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

Feb 14, 2016

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 $f \left(x\right) = {x}^{3} - 3 x + 1$ is $f ' \left(x\right) = 3 {x}^{2} - 3$.

The critical values are when $3 {x}^{2} - 3 = 0$, which simplifies to be $x = \pm 1$. 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.

$f \left(0\right) = 1$
$f \left(1\right) = - 1$
$f \left(3\right) = 19$

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]}