# Proving the convexity of a real function given that $f'$ is increasing and f is differentiable.

I'm a little stuck on this question. I got to the point where I stated that for some $c \in [x,z]$ and $d \in [z,y]$, $f'(d) \geq f'(c)$. I don't know how to proceed from hereon. Thanks guys.

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This can't work. There are functions that are increasing and differentiable, but not convex. – Hagen von Eitzen Apr 9 '13 at 21:26
@HagenvonEitzen The original title was hard to read, I think its fixed now. – muzzlator Apr 9 '13 at 21:30

We know there is a $c \in (x,y)$ such that $f'(c)$ is the average value $\frac{f(y)-f(x)}{y-x}$. Suppose there is a $d$ which increases this average value from $x$. Then this $d$ must be greater than $c$ since $f'(x)$ is increasing and mean value theorem on $[x,d]$ would be a contradiction. But then mean value theorem on $[d,y]$ produces a contradiction.

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