Minimum, maximum and $||f_n||_\infty$ of $f_n(x) = x^{\frac{n}{2}}e^{-nx}$

Question

$n \in \mathbb{N}, f_n: D=[0, +\infty)\rightarrow \mathbb{R}$
$f_n(x) = x^{\frac{n}{2}}e^{-nx}$

I need to determine the minimium and maximum of $f_n(x)$. I would say the minimum is 0 and the maximum is $f_1(0.5)$. Is that correct? Furthermore I need to determine $||f_n||_\infty$. Shouldn't that be the same as the maximum?

Thanks for your help!

Answer 1

$$ f'_n\left(x\right)=\frac{n}{2}x^{(n-2)/2}e^{-nx}+x^{n/2}\left(-ne^{-nx}\right)=e^{-nx}\left(\frac{n}{2}x^{(n-2)/2}-nx^{n/2}\right)$$ Hence for all $x \in \mathbb{R}^{+}$ $$ f'_n\left(x\right)=ne^{-nx}x^{n/2-1}\left(\frac{1}{2}-x\right)$$ $$ f'_n\left(x\right)=0 \Leftrightarrow x=\frac{1}{2}$$ Then

$$ \left\|f_n\right\|_{\infty,\mathbb{R}^{+}}=f_n\left(\frac{1}{2}\right)=\frac{1}{\left(2e\right)^{n/2}} $$

So i think you were right, however checking for $f_1$ could not be the right solution because the maximum possibly depends on $n$. ( it could have been reached for example when $x=n/2$ ).

As you stated, the minimum is $0$ which is reached when $x=0$ and is also the limit of $f_n$ as $x \rightarrow +\infty$.