. . . . . .

Section4.2TheMeanValueTheorem

V63.0121.034, CalculusI

November9, 2009

Announcements

I Quizthisweekon§§3.1–3.5

..Imagecredit: Jimmywayne22

. . . . . .

Outline

Review: TheClosedIntervalMethod

Rolle’sTheorem

TheMeanValueTheoremApplications

WhytheMVT istheMITC

. . . . . .

FlowchartforplacingextremaThankstoFermat

Suppose f isacontinuousfunctionontheclosed, boundedinterval [a,b], and c isaglobalmaximumpoint.

..start

.Is c an

endpoint?

. c = a orc = b

.c is a

local max

.Is f diff’ble

at c?

.f is notdiff at c

.f′(c) = 0

.no

.yes

.no

.yes

. . . . . .

TheClosedIntervalMethod

Thismeanstofindthemaximumvalueof f on [a,b], weneedto:I Evaluate f atthe endpoints a and bI Evaluate f atthe criticalpoints x whereeither f′(x) = 0 or f isnotdifferentiableat x.

I Thepointswiththelargestfunctionvaluearetheglobalmaximumpoints

I Thepointswiththesmallestormostnegativefunctionvaluearetheglobalminimumpoints.

. . . . . .

Outline

Review: TheClosedIntervalMethod

Rolle’sTheorem

TheMeanValueTheoremApplications

WhytheMVT istheMITC

. . . . . .

HeuristicMotivationforRolle’sTheorem

Ifyoubikeupahill, thenbackdown, atsomepointyourelevationwasstationary.

.

.Imagecredit: SpringSun

. . . . . .

MathematicalStatementofRolle’sTheorem

Theorem(Rolle’sTheorem)Let f becontinuouson [a,b]anddifferentiableon (a,b).Suppose f(a) = f(b). Thenthereexistsapoint c in (a,b)suchthat f′(c) = 0. . .

.a..b

..c

. . . . . .

MathematicalStatementofRolle’sTheorem

Theorem(Rolle’sTheorem)Let f becontinuouson [a,b]anddifferentiableon (a,b).Suppose f(a) = f(b). Thenthereexistsapoint c in (a,b)suchthat f′(c) = 0. . .

.a..b

..c

. . . . . .

ProofofRolle’sTheoremProof.

I BytheExtremeValueTheorem f mustachieveitsmaximumvalueatapoint c in [a,b].

I If c isin (a,b), great: it’salocalmaximumandsobyFermat’sTheorem f′(c) = 0.

I Ontheotherhand, if c = a or c = b, trywiththeminimum.Theminimumof f on [a,b] mustbeachievedatapoint d in[a,b].

I If d isin (a,b), great: it’salocalminimumandsobyFermat’sTheorem f′(d) = 0. Ifnot, d = a or d = b.

I Ifwestillhaven’tfoundapointintheinterior, wehavethatthemaximumandminimumvaluesof f on [a,b] occuratbothendpoints. Butwealreadyknowthat f(a) = f(b). Ifthesearethemaximumandminimumvalues, f is constanton [a,b] andanypoint x in (a,b) willhave f′(x) = 0.

. . . . . .

ProofofRolle’sTheoremProof.

I BytheExtremeValueTheorem f mustachieveitsmaximumvalueatapoint c in [a,b].

I If c isin (a,b), great: it’salocalmaximumandsobyFermat’sTheorem f′(c) = 0.

I Ontheotherhand, if c = a or c = b, trywiththeminimum.Theminimumof f on [a,b] mustbeachievedatapoint d in[a,b].

I If d isin (a,b), great: it’salocalminimumandsobyFermat’sTheorem f′(d) = 0. Ifnot, d = a or d = b.

I Ifwestillhaven’tfoundapointintheinterior, wehavethatthemaximumandminimumvaluesof f on [a,b] occuratbothendpoints. Butwealreadyknowthat f(a) = f(b). Ifthesearethemaximumandminimumvalues, f is constanton [a,b] andanypoint x in (a,b) willhave f′(x) = 0.

. . . . . .

ProofofRolle’sTheoremProof.

I BytheExtremeValueTheorem f mustachieveitsmaximumvalueatapoint c in [a,b].

I If c isin (a,b), great: it’salocalmaximumandsobyFermat’sTheorem f′(c) = 0.

I Ontheotherhand, if c = a or c = b, trywiththeminimum.Theminimumof f on [a,b] mustbeachievedatapoint d in[a,b].

