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I don't know if what I'm doing is correct.

Let $f:\mathbb{R} \to \mathbb{R}$ be a function, $f(x)=e^{x^4-3x^2}$. Choose the correct answer:

  1. $f$ has an absolute maximum at $x=0$.
  2. $f$ has a relative maximum at $x=0$ which isn't absolute.
  3. $f$ has a absolute minimum at $x=1$ and doesn't have any relative minimum. (Does this even make sense?)
  4. $f$ doesn't have relative maximum.

What I've been doing:

Since the question doesn't tell me the interval where I have to find the relative extrema, by the options the gave me I suppose that's $I=[0,1]$.

I know that when I have to find the relative extrema, I have to find the critical points for $f$ at the given interval, but this function doesn't have any. So how do I find it? (Algebraiclly).

When I look at the function it clearly does have a relative maximum at $x=0$ (Option 2), and it's clearly not absolute as $\lim_{x \to +\infty} f(x) = +\infty$ (is finding the limit as $x \to +\infty$ correct when trying to find the absolute extrema?).

So, how do I prove that option 2 is correct without looking at the graphic?

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  • $\begingroup$ I don't understand you approach. First, you tell us that the domain of $f$ is $\mathbb R$. Then you write that “the question doesn't tell me the interval where I have to find the relative extrema”. Finally, you decree that that interval is $[0,1]$. $\endgroup$ – José Carlos Santos Nov 22 '18 at 16:49
  • $\begingroup$ Yes it's kinda confussing, I thought that the interval was $[0,1]$ by the options they gave me. So I thought that I had to find the relative extrema at that interval. $\endgroup$ – Moria Nov 22 '18 at 16:52
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HINT

We have that

$$f(x)=e^{x^4-3x^2}>0 \implies f'(x)=(4x^3-6x)e^{x^4-3x^2}=0 \implies x=0 \,\lor \, x=\pm \sqrt{\frac23}$$

then consider the sign of $f'(x)$.

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