Showing $f(x) = xe^{-nx^2}$ is a bounded function

Question

Let $f: \Bbb{R} \rightarrow \Bbb{R}$ be defined as

$f(x) = xe^{-nx^2}$ for some natural number.

Then how can I show that $f$ is bounded.

So I need to find a real number $M$ such that $|f(x)| \le M$ for all $x \in \Bbb{R}$.

I can see the bound for $e^{-nx^2}$ is $1$.

But how do I find the bound of the whole function?

Answer 1

It is easy to see that $$\lim_{x\to\pm\infty}f(x)=0.$$ Hence, there is $x_0>0$ such that whenver $|x|>x_0$, we have $f(x)<1$. Moreover, $f\vert_{[-x_0,x_0]}$ is continuous and defined on a compact set, hence reaches some maximal value $M$. Thus $M+1$ is an upper bound for $f$.

Answer 2

Observe, for example with L'Hospital, that

$$\lim_{x\to\pm\infty}\frac x{e^{nx^2}}\stackrel{L'H}=\lim_{x\to\pm\infty}\frac1{2nxe^{nx^2}}=0$$

and since the function is continuous in $\;\Bbb R\;$ , then in any interval of the form $\;[-a,a]\;,\;\;0<a\in\Bbb R\;$, it is bounded and gets there its maximum (and minimum), by Weierstrass theorems 1 and 2.

Well, now do a little more mathematics here and deduce your result.

Answer 3

Note that $f'(x)=(1-2nx^2)e^{-nx^2}$. So, $f'(x)>0$ if $x\in\left(-\frac1{\sqrt{2n}},\frac1{\sqrt{2n}}\right)$ and $f'(x)<0$ if $x\notin\left[-\frac1{\sqrt{2n}},\frac1{\sqrt{2n}}\right]$. But then, since we also have $\lim_{x\to\pm\infty}f(x)=0$,$$\max f(x)=f\left(\frac1{\sqrt{2n}}\right)=\frac1{\sqrt{2en}}$$and$$\min f(x)=f\left(-\frac1{\sqrt{2n}}\right)=-\frac1{\sqrt{2en}}.$$