Question

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

Answer 1

Absolute minimum of `-1` at `x=1` and an absolute maximum of `19` at `x=3`.

Explanation:

There are two candidates for the absolute extrema of an interval. They are the endpoints of the interval (here, `0` and `3`) and the critical values of the function located within the interval.

The critical values can be found by finding the function's derivative and finding for which values of `x` it equals `0`.

We can use the power rule to find that the derivative of `f(x)=x^3-3x+1` is `f'(x)=3x^2-3`.

The critical values are when `3x^2-3=0`, which simplifies to be `x=+-1`. However, `x=-1` is not in the interval so the only valid critical value here is the one at `x=1`. We now know that the absolute extrema could occur at `x=0,x=1,` and `x=3`.

To determine which is which, plug them all into the original function.

`f(0)=1`
`f(1)=-1`
`f(3)=19`

From here we can see that there is an absolute minimum of `-1` at `x=1` and an absolute maximum of `19` at `x=3`.

Check the function's graph:

graph{x^3-3x+1 [-0.1, 3.1, -5, 20]}