# Darboux's theorem

In Dunham's "Calculus Gallery", it introduces Darboux's theorem:

If $f$ is differentiable on $[a,b]$, and if r is any number for which $f'(a)<r<f'(b)$ then there exists a c in (a,b) such that $f'(r)=c$

Proof: [summarized] Introduce $g(x)=f(x)-rx$. There is a point $c$ in $[a,b]$ where g takes a minimum. Because $g'(a)=f'(a)-r<0$ and $g'(b)=f'(b)-r>0$ we see that a minimum cannot occur at a or b, and so c lies in (a,b)...

I don't see why the minimum can't occur at a or b. Consider $f(x)=x^2, r=1.5,[a,b]=[1,2]$. The minimum value of $g'(x)=2x-1.5$ in $[1,2]$ does indeed seem to be $x=1$ or $x=a$. Does he mean $min(|g'(x)|)$?

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For your example, $f'(a) = 2$ and $f'(b) = 4$, so $1.5$ isn't in between them. – Jason DeVito Jul 18 '12 at 14:13

In your example, $f'(x)=2x$, so $f'(1)=2\gt r$.
The minimum can't occur at $a$ or $b$ because $g'(a)\lt0$ and $g'(b)\gt0$, so $g$ decreases from the boundary inwards on both sides.
Oh, I think I see. It's the minimum of $g$, not of $g'$ that we're looking for? – Xodarap Jul 18 '12 at 14:16
@Xodarap: Yes, it says "where $g$ takes a minimum". – joriki Jul 18 '12 at 14:17