Download - Lesson 17: The Mean Value Theorem and the shape of curves

Transcript
Page 1: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Section4.3TheMeanValueTheoremandtheshapeofcurves

Math1a

March14, 2008

Announcements

◮ Midtermisgraded.◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323

..Image: Flickruser Jimmywayne32

Page 2: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Announcements

◮ Midtermisgraded◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323

Page 3: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

HappyPiDay!

3:14PM Digitrecitationcontest! Reciteallthedigitsyouknowof π(inorder, please). Pleaseletusknowinadvanceifyou’llrecite π inabaseotherthan10(theusualchoice), 2, or16.Onlypositiveintegerbasesallowed–nofairtomemorize πinbase π/(π − 2)...

4PM —Pi(e)eatingcontest! Cornbreadaresquare; pieareround.Youhave3minutesand14secondstostuffyourselfwithasmuchpieasyoucan. Theleftoverswillbeweighedtocalculatehowmuchpieyouhaveeaten.

ConteststakeplaceinthefourthfloorloungeoftheMathDepartment. .

.Image: FlickruserPaulAdamSmith

Page 4: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Outline

Recall: Fermat’sTheoremandtheClosedIntervalMethod

TheMeanValueTheoremRolle’sTheorem

WhytheMVT istheMITC

TheIncreasing/DecreasingTestUsingthederivativetosketchthegraph

TestsforextemityTheFirstDerivativeTestTheSecondDerivativeTest

Page 5: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Fermat’sTheorem

DefinitionLet f bedefinednear a. a isa localmaximum of f if

f(x) ≤ f(a)

forall x inanopenintervalcontaining a.

TheoremLet f havealocalmaximumat a. If f isdifferentiableat a, thenf′(a) = 0.

Page 6: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

TheClosedIntervalMethod

Let f beacontinuousfunctiondefinedonaclosedinterval [a,b].Weareinsearchofitsglobalmaximum, callit c. Then:

◮ Either themaximumoccursatanendpointoftheinterval, i.e., c = aor c = b,

◮ Or themaximumoccursinside (a,b). Inthiscase, c isalsoalocalmaximum.

◮ Either f isdifferentiableat c, inwhichcase f′(c) = 0byFermat’sTheorem.

◮ Or f isnotdifferentiableat c.

Thismeanstofindthemaximumvalueof f on [a,b],weneedtocheck:

◮ a and b◮ Points x where f′(x) = 0◮ Points x where f isnot

differentiable.

Thelattertwoarebothcalledcriticalpoints of f. ThistechniqueiscalledtheClosedIntervalMethod.

Page 7: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

TheClosedIntervalMethod

Let f beacontinuousfunctiondefinedonaclosedinterval [a,b].Weareinsearchofitsglobalmaximum, callit c. Then:

◮ Either themaximumoccursatanendpointoftheinterval, i.e., c = aor c = b,

◮ Or themaximumoccursinside (a,b). Inthiscase, c isalsoalocalmaximum.

◮ Either f isdifferentiableat c, inwhichcase f′(c) = 0byFermat’sTheorem.

◮ Or f isnotdifferentiableat c.

Thismeanstofindthemaximumvalueof f on [a,b],weneedtocheck:

◮ a and b◮ Points x where f′(x) = 0◮ Points x where f isnot

differentiable.

Thelattertwoarebothcalledcriticalpoints of f. ThistechniqueiscalledtheClosedIntervalMethod.

Page 8: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

MeettheMathematician: PierredeFermat

◮ 1601–1665◮ Lawyerandnumber

theorist◮ Provedmanytheorems,

didn’tquiteprovehislastone

Page 9: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Outline

Recall: Fermat’sTheoremandtheClosedIntervalMethod

TheMeanValueTheoremRolle’sTheorem

WhytheMVT istheMITC

TheIncreasing/DecreasingTestUsingthederivativetosketchthegraph

TestsforextemityTheFirstDerivativeTestTheSecondDerivativeTest

Page 10: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Rolle’sTheorem

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

.a.•.b

.•.c

Proof.If f isnotconstant, ithasalocalmaximumorminimumin (a,b).Callthispoint c. ThenbyFermat’sTheorem f′(c) = 0.

Page 11: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Rolle’sTheorem

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

.a.•.b

.•.c

Proof.If f isnotconstant, ithasalocalmaximumorminimumin (a,b).Callthispoint c. ThenbyFermat’sTheorem f′(c) = 0.

Page 12: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Rolle’sTheorem

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

.a.•.b

.•.c

Proof.If f isnotconstant, ithasalocalmaximumorminimumin (a,b).Callthispoint c. ThenbyFermat’sTheorem f′(c) = 0.

Page 13: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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

Page 14: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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

Page 15: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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

Page 16: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ProofoftheMVT

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(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). Sothereexistsapoint c ∈ (a,b) suchthat

0 = g′(c) = f′(c) − f(b) − f(a)b− a

.

Page 17: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

QuestionOnatollroadadrivertakesatimestampedtoll-cardfromthestartingboothanddrivesdirectlytotheendofthetollsection.Afterpayingtherequiredtoll, thedriverissurprisedtoreceiveaspeedingticketalongwiththetollreceipt. Whichofthefollowingbestdescribesthesituation?

