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1x on [-1,1f(x) 2x2 cos

Question

Determine the location and value of extreme values of f on the given interval, if they exist.

 

Type exact answers, using pi as needed.

1
x on [-1,1
f(x) 2x2 cos
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1 x on [-1,1 f(x) 2x2 cos

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check_circleAnswer
Step 1

Given information:

The given function is  

-1x
f(x)= 2x2+2coslx
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-1x f(x)= 2x2+2coslx

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Step 2

Calculation:

First, we have to find the derivative of the given function

 

d
f (x)=212 +2cos1y
dx
4x+2cos(x)
dx
2
=4x
VI-2
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d f (x)=212 +2cos1y dx 4x+2cos(x) dx 2 =4x VI-2

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Step 3

Now, we have to find the critical ...

f '(x)=0
2
= 0
4x-
1-x
41
-xx-1= 0
2
=12
41
16x2-1641
Assume u= x*
-16u16u-1=0
+,/162-4(-16)(-1)
-16t
,2
2(-16
2- 3
2+3
4
4
-3
2-3
2+
v2+y3
2
=-
=-
2
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f '(x)=0 2 = 0 4x- 1-x 41 -xx-1= 0 2 =12 41 16x2-1641 Assume u= x* -16u16u-1=0 +,/162-4(-16)(-1) -16t ,2 2(-16 2- 3 2+3 4 4 -3 2-3 2+ v2+y3 2 =- =- 2

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Calculus

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