# Finding absolute and relative extrema of a function

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?

• 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 $[0,1]$. – José Carlos Santos Nov 22 '18 at 16:49
• 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. – Moria Nov 22 '18 at 16:52

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