..

The The right endpointright endpoint bb of the interval of the interval [a, b][a, b] gives rise to the gives rise to the absolute maximumabsolute maximum value value ff((bb)) of of ff..

xx

yy

Relative Relative minimumminimum

Relative Relative maximummaximum

Absolute Absolute minimumminimum

Absolute Absolute maximummaximum

aa xx11 xx22 xx33bb

Relative Relative maximummaximum

Page 81: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Finding Absolute ExtremaFinding Absolute Extrema

To find the To find the absolute extremaabsolute extrema of of f f on a on a closed intervalclosed interval [[aa, , bb]]..

1.1. Find the Find the critical numberscritical numbers of of ff that lie on that lie on ((aa, , bb))..

2.2. Compute the Compute the valuevalue of of f f at each at each critical numbercritical number found in found in step 1step 1 and compute and compute ff((aa)) and and ff((bb))..

3.3. The The absolute maximumabsolute maximum value and value and absolute minimumabsolute minimum value of value of ff will correspond to the will correspond to the largestlargest and and smallestsmallest numbers, respectively, found in numbers, respectively, found in step 2step 2..

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ExamplesExamples

Find the absolute extrema of the function Find the absolute extrema of the function FF((xx) = ) = xx22 defined defined on the interval on the interval [–1, 2][–1, 2]..

SolutionSolution The function The function FF is is continuouscontinuous on the closed interval on the closed interval [–1, 2][–1, 2]

and and differentiabledifferentiable on the open interval on the open interval (–1, 2)(–1, 2). . Setting Setting FF ′′ = 0 = 0, we get , we get FF ′′((xx) = 2) = 2x x = 0= 0, so there is , so there is only one only one

critical pointcritical point at at x x = 0 = 0.. So, So, FF(–1) = (–1)(–1) = (–1)22 = 1= 1, , FF(0) = (0)(0) = (0)22 = 0= 0, and , and FF(2) = (2)(2) = (2)22 = 4= 4.. It follows that It follows that 00 is the is the absolute minimumabsolute minimum of of FF, and , and 44 is the is the

absolute maximumabsolute maximum of of FF..

Example 1, page 300

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––22 –1 –1 11 22

ExamplesExamples

Find the absolute extrema of the function Find the absolute extrema of the function FF((xx) = ) = xx22 defined defined on the interval on the interval [–1, 2][–1, 2]..

SolutionSolution

Absolute Absolute minimumminimum

44

3 3

22

1 1

xx

yy

2( )f x x 2( )f x x

Absolute Absolute maximummaximum

Example 1, page 300

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ExamplesExamples

Find the absolute extrema of the function Find the absolute extrema of the function

defined on the interval defined on the interval [0, 3][0, 3]..

SolutionSolution The function The function ff is is continuouscontinuous on the closed interval on the closed interval [0, 3][0, 3]

and and differentiabledifferentiable on the open interval on the open interval (0, 3)(0, 3). . Setting Setting f f ′′ = 0 = 0, we get , we get

which gives which gives two critical pointstwo critical points at at x x = – = – 2/32/3 and and xx = 2 = 2.. We drop We drop xx = – = – 2/32/3 since it since it lies outside the intervallies outside the interval [0, 3][0, 3].. So, So, ff(0) = 4(0) = 4, , ff(2) = –(2) = – 44, and , and ff(3) = 1(3) = 1.. It follows that It follows that –– 44 is the is the absolute minimumabsolute minimum of of ff, and , and 44 is is

the the absolute maximumabsolute maximum of of ff..

3 2( ) 2 4 4f x x x x 3 2( ) 2 4 4f x x x x

2( ) 3 4 4 (3 2)( 2) 0f x x x x x 2( ) 3 4 4 (3 2)( 2) 0f x x x x x

Example 2, page 300

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ExamplesExamples

Find the absolute extrema of the function Find the absolute extrema of the function

defined on the interval defined on the interval [0, 3][0, 3]..

