Lagrange's Mean Value Theorem: Statement, Proof & Solved Examples

The statement

Before you start: this theorem is a tilted version of Rolle's theorem. Read that lesson first if it's new to you.

Lagrange's mean value 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 \((a, b)\).

Conclusion

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

There are only two conditions. Compared with Rolle's theorem, the condition \(f(a) = f(b)\) has been dropped.

Two other forms you will meet

Multiply out: \(f(b) = f(a) + (b - a)\,f'(c)\).
Put \(b = a + h\) and write \(c = a + \theta h\) with \(0 < \theta < 1\): \[f(a + h) = f(a) + h\,f'(a + \theta h)\] This "\(\theta\) form" is the one used to prove Taylor's theorem.

The idea in one picture

The fraction \(\dfrac{f(b) - f(a)}{b - a}\) is the slope of the chord joining \(A(a, f(a))\) and \(B(b, f(b))\). The theorem says that somewhere between \(a\) and \(b\) the curve has a tangent with exactly that slope: a tangent parallel to the chord.

The green chord \(AB\) has slope \(\dfrac{f(b) - f(a)}{b - a}\). At \(c\), the orange tangent has exactly the same slope, so it is parallel to the chord. Here \(f(x) = \sqrt{x}\) on \([0.25, 4]\), and \(c = 1.5625\).

A real-life version: you drive 120 km in 2 hours, so your average speed is 60 km/h. The theorem says that at some instant your speedometer showed exactly 60 km/h. You can't average 60 without passing through 60 at least once. Average-speed cameras on highways use exactly this idea: if your average speed between two cameras is above the limit, you must have crossed the limit at some instant.

If \(f(a) = f(b)\), the chord is horizontal and the theorem becomes Rolle's theorem. Rolle's theorem is the special case; Lagrange's theorem is the general one.

Proof (using Rolle's theorem)

"State and prove" is a common exam question, and the proof is short. The trick is to subtract the chord from the curve so that both ends become equal.

Let \(k = \dfrac{f(b) - f(a)}{b - a}\) and define \[\phi(x) = f(x) - f(a) - k\,(x - a)\]
\(\phi\) is continuous on \([a, b]\) and differentiable on \((a, b)\), because \(f\) is and \(k(x - a)\) is a polynomial.
\(\phi(a) = 0\) and \(\phi(b) = f(b) - f(a) - k(b - a) = 0\). So \(\phi(a) = \phi(b)\).
By Rolle's theorem there is \(c \in (a, b)\) with \(\phi'(c) = 0\), that is \(f'(c) - k = 0\), so \(f'(c) = \dfrac{f(b) - f(a)}{b - a}\). \(\blacksquare\)

Two useful consequences

If \(f'(x) = 0\) for every \(x\) in an interval, then \(f\) is constant on that interval.
If \(f'(x) > 0\) on an interval, then \(f\) is increasing there; if \(f'(x) < 0\), it is decreasing.

Both follow directly: for any two points \(x_1 < x_2\), \(f(x_2) - f(x_1) = (x_2 - x_1)\,f'(c)\), and the sign of \(f'(c)\) decides the sign of the difference.

How to verify the theorem

Continuity on \([a, b]\), with a one-line reason.
Differentiability on \((a, b)\), with a one-line reason.
Slope of the chord: calculate \(\dfrac{f(b) - f(a)}{b - a}\).
Find \(c\): solve \(f'(c) = \) that slope, and keep only values strictly inside \((a, b)\).

Solved examples

Six examples, including two "prove an inequality" questions, which are the favourite exam use of this theorem.

Example 1: A quadratic

Easy

Verify Lagrange's mean value theorem for \(f(x) = x^2\) on \([1, 3]\).

\(f\) is a polynomial, so it is continuous on \([1, 3]\) and differentiable on \((1, 3)\).
Slope of the chord: \(\dfrac{f(3) - f(1)}{3 - 1} = \dfrac{9 - 1}{2} = 4\).
\(f'(c) = 2c = 4\) gives \(c = 2\), which lies in \((1, 3)\).

Answer\(c = 2\)

Example 2: A cubic

Easy

Verify the theorem for \(f(x) = x^3\) on \([0, 2]\).

Polynomial, so both conditions hold.
Slope of the chord: \(\dfrac{8 - 0}{2 - 0} = 4\).
\(3c^2 = 4\) gives \(c = \pm\dfrac{2}{\sqrt 3}\). Only \(c = \dfrac{2}{\sqrt 3} \approx 1.155\) lies in \((0, 2)\).

