Mean Value Theorem/Proof 2

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.


Then:

$\exists \xi \in \openint a b: \map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$


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:

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

$\blacksquare$