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

Question

3495 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-05T13:56:01

The only critical number is $5/3$

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

For $f(x) = (4x-6)e^(-6x)$ the domain is $(-oo,oo)$ .

$f'(x) = 4e^(-6x) + (4x-6)(-6e^(-6x))$

$= 4e^(-6x)-24xe^(-6x)+36e^(-6x)$

$= -24xe^(-6x)+40e^(-6x)$

$= -8e^(-6x)(3x-5)$

$f'(x)$ is defined for all real $x$ and $f'(x) = 0$ at $x=5/3$

The only critical number is $5/3$

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