Calculate all extremums and decide whether they are global or not: $f(x)=x^{2} \cdot e^{-x}$ where $x \in \mathbb{R}$
I have of course started with derivatives (I have checked and they are correct!):
$$f'(x) = e^{-x}(2x-x^{2})$$
$$f''(x) = e^{-x}(x^{2}-4x+2)$$
$f'(x)=0$
$0=e^{-x}(2x-x^{2})$
$0=2x-x^{2}$
$0=x(2-x)$
$x_{1}=0$ $,$ $f''(0)=2$ $\Rightarrow$ minimum at $P(0|0)$
$x_{2}=2$ $,$ $f''(2)=-0,27$ $\Rightarrow$ maximum $Q(2|\frac{4}{e^{2}})$
So now comes the exciting part for me (I believe till here it is correct):
$$\lim_{x\rightarrow\infty}x^{2} \cdot e^{-x}= 0$$
$$\lim_{x\rightarrow -\infty}x^{2} \cdot e^{-x}= \infty$$
From this I can see that the minimum we calculated must be global minimum.
The maximum we calculated is local (so not global) because we are in $x \in \mathbb{R}$ and we know from above that there are greater maximums.
Please tell me if I did it right, I need to do that in the exam and I would do it like here?
You are correct. You can also note that $f(x)\ge 0 \quad \forall x\in \mathbb{R}$, so $f(0)=0$ is a global minimum.