# 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 $g'(b)<0$ ?

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What is $g$? The wikipedia page uses $f$ and $\phi$... – Matthew Conroy Sep 7 '12 at 16:35
I can't understand what you mean by "...only a maximum". – Siminore Sep 7 '12 at 17:09

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 $b$.