For What Valus of $a \in \mathbb{N}$ Does $y=x^a e^x$ Has Only One Extreme Point?

Question

The solution is every odd $a$, but I have a little bit of trouble getting it.

Here's what I got: $$y=x^a e^x\\ y'= ax^{a-1}e^x+x^ae^x \\ y' = e^x(ax^{a-1}+x^a) = 0\\ ax^{a-1}+x^a=0\\ \frac{ax^a}{x}+x^a=0\\ x^a(\frac{a}{x}+1)=0\\ $$ $x=0, -a$ are the suspicious points. It seemes that the "only one extreme point" is $0$, which means that $x=-a$ should not be an extreme point in some cases.

This itself is weird because $y'$ is not defined at $0$...

How do I continue from there? Help would be appreciated.

Thank you.

Edit: in "extreme point" I mean local maximum or minimum and not inflection point.

Answer 1

Better if you write $y' =e^xx^{a-1}(a+x) $.

Then $y' = 0$ at $x = -a$.

For $y'$ to have another root, it must be at $x = 0$ with $a > 1$.

I don't know where your answer of "every odd $a$" came from, but consider $a = 2$.

Then $y = x^2e^x$ so $y' =e^xx(x+2) $ which is zero at $x = -2$ and $x = 0$