Rolle's Theorem: Statement, Conditions & Solved Examples

Before you start: you need to be comfortable with differentiation. Revise with the Derivatives guide if it feels rusty.

The statement

Rolle's theorem

Let \(f\) be a real function such that

1. Continuity

\(f\) is continuous on the closed interval \([a, b]\);

2. Differentiability

\(f\) is differentiable on the open interval \((a, b)\);

3. Equal end values

\(f(a) = f(b)\).

Conclusion

Then there exists at least one \(c \in (a, b)\) such that \[f'(c) = 0\]

Notice the brackets. Continuity is needed on the closed interval, including the end points. Differentiability is only needed on the open interval, and the point \(c\) is always strictly between \(a\) and \(b\), never at an end.

The idea in one picture

Picture a curve that starts at point \(A\) and ends at point \(B\) at exactly the same height. If it goes up, it must come back down; if it goes down, it must come back up. At every turning point the tangent is horizontal, so its slope \(f'(c)\) is zero.

The curve starts and ends at the same height, \(f(a) = f(b)\). Here it has two points, \(c_1\) and \(c_2\), where the tangent is horizontal. Rolle's theorem promises at least one.

A real-life version: you leave home on a straight road and come back home later. At some instant you must have turned around, and at that instant your velocity was exactly zero. Position is \(f\), velocity is \(f'\), and leaving from and returning to the same place is \(f(a) = f(b)\).

Because the chord \(AB\) is horizontal, you can also say it this way: somewhere on the curve, the tangent is parallel to the chord \(AB\). Keep that sentence in mind, because it is exactly what Lagrange's mean value theorem generalises.

Why each condition matters

Drop any one condition and the conclusion can fail. These three counterexamples are worth remembering.

1. Not continuous on \([a, b]\)

Take \(f(x) = x\) for \(0 \le x < 1\) and \(f(1) = 0\). Then \(f(0) = f(1) = 0\), but the graph jumps at \(x = 1\). Inside the interval \(f'(x) = 1\), which is never zero.

2. Not differentiable on \((a, b)\)

Take \(f(x) = |x|\) on \([-1, 1]\). It is continuous and \(f(-1) = f(1) = 1\), but it has a sharp corner at \(x = 0\). The slope is \(-1\) on the left and \(+1\) on the right, and never zero.

3. \(f(a) \ne f(b)\)

Take \(f(x) = x\) on \([0, 1]\). It is continuous and differentiable, but \(f(0) = 0 \ne 1 = f(1)\), and \(f'(x) = 1\) is never zero.

Vipul Sir's tip

The conditions are sufficient, not necessary. If they hold, a \(c\) is guaranteed. If one fails, Rolle's theorem simply says nothing: a suitable \(c\) may or may not exist. For example, \(f(x) = x^2\) on \([-1, 2]\) has \(f(-1) \ne f(2)\), yet \(f'(0) = 0\).

How to verify Rolle's theorem

"Verify Rolle's theorem for \(f(x)\) on \([a, b]\)" is the most common question. Always follow the same four steps and write each one down:

Continuity on \([a, b]\). Polynomials, \(\sin x\), \(\cos x\) and \(e^x\) are continuous everywhere. Watch out for denominators that become zero, \(\log\) of zero or negative numbers, \(\tan x\) at odd multiples of \(\tfrac{\pi}{2}\), and \(|x|\).
Differentiability on \((a, b)\). Watch for corners (\(|x|\)) and fractional powers such as \(x^{2/3}\), whose derivative blows up.
Equal end values. Calculate \(f(a)\) and \(f(b)\) and show they are equal.
Find \(c\). Solve \(f'(c) = 0\) and keep only the values strictly inside \((a, b)\).

If you find at least one such \(c\), Rolle's theorem is verified.

Solved examples

Six examples, from warm-up to university-exam level. The last one is a case where the theorem cannot be applied.

Example 1: A quadratic

Easy

Verify Rolle's theorem for \(f(x) = x^2 - 5x + 6\) on \([2, 3]\).

\(f\) is a polynomial, so it is continuous on \([2, 3]\) and differentiable on \((2, 3)\).
\(f(2) = 4 - 10 + 6 = 0\) and \(f(3) = 9 - 15 + 6 = 0\), so \(f(2) = f(3)\).
\(f'(x) = 2x - 5 = 0\) gives \(x = \tfrac{5}{2}\), which lies in \((2, 3)\).

Answer\(c = \tfrac{5}{2}\). Rolle's theorem is verified.

Example 2: A trigonometric function

Easy

Verify Rolle's theorem for \(f(x) = \sin x + \cos x\) on \(\left[0, \tfrac{\pi}{2}\right]\).

\(\sin x\) and \(\cos x\) are continuous and differentiable everywhere, so both conditions hold.
\(f(0) = 0 + 1 = 1\) and \(f\left(\tfrac{\pi}{2}\right) = 1 + 0 = 1\).
\(f'(x) = \cos x - \sin x = 0\) gives \(\tan x = 1\), so \(x = \tfrac{\pi}{4}\), which lies in \(\left(0, \tfrac{\pi}{2}\right)\).

