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

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Answer 1 · 2017-05-05T13:56:01

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

A critical number for $f$ $f$ is a number, $c$ $c$ , in the domain of $f$ $f$ 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)=(4x−6)e−6xf(x) = (4 x - 6)e^{-6 x} ?