Download - Lesson 20: The Mean Value Theorem
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. . . . . .
Section4.2TheMeanValueTheorem
V63.0121.027, CalculusI
November10, 2009
Announcements
I Quizthisweekon§§3.1–3.5
..Imagecredit: Jimmywayne22
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. . . . . .
Outline
Review: TheClosedIntervalMethod
Rolle’sTheorem
TheMeanValueTheoremApplications
WhytheMVT istheMITC
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. . . . . .
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
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. . . . . .
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.
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. . . . . .
Outline
Review: TheClosedIntervalMethod
Rolle’sTheorem
TheMeanValueTheoremApplications
WhytheMVT istheMITC
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. . . . . .
HeuristicMotivationforRolle’sTheorem
Ifyoubikeupahill, thenbackdown, atsomepointyourelevationwasstationary.
.
.Imagecredit: SpringSun
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. . . . . .
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
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. . . . . .
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
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
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.
![Page 14: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/14.jpg)
. . . . . .
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.
![Page 15: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/15.jpg)
. . . . . .
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.
![Page 16: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/16.jpg)
. . . . . .
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
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. . . . . .
Outline
Review: TheClosedIntervalMethod
Rolle’sTheorem
TheMeanValueTheoremApplications
WhytheMVT istheMITC
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. . . . . .
HeuristicMotivationforTheMeanValueTheorem
Ifyoudrivebetweenpoints A and B, atsometimeyourspeedometerreadingwasthesameasyouraveragespeedoverthedrive.
.
.Imagecredit: ClintJCL
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. . . . . .
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
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. . . . . .
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
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. . . . . .
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
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. . . . . .
Rollevs. MVT
f′(c) = 0f(b) − f(a)
b− a= f′(c)
. ..a
..b
..c
. ..a
..b
..c
Ifthe x-axisisskewedthepictureslookthesame.
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. . . . . .
Rollevs. MVT
f′(c) = 0f(b) − f(a)
b− a= f′(c)
. ..a
..b
..c
. ..a
..b
..c
Ifthe x-axisisskewedthepictureslookthesame.
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. . . . . .
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
.
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. . . . . .
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
.
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. . . . . .
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
.
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. . . . . .
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
.
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. . . . . .
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
.
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
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.
![Page 32: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/32.jpg)
. . . . . .
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.
![Page 33: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/33.jpg)
. . . . . .
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|
![Page 34: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/34.jpg)
. . . . . .
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|
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. . . . . .
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))
![Page 36: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/36.jpg)
. . . . . .
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))
![Page 37: Lesson 20: The Mean Value Theorem](https://reader033.vdocuments.mx/reader033/viewer/2022052523/5561b5e0d8b42a46138b4af2/html5/thumbnails/37.jpg)
. . . . . .
QuestionA drivertravelsalongtheNewJerseyTurnpikeusingEZ-Pass. ThesystemtakesnoteofthetimeandplacethedriverentersandexitstheTurnpike. A weekafterhistrip, thedrivergetsaspeedingticketinthemail. Whichofthefollowingbestdescribesthesituation?
(a) EZ-Passcannotprovethatthedriverwasspeeding
(b) EZ-Passcanprovethatthedriverwasspeeding
(c) Thedriver’sactualmaximumspeedexceedshisticketedspeed
(d) Both(b)and(c).
Bepreparedtojustifyyouranswer.
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. . . . . .
QuestionA drivertravelsalongtheNewJerseyTurnpikeusingEZ-Pass. ThesystemtakesnoteofthetimeandplacethedriverentersandexitstheTurnpike. A weekafterhistrip, thedrivergetsaspeedingticketinthemail. Whichofthefollowingbestdescribesthesituation?
(a) EZ-Passcannotprovethatthedriverwasspeeding
(b) EZ-Passcanprovethatthedriverwasspeeding
(c) Thedriver’sactualmaximumspeedexceedshisticketedspeed
(d) Both(b)and(c).
Bepreparedtojustifyyouranswer.
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. . . . . .
Outline
Review: TheClosedIntervalMethod
Rolle’sTheorem
TheMeanValueTheoremApplications
WhytheMVT istheMITC
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. . . . . .
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
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. . . . . .
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
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. . . . . .
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
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. . . . . .
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
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. . . . . .
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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. . . . . .
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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. . . . . .
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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. . . . . .
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)
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. . . . . .
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)
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. . . . . .
MVT anddifferentiability
ExampleLet
f(x) =
{−x if x ≤ 0
x2 if x ≥ 0
Is f differentiableat 0?
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
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!
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. . . . . .
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!
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
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.
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. . . . . .
Whathavewelearnedtoday?
I Rolle’sTheorem: thereisastationarypointI MeanValueTheorem: atsomepointtheinstantaneousrateofchangeequalstheaveragerateofchange(TheMostImportantTheoreminCalculus)
I Onlyconstantfunctionshaveaderivativeofzero.