Rolle's Theorem
Suppose that \(a,b \in \mathbb{R}\) with \(a < b\). If
  1. \(f\) is continuous on \([a,b]\)
  2. \(f\) is differentiable on \((a,b)\)
  3. \(f(a) = f(b)\)
Then \(f'(c)=0\) for some \(c \in (a,b)\).

Proof

Since \(f\) is continuous on a closed and bounded interval \([a,b]\) then by the Extreme Value Theorem. \(f\) has a finite maximum \(M\) and a finite minimum \(m\) on \([a,b]\). If \(m = M\), then \(f\) is constant (the maximum point is the same as the minimum point). Thus, \(f'(x)=0\) for all \(x \in (a,b)\).

Otherwise suppose \(m \neq M\). Since \(f(a) = f(b)\), then either \(m\) or \(M\) is attained in the open interval \((a,b)\) (Since it cannot happen that both the maximum and minimum occur only at the endpoints). Suppose without the loss of generality that \(M\) is attained inside \((a,b)\). Hence, there exists a \(c\) such that \(f(c) = M\). Now, since \(M\) is the maximum of \(f\), then

$$ \begin{align*} f(c+h) - f(c) \leq 0 \end{align*} $$

for all \(h\) such that \(c+h \in (a,b)\). In the case when \(h > 0\), then this implies that

$$ \begin{align*} f'(c) = \lim_{h \to 0+} \frac{f(c+h) - f(c)}{h} \leq 0, \end{align*} $$

and in the case when \(h < 0\), this implies that

$$ \begin{align*} f'(c) = \lim_{h \to 0-} \frac{f(c+h) - f(c)}{h} \geq 0, \end{align*} $$

Since \(f\) is differentiable at \(c\), then the left and right limits must be equal. Hence, \(f'(c)=0\). \(\ \blacksquare\)

What happens when the continuity condition is relaxed?

If the continuity condition is relaxed even at point in \([a,b]\), then we can construct a function such that

$$ \begin{align} f(x) = \begin{cases} x, & \text{if } x \in [0,1), \\ 0, & \text{if } x = 1. \end{cases} \end{align} $$

This function is continuous and differentiable on \([0,1)\). We also have that \(f(0)=f(1)\). However it is not continuous at \(1\) since

$$ \begin{align*} \lim_{x \to 1-} = f(x) = 1 \neq f(1) = 0 \end{align*} $$

Note here that \(f'(x)\) is never zero.

What happens when the differentiability condition is relaxed?

Take \(f(x) = |x|\). It is continuous on \([-1,1]\) and differentiable on \((-1,1)\) except at \(x = 0\). We also have that \(f(1)=f(-1)\). But again \(f'(x)\) is never zero.

References