Answer\(c = \dfrac{2}{\sqrt 3}\)

Example 3: A logarithm

Medium

Verify the theorem for \(f(x) = \log x\) on \([1, e]\).

\(\log x\) is continuous and differentiable for \(x > 0\), so both conditions hold on \([1, e]\).
Slope of the chord: \(\dfrac{\log e - \log 1}{e - 1} = \dfrac{1}{e - 1}\).
\(f'(c) = \dfrac{1}{c} = \dfrac{1}{e - 1}\) gives \(c = e - 1 \approx 1.718\), which lies in \((1, e)\).

Answer\(c = e - 1\)

Example 4: A classic with an awkward root

Medium

Verify the theorem for \(f(x) = x(x - 1)(x - 2)\) on \(\left[0, \tfrac{1}{2}\right]\).

Expand: \(f(x) = x^3 - 3x^2 + 2x\), a polynomial. \(f(0) = 0\) and \(f\left(\tfrac{1}{2}\right) = \tfrac{1}{2}\cdot\left(-\tfrac{1}{2}\right)\cdot\left(-\tfrac{3}{2}\right) = \tfrac{3}{8}\).
Slope of the chord: \(\dfrac{3/8}{1/2} = \dfrac{3}{4}\).
\(f'(c) = 3c^2 - 6c + 2 = \dfrac{3}{4}\) gives \(12c^2 - 24c + 5 = 0\), so \(c = 1 \pm \dfrac{\sqrt{21}}{6}\).
\(1 + \dfrac{\sqrt{21}}{6} \approx 1.76\) is outside the interval. \(1 - \dfrac{\sqrt{21}}{6} \approx 0.236\) lies in \(\left(0, \tfrac{1}{2}\right)\).

Answer\(c = 1 - \dfrac{\sqrt{21}}{6}\)

Example 5: Proving an inequality

Exam level

Using Lagrange's mean value theorem, prove that \(\dfrac{b - a}{1 + b^2} < \tan^{-1} b - \tan^{-1} a < \dfrac{b - a}{1 + a^2}\) for \(0 < a < b\). Hence show that \(\dfrac{\pi}{4} + \dfrac{3}{25} < \tan^{-1}\dfrac{4}{3} < \dfrac{\pi}{4} + \dfrac{1}{6}\).

Apply the theorem to \(f(x) = \tan^{-1} x\) on \([a, b]\): \(\dfrac{\tan^{-1} b - \tan^{-1} a}{b - a} = \dfrac{1}{1 + c^2}\) for some \(a < c < b\).
Since \(0 < a < c < b\), we have \(1 + a^2 < 1 + c^2 < 1 + b^2\), so \(\dfrac{1}{1 + b^2} < \dfrac{1}{1 + c^2} < \dfrac{1}{1 + a^2}\).
Multiply through by \(b - a > 0\) to get the required inequality.
Now put \(a = 1\), \(b = \tfrac{4}{3}\): \(b - a = \tfrac{1}{3}\), \(1 + b^2 = \tfrac{25}{9}\), \(1 + a^2 = 2\), and \(\tan^{-1} 1 = \tfrac{\pi}{4}\). The bounds become \(\tfrac{3}{25}\) and \(\tfrac{1}{6}\).

Answer\(\dfrac{\pi}{4} + \dfrac{3}{25} < \tan^{-1}\dfrac{4}{3} < \dfrac{\pi}{4} + \dfrac{1}{6}\)

Example 6: When the theorem cannot be applied

Exam level

Can Lagrange's mean value theorem be applied to \(f(x) = |x|\) on \([-1, 2]\)?

\(|x|\) is continuous on \([-1, 2]\).
But it has a corner at \(x = 0\), which lies inside \((-1, 2)\), so it is not differentiable there.
Check: the chord slope is \(\dfrac{2 - 1}{3} = \dfrac{1}{3}\), while \(f'(x)\) is only ever \(-1\) or \(+1\). No \(c\) works.

AnswerNo. The differentiability condition fails at \(x = 0\).

Vipul Sir's tip

For "prove an inequality" questions, the pattern is always the same: pick the function whose difference appears in the middle, apply the theorem, then use \(a < c < b\) to bound \(f'(c)\) from both sides.

Common mistakes

1. Mixing up the order in the slope

Write \(\dfrac{f(b) - f(a)}{b - a}\), with the same order on top and bottom. \(\dfrac{f(b) - f(a)}{a - b}\) gives the wrong sign.

2. Keeping a value of \(c\) outside the interval

Equations for \(c\) often have two roots (Examples 2 and 4). Always check which ones lie strictly inside \((a, b)\).

