Section 4.2 Derivatives and the Shapes of...
Transcript of Section 4.2 Derivatives and the Shapes of...
. . . . . .
Section4.2DerivativesandtheShapesofCurves
V63.0121.027, CalculusI
November12, 2009
Announcements
I FinalExamFriday, December18, 2:00–3:50pm
..Imagecredit: cobalt123
. . . . . .
Outline
Recall: TheMeanValueTheorem
MonotonicityTheIncreasing/DecreasingTestFindingintervalsofmonotonicityTheFirstDerivativeTest
ConcavityDefinitionsTestingforConcavityTheSecondDerivativeTest
. . . . . .
Recall: TheMeanValueTheorem
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
. . . . . .
Recall: TheMeanValueTheorem
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
. . . . . .
Recall: TheMeanValueTheorem
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
. . . . . .
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.
. . . . . .
Outline
Recall: TheMeanValueTheorem
MonotonicityTheIncreasing/DecreasingTestFindingintervalsofmonotonicityTheFirstDerivativeTest
ConcavityDefinitionsTestingforConcavityTheSecondDerivativeTest
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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,∞)
. . . . . .
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,∞)
. . . . . .
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,∞)
. . . . . .
FindingintervalsofmonotonicityII
ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.
Solution
I f′(x) = 2x, whichispositivewhen x > 0 andnegativewhen xis.
I Wecandrawanumberline:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f isdecreasingon (−∞,0) andincreasingon (0,∞).
I Infactwecansay f isdecreasingon (−∞, 0] andincreasingon [0,∞)
. . . . . .
FindingintervalsofmonotonicityII
ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.
Solution
I f′(x) = 2x, whichispositivewhen x > 0 andnegativewhen xis.
I Wecandrawanumberline:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f isdecreasingon (−∞,0) andincreasingon (0,∞).
I Infactwecansay f isdecreasingon (−∞, 0] andincreasingon [0,∞)
. . . . . .
FindingintervalsofmonotonicityII
ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.
Solution
I f′(x) = 2x, whichispositivewhen x > 0 andnegativewhen xis.
I Wecandrawanumberline:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f isdecreasingon (−∞,0) andincreasingon (0,∞).I Infactwecansay f isdecreasingon (−∞, 0] andincreasing
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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.
. . . . . .
FindingintervalsofmonotonicityII
ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.
Solution
I f′(x) = 2x, whichispositivewhen x > 0 andnegativewhen xis.
I Wecandrawanumberline:
. .f′
.f
.−.↘
..0.0 .+
.↗
.min
I So f isdecreasingon (−∞,0) andincreasingon (0,∞).I Infactwecansay f isdecreasingon (−∞, 0] andincreasing
on [0,∞)
. . . . . .
FindingintervalsofmonotonicityII
ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.
Solution
I f′(x) = 2x, whichispositivewhen x > 0 andnegativewhen xis.
I Wecandrawanumberline:
. .f′
.f
.−.↘
..0.0 .+
.↗.min
I So f isdecreasingon (−∞,0) andincreasingon (0,∞).I Infactwecansay f isdecreasingon (−∞, 0] andincreasing
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
. . . . . .
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
. . . . . .
Outline
Recall: TheMeanValueTheorem
MonotonicityTheIncreasing/DecreasingTestFindingintervalsofmonotonicityTheFirstDerivativeTest
ConcavityDefinitionsTestingforConcavityTheSecondDerivativeTest
. . . . . .
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
. . . . . .
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)
. . . . . .
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)
. . . . . .
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)
. . . . . .
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)
. . . . . .
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)
. . . . . .
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
. . . . . .
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
. . . . . .
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 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
. . . . . .
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 factor
I So f isconcavedownon (−∞, 0], concavedownon [0, 2/5),concaveupon (2/5,∞), andhasaninflectionpointwhenx = 2/5
. . . . . .
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
. . . . . .
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.
. . . . . .
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.
. . . . . .
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 + 2
I Since f′′(−2/3) = −2 < 0, −2/3 isalocalmaximum.I Since f′′(0) = 2 > 0, 0 isalocalminimum.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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’tcatchthelocalminimum x = 0 since f isnotdifferentiablethere.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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.
. . . . . .
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
. . . . . .
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.
. . . . . .
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!