# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for f (x) = x^3 + 6x^2?

Dec 20, 2015

Find when $f ' \left(x\right) = 0$ or $\text{DNE}$.

$f ' \left(x\right) = 3 {x}^{2} + 12 x = 3 x \left(x + 4\right)$

$f ' \left(x\right) = 0$ when $x = - 4 , 0$.
$f ' \left(x\right)$ never $\text{DNE}$.

Now, use a sign chart with $- 4 , 0$.

$f ' \left(x\right) \textcolor{w h i t e}{\times \times \times \times} - 4 \textcolor{w h i t e}{\times \times \times \times \times \times x} 0$
$\leftarrow - - - - - - - - - - - - - - - - - - \rightarrow$
$\textcolor{w h i t e}{\times \times} \text{POSITIVE"color(white)(xxxxx)"NEGATIVE"color(white)(xxxxxx)"POSITIVE}$

$f$ is increasing whenever $f ' \left(x\right) > 0$.
$f$ is decreasing whenever $f ' \left(x\right) < 0$.

Thus,

$f$ is increasing on $\left(- \infty , - 4\right) \cup \left(0 , + \infty\right)$.
$f$ is decreasing on $\left(- 4 , 0\right)$.

A relative maximum occurs whenever $f '$ switches from positive to negative.
A relative minimum occurs whenever $f '$ switches from negative to positive.

Thus,

There is a relative maximum when $x = - 4$.
There is a relative minimum when $x = 0$.

graph{x^3+6x^2 [-51.76, 65.27, -14.2, 44.35]}