# How do you find the local max and min for x^3-2x^2-8x?

Jan 17, 2016

Take a look at the derivative.

#### Explanation:

Here, $f \left(x\right) = {x}^{3} - 2 {x}^{2} - 8 x$ so $f ' \left(x\right) = 3 {x}^{2} - 4 x - 8$.

The local max and min of $f$ will be given by the roots of $f '$.

We first calculate $\Delta = 16 - 4 \cdot 3 \cdot \left(- 8\right) = 112$ so $f '$ has 2 real roots.

By the quadratic formula, the zeros of $f '$ are given by $\frac{- b \pm \sqrt{\Delta}}{2} a$, which are here $\frac{4 \pm \sqrt{112}}{6}$.

So the critical points are $f \left(\frac{4 + \sqrt{112}}{6}\right)$ and $f \left(\frac{4 - \sqrt{112}}{6}\right)$. By evaluating them, you will see which one is a local max and which one is the local min.