The Mean Value Theorem is arguably one of the most important theorems from real analysis. It’s used to prove many well known results, such as the Fundamental Theorem of Calculus, L’Hospital’s rule, and even the symmetry of second derivatives in higher dimensions. Today, let’s take a look at its proof.
Theorem: Let $f : [a,b] \to \RR$ be a continuous function on the closed interval $[a,b]$, and differentiable on the open interval $(a,b)$, where $a < b$. Then there exists some $c$ in $(a,b)$ such that $$ f'(c) = \frac{f(b)-f(a)}{b-a}. $$
Proof: Define $g(x) = f(x) - mx$ where $m = \frac{f(b)-f(a)}{b-a}$. Note that with this choice of $m$, $g(a) = g(b)$. Futhermore, since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, $g$ also shares the same properties. Hence by Rolle’s Theorem, there exists $c \in (a,b)$ such that $g'(c) = 0$. Finally, since $g'(x) = f'(x) - m$, it follows that $$ g'(c) = f'(c) - m = 0 \quad \implies \quad f'(c) = m = \frac{f(b)-f(a)}{b-a} $$