please help. Thanks for all the help rendered before.

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embizze eNotes educator| Certified Educator

We are asked to find the relative extrema for f(x)=x^4-8x^3+8:

This function is everywhere continuous and infinitely differentiable, so the extrema occur where the first derivative is zero.




x=0 or 6

For x<0 f'(x)<0

For 0<x<6 f'(x)<0

For x>6 f'(x)>0

The only sign change is at x=6; since the derivative changes from negative to positive there is an absolute minimum at x=6.

There is an inflection point at x=0: f''(x)=12x^2-48x



x=0 or 4. The sign of the second derivative is positive for x<0, negative for 0<x<4, and positive for x>4.


There is an absolute minimum at x=6; there are inflection points when x=0 and x=4.


The graph:

gsarora17 eNotes educator| Certified Educator

Answer to the Question 2 

Absolute extrema of the function that is continuous on a closed interval , can be found by finding the critical numbers that are in the interval and then evaluating the function at the end points and at the critical numbers.



So to find the critical numbers,


`sinx=0` General solution for `sinx=0 : x=0+2npi , x=pi+2npi`

`x=2pin , x=pi+2pin`

Solutions for the range `0<=x<=2pi : x=0 , x=pi , x=2pi`

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`y(0)=5 , y(pi)=-5 , y(2pi)=5`

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Absolute maxima (0,5) , (2pi,5)

Absolute minima (pi,-5)


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