Given $f\colon\mathbb{R} \to \mathbb{R}$, $f$ is differentiable on $\mathbb{R}$ and the $\lim_{x \to \infty}f(x)$ does not exists . show/prove formally that there exists $x_0 \in \mathbb{R}$ such that $f'(x_0)=0$

My strategy is showing that $f$ is not monotonic function because all monotonic functions have limits when $x \to \infty$ ( $\lim_{x \to \infty}f(x)=l$ while $l \in \mathbb{R}$ or $l=+/- \infty$ ) now i can say that there exists $x_1$ and $x_2 \in \mathbb{R}$ such that $f'(x_1)<0$ and $f'(x_2)>0$ and use the mean value for derivative function (Darboux's theorem)

Can i really say that a continuous function is not monotonic just because $\lim_{x \to \infty}f(x)$ does not exists ? or its a "one way" statement ?

does all non-monotonic continues functions have $x_0 \in \mathbb{R}$ such that $f'(x_0)=0$ ?

  • $\begingroup$ Edited because $f(x_0)'$ is the derivative of the constant $f(x_0)$. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 26 '14 at 11:11
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    $\begingroup$ What about $f(x)=2x+\sin x$? (This is with regards to the question in the first paragraph not the one in the title). $\endgroup$ – Hagen von Eitzen Feb 26 '14 at 11:11
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    $\begingroup$ @Boris, in your definition, $\lim=\infty$ counts as $\exists$ or not? $\endgroup$ – Martín-Blas Pérez Pinilla Feb 26 '14 at 11:13
  • $\begingroup$ @Martín-Blas Pérez Pinilla yes it does . so i guess its not a counterexample. and thanks for editing $\endgroup$ – Boris Morozov Feb 26 '14 at 11:14
  • $\begingroup$ This is a subquestion of question math.stackexchange.com/questions/691042/… asked only a little while ago. $\endgroup$ – Andrea Mori Feb 26 '14 at 11:21

Suppose that $f'(x) \neq 0$ for all $x$. Then $f'(x)$ is of constant sign so that either $f(x)$ is strictly increasing (in which case $\lim_{x\to\infty}f(x) = L, \infty$) or $f(x)$ is strictly decreasing (in which case $\lim_{x \to \infty}f(x) = L, -\infty$). You get a contradiction in any case. Also see my answer to one of your earlier questions.

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  • $\begingroup$ so my version of the proof is more or less ok ? and we can say that all continues non-monotonic functions have a point at which the derivative is zero ? thanks . $\endgroup$ – Boris Morozov Feb 26 '14 at 11:47
  • $\begingroup$ yeah i think so... $\endgroup$ – Paramanand Singh Feb 26 '14 at 11:52
  • $\begingroup$ @BorisMorozov Unless we assume differentiability of $f$, it is not true. See for example $\arcsin( \sin x )$. $\endgroup$ – Adayah May 25 '18 at 18:19
  • $\begingroup$ @Adayah: differentiability of $f$ is a given hypothesis in question. See first sentence of the question. $\endgroup$ – Paramanand Singh May 26 '18 at 2:51
  • $\begingroup$ @ParamanandSingh I was answering the question in the above comment, not the question in the Original Post. The differentiability assumption (nor any other) doesn't implicitly carry over. $\endgroup$ – Adayah May 26 '18 at 10:49

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