Proof of Mean Value Theorem in One Dimension

Define

$$m=\frac{f(b)-f(a)}{b-a}$$

and set $\phi:I\to\mathbb{R}$ by $\phi(x)=f(x)-mx$. Since $f$ is continuous on $[a,b]$ and differentiable at every point of $(a,b)$ by hypothesis, the same is true for $\phi$. Also,

$$\phi(b)-\phi(a)=\bigl(f(b)-mb\bigr)-\bigl(f(a)-ma\bigr)=f(b)-f(a)-m(b-a)=0,$$

so $\phi(a)=\phi(b)$. Therefore Rolle's Theorem in One Dimension applies to $\phi$ on $[a,b]$, and there exists $c\in(a,b)$ such that $\phi'(c)=0$. Because

$$\phi'(c)=f'(c)-m,$$

we obtain

$$f'(c)=m=\frac{f(b)-f(a)}{b-a}.$$

This is the desired conclusion.