# `f(x) = x^2 e^(-3x)` Find the critical numbers of the function

You need to evaluate the critical numbers of the function, hence, you need to evaluate the soutions to the first derivative, such that:

`f'(x) = 0`

You need to find the first derivative using the product rule:

`f'(x) = (x^2)'*(e^(-3x)) + x^2*(e^(-3x))'`

`f'(x) = 2x*(e^(-3x)) + x^2*(e^(-3x))*(-3x)'`

`f'(x) = 2x*(e^(-3x))...

## Unlock This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

You need to evaluate the critical numbers of the function, hence, you need to evaluate the soutions to the first derivative, such that:

`f'(x) = 0`

You need to find the first derivative using the product rule:

`f'(x) = (x^2)'*(e^(-3x)) + x^2*(e^(-3x))'`

`f'(x) = 2x*(e^(-3x)) + x^2*(e^(-3x))*(-3x)'`

`f'(x) = 2x*(e^(-3x)) - 3x^2*(e^(-3x))`

You need to factor out `(e^(-3x)):`

`f'(x) = (e^(-3x))*(2x - 3x^2)`

You need to solve for x the equation f'(x) = 0:

`(e^(-3x))*(2x - 3x^2) = 0`

Since `e^(-3x) >0,` then `2x - 3x^2 = 0.`

You need to factor out x, such that:

`x(2 - 3x) = 0 => x = 0`

`2 - 3x = 0 => 2 = 3x => x = 2/3`

Hence, evaluating the critical values of the function yields `x = 0, x = 2/3.`

Approved by eNotes Editorial Team