What is Rolle's Theorem in calculus?

Rolle's Theorem states that if a function f(x) is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and satisfies f(a) = f(b), then there is at least one number c in (a, b) such that f'(c) = 0.

Rolle's Theorem Questions and Examples

Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.

In what follows, \( C \) represents the constant of integration where applicable.

Rolle's Theorem

Rolle's theorem is a direct consequence of the mean value theorem where, under the conditions that \( f(x) \) is a continuous function on the interval \( [a, b] \) and differentiable on the open interval \( (a, b) \), there exists at least one value \( c \) of \( x \) such that:

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

Now, if the condition \( f(a) = f(b) \) is also satisfied, then the above simplifies to: \( f'(c) = 0 \).

In other words, under these three conditions, we can always find a tangent to the curve of \( f \) that is horizontal (slope = \( f'(c) = 0 \)).

Rolle's Theorem

If \( f(x) \) is:
1) a continuous function on the interval \( [a, b] \)
2) differentiable on the open interval \( (a, b) \)
3) and \( f(a) = f(b) \)

then there is at least one value \( c \) of \( x \) in the interval \( (a, b) \) such that: \[ f'(c) = 0 \]

Examples with Detailed Solutions

Click on each example to view its detailed step-by-step solution.

Example 1

The graph of \( f(x) = -x^2 + 6x - 6 \) for \( 1 \le x \le 5 \) is shown below. \( f(1) = f(5) = -1 \) and \( f \) is continuous on \( [1, 5] \) and differentiable on \( (1, 5) \), hence, according to Rolle's theorem, there exists at least one value of \( x = c \) such that \( f'(c) = 0 \).

Show Solution to Example 1

\[ f'(x) = -2x + 6 \] \[ f'(c) = -2c + 6 = 0 \]

Solve the above equation to obtain:

\[ c = 3 \]

Therefore, at \( x = 3 \), there is a tangent to the graph of \( f \) that has a slope equal to zero (horizontal line) as shown in Figure 1 below.

use of Rolle's theorem example 1 — Figure 1. Rolle's Theorem, Example 1

Example 2

The graph of \( f(x) = \sin(x) + 2 \) for \( 0 \le x \le 2\pi \) is shown below. \( f(0) = f(2\pi) = 2 \) and \( f \) is continuous on \( [0, 2\pi] \) and differentiable on \( (0, 2\pi) \), hence, according to Rolle's theorem, there exists at least one value (there may be more than one!) of \( x = c \) such that \( f'(c) = 0 \).

Show Solution to Example 2

\[ f'(x) = \cos(x) \] \[ f'(c) = \cos(c) = 0 \]

The above equation has two solutions on the interval \( [0, 2\pi] \):

\[ c_1 = \dfrac{\pi}{2} \quad \text{and} \quad c_2 = \dfrac{3\pi}{2} \]

Therefore, at both \( x = \dfrac{\pi}{2} \) and \( x = \dfrac{3\pi}{2} \), there are tangents to the graph that have a slope equal to zero (horizontal lines) as shown in Figure 2 below.

use of Rolle's theorem example 2, two tangents — Figure 2. Rolle's Theorem, Example 2 with two tangents

Example 3

Function \( f \) in Figure 3 does not satisfy Rolle's theorem: although it is continuous and \( f(-1) = f(3) \), the function is not differentiable at \( x = 1 \), and therefore \( f'(c) = 0 \) with \( c \) in the interval \( (-1, 3) \) is not guaranteed. In fact, it is easy to see that there is no horizontal tangent to the graph of \( f \) on the interval \( (-1, 3) \).

Graphs of a function with at least one of the conditions not satisfied — Figure 3. Graph of function that does not satisfy the condition of differentiability in Rolle's Theorem

Practice Questions with Detailed Solutions

Click on each question to view its detailed step-by-step solution.

