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Let $f\colon\mathbb R\to\mathbb R$ be continuously differentiable and let's say, for simplicity, that $f(0)=0$. Then by mean value theorem it's $$f(x)=f'(\xi)\cdot x \,\text{ for some } \xi \in (0, x)$$

What I wondered is: What can we tell about the $\xi$ as we change $x$? My intuition says we should at least be able to find some $\xi\equiv \xi(x)$ that varies continuously with respect to $x$.

Or isn't this necessarily the case? Thanks for any ideas.

Dario
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1 Answers1

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No. Here is a counterexample: Choose $f$ such that $$f'(x)=\begin{cases} 2x-2 & x \le 1 \\ 0 & 1 \le x \le 2 \\ 2x-4 & x\ge 2\end{cases}$$ so $$f(x)=\begin{cases} x^2-2x & x\le 1 \\ -1 & 1\le x \le 2 \\ x^2-4x+3 & x\ge 2\end{cases}$$ Then $f(3)=0$, and for $0<x<3$, $f(x)<0$, and $\xi$ has to be chosen to be $\le 1$, and for $x>3$, $f(x)>0$, so $\xi\ge 2$ for these $x$.

Florian
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  • Ah, thank you for the counterexample – Dario Jun 10 '12 at 19:47
  • What makes the counterexample work is that $f'(\xi)=0$ for some $\xi$. Now what if we exclude this and assume that $f'(\xi)\neq0$ for all $\xi$? – Oskar Limka Dec 08 '15 at 09:26
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    @OskarLimka, if $f'(\xi)\neq0$ for all $\xi$ you can apply the implicit function theorem. It guarantees the existence of a function which is at least locally continuous. – Philipp Nov 14 '21 at 22:51
  • @Philipp thanks, that's what I thought, I was trying to stimulate a positive response to the OP, i.e., adding conditions that will make their conclusion correct. My hunch is that if $f$ is analytic (i.e., expressible as a convergent power series on some open interval containing 0) then we can continuously track $\xi$ as a (possibly multivalued) function of $x$. (I would start with a polynomial $f$.) – Oskar Limka Nov 19 '21 at 15:12