3. Adding the condition \(f(a) = f(b)\)

That condition belongs to Rolle's theorem. Lagrange's theorem needs only continuity and differentiability.

4. Skipping the condition checks

Examiners give marks for stating both conditions with reasons. A correct \(c\) without them loses marks.

How it's asked in exams

Verify the theorem for a given function and interval, and find \(c\).
State and prove the theorem, often with its geometric meaning. Use the proof in section 03 and the diagram in section 02.
Prove an inequality, such as the \(\tan^{-1}\) bound in Example 5 or \(\dfrac{x}{1 + x} < \log(1 + x) < x\).
Find \(\theta\) in \(f(a + h) = f(a) + h f'(a + \theta h)\) for a given function.

Practice questions

Try each one on paper before you open the answer.

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

\(f(1) = 2\), \(f(4) = 11\), so the chord slope is \(\dfrac{9}{3} = 3\). \(2c - 2 = 3\) gives \(c = \dfrac{5}{2}\).

Q2Verify the theorem for \(f(x) = \sqrt{x}\) on \([1, 4]\).

Chord slope \(= \dfrac{2 - 1}{3} = \dfrac{1}{3}\). \(\dfrac{1}{2\sqrt c} = \dfrac{1}{3}\) gives \(\sqrt c = \dfrac{3}{2}\), so \(c = \dfrac{9}{4}\).

Q3Verify the theorem for \(f(x) = \dfrac{1}{x}\) on \([1, 3]\).

Chord slope \(= \dfrac{1/3 - 1}{2} = -\dfrac{1}{3}\). \(-\dfrac{1}{c^2} = -\dfrac{1}{3}\) gives \(c = \sqrt 3\).

Q4Verify the theorem for \(f(x) = e^x\) on \([0, 1]\).

Chord slope \(= e - 1\). \(e^c = e - 1\) gives \(c = \log(e - 1) \approx 0.541\), which lies in \((0, 1)\).

Q5Show that for \(f(x) = px^2 + qx + r\), the value of \(c\) on any interval \([a, b]\) is the midpoint \(\dfrac{a + b}{2}\).

Chord slope \(= \dfrac{p(b^2 - a^2) + q(b - a)}{b - a} = p(a + b) + q\). Setting \(f'(c) = 2pc + q\) equal to this gives \(c = \dfrac{a + b}{2}\).

Q6Prove that \(\dfrac{x}{1 + x} < \log(1 + x) < x\) for \(x > 0\).

Apply the theorem to \(f(t) = \log(1 + t)\) on \([0, x]\): \(\log(1 + x) = \dfrac{x}{1 + c}\) for some \(0 < c < x\).

Since \(1 < 1 + c < 1 + x\), we get \(\dfrac{x}{1 + x} < \dfrac{x}{1 + c} < x\).

Q7Prove that \(|\sin b - \sin a| \le |b - a|\) for all real \(a, b\).

By the theorem, \(\sin b - \sin a = (b - a)\cos c\) for some \(c\) between \(a\) and \(b\). Since \(|\cos c| \le 1\), the result follows. (If \(a = b\), both sides are 0.)

Q8Challenge. For \(f(x) = x^2\), find \(\theta\) in \(f(a + h) = f(a) + h\,f'(a + \theta h)\).

\((a + h)^2 = a^2 + h \cdot 2(a + \theta h)\) gives \(2ah + h^2 = 2ah + 2\theta h^2\), so \(\theta = \dfrac{1}{2}\) for every \(a\) and \(h\).

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Frequently asked questions

What is Lagrange's mean value theorem in simple words?

If a curve is smooth and unbroken between two points, then somewhere in between its tangent is parallel to the straight line joining those two points.

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

Rolle's theorem needs the extra condition \(f(a) = f(b)\) and concludes \(f'(c) = 0\). Lagrange's 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.

Why is it called the mean value theorem?

Because \(\dfrac{f(b) - f(a)}{b - a}\) is the average (mean) rate of change of \(f\) over \([a, b]\). The theorem says the instantaneous rate \(f'(c)\) equals this mean value at some point.

What is the geometric meaning of Lagrange's mean value theorem?

There is at least one point on the curve, strictly between \(A\) and \(B\), where the tangent is parallel to the chord \(AB\).

Is c unique?

Not necessarily. The theorem guarantees at least one value. A wavy curve can have several points where the tangent is parallel to the chord.

Where Lagrange's theorem leads next

← All 33 Calculus topics

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.

Lagrange's mean value theorem: the tangent parallel to the chord.