Answer\(c = \tfrac{\pi}{4}\)

Example 3: Throw away the end point

Medium

Verify Rolle's theorem for \(f(x) = x(x-3)^2\) on \([0, 3]\).

Polynomial, so continuous and differentiable. \(f(0) = 0\) and \(f(3) = 0\).
Product rule: \(f'(x) = (x-3)^2 + 2x(x-3) = (x-3)(3x-3) = 3(x-3)(x-1)\).
\(f'(x) = 0\) gives \(x = 3\) or \(x = 1\). But \(x = 3\) is an end point, not inside \((0, 3)\), so reject it.

Answer\(c = 1\)

Example 4: Exponential times sine

Medium

Verify Rolle's theorem for \(f(x) = e^x \sin x\) on \([0, \pi]\).

Product of functions that are continuous and differentiable everywhere, so both conditions hold.
\(f(0) = e^0 \sin 0 = 0\) and \(f(\pi) = e^{\pi}\sin\pi = 0\).
\(f'(x) = e^x(\sin x + \cos x)\). Since \(e^x \ne 0\), we need \(\sin x + \cos x = 0\), that is \(\tan x = -1\), so \(x = \tfrac{3\pi}{4}\), which lies in \((0, \pi)\).

Answer\(c = \tfrac{3\pi}{4}\)

Example 5: A classic university question

Exam level

Verify Rolle's theorem for \(f(x) = \log\left(\dfrac{x^2 + ab}{(a+b)\,x}\right)\) on \([a, b]\), where \(0 < a < b\).

For \(x\) in \([a, b]\), \(x > 0\), so the expression inside the log is positive. Hence \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\).
\(f(a) = \log\dfrac{a^2 + ab}{(a+b)a} = \log\dfrac{a(a+b)}{a(a+b)} = \log 1 = 0\). In the same way, \(f(b) = \log 1 = 0\).
Write \(f(x) = \log(x^2 + ab) - \log(a+b) - \log x\). Then \[f'(x) = \frac{2x}{x^2 + ab} - \frac{1}{x} = \frac{x^2 - ab}{x\,(x^2 + ab)}\]
\(f'(c) = 0\) gives \(c^2 = ab\), so \(c = \sqrt{ab}\) (taking the positive root). Since \(a < b\), the geometric mean \(\sqrt{ab}\) lies strictly between \(a\) and \(b\).

Answer\(c = \sqrt{ab}\)

Example 6: When the theorem cannot be applied

Exam level

Can Rolle's theorem be applied to \(f(x) = 1 - (x-1)^{2/3}\) on \([0, 2]\)? Justify.

\(f(0) = 1 - (-1)^{2/3} = 1 - 1 = 0\) and \(f(2) = 1 - 1^{2/3} = 0\), so the end values are equal. \(f\) is also continuous on \([0, 2]\).
But \(f'(x) = -\dfrac{2}{3\,(x-1)^{1/3}}\), which does not exist at \(x = 1\), a point inside \((0, 2)\). The graph has a sharp point (a cusp) there.
So \(f\) is not differentiable on \((0, 2)\), and the second condition fails. Notice also that \(f'(x)\) is never zero.

AnswerNo. Rolle's theorem does not apply, because \(f\) is not differentiable at \(x = 1\).

Common mistakes

1. Keeping a value of \(c\) that isn't inside the interval

\(c\) must lie strictly between \(a\) and \(b\). In Example 3, \(f'(x) = 0\) at \(x = 1\) and \(x = 3\), but only \(c = 1\) counts. Always check and state that \(c \in (a, b)\).

2. Skipping the continuity check

\(f(x) = \tan x\) on \([0, \pi]\) has \(f(0) = f(\pi) = 0\), but \(\tan x\) is not continuous at \(x = \tfrac{\pi}{2}\). And \(f'(x) = \sec^2 x\) is never zero. Checking only \(f(a) = f(b)\) would give a wrong answer.

3. Thinking there is exactly one \(c\)

The theorem says at least one. In practice question 2 below there are two valid values, and you must give both.

4. Reading the theorem backwards

If a condition fails, you can only say the theorem does not apply. You cannot conclude that no \(c\) exists (see the tip in section 03).

How it's asked in exams

Verify Rolle's theorem for a given function on a given interval, and find \(c\). This is the most common form.
Can Rolle's theorem be applied? Justify. Here you must find the condition that fails, as in Example 6.
State Rolle's theorem and give its geometric meaning. State all three conditions and draw the diagram from section 02.
Use Rolle's theorem to prove a result, for example that an equation has exactly one real root (practice question 7).

Exam tip

Write the three conditions as three separate, numbered lines, each with a one-line reason. Examiners look for all three, and many students lose marks simply by skipping the continuity or differentiability line.

