# How do use the first derivative test to determine the local extrema f(x)=x^3-2x +pi ?

Sep 15, 2015

See the explanation.

#### Explanation:

Find the critical numbers for $f$.

$f \left(x\right) = {x}^{3} - 2 x + \pi$

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

$f '$ is never undefined and is $f ' \left(x\right) = 0$ at $x = \pm \frac{\sqrt{6}}{3}$

On $\left(- \infty , - \frac{\sqrt{6}}{3}\right)$, we get $f ' \left(x\right)$ is positive
on $\left(- \frac{\sqrt{6}}{3} , \frac{\sqrt{6}}{3}\right)$, we get $f ' \left(x\right)$ is negative.

So $f$ changes from increasing to decreasing as we move to the left past $x = - \frac{\sqrt{6}}{3}$.
Therefore, $f \left(- \frac{\sqrt{6}}{3}\right)$ is a local maximum.

Recall: on $\left(- \frac{\sqrt{6}}{3} , \frac{\sqrt{6}}{3}\right)$, we get $f ' \left(x\right)$ is negative.
now, on $\left(\frac{\sqrt{6}}{3} , \infty\right)$, we get $f ' \left(x\right)$ is positive.
So $f$ changes from decreasing to increasing as we move to the left past $x = \frac{\sqrt{6}}{3}$.
Therefore, $f \left(\frac{\sqrt{6}}{3}\right)$ is a local minimum.

Calculating (simplifying) $f \left(- \frac{\sqrt{6}}{3}\right)$ and $f \left(\frac{\sqrt{6}}{3}\right)$ is left to the student.