# What are the critical values, if any, of f(x)= x^6ln(x) ?

Jul 4, 2018

$x = 0$ or $x = {e}^{- \frac{1}{6}}$

#### Explanation:

We use the product rule

$\left(u v\right) ' = u ' v + u v '$
and the fact that

$\left(\ln \left(x\right)\right) ' = \frac{1}{x}$

so we get

$f ' \left(x\right) = 6 {x}^{5} + {x}^{6} \cdot \frac{1}{x}$

$f ' \left(x\right) = {x}^{5} \left(6 \ln \left(x\right) + 1\right)$
solving $f ' \left(x\right) = 0$
we get

$x = 0$ or $x = {e}^{- \frac{1}{6}}$