Find $f:\mathbb{Q} \rightarrow \mathbb{R}$ where the mean value theorem fails

Question

Find a differentiable function $f:\mathbb{Q} \rightarrow \mathbb{R}$ that is not monotonically increasing, but still fullfills $f'(x) \geq 0$ for all $x \in \mathbb{Q}$.

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The background of this problem is that functions from the rationals to the reals have different properties than the ones analyzed in real analysis. According to the exercise sheet, differentiability still implies continuity in a non-complete vector space, but the mean value theorem does not hold (hence the problem above).

I blindly tried few functions, but I don't quite get why the mean value theorem shouldn't hold. Thank you for helping me.

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Source: This problem is taken from my analysis course. However, this was not homework and I'm going back the exercise sheets to prepare for the exam. Link for anyone interested (German), the problem is "Aufgabe 1.B (a)": https://www.mathematik.hu-berlin.de/~wendl/Sommer2019/Analysis2/Blatt1.pdf

Answer 1

Just take $f(x)=\dfrac1{\sqrt2-x}$. That will work.

Answer 2

The easiest example might be

$$ f(x)=\begin{cases} 1, & x^2<2,\\ 0, & x^2>2.\end{cases}$$

Answer 3

Try $$f(x) = \begin{cases} 1 & \text{if $x < \sqrt{2}$} \\ 0 & \text{if $x > \sqrt{2}$} \end{cases} $$ This not only satisfies your criterion, but $f'(x)=0$ for all $x \in \mathbb Q$, and yet $f$ is not constant!.