# How do use the first derivative test to determine the local extrema f(x) = 3(x-4)^(2/3) +6?

Aug 28, 2015

$6$ is a local minimum. (It occurs at $4$.) There are no other local extrema.

#### Explanation:

$f \left(x\right) = 3 {\left(x - 4\right)}^{\frac{2}{3}} + 6$

$f ' \left(x\right) = 2 {\left(x - 4\right)}^{- \frac{1}{3}} = \frac{2}{\sqrt[3]{x - 4}}$

$f '$ is never $0$ and is not defined at $x = 4$.

$4$ is in the domain of $f$, so $4$ is a critical number for $f$.

First derivative test:

on the interval $\left(- \infty , 4\right)$, we have $f ' \left(x\right) < 0$ and

on the interval $\left(4 , \infty\right)$, we have $f ' \left(x\right) > 0$.

Therefore, $f \left(4\right)$ is a local minimum..

$f \left(4\right) = 3 {\left(\left(4\right) - 4\right)}^{\frac{2}{3}} + 6 = 6$

Because there are no other critical numbers, there are no other local extrema.