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.

f'(x)=4x^3-24x^2

4x^3-24x^2=0

4x^2(x-6)=0

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

12x^2-48x=0

12x(x-4)=0

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.

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

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The graph:

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.

`y=5cosx`

`y'=-5sinx`

So to find the critical numbers,

`-5sinx=0`

`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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