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If $f(x)$ is differentiable on $(a,b)$, please prove: there exists $(\alpha,\beta)\subset (a,b)$ such that $f'(x)$ is bounded on $(\alpha,\beta)$.

Suppose $f'(x)$ is unbounded on every subintervals of $(a,b)$, I need to derive a contradiction. But the only thing I know is that $f(x)$ is differentiable. The derivative of $f(x)$ satisfies the intermediate value theorem: If $u<v, f'(u)\ne f'(v)$, $\exists w\in (u,v)$ s.t. $f'(w)\in (f'(u),f'(v))$ or $f'(w)\in (f'(v),f'(u))$. Then I don't know what to do next.

Arctic Char
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Yan Peng
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  • The continuity set of a derivative must be dense — see (https://math.stackexchange.com/q/112067/148510 ) — and continuity at a point implies there is an interval where the function is bounded. – RRL Sep 23 '20 at 11:17
  • It is proved as part of this answer: https://math.stackexchange.com/a/3807401. – Martin R Sep 23 '20 at 11:20

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