# Question #5faa3

Feb 24, 2015

I'm going to assume you mean the open intervals
for which
$f \left(x\right)$ is increasing, and
$f \left(x\right)$ is decreasing.
(The open interval of $f \left(x\right)$ is ($- \infty , + \infty$)

If $f \left(x\right) = 12 x - {x}^{3}$
then the critical points (where the function changes from increasing to decreasing, or visa versa) occur when
$\frac{d f \left(x\right)}{\mathrm{dx}} = 0$

$\frac{d \left(12 x - {x}^{3}\right)}{\mathrm{dx}} = 12 - 3 {x}^{2}$

So the critical points occur when
$3 {x}^{2} = 12$ or $x = \pm 2$

When $x = - 2$
$f \left(x\right) = 12 \left(2\right) - {\left(- 2\right)}^{3} = 32$

When $x = 2$
$f \left(x\right) = 12 \left(2\right) - {\left(2\right)}^{3} = 16$

Therefore, for $x$ in the open interval ($- 2 , + 2$), f(x) is decreasing.

It follows that for $x$ in the open intervals ($- \infty , - 2$) and ($2 , + \infty$), f(x) is increasing.

Hope this is what you were looking for.