Lesson 21: Derivatives and the Shapes of Curves

. . . . . .

Section4.2DerivativesandtheShapesofCurves

V63.0121.027, CalculusI

November12, 2009

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I FinalExamFriday, December18, 2:00–3:50pm

..Imagecredit: cobalt123

http://www.flickr.com/photos/cobalt/401656972/

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Outline

Recall: TheMeanValueTheorem

MonotonicityTheIncreasing/DecreasingTestFindingintervalsofmonotonicityTheFirstDerivativeTest

ConcavityDefinitionsTestingforConcavityTheSecondDerivativeTest

. . . . . .


Theorem(TheMeanValueTheorem)Let f becontinuouson [a,b]anddifferentiableon (a,b).Thenthereexistsapoint c in(a,b) suchthat

f(b) − f(a)b− a

= f′(c). . ..a

..b

..c

. . . . . .



f(b) − f(a)b− a

= f′(c). . ..a

..b

..c

. . . . . .



f(b) − f(a)b− a

= f′(c). . ..a

..b

..c

. . . . . .

WhytheMVT istheMITCMostImportantTheoremInCalculus!

TheoremLet f′ = 0 onaninterval (a,b). Then f isconstanton (a,b).

Proof.Pickanypoints x and y in (a,b) with x < y. Then f iscontinuouson [x, y] anddifferentiableon (x, y). ByMVT thereexistsapointz in (x, y) suchthat

f(y) − f(x)y− x

= f′(z) = 0.

So f(y) = f(x). Sincethisistrueforall x and y in (a,b), then f isconstant.

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Outline




. . . . . .

Whatdoesitmeanforafunctiontobeincreasing?

DefinitionA function f is increasing on (a,b) if

f(x) < f(y)

whenever x and y aretwopointsin (a,b) with x < y.

I Anincreasingfunction“preservesorder.”I Writeyourowndefinition(mutatismutandis)of decreasing,

nonincreasing, nondecreasing

. . . . . .

Whatdoesitmeanforafunctiontobeincreasing?

DefinitionA function f is increasing on (a,b) if

f(x) < f(y)

whenever x and y aretwopointsin (a,b) with x < y.

I Anincreasingfunction“preservesorder.”I Writeyourowndefinition(mutatismutandis)of decreasing,

nonincreasing, nondecreasing

. . . . . .

TheIncreasing/DecreasingTest

Theorem(TheIncreasing/DecreasingTest)If f′ > 0 on (a,b), then f isincreasingon (a,b). If f′ < 0 on (a,b),then f isdecreasingon (a,b).

Proof.Itworksthesameasthelasttheorem. Picktwopoints x and y in(a,b) with x < y. Wemustshow f(x) < f(y). ByMVT thereexistsapoint c in (x, y) suchthat

f(y) − f(x)y− x

= f′(c) > 0.

Sof(y) − f(x) = f′(c)(y− x) > 0.

. . . . . .

TheIncreasing/DecreasingTest

Theorem(TheIncreasing/DecreasingTest)If f′ > 0 on (a,b), then f isincreasingon (a,b). If f′ < 0 on (a,b),then f isdecreasingon (a,b).

Proof.Itworksthesameasthelasttheorem. Picktwopoints x and y in(a,b) with x < y. Wemustshow f(x) < f(y). ByMVT thereexistsapoint c in (x, y) suchthat

f(y) − f(x)y− x

= f′(c) > 0.

Sof(y) − f(x) = f′(c)(y− x) > 0.

. . . . . .

FindingintervalsofmonotonicityI

ExampleFindtheintervalsofmonotonicityof f(x) = 2x− 5.

Solutionf′(x) = 2 isalwayspositive, so f isincreasingon (−∞,∞).

ExampleDescribethemonotonicityof f(x) = arctan(x).

SolutionSince f′(x) =

11 + x2

isalwayspositive, f(x) isalwaysincreasing.

. . . . . .






11 + x2


. . . . . .






11 + x2


. . . . . .






11 + x2


. . . . . .

FindingintervalsofmonotonicityII

ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.

Solution

I f′(x) = 2x, whichispositivewhen x > 0 andnegativewhen xis.

I Wecandrawanumberline:

. .f′.− ..0.0 .+

.min

I So f isdecreasingon (−∞,0) andincreasingon (0,∞).I Infactwecansay f isdecreasingon (−∞, 0] andincreasing

on [0,∞)

. . . . . .