SolutionSolution

3 2( ) 2 4 4f x x x x 3 2( ) 2 4 4f x x x x

––2 2 22 44

Absolute Absolute minimumminimum

44

2 2

––2 2

––44

xx

yy

Absolute Absolute maximummaximum

3 2( ) 2 4 4f x x x x 3 2( ) 2 4 4f x x x x

Example 2, page 300

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Applied Example:Applied Example: Maximizing Profits Maximizing Profits

Acrosonic’s total profit (in dollars) from manufacturing Acrosonic’s total profit (in dollars) from manufacturing and selling and selling xx units of their model F speakers is given by units of their model F speakers is given by

How many unitsHow many units of the loudspeaker system must Acrosonic of the loudspeaker system must Acrosonic produce to produce to maximize profitsmaximize profits??

SolutionSolution To find the To find the absolute maximumabsolute maximum of of PP on on [0, 20,000][0, 20,000], first find , first find

the the stationary pointsstationary points of of PP on the interval on the interval (0, 20,000)(0, 20,000).. Setting Setting f f ′′ = 0 = 0, we get , we get

which gives us which gives us only one only one stationary pointstationary point at at xx = 7500 = 7500..

2( ) 0.02 300 200,000 (0 20,000) P x x x x 2( ) 0.02 300 200,000 (0 20,000) P x x x x

( ) 0.04 300 0

3007500

0.04

P x x

x

( ) 0.04 300 0

3007500

0.04

P x x

x

Applied Example 4, page 301

Page 87: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Applied Example:Applied Example: Maximizing Profits Maximizing Profits

Acrosonic’s total profit (in dollars) from manufacturing Acrosonic’s total profit (in dollars) from manufacturing and selling and selling xx units of their model F speakers is given by units of their model F speakers is given by

How many unitsHow many units of the loudspeaker system must Acrosonic of the loudspeaker system must Acrosonic produce to produce to maximize profitsmaximize profits??

SolutionSolution Evaluating the only Evaluating the only stationary pointstationary point we get we get

PP(7500)(7500) = 925,000= 925,000 Evaluating the Evaluating the endpointsendpoints we get we get

PP(0)(0) = –200,000= –200,000 PP(20,000)(20,000) = –2,200,000= –2,200,000

Thus, Acrosonic will realize the Thus, Acrosonic will realize the maximum profitmaximum profit of of $925,000$925,000 by by producingproducing 7500 7500 speakers.speakers.

2( ) 0.02 300 200,000 (0 20,000) P x x x x 2( ) 0.02 300 200,000 (0 20,000) P x x x x

Applied Example 4, page 301

Page 88: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Applied Example:Applied Example: Maximizing Profits Maximizing Profits

Acrosonic’s total profit (in dollars) from manufacturing Acrosonic’s total profit (in dollars) from manufacturing and selling and selling xx units of their model F speakers is given by units of their model F speakers is given by

How many unitsHow many units of the loudspeaker system must Acrosonic of the loudspeaker system must Acrosonic produce to produce to maximize profitsmaximize profits??

SolutionSolution

2( ) 0.02 300 200,000 (0 20,000) P x x x x 2( ) 0.02 300 200,000 (0 20,000) P x x x x

10001000

800800

600600

400400

200200

––200200

xx

yy

22 44 66 88 1010 1212 1414 1616

(Thousands of speakers)(Thousands of speakers)

(Tho

usan

ds o

f d

olla

rs)

(Tho

usan

ds o

f d

olla

rs)

Maximum Maximum ProfitProfit

( )P x( )P x

Applied Example 4, page 301

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4.54.5Optimization IIOptimization II

xx

Inve

ntor

y L

evel

Inve

ntor

y L

evel

TimeTime

2x2x

Average Average inventoryinventory

Page 90: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Guidelines for Solving Optimization ProblemsGuidelines for Solving Optimization Problems

1.1. Assign a Assign a letterletter to each to each variablevariable mentioned in the problem. mentioned in the problem.If appropriate, draw and If appropriate, draw and labellabel a figure. a figure.

2.2. Find an Find an expressionexpression for the quantity to be optimized. for the quantity to be optimized.3.3. Use the conditions given in the problem to write the Use the conditions given in the problem to write the

quantity to be optimized quantity to be optimized as a functionas a function ff of one variable. of one variable.Note any Note any restrictionsrestrictions to be placed on the to be placed on the domaindomain of of ff from from physical considerations of the problem.physical considerations of the problem.

4.4. OptimizeOptimize the function the function ff over its domain using the methods over its domain using the methods of of Section 4.4Section 4.4..

