Question

What are the absolute extrema of `f(x) =x^4 − 8x^2 − 12 in[-3,-1]`?

Answer 1

`-3` (occurring at `x=-3`) and `-28` (occurring at `x=-2`)

Explanation:

Absolute extrema of a closed interval occur at the endpoints of the interval or at `f'(x)=0`.

That means we'll have to set the derivative equal to `0` and see what `x`-values that gets us, and we'll have to use `x=-3` and `x=-1` (because these are the endpoints).

So, starting with taking the derivative:
`f(x)=x^4-8x^2-12`
`f'(x)=4x^3-16x`

Setting it equal to `0` and solving:
`0=4x^3-16x`
`0=x^3-4x`
`0=x(x^2-4)`
`x=0` and `x^2-4=0`
Thus the solutions are `0,2,` and `-2`.

We immediately get rid of `0` and `2` because they are not on the interval `[-3,-1]`, leaving only `x=-3,-2,` and `-1` as the possible places where extrema can occur.

Finally, we evaluate these one by one to see what the absolute min and max are:
`f(-3)=-3`
`f(-2)=-28`
`f(-1)=-19`

Therefore `-3` is the absolute maximum and `-28` is the absolute minimum on the interval `[-3,-1]`.