# How do you find the critical numbers of f(x) = (4 x - 6)e^{-6 x} ?

May 5, 2017

The only critical number is $\frac{5}{3}$

#### Explanation:

A critical number for $f$ is a number, $c$, in the domain of $f$ with $f ' \left(c\right) = 0$ or f'(c)# does not exist.

For $f \left(x\right) = \left(4 x - 6\right) {e}^{- 6 x}$ the domain is $\left(- \infty , \infty\right)$.

$f ' \left(x\right) = 4 {e}^{- 6 x} + \left(4 x - 6\right) \left(- 6 {e}^{- 6 x}\right)$

$= 4 {e}^{- 6 x} - 24 x {e}^{- 6 x} + 36 {e}^{- 6 x}$

$= - 24 x {e}^{- 6 x} + 40 {e}^{- 6 x}$

$= - 8 {e}^{- 6 x} \left(3 x - 5\right)$

$f ' \left(x\right)$ is defined for all real $x$ and $f ' \left(x\right) = 0$ at $x = \frac{5}{3}$

The only critical number is $\frac{5}{3}$