Rolle's Theorem

If a differentiable function starts and ends at the same height, its graph must have at least one point where the tangent is horizontal. This is Rolle's Theorem, and it is the key step in proving the Mean Value Theorem.

Condition	Conclusion
$f$ continuous on $[a, b]$
f'(x) exists on $(a, b)$	There exists $c \in (a, b)$ with f'(c)=0
$f (a) = f (b)$

Statement of Rolle's Theorem

Graph of a function with equal values at a and b, and a horizontal tangent at an interior point c — If $f (a) = f (b)$ , there is at least one point $c$ between $a$ and $b$ such that f'(c)=0.

Let $f (x)$ be a function with the following properties:

$f$ is continuous everywhere on the closed interval $[a, b]$ .
f'(x) exists at each point of the open interval $(a, b)$ .

If $f (a) = f (b)$ , then there is at least one point $c$ in the open interval $(a, b)$ such that f'(c)=0.

This theorem is intuitively clear: as $x$ increases from $a$ to $b$ , $f (x)$ cannot always increase or always decrease, since $f (a) = f (b)$ . Hence, for at least one value of $x$ between $a$ and $b$ , the function must stop increasing and begin decreasing (or vice versa). By Fermat's Theorem, the derivative must be zero at that turning point.

The conditions require:

At least one point $c \in (a, b)$ where f'(c)=0, but there may be more than one.

Graph showing three interior points where the derivative is zero, all consistent with Rolle's Theorem — There can be more than one point where the derivative is zero. This graph has three such points.

Roots of Derivatives

In a special case where $f (a) = f (b) = 0$ , Rolle's Theorem states that between any pair of roots of $f (x)$ lies a root of f'(x).

Example. Let $f (x) = (x - 3) (x - 2) (x + 1) (x + 2) (x + 4)$ . This function has zeros at $x = 3, 2, - 1, - 2, - 4$ . Since $f$ is a polynomial, it is continuous and differentiable everywhere. Rolle's Theorem guarantees that f'(x) has at least one zero in each of the intervals $(2,3)$, $(- 1, 2)$ , $(- 2, - 1)$ , and $(- 4, - 2)$ .

Graph showing roots of f(x) and the guaranteed roots of f'(x) between them — The derivative f'(x) has a root between each pair of consecutive roots of $f (x)$ .

Another example. Consider $g (x) = x^{4} - x^{2} - 2$ , which has zeros at $\pm \sqrt{2}$ . Rolle's Theorem guarantees g'(x)=4x^3-2x=2x(2x^2-1) has at least one zero between $- \sqrt{2}$ and $\sqrt{2}$ . In fact, g' has three roots: $x = 0$ and $x = \pm 1 / \sqrt{2}$ .

Graph of g(x) = x^4 - x^2 - 2 and its derivative g'(x), showing three roots of g' between the two roots of g — The derivative g'(x) has three roots between the two roots of $g (x)$ .

Examples

Show that the equation $x^{3} + x - 1 = 0$ has exactly one real solution.

Solution

Let

f (x) = x^{3} + x - 1

. Since

f

is a polynomial, it is continuous everywhere. Because

f (- 1) = - 3 < 0

and

f (1) = 1 > 0

, by Bolzano's Theorem,

f

has at least one zero between

- 1

and $1$. If

f

had more than one zero, Rolle's Theorem would guarantee the existence of a zero of f'(x). However: f'(x)=3x^2+1>0\quad\text{for all }x. Since f' has no zeros,

f

cannot have more than one root.

Graph of f(x) = x^3 + x - 1 crossing the x-axis exactly once

When Rolle's Theorem Does Not Apply

The function $f (x)$ must be continuous at all points in the closed interval $[a, b]$ , including the endpoints, but its derivative does not need to exist at the endpoints. If $f (x)$ is discontinuous even at one point of the closed interval, or if f'(x) does not exist at an interior point, then Rolle's Theorem does not apply.

Three graphs showing cases where Rolle's Theorem fails: discontinuous f, f' undefined at an interior point with cusp, and another case — Rolle's Theorem does not apply if either $f$ is discontinuous or f' is not defined everywhere between $a$ and $b$ .

Consider $y = f (x) = | x |$ on $[- 2, 2]$ . Can we apply Rolle's Theorem?

Solution

This function is continuous on

[- 2, 2]

and

f (2) = f (- 2) = 2

. The derivative is: f'(x)=\begin{cases}1 & \text{if }0 Since f'(x) does not exist at

x = 0

(an interior point), Rolle's Theorem does not apply. Indeed, f'(x) never equals zero in

(- 2, 2)

Graph of y = |x| on [-2, 2] showing the corner at x = 0 where f' is undefined

Consider $y = f (x) = \sqrt{1 - x^{2}}$ on $[- 1, 1]$ . Can we apply Rolle's Theorem?

Solution

The function

f (x) = \sqrt{1 - x^{2}}

is continuous on

[- 1, 1]

and

f (- 1) = f (1) = 0

. Its derivative: f'(x)=\frac{-x}{\sqrt{1-x^{2}}} exists everywhere on

(- 1, 1)

except at the endpoints

x = \pm 1

. The conditions of Rolle's Theorem do not require differentiability at endpoints. Indeed, f'(0)=0, confirming the theorem's conclusion.

Graph of y = sqrt(1 - x^2), a semicircle, with a horizontal tangent at the top

A Physical Interpretation

Consider a train moving along a straight track. If we observe the same train at the same station at 10 am and again at 4 pm, we know its speed must have been zero at some point between those times. If it never left the station, its speed was zero throughout. If it left and returned, it had to stop (even briefly) to reverse direction. This is exactly what Rolle's Theorem states.

Frequently Asked Questions

Can there be more than one point where f'(c) = 0?

Yes. Rolle's Theorem only guarantees **at least one** such point. There can be many. For example,

f (x) = \sin x

[0, 2 π]

satisfies

f (0) = f (2 π) = 0

, and f'(x)=\cos x vanishes at both

x = π / 2

and

x = 3 π / 2

Is differentiability required at the endpoints?

No. The hypothesis requires f'(x) to exist at each point of the **open** interval

(a, b)

, not at the closed interval. The function

f (x) = \sqrt{1 - x^{2}}

[- 1, 1]

illustrates this: f' does not exist at

\pm 1

, yet Rolle's Theorem applies.

How is Rolle's Theorem used in practice?

One important application is proving that a polynomial (or other function) has exactly one root in an interval: show it has at least one root (by the Intermediate Value Theorem) and then show its derivative has no roots (so, by Rolle's Theorem, the function cannot have two roots).

Statement of Rolle's Theorem

Roots of Derivatives

Examples

When Rolle's Theorem Does Not Apply

A Physical Interpretation

Frequently Asked Questions

Quick Links

Social Media