Solution



. .f′.− ..0.0 .+

.min


on [0,∞)

. . . . . .



Solution



. .f′.− ..0.0 .+

.min


on [0,∞)

. . . . . .



Solution



. .f′

.f

.−.↘

..0.0 .+

.↗

.min

I So f isdecreasingon (−∞,0) andincreasingon (0,∞).

I Infactwecansay f isdecreasingon (−∞, 0] andincreasingon [0,∞)

. . . . . .



Solution



. .f′

.f

.−.↘

..0.0 .+

.↗

.min

I So f isdecreasingon (−∞,0) andincreasingon (0,∞).

I Infactwecansay f isdecreasingon (−∞, 0] andincreasingon [0,∞)

. . . . . .



Solution



. .f′

.f

.−.↘

..0.0 .+

.↗

.min


on [0,∞)

. . . . . .

FindingintervalsofmonotonicityIIIExampleFindtheintervalsofmonotonicityof f(x) = x2/3(x + 2).

Solution

f′(x) = 23x

−1/3(x + 2) + x2/3 = 13x

−1/3 (5x + 4)

Thecriticalpointsare 0 andand −4/5.

. .x−1/3..0.×.− .+

.5x + 4..−4/5

.0.− .+

.f′(x)

.f(x).

.−4/5

.0 ..0.×

.+

.↗.−.↘

.+

.↗.max .min

. . . . . .


Solution

f′(x) = 23x

−1/3(x + 2) + x2/3 = 13x

−1/3 (5x + 4)


. .x−1/3..0.×.− .+

.5x + 4..−4/5

.0.− .+

.f′(x)

.f(x).

.−4/5

.0 ..0.×

.+

.↗.−.↘

.+

.↗.max .min

. . . . . .


Solution

f′(x) = 23x

−1/3(x + 2) + x2/3 = 13x

−1/3 (5x + 4)


. .x−1/3..0.×.− .+

.5x + 4..−4/5

.0.− .+

.f′(x)

.f(x).

.−4/5

.0 ..0.×

.+

.↗.−.↘

.+

.↗.max .min

. . . . . .


Solution

f′(x) = 23x

−1/3(x + 2) + x2/3 = 13x

−1/3 (5x + 4)


. .x−1/3..0.×.− .+

.5x + 4..−4/5

.0.− .+

.f′(x)

.f(x).

.−4/5

.0 ..0.×.+

.↗.−.↘

.+

.↗

.max .min

. . . . . .

TheFirstDerivativeTest

Theorem(TheFirstDerivativeTest)Let f becontinuouson [a,b] and c acriticalpointof f in (a,b).

I If f′(x) > 0 on (a, c) and f′(x) < 0 on (c,b), then c isalocalmaximum.

I If f′(x) < 0 on (a, c) and f′(x) > 0 on (c,b), then c isalocalminimum.

I If f′(x) hasthesamesignon (a, c) and (c,b), then c isnotalocalextremum.

. . . . . .



Solution



. .f′

.f

.−.↘

..0.0 .+

.↗

.min


on [0,∞)

. . . . . .



Solution



. .f′

.f

.−.↘

..0.0 .+

.↗.min


on [0,∞)

. . . . . .


Solution

f′(x) = 23x

−1/3(x + 2) + x2/3 = 13x

−1/3 (5x + 4)


. .x−1/3..0.×.− .+

.5x + 4..−4/5

.0.− .+

.f′(x)

.f(x).

.−4/5

.0 ..0.×.+

.↗.−.↘

.+

.↗

.max .min

. . . . . .


Solution

f′(x) = 23x

−1/3(x + 2) + x2/3 = 13x

−1/3 (5x + 4)


. .x−1/3..0.×.− .+

.5x + 4..−4/5

.0.− .+

.f′(x)

.f(x).

.−4/5

.0 ..0.×.+

.↗.−.↘

.+

.↗.max .min

. . . . . .

Outline




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DefinitionThegraphof f iscalled concaveup onandinterval I ifitliesaboveallitstangentson I. Thegraphof f iscalled concavedownon I ifitliesbelowallitstangentson I.

.

concaveup

.

concavedownWesometimessayaconcaveupgraph“holdswater”andaconcavedowngraph“spillswater”.

. . . . . .