Page 91: Applications of the First Derivative Applications of the Second Derivative Curve Sketching

Applied Maximization Problem:Applied Maximization Problem: Packaging Packaging

By cutting away By cutting away identical squaresidentical squares from each from each cornercorner of a of a rectangularrectangular piece of cardboard and folding up the resulting piece of cardboard and folding up the resulting flaps, the cardboard may be turned into an flaps, the cardboard may be turned into an open boxopen box..

If the cardboard is If the cardboard is 1616 inches long and inches long and 1010 inches wide, find the inches wide, find the dimensionsdimensions of the box that will yield the of the box that will yield the maximum volumemaximum volume..

16

10

Applied Example 2, page 314

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Applied Maximization Problem:Applied Maximization Problem: Packaging Packaging

SolutionSolution1.1. Let Let xx denote the denote the lengthlength in inches of in inches of oneone sideside of each of the of each of the

identical squaresidentical squares to be cut out of the cardboard. to be cut out of the cardboard.The The dimensionsdimensions of the box are of the box are (16 – 2(16 – 2xx)) by by (10 – 2(10 – 2xx)) by by xx in. in.

x

16 – 2x

10

16

x

x x

10 – 2x

Applied Example 2, page 314

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Applied Maximization Problem:Applied Maximization Problem: Packaging Packaging

SolutionSolution2.2. Let Let VV denote the denote the volumevolume (in cubic inches) of the resulting (in cubic inches) of the resulting

box. The volume, box. The volume,

is the quantity to be is the quantity to be maximizedmaximized..

x

16 – 2x10 – 2x

3 2

(16 2 )(10 2 )4( 13 40 )

V x x xx x x

3 2

(16 2 )(10 2 )4( 13 40 )

V x x xx x x

Applied Example 2, page 314

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Applied Maximization Problem:Applied Maximization Problem: Packaging Packaging

SolutionSolution3.3. Each side of the box must be Each side of the box must be nonnegativenonnegative, so , so xx must satisfy must satisfy

the inequalities the inequalities xx 0 0,, 16 – 2 16 – 2xx 0 0, and, and 10 – 2 10 – 2xx 00..✦ All these inequalities are satisfied if All these inequalities are satisfied if 0 0 xx 5 5..✦ Therefore, the problem at hand is equivalent to Therefore, the problem at hand is equivalent to findingfinding

the the absolute maximumabsolute maximum of of

on the on the closed intervalclosed interval [0, 5][0, 5]..

3 2( ) 4( 13 40 )V f x x x x 3 2( ) 4( 13 40 )V f x x x x

Applied Example 2, page 314

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Applied Maximization Problem:Applied Maximization Problem: Packaging Packaging

SolutionSolution4.4. ff is continuous on is continuous on [0, 5][0, 5]. Setting . Setting f f ′′((xx) = 0) = 0 we get we get

which yields the which yields the critical numberscritical numbers xx = 20/3 = 20/3 and and xx = 2 = 2..We discard We discard xx = 20/3 = 20/3 for being outside the interval for being outside the interval [0, 5][0, 5]..We evaluate We evaluate f f at the at the critical pointcritical point and at the and at the endpointsendpoints::

ff(0) = 0(0) = 0 ff(2) = 144(2) = 144 ff(5) = 0(5) = 0Thus, the Thus, the volumevolume of the box is of the box is maximizedmaximized by taking by taking xx = 2 = 2..The resulting The resulting dimensions of the boxdimensions of the box are are 1212″″ ☓☓ 6 6″″ ☓☓ 2 2″″..

2( ) 4(3 26 40)

4(3 20)( 2)

f x x x

x x

2( ) 4(3 26 40)

4(3 20)( 2)

f x x x

x x

Applied Example 2, page 314

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Applied Minimization Problem:Applied Minimization Problem: Inventory Control Inventory Control

Dixie’s Import-Export is the sole seller of the Excalibur Dixie’s Import-Export is the sole seller of the Excalibur 250 cc motorcycle.250 cc motorcycle.

Management estimates that the Management estimates that the demanddemand for these for these motorcycles will be motorcycles will be 10,00010,000 for the coming year and that for the coming year and that they will sell at a they will sell at a uniformuniform raterate throughout the year. throughout the year.