I If d isin (a,b), great: it’salocalminimumandsobyFermat’sTheorem f′(d) = 0. Ifnot, d = a or d = b.

I Ifwestillhaven’tfoundapointintheinterior, wehavethatthemaximumandminimumvaluesof f on [a,b] occuratbothendpoints. Butwealreadyknowthat f(a) = f(b). Ifthesearethemaximumandminimumvalues, f is constanton [a,b] andanypoint x in (a,b) willhave f′(x) = 0.

. . . . . .

ProofofRolle’sTheoremProof.

I BytheExtremeValueTheorem f mustachieveitsmaximumvalueatapoint c in [a,b].

I If c isin (a,b), great: it’salocalmaximumandsobyFermat’sTheorem f′(c) = 0.

I Ontheotherhand, if c = a or c = b, trywiththeminimum.Theminimumof f on [a,b] mustbeachievedatapoint d in[a,b].

I If d isin (a,b), great: it’salocalminimumandsobyFermat’sTheorem f′(d) = 0. Ifnot, d = a or d = b.

I Ifwestillhaven’tfoundapointintheinterior, wehavethatthemaximumandminimumvaluesof f on [a,b] occuratbothendpoints. Butwealreadyknowthat f(a) = f(b). Ifthesearethemaximumandminimumvalues, f is constanton [a,b] andanypoint x in (a,b) willhave f′(x) = 0.

. . . . . .

ProofofRolle’sTheoremProof.

I BytheExtremeValueTheorem f mustachieveitsmaximumvalueatapoint c in [a,b].

I If c isin (a,b), great: it’salocalmaximumandsobyFermat’sTheorem f′(c) = 0.

I Ontheotherhand, if c = a or c = b, trywiththeminimum.Theminimumof f on [a,b] mustbeachievedatapoint d in[a,b].

I If d isin (a,b), great: it’salocalminimumandsobyFermat’sTheorem f′(d) = 0. Ifnot, d = a or d = b.

I Ifwestillhaven’tfoundapointintheinterior, wehavethatthemaximumandminimumvaluesof f on [a,b] occuratbothendpoints.

Butwealreadyknowthat f(a) = f(b). Ifthesearethemaximumandminimumvalues, f is constanton [a,b] andanypoint x in (a,b) willhave f′(x) = 0.

. . . . . .

ProofofRolle’sTheoremProof.

I BytheExtremeValueTheorem f mustachieveitsmaximumvalueatapoint c in [a,b].

I If c isin (a,b), great: it’salocalmaximumandsobyFermat’sTheorem f′(c) = 0.

I Ontheotherhand, if c = a or c = b, trywiththeminimum.Theminimumof f on [a,b] mustbeachievedatapoint d in[a,b].

I If d isin (a,b), great: it’salocalminimumandsobyFermat’sTheorem f′(d) = 0. Ifnot, d = a or d = b.

I Ifwestillhaven’tfoundapointintheinterior, wehavethatthemaximumandminimumvaluesof f on [a,b] occuratbothendpoints. Butwealreadyknowthat f(a) = f(b).

Ifthesearethemaximumandminimumvalues, f is constanton [a,b] andanypoint x in (a,b) willhave f′(x) = 0.

. . . . . .

ProofofRolle’sTheoremProof.

I BytheExtremeValueTheorem f mustachieveitsmaximumvalueatapoint c in [a,b].

I If c isin (a,b), great: it’salocalmaximumandsobyFermat’sTheorem f′(c) = 0.

I Ontheotherhand, if c = a or c = b, trywiththeminimum.Theminimumof f on [a,b] mustbeachievedatapoint d in[a,b].

I If d isin (a,b), great: it’salocalminimumandsobyFermat’sTheorem f′(d) = 0. Ifnot, d = a or d = b.

I Ifwestillhaven’tfoundapointintheinterior, wehavethatthemaximumandminimumvaluesof f on [a,b] occuratbothendpoints. Butwealreadyknowthat f(a) = f(b). Ifthesearethemaximumandminimumvalues, f is constanton [a,b] andanypoint x in (a,b) willhave f′(x) = 0.

. . . . . .

FlowchartproofofRolle’sTheorem

.

.

..Let c be

the max pt.

.Let d bethe min pt

.

.endpointsare maxand min

.

..is c an

endpoint?.

.is d an

endpoint?.

.f is

constanton [a,b]

..f′(c) = 0 ..

f′(d) = 0 ..f′(x) ≡ 0on (a,b)

.no .no

.yes .yes

. . . . . .

