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)
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 $min(|g'(x)|)$?