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March 29, 2015

March 29, 2015

Posted by **Cady** on Tuesday, April 5, 2011 at 11:07pm.

f(x)=(x−2)(x−5)^3+11 on each of the indicated intervals.

Enter -1000 for any absolute extrema that does not exist.

(A) Interval = [1,4]

Absolute maximum =

Absolute minimum =

(B) Interval = [1,8] .

Absolute maximum =

Absolute minimum =

(C) Interval = [4,9] .

Absolute maximum =

Absolute minimum =

My answers are:

(A) Interval = [1,4]

Absolute maximum = 9

Absolute minimum = -5

(B) Interval = [1,8] .

Absolute maximum = 173

Absolute minimum = -5

(C) Interval = [4,9] .

Absolute maximum = 459

Absolute minimum = -1000

Only the max for the second one and the max for the third one are right.... I don't know what to do....

- Calculus (pleas help!!!) -
**Mgraph**, Wednesday, April 6, 2011 at 11:05amf'(x)=(x-5)^3+(x-2)3(x-5)^2=(4x-11)(x-5)^2

(A) Absmax=f(1)=64+11=75

Absmin=f(11/4)=-3^7/2^8+11=-2187/256+

+11=629/256

(B) Absmax=f(8)=173

Absmin=f(11/4)=629/256

(C) Absmax=f(9)=459

Absmin=f(4)=9

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