Calculate all local and global extremums: $f(x) = (x^{2}+x+1) \cdot e^{-x}$ where $f: \mathbb{R} \rightarrow \mathbb{R}$
I'm pretty sure that I have created the derivatives and also calculated the local extremums correctly. But I'm not sure at all about the global ones. This means I'm especially / mainly interested in knowing if I did the calculation for the global ones correctly.
$$f'(x) = e^{-x}(x-x^{2})$$
$$f'(x) = 0$$
$$e^{-x}(x-x^{2}) = 0 \Rightarrow x_{1} = 0, x_{2}= 1$$
$$f''(x) = e^{-x}(x^{2}-3x+1)$$
$$f''(0) = 1 > 0$$
$$f''(1)= -e < 0$$
$\Rightarrow$
Local minimum at $P_{1}(0|1)$
Local maximum at $P_{2}(1|\frac{3}{e})$
Now for global ones, I would say we need to check the limits for $+- \infty$:
$\lim_{x\rightarrow \infty}e^{-x}(x^{2}+x+1)= 0$ $\Rightarrow$ There is a global minimum
$\lim_{x\rightarrow -\infty}e^{-x}(x^{2}+x+1)= \infty$ $\Rightarrow$ There is a global maximum
actually it is $$f''(1)=-e^{-1}$$ your other answers are ok, the extrema are called infimum or supremum