(a) Theboothattendantdoesnothaveenoughinformationtoprovethatthedriverwasspeeding.

(b) Theboothattendantcanprovethatthedriverwasspeedingduringhistrip.

(c) Thedriverwillgetaticketforalowerspeedthanhisactualmaximumspeed.

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

Bepreparedtojustifyyouranswer.

Answer(b)and(c).

Page 18: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

QuestionOnatollroadadrivertakesatimestampedtoll-cardfromthestartingboothanddrivesdirectlytotheendofthetollsection.Afterpayingtherequiredtoll, thedriverissurprisedtoreceiveaspeedingticketalongwiththetollreceipt. Whichofthefollowingbestdescribesthesituation?

(a) Theboothattendantdoesnothaveenoughinformationtoprovethatthedriverwasspeeding.

(b) Theboothattendantcanprovethatthedriverwasspeedingduringhistrip.

(c) Thedriverwillgetaticketforalowerspeedthanhisactualmaximumspeed.

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

Bepreparedtojustifyyouranswer.

Answer(b)and(c).

Page 19: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Outline

Recall: Fermat’sTheoremandtheClosedIntervalMethod

TheMeanValueTheoremRolle’sTheorem

WhytheMVT istheMITC

TheIncreasing/DecreasingTestUsingthederivativetosketchthegraph

TestsforextemityTheFirstDerivativeTestTheSecondDerivativeTest

Page 20: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

WhytheMVT istheMITC

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 ∈ (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.

Page 21: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

WhytheMVT istheMITC

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 ∈ (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.

Page 22: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

WhytheMVT istheMITC

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 ∈ (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.

Page 23: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Outline

Recall: Fermat’sTheoremandtheClosedIntervalMethod

TheMeanValueTheoremRolle’sTheorem

WhytheMVT istheMITC

TheIncreasing/DecreasingTestUsingthederivativetosketchthegraph

TestsforextemityTheFirstDerivativeTestTheSecondDerivativeTest

Page 24: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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 ∈ (x, y) suchthat

f(y) − f(x)y− x

= f′(c) > 0.

Sof(y) − f(x) = f′(c)(y− x) > 0.

Page 25: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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 ∈ (x, y) suchthat

f(y) − f(x)y− x

= f′(c) > 0.

Sof(y) − f(x) = f′(c)(y− x) > 0.

Page 26: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ExampleFindtheintervalsofmonotonicityof f(x) = 2/3x− 5.

Solutionf′(x) = 2/3 isalwayspositive, so f isincreasingon (−∞,∞).

Page 27: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ExampleFindtheintervalsofmonotonicityof f(x) = 2/3x− 5.

Solutionf′(x) = 2/3 isalwayspositive, so f isincreasingon (−∞,∞).

Page 28: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.

Solutionf′(x) = 2x, whichispositivewhen x > 0 andnegativewhen x is.Wecandrawanumberline:

. .f′.− ..0.0 .+

Page 29: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.

Solutionf′(x) = 2x, whichispositivewhen x > 0 andnegativewhen x is.

Wecandrawanumberline:

. .f′.− ..0.0 .+

Page 30: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.

Solutionf′(x) = 2x, whichispositivewhen x > 0 andnegativewhen x is.Wecandrawanumberline:

. .f′.− ..0.0 .+

Page 31: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ExampleFindtheintervalsofmonotonicityof f(x) = x2 − 1.

Solutionf′(x) = 2x, whichispositivewhen x > 0 andnegativewhen x is.Wecandrawanumberline:

. .f′

.f

.−.↘

..0.0 .+

.↗

Page 32: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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

SolutionWrite f(x) = x5/3 + 2x2/3. Then

f′(x) = 53x

2/3 + 43x

−1/3

= 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.×.+

.↗.−.↘

.+

.↗

Page 33: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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

SolutionWrite f(x) = x5/3 + 2x2/3. Then

f′(x) = 53x

2/3 + 43x

−1/3

= 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.×.+

.↗.−.↘

.+

.↗

Page 34: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

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

SolutionWrite f(x) = x5/3 + 2x2/3. Then

f′(x) = 53x

2/3 + 43x

−1/3

= 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.×.+

.↗.−.↘

.+

.↗

Page 35: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Outline

Recall: Fermat’sTheoremandtheClosedIntervalMethod

TheMeanValueTheoremRolle’sTheorem

WhytheMVT istheMITC

TheIncreasing/DecreasingTestUsingthederivativetosketchthegraph

TestsforextemityTheFirstDerivativeTestTheSecondDerivativeTest

Page 36: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

TheFirstDerivativeTest

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

Theorem

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

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

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

Page 37: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

TheSecondDerivativeTest

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

Theorem

◮ If f′′(c) < 0, then f(c) isalocalmaximum.◮ If f′′(c) > 0, then f(c) isalocalminimum.◮ If f′′(c) = 0, thesecondderivativeisinconclusive(thisdoes

notmean c isneither; wejustdon’tknowyet).

Page 38: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

ExampleFindthelocalextremaof f(x) = x3 − x.

Page 39: Lesson 17: The Mean Value Theorem and the shape of curves

. . . . . .

Nexttime: graphingfunctions