Mean Value Theorem/Proof 2

Proof
Let $g : \closedint a b \to \R$ be a real function with:


 * $\map g x = x$

for all $x \in \closedint a b$.

By Power Rule for Derivatives, we have:


 * $g$ is differentiable with $\map {g'} x = 1$ for all $x \in \closedint a b$.

Note that in particular:


 * $\map {g'} x \ne 0$ for all $x \in \openint a b$.

Since $f$ is continuous on $\closedint a b$ and differentiable on $\openint a b$, we can apply the Cauchy Mean Value Theorem.

We therefore have that there exists $\xi \in \openint a b$ such that:


 * $\dfrac {\map {f'} \xi} {\map {g'} \xi } = \dfrac {\map f b - \map f a} {\map g b - \map g a}$

Note that:


 * $\map {g'} \xi = 1$

and:


 * $\map g b - \map g a = b - a$

so this can be rewritten:


 * $\map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$