DefinitionA point P onacurve y = f(x) iscalledan inflectionpoint if f iscontinuousthereandthecurvechangesfromconcaveupwardtoconcavedownwardat P (orviceversa).

..concavedown

.concaveup

..inflectionpoint

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Theorem(ConcavityTest)

I If f′′(x) > 0 forall x inaninterval I, thenthegraphof f isconcaveupwardon I

I If f′′(x) < 0 forall x in I, thenthegraphof f isconcavedownwardon I

Proof.Suppose f′′(x) > 0 on I. Thismeans f′ isincreasingon I. Let a andx bein I. Thetangentlinethrough (a, f(a)) isthegraphof

L(x) = f(a) + f′(a)(x− a)

ByMVT,thereexistsa c between a and x withf(x) − f(a)

x− a= f′(c).

So

f(x) = f(a) + f′(c)(x− a) ≥ f(a) + f′(a)(x− a) = L(x)

. . . . . .

Theorem(ConcavityTest)

I If f′′(x) > 0 forall x inaninterval I, thenthegraphof f isconcaveupwardon I

I If f′′(x) < 0 forall x in I, thenthegraphof f isconcavedownwardon I

Proof.Suppose f′′(x) > 0 on I. Thismeans f′ isincreasingon I. Let a andx bein I. Thetangentlinethrough (a, f(a)) isthegraphof

L(x) = f(a) + f′(a)(x− a)

ByMVT,thereexistsa c between a and x withf(x) − f(a)

x− a= f′(c).

So

f(x) = f(a) + f′(c)(x− a) ≥ f(a) + f′(a)(x− a) = L(x)

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ExampleFindtheintervalsofconcavityforthegraphof f(x) = x3 + x2.

Solution

I Wehave f′(x) = 3x2 + 2x, so f′′(x) = 6x + 2.I Thisisnegativewhen x < −1/3, positivewhen x > −1/3, and0 when x = −1/3

I So f isconcavedownon (−∞,−1/3), concaveupon(1/3,∞), andhasaninflectionpointat (−1/3, 2/27)

. . . . . .


Solution

I Wehave f′(x) = 3x2 + 2x, so f′′(x) = 6x + 2.

I Thisisnegativewhen x < −1/3, positivewhen x > −1/3, and0 when x = −1/3


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Solution



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Solution



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ExampleFindtheintervalsofconcavityofthegraphof f(x) = x2/3(x + 2).

Solution

I f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2)

I Thesecondderivative f′′(x) isnotdefinedat 0I Otherwise, x−4/3 isalwayspositive, sotheconcavityis

determinedbythe 5x− 2 factorI So f isconcavedownon (−∞, 0], concavedownon [0, 2/5),

concaveupon (2/5,∞), andhasaninflectionpointwhenx = 2/5

. . . . . .


Solution

I f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2)




. . . . . .


Solution

I f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2)

I Thesecondderivative f′′(x) isnotdefinedat 0

I Otherwise, x−4/3 isalwayspositive, sotheconcavityisdeterminedbythe 5x− 2 factor

I So f isconcavedownon (−∞, 0], concavedownon [0, 2/5),concaveupon (2/5,∞), andhasaninflectionpointwhenx = 2/5

. . . . . .


Solution

I f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2)


determinedbythe 5x− 2 factor

I So f isconcavedownon (−∞, 0], concavedownon [0, 2/5),concaveupon (2/5,∞), andhasaninflectionpointwhenx = 2/5

. . . . . .


Solution

I f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2)




. . . . . .

TheSecondDerivativeTest

Theorem(TheSecondDerivativeTest)Let f, f′, and f′′ becontinuouson [a,b]. Let c bebeapointin(a,b) with f′(c) = 0.

I If f′′(c) < 0, then f(c) isalocalmaximum.I If f′′(c) > 0, then f(c) isalocalminimum.

Remarks

I If f′′(c) = 0, thesecondderivativetestis inconclusive (thisdoesnotmean c isneither; wejustdon’tknowyet).

I Welookforzeroesof f′ andplugtheminto f′′ todetermineiftheir f valuesarelocalextremevalues.

. . . . . .

TheSecondDerivativeTest

Theorem(TheSecondDerivativeTest)Let f, f′, and f′′ becontinuouson [a,b]. Let c bebeapointin(a,b) with f′(c) = 0.

I If f′′(c) < 0, then f(c) isalocalmaximum.I If f′′(c) > 0, then f(c) isalocalminimum.

