Multiple answer question for the function $f(x)=e^{2-x}x^2$ Considering that I don't use calculators, graphic programs that draw functions, books, let the function $f\colon \mathbb R \to \mathbb R$ defined by law
$$f(x)=e^{2-x}x^2=\frac{e^2x^2}{e^{x}} \tag 1$$
Which of the following assertions is TRUE? I have these 5 options: A: $f$ has no relative extremes;
B: $f$ restricted to $]-∞, 0[$ is increasing;
C: $f$ has two points of inflection;
D: $f$ has no asymptotes;
E: None of the previous answers is true.

This is my fast solution hoping that I not make any mistakes: for my eyes and my mind the A is false, because $f$ will have relative extremes calculating the $f'(x)$. I think that there is almost a root when $f'(x)=0$. Obviously I haven't done the calculations and put it on stand-by. The D, for my humble opinion can be true: generally I remember that if is $f$ it the typology (1) may have oblique asymptotes. It has no horizontal or vertical asymptotes. Excluding the answers that include calculations I think that the B is true: in fact
$$\lim_{x\to-\infty}f(x)=+\infty$$
and imagining to do the calculations in the mind using the Hopital theorem
$$\lim_{x\to+\infty}f(x)=0$$
Hence I not consider the C and E.
A: Note that we have that $f(0)=0$ and as you noticed $\lim_{x\to-\infty}f(x)=+\infty$, therefore of course also $B$ is false (by MVT).
Indeed since $\lim_{x\to-\infty}f(x)=+\infty$, exists $-a<0$ such that $f(-a)>0$ then $$\frac{f(0)-f(-a)}{0-(-a)}=f'(\xi)<0$$ for some $\xi \in(-a,0)$.
To analize $A$ and $C$ I think we can't avoid to take the derivatives.
