Mean_Value_Theorem

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Definition

If $f:[a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ where

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

Proof

Define the auxiliary function

$$ g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a) $$

Then $g(a) = f(a)$ and $g(b) = f(a)$, so $g(a) = g(b)$.

By Rolle’s Theorem (a special case of MVT where $f(a) = f(b)$), there exists $c \in (a,b)$ where $g'(c) = 0$.

Computing $g'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} = 0$ gives the result.