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I am looking for an example of a function $f$ such that:

$f$ that is continuous on $\mathbb{R}$, differentiable on $\mathbb{R} \backslash\{0\}$, and does not satisfy the mean value theorem on $\mathbb{R}$.

So far, I am not quite sure how to handle the last part of that sentence, and what it implies.

Does this mean,I can't use $|x|$ or $x^\frac{2}{3}$ ?

Why/why not?

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Take the function $|x|$. Now consider the interval $[-1,1]$, if we where to attempt to naively use the mean value theorem it would tell us there is a point $x$ in $[-1,1]$ where the derivative is $0$. This is false since all the points which come with a derivative have either derivative $1$ or $-1$.

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  • $\begingroup$ is the part "does not satisfy the mean value theorem on $\mathbb{R}$", a trivial characteristic to include? $\endgroup$ – piman314 May 14 '15 at 3:51
  • $\begingroup$ It does not satisfy the mean value theorem on $\mathbb R$ because if it did then there would be a point in the interval $[-1,1]$ with derivative zero. Since this does not happen it does not satisfy the mean value theorem. $\endgroup$ – Jorge Fernández Hidalgo May 14 '15 at 3:52

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