minimum point of $x^2e^{-x}$ I have looking for the minimum point of $$f(x)=x^2e^{-x}$$
I differentiated once and got $f'(x)=-e^{-x}(-2+x)x$ so $x=2$ and $x=0$ can be min/max points.
I have differentiated again and got $f''(x)=e^{-x}(2-4x+x^2)$ then I plugged in $x=2$ and got $\frac{-2}{e^2}$ so it is maximum.
But in the graph it does not seems like a maximum point.
moreover How do I prove that the function is not bounded from above?
 A: For all $x\neq 0$, $$f(x)=x^2e^{-x}>0,$$
and $f(0)=0$, therefore it's the minimum ;-)
A: If $f''$ is negative at a point where $f'$ is $0$, then that is a local maximum point.
The reason for that should appeal to common sense: $f''<0$ means $f'$ is decreasing.  At a point where $f'=0$, that means $f'$ changes from positive to negative.  Think about what a curve with a positive slope looks like, and what a curve with a negative slope looks like, and you'll see it.
You can get all this (at least in the case of this particular function) from $f'$ alone.  You have $f'(x)=e^{-x}x(2-x)$.  Since $e$ raised to a real poser is always positive, look at $x(2-x)$ and see that it's positive when $x$ is between $0$ and $2$, and negative when $x<0$ or $x>2$.  That means $f$ decreases on $(-\infty,0]$, increases on $[0,2]$, and decreases on $[2,\infty)$. Therefore you have a minimum at $x=0$ and a maximum at $x=2$.
Since $f(0)=0$ and $f(x)>0$ elsewhere, that is a global minimum (and in fact you can see that without even using derivatives).  With a bit more work, you can see that $f(x)\to+\infty$ as $x\to-\infty$, so there is no global maximum.
A: How do I prove that the function is not bounded from above?
One answer (probably, the simplest one) was given in the @PrinceM's comment. Here is other approach:
Let $g(x)=e^x$ and $h(x)=-x$. Then, $g'(x)=e^x>0$ and $h(x)=-1<0$ for all $x$. So, $g$ is strictly increasing and $h$ is strictly decreasing. Since 
$g(-1)=\frac{1}{e}<1=h(-1)$ we conclude that
$$\begin{align}
g(x)&<h(x),\qquad &&\forall\ x<-1\\\\
e^x&<-x,\qquad &&\forall\ x<-1\\\\
1&<-xe^{-x},\qquad &&\forall\ x<-1\\\\
x&>-x^2e^{-x},\qquad &&\forall\ x<-1\\\\
-x&<x^2e^{-x},\qquad &&\forall\ x<-1\\\\
h(x)&<f(x),\qquad &&\forall\ x<-1\\\\
\end{align}$$
Since $h$ is not bounded from above on $(-\infty,-1)$, the last inequality implies that $f$ is not bounded from above.
