Question

How do you find the critical points for `f(x)=x^(8)` and the local max and min?

Answer 1

Check below.

Explanation:

`f(x)=x^8`

`D_f=RR`

`f` is continuous in `RR`

`f'(x)=8x^7` and

`f'(x)=0 <=> x=0`
`x_0=0` is a critical point.
There is no other critical point because `f'` is defined in `RR`.

If `x<0` then `f'(x)<0` so `f` is decreasing in `(-oo,0]`
If `x>0` then `f'(x)>0` so `f` is increasing in `[0,+oo)`

• `0` `<` `x` `<=> f(x)>f(0)=0`

• `x>=0` `<=> f(x)>=f(0)=0`

So for `x` `in` `RR`: `f(x)>=f(0)`

Therefore `f` has global minimum at `x_0=0` , `f(0)=0`