Outline

Review: TheClosedIntervalMethod

Rolle’sTheorem

TheMeanValueTheoremApplications

WhytheMVT istheMITC

. . . . . .

HeuristicMotivationforTheMeanValueTheorem

Ifyoudrivebetweenpoints A and B, atsometimeyourspeedometerreadingwasthesameasyouraveragespeedoverthedrive.

.

.Imagecredit: ClintJCL

. . . . . .

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

. . . . . .

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

. . . . . .

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

. . . . . .

Rollevs. MVT

f′(c) = 0f(b) − f(a)

b− a= f′(c)

. ..a

..b

..c

. ..a

..b

..c

Ifthe x-axisisskewedthepictureslookthesame.

. . . . . .

Rollevs. MVT

f′(c) = 0f(b) − f(a)

b− a= f′(c)

. ..a

..b

..c

. ..a

..b

..c

Ifthe x-axisisskewedthepictureslookthesame.

. . . . . .

ProofoftheMeanValueTheorem

Proof.Thelineconnecting (a, f(a)) and (b, f(b)) hasequation

y− f(a) =f(b) − f(a)

b− a(x− a)

ApplyRolle’sTheoremtothefunction

g(x) = f(x) − f(a) − f(b) − f(a)b− a

(x− a).

Then g iscontinuouson [a,b] anddifferentiableon (a,b) since fis. Also g(a) = 0 and g(b) = 0 (checkboth) SobyRolle’sTheoremthereexistsapoint c in (a,b) suchthat

0 = g′(c) = f′(c) − f(b) − f(a)b− a

.

. . . . . .

ProofoftheMeanValueTheorem

Proof.Thelineconnecting (a, f(a)) and (b, f(b)) hasequation

y− f(a) =f(b) − f(a)

b− a(x− a)

ApplyRolle’sTheoremtothefunction

g(x) = f(x) − f(a) − f(b) − f(a)b− a

(x− a).

Then g iscontinuouson [a,b] anddifferentiableon (a,b) since fis. Also g(a) = 0 and g(b) = 0 (checkboth) SobyRolle’sTheoremthereexistsapoint c in (a,b) suchthat

0 = g′(c) = f′(c) − f(b) − f(a)b− a

.

. . . . . .

ProofoftheMeanValueTheorem

Proof.Thelineconnecting (a, f(a)) and (b, f(b)) hasequation

y− f(a) =f(b) − f(a)

b− a(x− a)

ApplyRolle’sTheoremtothefunction

g(x) = f(x) − f(a) − f(b) − f(a)b− a

(x− a).

Then g iscontinuouson [a,b] anddifferentiableon (a,b) since fis.

Also g(a) = 0 and g(b) = 0 (checkboth) SobyRolle’sTheoremthereexistsapoint c in (a,b) suchthat

0 = g′(c) = f′(c) − f(b) − f(a)b− a

.

. . . . . .

ProofoftheMeanValueTheorem

Proof.Thelineconnecting (a, f(a)) and (b, f(b)) hasequation

y− f(a) =f(b) − f(a)

b− a(x− a)

ApplyRolle’sTheoremtothefunction

g(x) = f(x) − f(a) − f(b) − f(a)b− a

(x− a).

Then g iscontinuouson [a,b] anddifferentiableon (a,b) since fis. Also g(a) = 0 and g(b) = 0 (checkboth)

SobyRolle’sTheoremthereexistsapoint c in (a,b) suchthat

0 = g′(c) = f′(c) − f(b) − f(a)b− a

.

. . . . . .

ProofoftheMeanValueTheorem

Proof.Thelineconnecting (a, f(a)) and (b, f(b)) hasequation

y− f(a) =f(b) − f(a)

b− a(x− a)

ApplyRolle’sTheoremtothefunction

g(x) = f(x) − f(a) − f(b) − f(a)b− a

(x− a).

Then g iscontinuouson [a,b] anddifferentiableon (a,b) since fis. Also g(a) = 0 and g(b) = 0 (checkboth) SobyRolle’sTheoremthereexistsapoint c in (a,b) suchthat

0 = g′(c) = f′(c) − f(b) − f(a)b− a

.

. . . . . .

UsingtheMVT tocountsolutions

ExampleShowthatthereisauniquesolutiontotheequation x3 − x = 100intheinterval [4, 5].

Solution

I BytheIntermediateValueTheorem, thefunctionf(x) = x3 − x musttakethevalue 100 atsomepointon c in(4, 5).

