# What are the critical values, if any, of  f(x)=x^3 + 6x^2 − 135x?

Nov 19, 2015

$- 9$ and $5$.

#### Explanation:

A critical number for a function $f$, is a number in the domain of $f$ at which either the derivative of $f$ is $0$ or it fails to exist.

For  f(x)=x^3 + 6x^2 − 135x, the domain is $\left(- \infty , \infty\right)$ and

$f ' \left(x\right) = 3 {x}^{2} + 12 x - 136$.

This $f ' \left(x\right)$ exists for all numbers.

$f ' \left(x\right) = 0$ at $- 9$ and at $5$, both of which are in the domain of $f$, so both are critical numbers for $f$.

To solve $f ' \left(x\right) = 0$, use the quadratic formula or factor:

$3 {x}^{2} + 12 x - 136 = 3 \left({x}^{2} + 4 x - 45\right) = 3 \left(x + 9\right) \left(x - 5\right)$