Rolle's Theorem for Differentiable Functions

# Rolle's Theorem for Differentiable Functions

Lemma 1: Let $f : [a, b] \to \mathbb{R}$. If $f$ is continuous on $[a, b]$, $f$ is differentiable on $(a, b)$, and $f$ attains a local maximum (or minimum) value at $c \in (a, b)$ then $f'(c) = 0$. |

**Proof:**If $f$ attains a local maximum at the value at $c \in (a, b)$ there exists an open interval $(w, z) \subseteq (a, b)$ containing $c$ for which $f(x) - f(c) \leq 0$ for all $x \in (w, z)$. Since $f$ is differentiable at $c \in (w, z) \subseteq (a, b)$ we have that the limit $f'(c)$ exists. Looking at the lefthand and righthand limits we see that:

\begin{align} \quad f'(c) = \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} \geq 0 \end{align}

(2)
\begin{align} \quad f'(c) = \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c} \leq 0 \end{align}

- Therefore $f'(c) = 0$. The proof is analogous when $f$ attains a local minimum at $c$. $\blacksquare$

Theorem 2 (Rolle's Theorem): Let $f : [a, b] \to \mathbb{R}$. If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$. Then there exists a point $c \in (a, b)$ such that $f'(c) = 0$. |

**Proof:**Since $f$ is continuous on the closed interval $[a, b]$ - it attains a maximum and minimum value. If $f(a) = f(b)$ is both its maximum and minimum value then $f$ is constant on $[a, b]$ and so $f'(x) = 0$ on $(a, b)$ and we are done.

- Suppose instead that $f(a) = f(b)$ is not both the maximum and minimum value of $f$ on $[a, b]$. Then there exists a point $c \in (a, b)$ for which $f(c)$ is either a maximum or minimum value.

**Case 1:**Suppose that $f(c)$ is a maximum value. Then $f(x) - f(c) \leq 0$ for all $x \in [a, b]$. Consider the derivative of $f$ at $c$ (which exists since $f$ is differentiable on $(a, b)$):

\begin{align} \quad \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \end{align}

- We look at the lefthand and righthand limits to see that:

\begin{align} \quad \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} \geq 0 \end{align}

(5)
\begin{align} \quad \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c} \leq 0 \end{align}

- But this implies that $f'(c) = 0$.

**Case 2:**Suppose that $f(c)$ is a minimum value. Then $f(x) - f(c) \geq 0$ for all $x \in [a, b]$. Consider the derivative of $f$ at $c$ (which exists since $f$ is differentiable on $(a, b)$):

\begin{align} \quad \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \end{align}

- We look at the lefthand and righthand limits to see that:

\begin{align} \quad \lim_{x \to c^-} \frac{f(x) - f(c)}{x - c} \leq 0 \end{align}

(8)
\begin{align} \quad \lim_{x \to c^+} \frac{f(x) - f(c)}{x - c} \geq 0 \end{align}

- But this also implies that $f'(c) = 0$.

- So there exists a point $c \in (a, b)$ for which $f'(c) = 0$. $\blacksquare$