I Ifthereweretwopoints c1 and c2 with f(c1) = f(c2) = 100,thensomewherebetweenthemwouldbeapoint c3betweenthemwith f′(c3) = 0.

I However, f′(x) = 3x2 − 1, whichispositiveallalong (4, 5).Sothisisimpossible.

. . . . . .

UsingtheMVT tocountsolutions

ExampleShowthatthereisauniquesolutiontotheequation x3 − x = 100intheinterval [4, 5].

Solution

I BytheIntermediateValueTheorem, thefunctionf(x) = x3 − x musttakethevalue 100 atsomepointon c in(4, 5).

I Ifthereweretwopoints c1 and c2 with f(c1) = f(c2) = 100,thensomewherebetweenthemwouldbeapoint c3betweenthemwith f′(c3) = 0.

I However, f′(x) = 3x2 − 1, whichispositiveallalong (4, 5).Sothisisimpossible.

. . . . . .

UsingtheMVT tocountsolutions

ExampleShowthatthereisauniquesolutiontotheequation x3 − x = 100intheinterval [4, 5].

Solution

I BytheIntermediateValueTheorem, thefunctionf(x) = x3 − x musttakethevalue 100 atsomepointon c in(4, 5).

I Ifthereweretwopoints c1 and c2 with f(c1) = f(c2) = 100,thensomewherebetweenthemwouldbeapoint c3betweenthemwith f′(c3) = 0.

I However, f′(x) = 3x2 − 1, whichispositiveallalong (4, 5).Sothisisimpossible.

. . . . . .

UsingtheMVT tocountsolutions

ExampleShowthatthereisauniquesolutiontotheequation x3 − x = 100intheinterval [4, 5].

Solution

I BytheIntermediateValueTheorem, thefunctionf(x) = x3 − x musttakethevalue 100 atsomepointon c in(4, 5).

I Ifthereweretwopoints c1 and c2 with f(c1) = f(c2) = 100,thensomewherebetweenthemwouldbeapoint c3betweenthemwith f′(c3) = 0.

I However, f′(x) = 3x2 − 1, whichispositiveallalong (4, 5).Sothisisimpossible.

. . . . . .

ExampleWeknowthat |sin x| ≤ 1 forall x. Showthat |sin x| ≤ |x|.

SolutionApplytheMVT tothefunction f(t) = sin t on [0, x]. Weget

sin x− sin 0x− 0

= cos(c)

forsome c in (0, x). Since |cos(c)| ≤ 1, weget∣∣∣∣sin xx∣∣∣∣ ≤ 1 =⇒ |sin x| ≤ |x|

. . . . . .

ExampleWeknowthat |sin x| ≤ 1 forall x. Showthat |sin x| ≤ |x|.

SolutionApplytheMVT tothefunction f(t) = sin t on [0, x]. Weget

sin x− sin 0x− 0

= cos(c)

forsome c in (0, x). Since |cos(c)| ≤ 1, weget∣∣∣∣sin xx∣∣∣∣ ≤ 1 =⇒ |sin x| ≤ |x|

. . . . . .

ExampleLet f beadifferentiablefunctionwith f(1) = 3 and f′(x) < 2 forall x in [0, 5]. Could f(4) ≥ 9?

Solution

ByMVT

f(4) − f(1)

4− 1= f′(c) < 2

forsome c in (1,4). Therefore

f(4) = f(1)+ f′(c)(3) < 3+2 ·3 = 9.

Sono, itisimpossiblethat f(4) ≥ 9.. .x

.y

..(1, 3)

..(4, 9)

..(4, f(4))

. . . . . .

ExampleLet f beadifferentiablefunctionwith f(1) = 3 and f′(x) < 2 forall x in [0, 5]. Could f(4) ≥ 9?

Solution

ByMVT

f(4) − f(1)

4− 1= f′(c) < 2

forsome c in (1,4). Therefore

f(4) = f(1)+ f′(c)(3) < 3+2 ·3 = 9.

Sono, itisimpossiblethat f(4) ≥ 9.. .x

.y

..(1, 3)

..(4, 9)

..(4, f(4))

. . . . . .

QuestionA drivertravelsalongtheNewJerseyTurnpikeusingEZ-Pass. ThesystemtakesnoteofthetimeandplacethedriverentersandexitstheTurnpike. A weekafterhistrip, thedrivergetsaspeedingticketinthemail. Whichofthefollowingbestdescribesthesituation?

