Proof of Mean Value Theorem in One Dimension

Define the secant slope $m=\dfrac{f(b)-f(a)}{b-a}$ and auxiliary function $\phi(x)=f(x)-mx$. Then $\phi$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $\phi(a)=\phi(b)$. By Rolle's Theorem in One Dimension, there exists $c\in(a,b)$ with $\phi'(c)=0$. Since $\phi'(x)=f'(x)-m$, we get $f'(c)=m=\dfrac{f(b)-f(a)}{b-a}$.