Practice questions

Try each one on paper before you open the answer.

Q1Verify Rolle's theorem for \(f(x) = x^2 - 4x + 3\) on \([1, 3]\).

\(f(1) = 0 = f(3)\). \(f'(x) = 2x - 4 = 0\) gives \(c = 2 \in (1, 3)\).

Q2Verify Rolle's theorem for \(f(x) = x^3 - 6x^2 + 11x - 6\) on \([1, 3]\).

\(f(1) = 1 - 6 + 11 - 6 = 0\) and \(f(3) = 27 - 54 + 33 - 6 = 0\). \(f'(x) = 3x^2 - 12x + 11 = 0\) gives \[x = \frac{12 \pm \sqrt{144 - 132}}{6} = 2 \pm \frac{1}{\sqrt{3}}\]

Both values (about 1.42 and 2.58) lie in \((1, 3)\), so there are two values of \(c\).

Q3Verify Rolle's theorem for \(f(x) = \sin 2x\) on \(\left[0, \tfrac{\pi}{2}\right]\).

\(f(0) = 0 = f\left(\tfrac{\pi}{2}\right)\). \(f'(x) = 2\cos 2x = 0\) gives \(2x = \tfrac{\pi}{2}\), so \(c = \tfrac{\pi}{4}\).

Q4Verify Rolle's theorem for \(f(x) = e^{-x}\sin x\) on \([0, \pi]\).

\(f(0) = 0 = f(\pi)\). \(f'(x) = e^{-x}(\cos x - \sin x) = 0\) gives \(\tan x = 1\), so \(c = \tfrac{\pi}{4}\).

Q5Can Rolle's theorem be applied to \(f(x) = |x - 2|\) on \([0, 4]\)?

No. \(f(0) = f(4) = 2\) and \(f\) is continuous, but it has a corner at \(x = 2\), so it is not differentiable on \((0, 4)\).

Q6Verify Rolle's theorem for \(f(x) = x(x+3)\,e^{-x/2}\) on \([-3, 0]\).

\(f(-3) = 0 = f(0)\). By the product rule, \[f'(x) = e^{-x/2}\left[(2x + 3) - \frac{x^2 + 3x}{2}\right] = -\frac{e^{-x/2}}{2}\,(x - 3)(x + 2)\]

So \(f'(x) = 0\) at \(x = 3\) or \(x = -2\). Only \(c = -2\) lies in \((-3, 0)\).

Q7Show that the equation \(x^3 + x - 1 = 0\) has exactly one real root.

At least one root: let \(f(x) = x^3 + x - 1\). Then \(f(0) = -1 < 0\) and \(f(1) = 1 > 0\), and \(f\) is continuous, so it crosses zero somewhere in \((0, 1)\).

At most one root: suppose there were two roots \(r_1 < r_2\). Then \(f(r_1) = f(r_2) = 0\), and by Rolle's theorem \(f'(c) = 3c^2 + 1 = 0\) for some \(c\). But \(3c^2 + 1 \ge 1\) for every \(c\), which is a contradiction. So there is exactly one real root.

Q8Challenge. For \(f(x) = (x-a)^m (x-b)^n\) on \([a, b]\), where \(m, n\) are positive integers, find \(c\).

\(f(a) = f(b) = 0\). Differentiating and taking out common factors, \[f'(x) = (x-a)^{m-1}(x-b)^{n-1}\big[m(x-b) + n(x-a)\big]\]

Inside \((a, b)\) the first two factors are non-zero, so \(m(c - b) + n(c - a) = 0\), giving \[c = \frac{mb + na}{m + n}\] This point divides \([a, b]\) internally in the ratio \(m : n\), so it always lies inside the interval.

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Frequently asked questions

What is Rolle's theorem in simple words?

If a smooth, unbroken curve starts and ends at the same height, then somewhere in between it has a flat (horizontal) tangent, where the derivative is zero.

What are the three conditions of Rolle's theorem?

\(f\) must be continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and satisfy \(f(a) = f(b)\).

What is the geometric meaning of Rolle's theorem?

There is at least one point on the curve between \(A\) and \(B\) where the tangent is parallel to the \(x\)-axis. Because \(f(a) = f(b)\), that tangent is also parallel to the chord \(AB\).

Can there be more than one value of c?

Yes. The theorem guarantees at least one, but there can be several. Practice question 2 has two.

How is Rolle's theorem different from Lagrange's mean value theorem?

Rolle's theorem needs \(f(a) = f(b)\) and concludes \(f'(c) = 0\). Lagrange's mean value theorem drops that condition and concludes \(f'(c) = \dfrac{f(b) - f(a)}{b - a}\). Rolle's theorem is the special case where the chord is horizontal, and it is used to prove Lagrange's theorem.

Where Rolle's theorem leads next

Written by Vipul Sir

Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.

Rolle's theorem: same height at both ends, a flat tangent in between.