Theorem – Mean value theorem

Contents
1. Theorem
2. Proof

1. Theorem

Let f be a continuous function on [a,b] that is differentiable on (a,b). Then, there exists at least one point c\in (a,b) such that

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

2. Proof

Firstly, let g(x) = \frac{f(b)-f(a)}{b-a} (x-a) +f(a). Then, the function f-g is continuous on [a,b] and differentiable on (a,b). Because f(a)-g(a) = f(b)-g(b) = 0, f-g there exists some point c\in (a,b) such that f'(c)-g'(c) = 0, and so

0=f'(c)-\frac{f(b)-f(a)}{b-a} \implies f'(c)=\frac{f(b)-f(a)}{b-a}

as was to be proven. \square