The The costcost incurred in incurred in orderingordering each shipment of each shipment of motorcycles is motorcycles is $10,000$10,000, and the , and the costcost per year of per year of storingstoring each motorcycle is each motorcycle is $200$200..

Dixie’s management faces the following problem: Dixie’s management faces the following problem: ✦ Ordering too manyOrdering too many motorcycles at one time motorcycles at one time increases increases

storage coststorage cost. . ✦ On the other hand, On the other hand, ordering too frequentlyordering too frequently increasesincreases the the

ordering costsordering costs.. How largeHow large should each order be, and should each order be, and how oftenhow often should should

orders be placed, to orders be placed, to minimize ordering and storage costsminimize ordering and storage costs??

Applied Example 5, page 317

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Applied Minimization Problem:Applied Minimization Problem: Inventory Control Inventory Control

SolutionSolution Let Let xx denote the denote the numbernumber of motorcycles of motorcycles in each orderin each order.. Assuming each shipment arrives just as the previous Assuming each shipment arrives just as the previous

shipment is sold out, the shipment is sold out, the average numberaverage number of motorcycles of motorcycles in in storagestorage during the year is during the year is xx/2/2, as shown below:, as shown below:

Thus, Dixie’s Thus, Dixie’s storage coststorage cost for the year is given by for the year is given by 200(200(xx/2)/2), , or or 100100xx dollars.dollars.

xx

Inve

ntor

y L

evel

Inve

ntor

y L

evel

TimeTime

2x2x

Average Average inventoryinventory

Applied Example 5, page 317

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Applied Minimization Problem:Applied Minimization Problem: Inventory Control Inventory Control

SolutionSolution Next, since the company Next, since the company requiresrequires 10,00010,000 motorcycles for the motorcycles for the

year and since year and since each ordereach order is for is for x x motorcycles, the motorcycles, the number number of ordersof orders required is required is

This gives an This gives an ordering costordering cost of of

dollars for the year.dollars for the year. Thus, the Thus, the total yearly costtotal yearly cost incurred by Dixie’s, including incurred by Dixie’s, including

ordering ordering andand storage costs storage costs, is given by, is given by

10,000

x

10,000

x

10,000 100,000,00010,000

x x

10,000 100,000,00010,000

x x

100,000,000( ) 100C x x

x

100,000,000( ) 100C x x

x

Applied Example 5, page 317

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Applied Minimization Problem:Applied Minimization Problem: Inventory Control Inventory Control

SolutionSolution The problem is reduced to The problem is reduced to findingfinding the the absolute minimumabsolute minimum of of

the function the function CC in the in the intervalinterval (0,10,000](0,10,000].. To accomplish this, we computeTo accomplish this, we compute

Setting Setting CC ′′((xx) =0) =0 and solving we obtain and solving we obtain xx = = ++10001000.. We We reject the negativereject the negative for being outside the domain. for being outside the domain. So we have So we have xx = 1000 = 1000 as the only as the only critical numbercritical number..

2

100,000,000( ) 100C x

x 2

100,000,000( ) 100C x

x

Applied Example 5, page 317

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Applied Minimization Problem:Applied Minimization Problem: Inventory Control Inventory Control

SolutionSolution So we have So we have xx = 1000 = 1000 as the only as the only critical numbercritical number.. Now we findNow we find

Since Since CC ″″(1000) > 0(1000) > 0, the , the second derivative testsecond derivative test implies that implies that xx = 1000 = 1000 is a is a relative minimumrelative minimum of of CC..

Also, since Also, since CC ″″(1000) > 0(1000) > 0 for all for all xx in in (0,10,000](0,10,000], the function , the function CC is is concave upwardconcave upward everywhereeverywhere so that also gives the so that also gives the absolute minimumabsolute minimum of of CC..

Thus, Thus, toto minimizeminimize the the orderingordering and and storage costsstorage costs, Dixie’s , Dixie’s should should placeplace 10,000/100010,000/1000, or , or 1010, , orders per yearorders per year, each for a , each for a shipment of shipment of 10001000 motorcycles. motorcycles.

3

200,000,000( )C x

x 3

200,000,000( )C x

x

Applied Example 5, page 317

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End of End of Chapter Chapter