# How do you use the first derivative test to find the local max and local min values of f(x)= -x^3 + 12x?

Apr 16, 2016

For local minimum and maximum values the derivative is equal to zero.

#### Explanation:

Given $f \left(x\right) = - {x}^{3} + 12 x$

The first derivative is
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = - 3 {x}^{2} + 12$

For local minimum and maximum values
$\textcolor{w h i t e}{\text{XXX}} f ' \left(x\right) = 0$

So
$\textcolor{w h i t e}{\text{XXX}} - 3 {x}^{2} + 12 = 0$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} - 4 = 0$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = - 2 \mathmr{and} x = + 2$

If $x = - 2$
$\textcolor{w h i t e}{\text{XXX}} f \left(x = - 2\right) = - {\left(- 2\right)}^{3} + 12 \left(- 2\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = 8 - 24$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = - 16$

If $x = + 2$
$\textcolor{w h i t e}{\text{XXX}} f \left(x = + 2\right) = - {\left(+ 2\right)}^{3} + 12 \left(+ 2\right)$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = - 8 + 24$
$\textcolor{w h i t e}{\text{XXXXXXXXXX}} = + 16$

So $x = - 2$ gives a local minimum
and $x = + 2$ give a local maximum

This can be verified by checking the graph of the original function:
graph{-x^3+12x [-41.1, 41.07, -20.53, 20.55]}