# Rolle's Theorem: Statement, Conditions and Worked Examples

> Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then f'(c) = 0 for some c. Graph, counterexamples, examples.

Source: https://practicecalculus.com/theorems/rolles-theorem
Updated: 2026-11-03

$$ f'(c) = 0 \text{ for some } c \in (a, b) $$

## At a glance

- **Hypotheses:** $f$ continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$
- **Conclusion:** There is a $c$ in $(a, b)$ with $f'(c) = 0$
- **Geometric meaning:** Somewhere between two equal heights the tangent line is horizontal
- **Generalization:** The mean value theorem
- **Used for:** Proving an equation has at most one root, proving the MVT
- **Course:** Calculus 1

## Key takeaways

- If a smooth function has the same value at both ends of an interval, its derivative is zero somewhere in between.
- All three hypotheses are needed: continuity on $[a, b]$, differentiability on $(a, b)$ and $f(a) = f(b)$.
- Rolle's theorem is the special case of the mean value theorem where the secant line is horizontal.
- Combined with the IVT, it proves an equation has exactly one solution.

**Rolle's theorem** says that if a smooth function starts and ends at the same height, then somewhere in between its graph has a horizontal tangent line. It is the simplest of the mean value theorems and the step used to prove the others.

## Statement of Rolle's theorem

Let $f$ be a function that satisfies all three hypotheses:

1. $f$ is **continuous** on the closed interval $[a, b]$;
2. $f$ is **differentiable** on the open interval $(a, b)$;
3. $f(a) = f(b)$.

Then there is at least one number $c$ in $(a, b)$ such that

$$f'(c) = 0$$

## What Rolle's theorem means

Throw a ball up and catch it at the same height. At some moment between the throw and the catch, its vertical velocity was exactly zero: the top of the arc. Rolle's theorem is that observation for any smooth function. If the graph goes up, it has to turn around to come back down, and at the turning point the slope is zero. If the function is constant, the slope is zero everywhere.

```graph
title: 'Rolle''s theorem: y = x³ − 4x on [−2, 2] has two horizontal tangents'
caption: 'Since $f(-2) = f(2) = 0$, Rolle''s theorem guarantees a horizontal tangent in $(-2, 2)$. Here there are two, at $x = \pm\frac{2}{\sqrt{3}} \approx \pm 1.15$.'
x: [-2.5, 2.5]
y: [-4.5, 4.5]
functions:
  - expr: x^3 - 4x
    label: y = x³ − 4x
  - expr: '3.079'
    label: horizontal tangents
    dashed: true
    domain: [-1.9, -0.4]
  - expr: '-3.079'
    dashed: true
    domain: [0.4, 1.9]
points:
  - [-2, 0, 'f(−2) = 0']
  - [2, 0, 'f(2) = 0']
  - [-1.1547, 3.079, 'c ≈ −1.15']
  - [1.1547, -3.079, 'c ≈ 1.15']
```

## Why each hypothesis matters

**Differentiability.** $f(x) = |x|$ on $[-1, 1]$ is continuous and $f(-1) = f(1) = 1$, but the graph is a sharp V. Its derivative is $-1$ on the left and $1$ on the right and is never $0$. The corner at $x = 0$, where $f$ is not differentiable, is what breaks the theorem.

**Continuity at the endpoints.** Let $f(x) = x$ for $0 \le x < 1$ and $f(1) = 0$. Then $f(0) = f(1) = 0$ and $f$ is differentiable on $(0, 1)$, but $f'(x) = 1$ everywhere inside. The jump at $x = 1$ breaks continuity on $[a, b]$.

**Equal endpoint values.** $f(x) = x$ on $[0, 1]$ is perfectly smooth, but $f(0) \ne f(1)$ and $f'(x) = 1$ is never zero. For unequal endpoints you need the [mean value theorem](/theorems/mean-value-theorem) instead.

