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If $f$ is continuous on $[a,b]$ and $f\;'> 0$ on $(a,b)$, then $f$ is monotonically increasing on $[a,b]$ and proof follows from mean value theorem. Now suppose we have $f \in C[a,b]$ and $f\;'(x+)> 0$ for all $x \in (a,b)$. Does it follow that $f$ is monotonically increasing on $[a,b]$?

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Suppose that $a<u<v<b$ with $f(u)>f(v)$. Let $C=\{x\in[u,v]:f(x)=f(u)\}$; $f$ is continuous, so $C$ is a closed subset of $[u,v]$ and therefore has a maximum element. Without loss of generality we may assume that this maximum element is $u$. The intermediate value theorem then ensures that $f(x)<f(u)$ for all $x\in(u,v]$. But $$0<f\;'(u+)=\lim_{h\to 0^+}\frac{f(u+h)-f(u)}{h},$$ so there is a $y\in(u,b)$ such that $f(x)>f(u)$ for every $x\in(u,y)$. This contradiction shows that $f$ must be monotone increasing.

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@Tigran: I don’t think that you understand the argument. I’m not using the mean value theorem anywhere. Of course $f$ is constant on $C$; so what? $C$ is used merely to find its maximum, $u$, so that I have an interval $(u,v]$ on which $f(x)<f(u)$. – Brian M. Scott Nov 2 '11 at 13:24
Now I got your argument. Thank you for your nice solution. – Tigran Hakobyan Nov 2 '11 at 13:36
@Tigran: You’re very welcome. – Brian M. Scott Nov 2 '11 at 13:52

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