Question

Let \(f(x)=x^2 e^{-x}\)

Answer 1

Find the critical points of \(f\) and identify as local min or local max. Find the inflection points.

Answer 2

I got \(\large f'(x)=-e^{-x}(x^2-2x)\)
Not sure if I solved it right tho..
I got x=0, 1, 2

Answer 3

@zepdrix ? c:

Answer 4

Your derivative looks good!
Hmm where did the x=1 solution come from?

I see the 0 and 2.

Answer 5

Meh, I don't know really either >.<

\[\large e^{-x}=0\]
Is that possible?

Answer 6

It is not! :)
The exponential function has no x-intercepts.

\(\Large\rm e^{-x} \to 0\) as \(\Large\rm x\to \infty\).
But that's not really useful here.
We're not thinking of "infinity" as an intercept.

Answer 7

So there's going to be a local min/max at 0 and 2?