## Worked examples

**Example 1: find the $c$ for $f(x) = x^2 - 4x + 3$ on $[1, 3]$.**

**Step 1.** Check the hypotheses. $f$ is a polynomial, so it is continuous and differentiable everywhere. $f(1) = 1 - 4 + 3 = 0$ and $f(3) = 9 - 12 + 3 = 0$, so $f(1) = f(3)$.

**Step 2.** Solve $f'(c) = 2c - 4 = 0$, giving $c = 2$, which is in $(1, 3)$. The vertex of the parabola is the guaranteed point.

**Example 2: $f(x) = x^3 - 4x$ on $[-2, 2]$.**

$f(-2) = -8 + 8 = 0$ and $f(2) = 8 - 8 = 0$, and $f$ is a polynomial. Solve

$$f'(c) = 3c^2 - 4 = 0 \implies c = \pm\frac{2}{\sqrt{3}} \approx \pm 1.155$$

Both values lie in $(-2, 2)$, so this function has two points where Rolle's conclusion holds. The theorem promised at least one.

**Example 3: prove $x^3 + x - 1 = 0$ has exactly one real solution.**

**Step 1 (at least one).** $f(x) = x^3 + x - 1$ is continuous, $f(0) = -1 < 0$ and $f(1) = 1 > 0$. By the [intermediate value theorem](/theorems/intermediate-value-theorem) there is a root in $(0, 1)$.

**Step 2 (at most one).** Suppose there were two roots $r_1 < r_2$. Then $f(r_1) = f(r_2) = 0$, and Rolle's theorem would give a $c$ between them with $f'(c) = 0$. But $f'(x) = 3x^2 + 1 \ge 1$ for every $x$, so $f'$ is never zero. That contradiction means there cannot be two roots.

So the equation has exactly one real solution. This IVT plus Rolle argument is the most common exam use of the theorem.

## How Rolle's theorem is proved

By the [extreme value theorem](/theorems/extreme-value-theorem), $f$ has a maximum and a minimum on $[a, b]$. If both occur at the endpoints, then since $f(a) = f(b)$ the function is constant and $f' = 0$ everywhere. Otherwise one of them occurs at an interior point $c$, and [Fermat's theorem](/theorems/fermats-theorem) gives $f'(c) = 0$.

## Common mistakes

1. **Not checking the hypotheses.** Exams often ask whether Rolle's theorem applies. Check continuity, differentiability and $f(a) = f(b)$ explicitly.
2. **Giving a $c$ outside $(a, b)$.** A solution of $f'(c) = 0$ outside the open interval does not count.
3. **Thinking the theorem finds every critical point.** It only guarantees at least one between $a$ and $b$.

## Related theorems

Tilt the picture so the endpoints are at different heights and you get the [mean value theorem](/theorems/mean-value-theorem). For a full treatment with applications, read the [mean value theorem guide](/blog/mean-value-theorem), and see [curve sketching with derivatives](/blog/curve-sketching-with-derivatives) for where horizontal tangents matter.

## Frequently asked questions

### What are the three conditions of Rolle's theorem?

The function must be continuous on the closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and have equal values at the endpoints, $f(a) = f(b)$. If all three hold, $f'(c) = 0$ for some $c$ in $(a, b)$.

### What is the difference between Rolle's theorem and the mean value theorem?

Rolle's theorem requires $f(a) = f(b)$ and concludes $f'(c) = 0$. The mean value theorem drops that requirement and concludes $f'(c) = \frac{f(b) - f(a)}{b - a}$. Rolle's theorem is the special case where that slope is zero.

### Can there be more than one value of c in Rolle's theorem?

Yes. The theorem guarantees at least one. For example $f(x) = x^3 - 4x$ on $[-2, 2]$ has two, $c = \pm\frac{2}{\sqrt{3}}$.

### Why does Rolle's theorem fail for |x| on [-1, 1]?

Because $|x|$ is not differentiable at $x = 0$. The endpoint values are equal, but the derivative is $-1$ or $1$ everywhere it exists, never $0$.