Question

How do you locate the absolute extrema of the function `f(x)=x^3-12x` on the closed interval [0,4]?

Answer 1

The absolute extrema of a function, `f`, that is continuous on a closed interval must occur at either a critical number for `f` in the interval, or at an endpoint of the interval.

Explanation:

So consider:

`f(x) =x^3-12x`

`f'(x) = 3x^2-12`.

`f'` is polynomial, so every critical number for `f` is a zero for `f'`.

The zeros of `f'` are: `3(x^2-4)=0`, `x=-2, 2`. But `-2` is not in the interval `[0,4]`.
So the only critical number in the interval is `2`.

The minimum and maximum must occur at one of the values

`x=` `0` or `2` or `4`.

To finish, evaluate `f` at each of these values.

`f(0) = 0`

`f(2) = 8-24 = -16`

`f(4) = 64-48 = 16`

The minimum is `-16` (it occurs at `2`)
The maximum is `16` (it occurs at `4`).

To understand what we've done it may help to see the graph:

graph{y=(x^3-12x)*sqrt(4-(x-2)^2)/(sqrt(4-(x-2)^2)) [-52.2, 40.28, -23.87, 22.4]}