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$$f(x) = x(2+x)e^{-x}$$

I have to find $\max_{\mathbb{Q}} \{ f(x) \}$ but I don't know how to proceed.

I started with $f'(x) = (-x^2+2)e^{-x}$ whence the stationary points are given when the $x$ coordinate is $x = \pm \sqrt{2}$

Yet those points do not belong to $\mathbb{Q}$. I thought like "I hav to find the closest possible point to $\sqrt{2}$, but it doesn't exist. I can always find a closer rational number from the one I found (call it $p$) and $\sqrt{2}$.

My conclusion would be that $f(x)$ has no maximum in $\mathbb{Q}$, but it does on $\mathbb{R}$ or if we want to be fancy, it even does on $\mathbb{R}\backslash\mathbb{Q}$.

Please correct my wrongs.

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  • $\begingroup$ You are correct - there is no maximum over $\mathbb{Q}$ of that function. $\endgroup$
    – Martin
    Jan 1 at 20:08

1 Answer 1

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Yes, that maximum does not exist. Your approach is good, but I think that it can be expressed in a better way.

Let $(q_n)_{n\in\Bbb N}$ be a sequence of rational numbers such that $\lim_{n\to\infty}q_n=\sqrt2$. Then\begin{align}\lim_{n\to\infty}f(q_n)&=f\left(\sqrt2\right)\\&=\left(2+2\sqrt2\right)e^{-\sqrt2}.\end{align}But, for each $q\in\Bbb Q$, $q\ne\sqrt2$, and the maximum of $f$ is attained at $\sqrt2$ and only at that point. Therefore\begin{align}f(q)&<f\left(\sqrt2\right)\\&=\left(2+2\sqrt2\right)e^{-\sqrt2}\\&=\lim_{n\to\infty}f(q_n).\end{align}So, there is some $n\in\Bbb N$ such that $f(q)<f(q_n)$. This proves that, for each rational number $q$, there is some rational number $q^\star$ such that $f(q)<f(q^\star)$, and therefore the set $\{f(q)\mid q\in\Bbb Q\}$ has no maximum.

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    $\begingroup$ Wow, that was really enlightening on how to write a solid proof. Thank you! $\endgroup$
    – Heidegger
    Jan 1 at 20:52
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    $\begingroup$ I'm glad I could help. $\endgroup$ Jan 1 at 21:04

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