proof of mean value theorem

Define $h(x)$ on $[a,b]$ by

$$h(x)=f(x)-f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)(x-a)$$

Clearly, $h$ is continuous on $[a,b]$, differentiable on $(a,b)$, and

Notice that $h$ satisfies the conditions of Rolle’s Theorem. Therefore, by Rolle’s Theorem there exists $c\in (a,b)$ such that $h^{\prime }(c)=0$.
However, from the definition of $h$ we obtain by differentiation that

$$h^{\prime }(x)=f^{\prime }(x)-\frac{f(b)-f(a)}{b-a}$$

Since $h^{\prime }(c)=0$, we therefore have

$$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$$

as required.