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$f:\mathbb R \to \mathbb R, \: f(x) =\int _0^x\:e^{-t^3}\left(t^3+t^2-t-1\right)dt$

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.

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    Hint: the minimum of $f(x)$ is somewhere where $f'(x) = 0$. What is $f'(x)$? – 2017-02-23
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    For the asymptotes, note that $\int_0^\infty t^ne^{-t^3} \, dt$ exists for all $n = 0, 1, 2, \ldots$. You can see this by bounding $t^n e^{-t^3} \leq t^n e^{-t}$. But, for $t \to -\infty$, $e^{-t^3}$ gets really big. – 2017-02-23

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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?

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    How can i make an idea of how the graph looks ? – 2017-02-23
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    @Liviu Typo, I meant the graph of the integrand. Let $g(t)$ be the integrand function. Then you know that $g(t) = 0$ if and only if $t = \pm 1$. Also you know that $g(t) < 0$ for $t \in (-\infty, -1) \cup (-1, 1)$ and $g(t) > 0$ for $t \in (1, +\infty)$. Finally, $\lim\limits_{t \to -\infty} g(t) = -\infty$, while $\lim\limits_{t \to +\infty} = 0$. Now, can you guess what the left asymptote is? – 2017-02-23
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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)

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    I actually don't know. $f'(x) = 0 => x = 1$ or $x = -1$ Why should the second derivative be positive? – 2017-02-23
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    http://math.stackexchange.com/questions/480711/second-derivative-criteria-for-maxima-and-minima see here. – 2017-02-23
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    Hope you get it now. – 2017-02-23
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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 $$ \lim_{x\to-\infty}f'(x)=-\infty $$

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    What connection is between $\lim_{x\to\infty}f'(x)$ and the limit of $f(x)$ ? – 2017-02-24
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    @Liviu You can conclude that $f$ has no asymptote at $-\infty$. If it has an asymptote at $\infty$, it must be horizontal. Now you have to show that $\lim_{x\to\infty}f(x)$ is finite; hint: the function is increasing from some point on, so you have to see whether it's upper bounded in some interval $[k,\infty)$. – 2017-02-24