# How do you solve x^3<9x?

Jul 24, 2016

$\textcolor{g r e e n}{x \in \left(0 , 3\right)}$ or $\textcolor{g r e e n}{x \in \left(- \infty , - 3\right)}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{3} < 9 x$

Note that the equation can not be true if $x = 0$

Case 1: $\textcolor{b l a c k}{x > 0}$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} < 9$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x < 3$
therefore $x \in \left(0 , 3\right)$

Case 2: $\textcolor{b l a c k}{x < 0}$
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} > 9$ (remember when dividing by a negative the inequality is reversed)
$\textcolor{w h i t e}{\text{XXX}} \left\mid x \right\mid > 3$
and since $x < 0$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow x < - 3$
therefore $x \in \left(- \infty , - 3\right)$