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4.4 Mean Value Theorem

Standards

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This was created by Keenan Xavier Lee, 2013. See my website for more information, lee-apcalculus.weebly.com.

§ Extreme Value Theorem

tffk ) is Continuous on the closed interval [ a ,D,

thenfklattainsitsAbs max

-- . . .

.

tabsmax&*snwmm* '

¥€¥t[ Exampkffindabsmnxtmiwfrftxtzs - Mx on E3D .

absm.ms#p:ftx1=zt.80xSz4.so=OfKHSz4.soS(tt- 161=0

z' -16=02-4=16

z=2,

-2.

Step# criticality endpoints

f (2) : -128 fly =3

f H)=128 ffD= -3

Slept:The abs Maxis 128 occurring at X=2& Nbs minis -128 occurringat ×=2 .

This was created by Keenan Xavier Lee, 2013. See my website for more information, lee-apcalculus.weebly.com.

n@NkanValueThemempkasenolIThemainreswttofthistpicistheMeanValueTheaom.Topovelunderstandthiscmoept.weneedtoKnWtheRoHe1sThewemBasicldeaofRo11eisthenem-lftheendpointva1uesoffareequahtslopeofthethenfmusthareanexbemevalneat-tungentlheissomenumberstricHybetnuntheM0iendpoints.p

wvidedthatfiscmhnnms•... ... .f... .• Mtheintemallfis differentiable

,# then f 'kj=o .

Ca b

Hsohol=

÷A slope might be Otniw( or more ) between numbers

•[[¥¥• hehqehowtheralnesoffare±a b ¢f'÷

•

Rolle 's Theorem F)Isuppose that I I•

§i¥EhEstM¥Eewn¥asYeEnBan⇐Then

,there must exist atleastanumberc

In [ aidsuohthatf 'H=O .

Rolle's Theorem witlleadustothe Mean Value Theorem.

This was created by Keenan Xavier Lee, 2013. See my website for more information, lee-apcalculus.weebly.com.

Meanvaluetheorenx

suppose that1

. ftx) is continuous on [a, D

2. fk ) is differentiable on (a , b)

then,

there exists a number c in ( a,b) such

that§ (d = flbtftfs

- a .

noted The Mean Value Theorem is a more general case of Rolle's Theorem

which has additional hypothesis that fk#b ) and in conclusion states

that there is a number c such that f 'H= 0 :

Example 1) Find a point c Satisfying the conclusion of MVT for fatty in

[2,8] .

FITIs f continuous on GD ? yesSecond Is f differentiable on (2 , 8) ? yes

Now, ftc) : f (b) -f

b - aLet a=2& b=8

.

so, ftfifztt = ¥6

Also,

fly = te Erik C't

.f '

Cx ) =- Ice = ¥

.

So now set f' 4 = ¥6 =) ¥ = I

- a = - 16c

2 = 16

C = ± 4.But since c= - 4 isn't in [2,8] , C= -4 is not the answer .

Therefore I

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Example D Find a point c Satisfying the conclusion of MVT frftx ) =pon [9,25] .

First Is f continuous on [0,253? yesSecond : Is f is differentiable on (9,25) ? yes

Now,

f 'C D= flb)-fCa÷

a.

Let a- 9,

b : 25. So

, ftp.#=z5f=sIAlso fH=F .

Now, f ' H Izi B

= Ifso

, Frets8=2 R

4¥Html16 = c

Therefore ,c-- 16 meeting MVT 's requirements

.