(a) EZ-Passcannotprovethatthedriverwasspeeding

(b) EZ-Passcanprovethatthedriverwasspeeding

(c) Thedriver’sactualmaximumspeedexceedshisticketedspeed

(d) Both(b)and(c).

Bepreparedtojustifyyouranswer.

. . . . . .

QuestionA drivertravelsalongtheNewJerseyTurnpikeusingEZ-Pass. ThesystemtakesnoteofthetimeandplacethedriverentersandexitstheTurnpike. A weekafterhistrip, thedrivergetsaspeedingticketinthemail. Whichofthefollowingbestdescribesthesituation?

(a) EZ-Passcannotprovethatthedriverwasspeeding

(b) EZ-Passcanprovethatthedriverwasspeeding

(c) Thedriver’sactualmaximumspeedexceedshisticketedspeed

(d) Both(b)and(c).

Bepreparedtojustifyyouranswer.

. . . . . .

Outline

Review: TheClosedIntervalMethod

Rolle’sTheorem

TheMeanValueTheoremApplications

WhytheMVT istheMITC

. . . . . .

FactIf f isconstanton (a,b), then f′(x) = 0 on (a,b).

I Thelimitofdifferencequotientsmustbe 0I Thetangentlinetoalineisthatline, andaconstantfunction’sgraphisahorizontalline, whichhasslope 0.

I Impliedbythepowerrulesince c = cx0

QuestionIf f′(x) = 0 is f necessarilyaconstantfunction?

I ItseemstrueI Butsofarnotheorem(thatwehaveproven)usesinformationaboutthederivativeofafunctiontodetermineinformationaboutthefunctionitself

. . . . . .

FactIf f isconstanton (a,b), then f′(x) = 0 on (a,b).

I Thelimitofdifferencequotientsmustbe 0I Thetangentlinetoalineisthatline, andaconstantfunction’sgraphisahorizontalline, whichhasslope 0.

I Impliedbythepowerrulesince c = cx0

QuestionIf f′(x) = 0 is f necessarilyaconstantfunction?

I ItseemstrueI Butsofarnotheorem(thatwehaveproven)usesinformationaboutthederivativeofafunctiontodetermineinformationaboutthefunctionitself

. . . . . .

FactIf f isconstanton (a,b), then f′(x) = 0 on (a,b).

I Thelimitofdifferencequotientsmustbe 0I Thetangentlinetoalineisthatline, andaconstantfunction’sgraphisahorizontalline, whichhasslope 0.

I Impliedbythepowerrulesince c = cx0

QuestionIf f′(x) = 0 is f necessarilyaconstantfunction?

I ItseemstrueI Butsofarnotheorem(thatwehaveproven)usesinformationaboutthederivativeofafunctiontodetermineinformationaboutthefunctionitself

. . . . . .

FactIf f isconstanton (a,b), then f′(x) = 0 on (a,b).

I Thelimitofdifferencequotientsmustbe 0I Thetangentlinetoalineisthatline, andaconstantfunction’sgraphisahorizontalline, whichhasslope 0.

I Impliedbythepowerrulesince c = cx0

QuestionIf f′(x) = 0 is f necessarilyaconstantfunction?

I ItseemstrueI Butsofarnotheorem(thatwehaveproven)usesinformationaboutthederivativeofafunctiontodetermineinformationaboutthefunctionitself

. . . . . .

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.

. . . . . .

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.

. . . . . .

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.

. . . . . .

TheoremSuppose f and g aretwodifferentiablefunctionson (a,b) withf′ = g′. Then f and g differbyaconstant. Thatis, thereexistsaconstant C suchthat f(x) = g(x) + C.

Proof.

I Let h(x) = f(x) − g(x)I Then h′(x) = f′(x) − g′(x) = 0 on (a,b)

I So h(x) = C, aconstantI Thismeans f(x) − g(x) = C on (a,b)

. . . . . .

TheoremSuppose f and g aretwodifferentiablefunctionson (a,b) withf′ = g′. Then f and g differbyaconstant. Thatis, thereexistsaconstant C suchthat f(x) = g(x) + C.

Proof.

I Let h(x) = f(x) − g(x)I Then h′(x) = f′(x) − g′(x) = 0 on (a,b)

I So h(x) = C, aconstantI Thismeans f(x) − g(x) = C on (a,b)

. . . . . .

MVT anddifferentiability

ExampleLet

f(x) =

{−x if x ≤ 0

x2 if x ≥ 0

Is f differentiableat 0?

