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

May 25, 2016

On $\left[0 , 3\right]$, 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.

$f \left(x\right) = {x}^{3} - 3 x + 1$ has derivative

$f ' \left(x\right) = 3 {x}^{2} - 3$.

$3 {x}^{2} - 3$ is never undefined and $3 {x}^{2} - 3 = 0$ at $x = \pm 1$.

Since $- 1$ is not in the interval $\left[0 , 3\right]$, we discard it.

The only critical number to consider is $1$.

$f \left(0\right) = 1$

$f \left(1\right) = - 1$ and

$f \left(3\right) = 19$.

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