Darboux Theorem Proof http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29
I'm having a bit of trouble understanding the proof of Darboux's Theorem on the IVP of derivatives.
Why should there be an extremum such that $g'(x) = 0$ from the fact that $g'(a)>0$ and $g'(b)<0$ ? 

Suppose that $g$ attains a local maximum at $a$. Then $$\lim_{x \to a+} \frac{g(x)-g(a)}{x-a} \leq 0.$$ Analogously, if $g$ attains a local maximum at $b$, then $$\lim_{x \to b-} \frac{g(x)-g(b)}{x-b}\geq 0.$$ But both contradict $g'(a)>0$ and $g'(b)<0$. Hence the maximum, which exists since $g$ is continuous on $[a,b]$, must lie inside $[a,b]$.
Actually, you can visualize the setting: $g'(a)>0$ means that $g$ leaves $a$ in an increasing way, and then reaches $b$ in a decreasing way. Therefore $g$ must attain a maximum somewhere between $a$ and $b$.
