# Find the absolute maximum and absolute minimum values of f(x) = x^(1/3) e^(−x^2/8) on the interval [−1, 4]?

Mar 17, 2017

The maximum is $\sqrt[6]{\frac{4}{3}} {e}^{- \frac{1}{6}} \approx 0.888$ (it occurs at $x = \sqrt{\frac{4}{3}}$) and the minimum is $- {e}^{- \frac{1}{8}} \approx - 0.882$ (at x=-1).

#### Explanation:

$f ' \left(x\right) = \frac{4 - 3 {x}^{2}}{12 {x}^{\frac{2}{3}} {e}^{{x}^{2} / 8}}$

$f '$ is undefined at $x = 0$ and $f ' \left(x\right) = 0$ for $x = \pm \sqrt{\frac{4}{3}}$

The critical numbers in $\left[- 1 , 4\right]$ are $0$, $\sqrt{\frac{4}{3}}$.

Evaluating, we find that

$f \left(- 1\right) \approx - 0.8825$

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

$f \left(\sqrt{\frac{4}{3}}\right) \approx 0.888$

$f \left(3\right) \approx 0.468$