# How do you find the local extremas for f(x)=x^(1/3)(x+8)?

May 16, 2016

Has a local minimum at $x = - 2$

#### Explanation:

The critical points are obtained by solving
$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{4 \left(2 + x\right)}{3 {x}^{\frac{2}{3}}} = 0$
Solving for $x$ we get $x = - 2$
The critical point qualification is done by calculating ${d}^{2} / {\left(\mathrm{dx}\right)}^{2} f \left(x\right) = \frac{4 \left(- 4 + x\right)}{9 {x}^{\frac{5}{3}}}$
So ${d}^{2} / {\left(\mathrm{dx}\right)}^{2} f \left(- 2\right) = 0.839947$
Then the critical point is a minimum.
If you where using a symbolic processor be aware of
${x}^{\frac{1}{3}} \equiv \frac{x}{\left\mid x \right\mid} {\left\mid x \right\mid}^{\frac{1}{3}}$ and ${x}^{\frac{5}{3}} \equiv \frac{x}{\left\mid x \right\mid} {\left\mid x \right\mid}^{\frac{5}{3}}$