. . . . . .

MVT anddifferentiability

ExampleLet

f(x) =

{−x if x ≤ 0

x2 if x ≥ 0

Is f differentiableat 0?

Solution(fromthedefinition)Wehave

limx→0−

f(x) − f(0)

x− 0= lim

x→0−

−xx

= −1

limx→0+

f(x) − f(0)

x− 0= lim

x→0+

x2

x= lim

x→0+x = 0

Sincetheselimitsdisagree, f isnotdifferentiableat 0.

. . . . . .

MVT anddifferentiability

ExampleLet

f(x) =

{−x if x ≤ 0

x2 if x ≥ 0

Is f differentiableat 0?

Solution(Sortof)If x < 0, then f′(x) = −1. If x > 0, then f′(x) = 2x. Since

limx→0+

f′(x) = 0 and limx→0−

f′(x) = −1,

thelimit limx→0

f′(x) doesnotexistandso f isnotdifferentiableat 0.

. . . . . .

I Thissolutionisvalidbutlessdirect.I Weseemtobeusingthefollowingfact: If lim

x→af′(x) doesnot

exist, then f isnotdifferentiableat a.I equivalently: If f isdifferentiableat a, then lim

x→af′(x) exists.

I Butthis“fact”isnottrue!

. . . . . .

Differentiablewithdiscontinuousderivative

Itispossibleforafunction f tobedifferentiableat a eveniflimx→a

f′(x) doesnotexist.

Example

Let f′(x) =

{x2 sin(1/x) if x ̸= 0

0 if x = 0. Thenwhen x ̸= 0,

f′(x) = 2x sin(1/x)+x2 cos(1/x)(−1/x2) = 2x sin(1/x)−cos(1/x),

whichhasnolimitat 0. However,

f′(0) = limx→0

f(x) − f(0)

x− 0= lim

x→0

x2 sin(1/x)x

= limx→0

x sin(1/x) = 0

So f′(0) = 0. Hence f isdifferentiableforall x, but f′ isnotcontinuousat 0!

. . . . . .

MVT totherescue

LemmaSuppose f iscontinuouson [a,b] and lim

x→a+f′(x) = m. Then

limx→a+

f(x) − f(a)x− a

= m.

Proof.Choose x near a andgreaterthan a. Then

f(x) − f(a)x− a

= f′(cx)

forsome cx where a < cx < x. As x → a, cx → a aswell, so:

limx→a+

f(x) − f(a)x− a

= limx→a+

f′(cx) = limx→a+

f′(x) = m.

. . . . . .

MVT totherescue

LemmaSuppose f iscontinuouson [a,b] and lim

x→a+f′(x) = m. Then

limx→a+

f(x) − f(a)x− a

= m.

Proof.Choose x near a andgreaterthan a. Then

f(x) − f(a)x− a

= f′(cx)

forsome cx where a < cx < x. As x → a, cx → a aswell, so:

limx→a+

f(x) − f(a)x− a

= limx→a+

f′(cx) = limx→a+

f′(x) = m.

. . . . . .

TheoremSuppose

limx→a−

f′(x) = m1 and limx→a+

f′(x) = m2

If m1 = m2, then f isdifferentiableat a. If m1 ̸= m2, then f isnotdifferentiableat a.

Proof.Weknowbythelemmathat

limx→a−

f(x) − f(a)x− a

= limx→a−

f′(x)

limx→a+

f(x) − f(a)x− a

= limx→a+

f′(x)

Thetwo-sidedlimitexistsif(andonlyif)thetworight-handsidesagree.

. . . . . .

TheoremSuppose

limx→a−

f′(x) = m1 and limx→a+

f′(x) = m2

If m1 = m2, then f isdifferentiableat a. If m1 ̸= m2, then f isnotdifferentiableat a.

Proof.Weknowbythelemmathat

limx→a−

f(x) − f(a)x− a

= limx→a−

f′(x)

limx→a+

f(x) − f(a)x− a

= limx→a+

f′(x)

Thetwo-sidedlimitexistsif(andonlyif)thetworight-handsidesagree.

. . . . . .

Whathavewelearnedtoday?

I Rolle’sTheorem: thereisastationarypointI MeanValueTheorem: atsomepointtheinstantaneousrateofchangeequalstheaveragerateofchange(TheMostImportantTheoreminCalculus)

I Onlyconstantfunctionshaveaderivativeofzero.