Proof of the Mean Value Theorem The equation of the secant through $(a,f(a))$ and $(b,f(b))$ is $y-f(a)=\frac{f(b)-f(a)}{b-a}(x-a)$ which we can rewrite as $y=\frac{f(b)-f(a)}{b-a}(x-a)+f(a).$ Let $g(x)=f(x)-\left[\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\right].$ Note that $g(a)=g(b)=0$. Also, $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ since $f$ is. So by Rolle's Theorem there exists $c$ in $(a,b)$ such that $g'(c)=0$. But $\displaystyle g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$, so $g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}=0.$ Therefore, $f'(c)=\frac{f(b)-f(a)}{b-a}$ and the proof is complete.