Question 1

Which of the functions given below satisfy all three conditions of Rolle's theorem?

a) \( f(x) = \cos(x) \), for \( x \) in \( [0, 2\pi] \)
b) \( g(x) = |x - 2| \), for \( x \) in \( [0, 4] \)
c) \( h(x) = \dfrac{1}{x^2} \), for \( x \) in \( [-1, 1] \)
d) \( k(x) = |\sin(x)| \), for \( x \) in \( [0, 2\pi] \)

Show Solution to Question 1

a) \( f(0) = 1 \) and \( f(2\pi) = 1 \), therefore \( f(0) = f(2\pi) \). \( f \) is continuous on \( [0, 2\pi] \) and differentiable in \( (0, 2\pi) \). Function \( f \) satisfies all conditions of Rolle's theorem.

b) Function \( g \) has a V-shaped graph with a vertex at \( x = 2 \) and is therefore not differentiable at \( x = 2 \). Function \( g \) does not satisfy all conditions of Rolle's theorem.

c) Function \( h \) is undefined at \( x = 0 \). Function \( h \) does not satisfy all conditions of Rolle's theorem.

d) \( k(x) = |\sin(x)| \) for \( x \) in \( [0, 2\pi] \). The graph of function \( k \) is shown below and reveals that function \( k \) is not differentiable at \( x = \pi \). Function \( k \) does not satisfy all conditions of Rolle's theorem.

Graph of k(x) = |sin(x)| , for x in [0 , 2π] — Figure 4. Graph of \( k(x) = |\sin(x)| \), for \( x \) in \( [0, 2\pi] \)

Question 2

Check that function \( f(x) = x^2 - 4x + 3 \) on the interval \( [1, 3] \) satisfies all conditions of Rolle's theorem, and then find all values of \( x = c \) such that \( f'(c) = 0 \).

Show Solution to Question 2

\( f \) is a polynomial function and is therefore continuous on the interval \( [1, 3] \) and differentiable on the interval \( (1, 3) \). Also, \( f(1) = f(3) = 0 \), and therefore function \( f \) satisfies all three conditions of Rolle's theorem, ensuring at least one value of \( x = c \) exists such that \( f'(c) = 0 \).

\[ f'(x) = 2x - 4 \] \[ f'(c) = 2c - 4 = 0 \implies c = 2 \]

The graph below shows \( f \), and the tangent at \( x = c = 2 \) is horizontal with \( f'(2) = 0 \).

use of Rolle's theorem question 2, f(x) = x^2 - 4x + 3 — Figure 5. Rolle's Theorem, Question 2, \( f(x) = x^2 - 4x + 3 \)

Question 3

Check that function \( g(x) = \cos(x) \) on the interval \( \left[-\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \) satisfies all conditions of Rolle's theorem, and then find all values \( x = c \) such that \( g'(c) = 0 \).

Show Solution to Question 3

Function \( g \) is a cosine function and is therefore continuous on the interval \( \left[-\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \) and differentiable on the interval \( \left(-\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right) \). Also:

\[ g\left(-\dfrac{\pi}{2}\right) = g\left(\dfrac{3\pi}{2}\right) = 0 \]

Therefore, function \( g \) satisfies all three conditions of Rolle's theorem, and there is at least one value of \( x = c \) such that \( g'(c) = 0 \).

\[ g'(x) = -\sin(x) \] \[ g'(c) = -\sin(c) = 0 \implies c = n\pi, \quad n = 0, \pm 1, \pm 2, \dots \]

Solutions on the interval \( \left[-\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] \) are:

\[ c_1 = 0 \quad \text{and} \quad c_2 = \pi \]

The graph below shows \( g \) and the tangents at \( x = c_1 = 0 \) and \( x = c_2 = \pi \) are horizontal, with \( g'(0) = 0 \) and \( g'(\pi) = 0 \).

use of Rolle's theorem question 3 , g(x) = cos(x) — Figure 6. Rolle's Theorem, Question 3, \( g(x) = \cos(x) \)