Local maxima in $f(x) = e^{-|x|}|x^2-1|$

Question

I am asked to determine whether this function has 0, 1, 2, or 3 local maximum points: $$f(x) = e^{-|x|}|x^2-1|$$

I took a look at the function and it seemed to me like it could have a local maximum in $x = 0$. My reasoning: $$f(0) = 1$$ and $$\lim_{x \to \pm\infty}f(x) = 0$$

so there had to be a maximum there. However, after looking at the plot, there turned out to be two more local maxima. I have no idea how I would have been supposed to find that. Anyone could please guide me on this? Thank you!

Answer 1

Let $g(x)=e^{-x}(x^2-1)$. Then $g'(x)=e^{-x}(1+2x-x^2)$, which is positive in $\left[0,1+\sqrt2\right)$ and negative in $\left(1+\sqrt2,\infty\right)$. So, $g$ increases in $\left[0,1+\sqrt2\right)$ and decreases in $\left(1+\sqrt2,\infty\right)$. But $\bigl(\forall x\in[0,\infty)\bigr):f(x)=\bigl\lvert g(x)\bigr\rvert$ and $g(x)$ is negative if $x\in[0,1)$ and positive if $x\in(1,\infty)$. So:

• $f(0)=1$;
• $f$ decreases on $[0,1)$;
• $f$ increases on $\left(1,1+\sqrt2\right)$;
• $f$ decreases on $\left(1+\sqrt2,\infty\right)$.

So, $f$ has two local maxima on $[0,\infty)$: at $0$ and at $1+\sqrt2$. Since it is an even fuction, it has three local maxima: at $\pm\left(1+\sqrt2\right)$ and at $0$:

Answer 2

A perfectly valid proof would be to plot it and count, since the question presents finitely many alternatives.