Studying properties of an integral $f:\mathbb R \to \mathbb R, \: f(x) =\int _0^x\:e^{-t^3}\left(t^3+t^2-t-1\right)dt$
Firstly, I have to find $min(f(x))$. The possible answers are:
A) $f(0)$
B) $f(1)$
C) $f(-1)$
D) $f(3)$
E) $f$ tends to $- \infty$
In order to do that, I have rewritten the function like this
$f\left(x\right)=\int _0^x\:e^{-t^3}\left(t+1\right)^2\left(t-1\right)dt$
and I studied the sign of the function $e^{-t^3}\left(t+1\right)^2\left(t-1\right)$ but there was nothing conclusive. Can someone teach me how to think this kind of problem ?
Secondly, for the same function, I have to mark the correct answer for" $G_f$ has":
A) a horisontal asymptote to $\infty$
B) asymptotes to $\infty$ and $-\infty$
C) a horisontal asymptote and an oblique asymptote
D) two oblique asymptotes
E) no asymptotes
Any hint to this problems would be greatly appreciated.
 A: Hint $1$: The fundamental theorem of calculus states that
$$\int_a^b g(x)\,\mathrm dx = G(b) - G(a)$$
for a primitive $G$ of $g$. Use it to find $f'(x)$. What can you conclude now?
Hint $2$: To find the asymptotes, visualize the graph of the integrand function. This should give you an idea for $x \to -\infty$. For the other side, it boils down to the improper integral $\lim\limits_{x \to +\infty} f(x)$. Does it converge?
A: Apply Newton-Leibnitz theorem. Then note that max or min is achieved at $0$ for derivative . Now differentiate the derivative to get second derivative by product rule. See for which of the root of first derivative the value of second derivative is positive(because thats the condition for minima. Hope you know why)
A: The derivative is $f'(x)=e^{-x^3}(x+1)^2(x-1)$ so it only vanishes at $1$ and $-1$.
However $f$ is decreasing on $(-\infty,1]$ and increasing on $[1,\infty)$, because the derivative is positive in a neighborhood of $-1$ (except only at $-1$).
For the question about asymptotes, note that
$$
\lim_{x\to\infty}f'(x)=0
$$
and
$$
\lim_{x\to-\infty}f'(x)=-\infty
$$
