Consequences of the Mean Value Theorem for Differentiable Functions

Corollary 1: If $f : [a, b] \to \mathbb{R}$ is continuous on $[a, b]$ and differentiable on $(a, b)$ then $f$ is constant on $[a, b]$ if and only if $f'(x) = 0$ on $(a, b)$.

• Proof: $\Rightarrow$ Suppose that $f$ is constant on $[a, b]$. Then $f(x) = M$ for some $M \in \mathbb{R}$ and $f'(x) = [M]' = 0$ on $(a, b)$.

• $\Leftarrow$ Suppose that $f'(x) = 0$ on $(a, b)$. By the Mean Value theorem, for each $x_0 \in [a, b]$ we have that there exists a $c \in (a, b)$ such that:

(1)

\begin{align} \quad f'(c) = \frac{f(b) - f(x_0)}{b - x_0} \end{align}

• But $f'(c) = 0$. So $f(b) = f(x_0)$ for all $x_0 \in [a, b]$. Hence $f$ is constant on $[a, b]$. $\blacksquare$

Corollary 2: If $f : [a, b] \to \mathbb{R}$ and $g : [a, b] \to \mathbb{R}$ are continuous on $[a, b]$, differentiable on $(a, b)$, and $f'(x) = g'(x)$ for all $x \in (a, b)$. Then there exists a constant $C \in \mathbb{R}$ for which $f(x) = g(x) + C$ on all of $I$.

• Proof: Consider the function $f - g$. It is continuous on $[a, b]$ and differentiable on $(a, b)$. Furthermore, $(f - g)'(x) = 0$ on $(a, b)$. By Corollary 1 we must have that $f - g$ is constant. So there exists a $C \in \mathbb{R}$ for which $f(x) - g(x) = C$ for all $x \in [a, b]$, i.e.:

(2)

\begin{align} \quad f(x) = g(x) + C \quad \blacksquare \end{align}