Finding all critical points of $g(x,y) = (x^2 + y^2)e^{-x}$

Question

Find all critical points of the equation: $$g(x,y) = (x^2 + y^2)e^{-x}$$

For this, I found the gradient of $g(x,y)$ which turned out to be $<(-x^2 + 2x -y^2)e^{-x}$ and $2ye^{-x}>$.

I set $2ye^{-x} = 0$ and I got $y = 0$. I'm not sure if I was supposed to do that but I plugged that into $$(-x^2 + 2x - y^2) e^{-x} = 0 \implies-x^2e^{-x} = -2xe^{-x} \implies x=2$$
I'm not sure if I did this right but I ended up getting $(2,0)$ as a critical point.

Just wanted to verify that I got this right or not.

Answer 1

First we can rewrite $g(x,y) =( x^2+y^2)e^{-x} = x^2e^{-x}+y^2e^{-x}.$

Now we compute the gradient of $g$: $\nabla g = (2xe^{-x}-x^2e^{-x}-y^2e^{-x},2ye^{-x})$ and we are looking for the points $(x,y)$ that make $\nabla g=\bar{0}$. So then

\begin{align} \nabla g = \bar{0} \iff &(i)\ \ 2xe^{-x}-x^2e^{-x}-y^2e^{-x}=0\\ &(ii) \ \ 2ye^{-x}=0 \end{align} But $e^{-x} \neq 0$ for all $x$, so then from (ii) we get that $y=0$. Now we substitute $y=0$ into (i) and get $$ 2xe^{-x}-x^2e^{-x}= 0 \iff e^{-x}x(2-x)=0 \iff x=0 \ \text{ or }\ x=2.$$

So the two critical points are $(0,0)$ and $(2,0)$.

Answer 2

Putting all the comments in an answer:

for $y=0$ the equation $(-x^2 + 2x - y^2) e^{-x} = 0 $ becomes $$ (-x^2 + 2x ) e^{-x} = 0 \quad \Rightarrow\quad (-x +2)xe^{-x}=0 \quad \Rightarrow \quad (-x +2)x=0 $$

so we have the two solutions $(0,0)$ and $(2,0)$.