The Mean Value Theorem for Differentiable Functions

The Mean Value Theorem for Differentiable Functions

Recall from the Rolle's Theorem for Differentiable Functions page that if $f : [a, b] \to \mathbb{R}$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$ then there exists a point $c \in (a, b)$ for which $f'(c) = 0$. We now state a more general version of Rolle's theorem known as the Mean-Value theorem.

 Theorem 1 (The Mean-Value Theorem): Let $f : [a, b] \to \mathbb{R}$. If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ then there exists a point $c \in (a, b)$ such that $\displaystyle{f'(c) = \frac{f(b) - f(a)}{b - a}}$.
• Proof: Define a function $\varphi$ on $[a, b]$ by:
(1)
\begin{align} \quad \varphi(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a} (x - a) \end{align}
• Observe that $\varphi$ is continuous on $[a, b]$, differentiable on $(a, b)$ and that $\varphi(b) = 0 = \varphi(a)$. By Rolle's theorem there exists a point $c \in (a, b)$ such that $\varphi'(c) = 0$. The derivative of $\varphi$ is:
(2)
\begin{align} \quad \varphi'(x) = f'(x) - \frac{f(b) - f(a)}{b - a} \end{align}
• Plugging in $x = c$ gives us:
(3)
\begin{align} \quad 0 = f'(c) - \frac{f(b) - f(a)}{b - a} \end{align}
• And so there exists a point $c \in (a, b)$ for which $\displaystyle{f'(c) = \frac{f(b) - f(a)}{b - a}}$. $\blacksquare$