Rolle's Theorem

The theorem

If $f$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and $f(a) = f(b)$ , then there exists at least one point $c \in (a,b)$ where $f'(c) = 0$ : intuitively, if a smooth curve starts and ends at the same height, it must have a flat point somewhere in between.

Worked example

Verify Rolle's theorem for $f(x) = x^2 + 2x - 8$ on $[-4, 2]$ .

Solution: $f$ is a polynomial, so it's continuous and differentiable everywhere. Checking the endpoints:

f(-4) = 16 - 8 - 8 = 0, \qquad f(2) = 4 + 4 - 8 = 0

Since $f(-4) = f(2)$ , Rolle's theorem guarantees a point $c \in (-4, 2)$ with $f'(c)=0$ . Since $f'(x) = 2x+2$ :

2c + 2 = 0 \;\Rightarrow\; c = -1

$c=-1$ does lie in $(-4,2)$ , so the theorem is verified.

f(x) = x² + 2x − 8 on [−4, 2]. The curve starts and ends at height 0, so somewhere between it must level off: at c = −1, where the tangent is horizontal.

When the theorem doesn't apply

All three conditions matter. $f(x) = |x|$ on $[-1, 1]$ has $f(-1) = f(1) = 1$ and is continuous, but it is not differentiable at $x = 0$ , and there is no point where $f'(c) = 0$ : the slope is $-1$ on the left and $+1$ on the right. So before using the theorem, check continuity on $[a, b]$ , differentiability on $(a, b)$ , and equal end values.

The mean value theorem

Rolle's theorem is a special case of the mean value theorem. Drop the condition $f(a) = f(b)$ , and some point still has a tangent parallel to the chord joining the endpoints: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , there is a $c \in (a, b)$ with

f'(c) = \dfrac{f(b) - f(a)}{b - a}

When $f(a) = f(b)$ , the right side is 0, which is exactly Rolle's theorem.

Rolle's theorem and the mean value theorem side by side

	Rolle's theorem	Mean value theorem
Continuous on	$[a, b]$	$[a, b]$
Differentiable on	$(a, b)$	$(a, b)$
End values	$f(a) = f(b)$	no condition
Conclusion	$f'(c) = 0$	$f'(c) = \dfrac{f(b) - f(a)}{b - a}$

Worked example: the mean value theorem

Find the value of $c$ guaranteed by the mean value theorem for $f(x) = \sqrt{x}$ on $[0, 4]$ .

Solution: $\sqrt{x}$ is continuous on $[0, 4]$ and differentiable on $(0, 4)$ . The chord's slope is $\dfrac{f(4) - f(0)}{4 - 0} = \dfrac{2}{4} = \dfrac{1}{2}$ . Setting the derivative equal to it:

\dfrac{1}{2\sqrt{c}} = \dfrac{1}{2} \;\Rightarrow\; \sqrt{c} = 1 \;\Rightarrow\; c = 1

$c = 1$ lies in $(0, 4)$ , as the theorem promises.

f(x) = √x on [0, 4]. The chord from (0, 0) to (4, 2) has slope 1/2, and the tangent at c = 1 is parallel to it.