Mean Value Theorem

The Mean Value Theorem (often abbreviated MVT) is considered one of the most important results in real analysis. An elegant proof of the Fundamental Theorem of Calculus can be given using Mean Value Theorem.

Statement

Let $f:[a,b]\rightarrow\mathbb R$ be differentiable function on the interval $(a,b)$ . The Mean Value Theorem states there exists a point $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}.$

Proof

Note that $\frac{f(b)-f(a)}{b-a}$ is the slope of the secant line which passes through the graph of $f$ at points $(a,f(a))$ and $(b,f(b))$ . Let $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)$ . As $g$ is continuous on $[a,b]$ since it is the linear combinatoin of the continuous functions $f(x)$ and $(x-a)$ , and that $g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}.$ Since $g$ is differentiable on $(a,b)$ and $g(a)=g(b)=f(a)$ , by Rolle's Theorem, there exists some $c\in(a,b)$ such that $g'(c)=0$ . Thus $f'(c)=g'(c)+\frac{f(b)-f(a)}{b-a}=\frac{f(b)-f(a)}{b-a}$ as desired. $\fbox{}$

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Mean Value Theorem

Statement

Proof