Remarks

I If f′′(c) = 0, thesecondderivativetestis inconclusive (thisdoesnotmean c isneither; wejustdon’tknowyet).

I Welookforzeroesof f′ andplugtheminto f′′ todetermineiftheir f valuesarelocalextremevalues.

. . . . . .

ExampleFindthelocalextremaof f(x) = x3 + x2.

Solution

I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 isalocalmaximum.I Since f′′(0) = 2 > 0, 0 isalocalminimum.

. . . . . .


Solution

I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.

I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 isalocalmaximum.I Since f′′(0) = 2 > 0, 0 isalocalminimum.

. . . . . .


Solution

I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2

I Since f′′(−2/3) = −2 < 0, −2/3 isalocalmaximum.I Since f′′(0) = 2 > 0, 0 isalocalminimum.

. . . . . .


Solution

I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 isalocalmaximum.

I Since f′′(0) = 2 > 0, 0 isalocalminimum.

. . . . . .


Solution

I f′(x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x + 2I Since f′′(−2/3) = −2 < 0, −2/3 isalocalmaximum.I Since f′′(0) = 2 > 0, 0 isalocalminimum.

. . . . . .

ExampleFindthelocalextremaof f(x) = x2/3(x + 2)

Solution

I Remember f′(x) =13x−1/3(5x + 4) whichiszerowhen

x = −4/5

I Remember f′′(x) =109x−4/3(5x− 2), whichisnegativewhen

x = −4/5

I So x = −4/5 isalocalmaximum.I NoticetheSecondDerivativeTestdoesn’tcatchthelocal

minimum x = 0 since f isnotdifferentiablethere.

. . . . . .


Solution


x = −4/5


x = −4/5



. . . . . .


Solution


x = −4/5


x = −4/5



. . . . . .


Solution


x = −4/5


x = −4/5

I So x = −4/5 isalocalmaximum.

I NoticetheSecondDerivativeTestdoesn’tcatchthelocalminimum x = 0 since f isnotdifferentiablethere.

. . . . . .


Solution


x = −4/5


x = −4/5



. . . . . .

Graph

Graphof f(x) = x2/3(x + 2):

. .x

.y

..(−4/5, 1.03413)

..(0,0)

..(2/5, 1.30292)

..(−2, 0)

. . . . . .

Whenthesecondderivativeiszero

I Atinflectionpoints c, if f′ isdifferentiableat c, then f′′(c) = 0I Isitnecessarilytrue, though?

Considertheseexamples:

f(x) = x4 g(x) = −x4 h(x) = x3

Allofthemhavecriticalpointsatzerowithasecondderivativeofzero. Butthefirsthasalocalminat 0, thesecondhasalocalmaxat 0, andthethirdhasaninflectionpointat 0. Thisiswhywesay2DT hasnothingtosaywhen f′′(c) = 0.

. . . . . .




f(x) = x4 g(x) = −x4 h(x) = x3


. . . . . .

Whenfirstandsecondderivativearezero

function derivatives graph type

f(x) = x4f′(x) = 4x3, f′(0) = 0

.min

f′′(x) = 12x2, f′′(0) = 0

g(x) = −x4g′(x) = −4x3, g′(0) = 0

.

maxg′′(x) = −12x2, g′′(0) = 0

h(x) = x3h′(x) = 3x2, h′(0) = 0

.infl.

h′′(x) = 6x, h′′(0) = 0

. . . . . .




f(x) = x4 g(x) = −x4 h(x) = x3


. . . . . .

Whathavewelearnedtoday?

I Concepts: MeanValueTheorem, monotonicity, concavityI Facts: derivativescandetectmonotonicityandconcavityI Techniquesfordrawingcurves: the Increasing/DecreasingTest andthe ConcavityTest

I Techniquesforfindingextrema: the FirstDerivativeTest andthe SecondDerivativeTest

Nextweek: Graphingfunctions!

. . . . . .

Whathavewelearnedtoday?

I Concepts: MeanValueTheorem, monotonicity, concavityI Facts: derivativescandetectmonotonicityandconcavityI Techniquesfordrawingcurves: the Increasing/DecreasingTest andthe ConcavityTest

I Techniquesforfindingextrema: the FirstDerivativeTest andthe SecondDerivativeTest

Nextweek: Graphingfunctions!

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Lesson 21: Derivatives and the Shapes of Curves