Find the critical numbers
x^2 e^(-4 x)

I'm not sure if I should use both chain rule and product rule?

Question

Find the critical numbers 
x^2 e^(-4 x) 

I'm not sure if I should use both chain rule and product rule?

ganeshie8 · Answer

u = x^2
v = e^(-4x)

the given function is a `product` of two functions

ganeshie8 · Answer

you must use product rule to start with, and then use chain rule to work u' and v'

anonymous · Answer

thank you!

ganeshie8 · Answer

np :) so have you found ur critical numbers ?

mathmale · Answer

Help others help you by typing in your work here.  
Since your function is indeed a product, the derivative should have the form

(    )(    ) + (    )(     ) = 0.  Solve for the critical numbers, x.

anonymous · Answer

So the product rule says f'(x)g(x) + f(x)g'(x). So first I will take the derivative of x^2 which is 2x and multiply by g(x) which is e^-4x? @ganeshie8 I'm still working on it:)

ganeshie8 · Answer

you're right

anonymous · Answer

So then the second part of product rule says add to f(x)g'(x). Now for chain rule, my whole function is f(x) right? so it will be (2x)(e^-4x) + (x^2e^-4x)(-4) ?

ganeshie8 · Answer

looks perfect ! set that equal to 0 and solve x

ganeshie8 · Answer

$$\large (2x)e^{-4x} -4 x^2e^{-4x} = 0$$

anonymous · Answer

ok thank you. Here I get kinda stuck. Do i factor out an x?

ganeshie8 · Answer

factor out everything u can

ganeshie8 · Answer

GCF of the terms

anonymous · Answer

Oh I see! So i factored out an x and e^-4x which makes my critical points 0 and 1/2?

anonymous · Answer

thank you!!!! @ganeshie8

ganeshie8 · Answer

yes, that looks good !