Question

What are the absolute extrema of `f(x)=x^3 - 3x + 1 in[0,3]`?

Answer 1

On `[0,3]`, the maximum is `19` (at `x=3`) and the minimum is `-1` (at `x=1`).

Explanation:

To find the absolute extrema of a (continuous) function on a closed interval, we know that the extrema must occur at either crtical numers in the interval or at the endpoints of the interval.

`f(x) = x^3-3x+1` has derivative

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

`3x^2-3` is never undefined and `3x^2-3=0` at `x=+-1`.

Since `-1` is not in the interval `[0,3]`, we discard it.

The only critical number to consider is `1`.

`f(0) = 1`

`f(1) = -1` and

`f(3) = 19`.

So, the maximum is `19` (at `x=3`) and the minimum is `-1` (at `x=1`).