Cauchy's Mean Value Theorem for Differentiable Functions

Theorem 1 (Cauchy's Mean Value Theorem): Let $f : [a, b] \to \mathbb{R}$ and $g : [a, b] \to \mathbb{R}$ be continuous on $[a, b]$, differentiable on $(a, b)$, and $g'(x) \neq 0$ for all $x \in (a, b)$. Then there exists a $c \in (a, b)$ such that $\displaystyle{\frac{f'(c)}{g'(x)} = \frac{f(b) - f(a)}{g(b) - g(a)}}$.

Note that when $g(x) = x$, Cauchy's Mean Value theorem reduces down to the regular Mean Value theorem.

• Proof: Define a function $h : [a, b] \to \mathbb{R}$ by:

(1)

\begin{align} \quad h(x) = \frac{f(b) - f(a)}{g(b) - g(a)}[g(x) - g(a)] - [f(x) - f(a)] \end{align}

• Since $f$ and $g$ are continuous on $[a, b]$ we have that $h$ is continuous on $[a, b]$. Furthermore, since $f$ and $g$ are differentiable on $(a, b)$ we have that $h$ is differentiable on $(a, b)$.

• Also observe that:

(2)

\begin{align} \quad h(b) &= \frac{f(b) - f(a)}{g(b) - g(a)}[g(b) - g(a)] - [f(b) - f(a)] = f(b) - f(a) - [f(b) - f(a)] = 0 \\ \quad h(a) &= \frac{f(b) - f(a)}{g(b) - g(a)}[g(a) - g(a)] - [f(a) - f(a)] = 0 \end{align}

• So $h(b) = h(a) = 0$. By Rolle's Theorem for Differentiable Functions we have that there exists a point $c \in (a, b)$ such that $h'(c) = 0$. The derivative of $h$ is:

(3)

\begin{align} \quad h'(x) = \frac{f(b) - f(a)}{g(b) - g(a)}g'(x) - f'(x) \end{align}

• Plugging in $x = c$ gives us:

(4)

\begin{align} \quad 0 &= h'(c) \\ &= \frac{f(b) - f(a)}{g(b) - g(a)}g'(c) - f'(c) \end{align}

• Rearranging the equation above gives us:

(5)

\begin{align} \quad f'(c) = \frac{f(b) - f(a)}{g(b) - g(a)} g'(c) \end{align}

• Since $g'(x) \neq 0$ for all $x \in (a, b)$ we divide both sides of the equation above by $g'(c)$ to get:

(6)

\begin{align} \quad \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} \quad \blacksquare \end{align}