Question

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

Answer 1

The only critical number is `5/3`

Explanation:

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`