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Suppose $f$ is a function, $a$ is a point in domain and $f'(a)>0$ and $f'(x)$ is continuous in $x=a$. How to prove $f$ is monotonically increasing in an interval including point $a$?

an4s
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saleh
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  • The statement isn't true. Take $f(x)=\tan(x)$ for rational $x$ and $f(x)=x$ otherwise for $a=0$. Then $f'(0)=1$ and $f'$ is continuous in $x=0$, but surely $f$ isn't monotonically increasing on any interval containing $0$. – Michael Hoppe Jan 04 '20 at 10:53
  • f' is not continuous in x=0 because f is not continuous near x=0 (except x=0) so derivative of f near 0 is not defined – saleh Jan 04 '20 at 11:28
  • Epsilon-Delta-Definition of continuity – whiteian motion Jan 04 '20 at 10:40
  • @saleh The only assumptions you made are: (1) $f$ is differentiable in one point and (2) $f'$ is continuous in that point. The example I gave satisfies both, doesn't it? – Michael Hoppe Jan 04 '20 at 12:05
  • See also https://math.stackexchange.com/questions/659558/counterexample-increasing-function/659584#659584, please. – Michael Hoppe Jan 04 '20 at 12:09

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If it's not, it means that there are sequences $x_n < y_n$ converging to $a$ such that $f(x_n)\ge f(y_n)$. By the mean value theorem, there is a sequence $z_n\in (x_n, y_n)$ such that $f'(z_n)\le 0$. Hence $f'(a) = \lim f'(z_n) \le 0$.

Of course this proof supposes that $f$ is differentiable in a neighborhood of $a$.

Gribouillis
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