The statement
Before you start: this theorem extends Lagrange's mean value theorem to two functions, and its proof uses Rolle's theorem.
Let \(f\) and \(g\) be real functions such that
1. Continuity
\(f\) and \(g\) are continuous on the closed interval \([a, b]\);
2. Differentiability
\(f\) and \(g\) are differentiable on the open interval \((a, b)\);
3. Non-zero derivative
\(g'(x) \ne 0\) for every \(x \in (a, b)\).
Conclusion
Then there exists at least one \(c \in (a, b)\) such that \[\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}\]
The third condition does two jobs. It keeps the right-hand side defined, and it also guarantees \(g(b) \ne g(a)\): if \(g(b) = g(a)\), Rolle's theorem would give a point where \(g' = 0\). So the left-hand side is never \(\tfrac{0}{0}\).
It is also called the generalised or extended mean value theorem.
Where it fits
The three mean value theorems form a family. Each one is a special case of the next:
| Theorem | Extra condition | Conclusion |
|---|---|---|
| Rolle's | \(f(a) = f(b)\) | \(f'(c) = 0\) |
| Lagrange's | none | \(f'(c) = \dfrac{f(b) - f(a)}{b - a}\) |
| Cauchy's | a second function \(g\), with \(g' \ne 0\) | \(\dfrac{f'(c)}{g'(c)} = \dfrac{f(b) - f(a)}{g(b) - g(a)}\) |
Put \(g(x) = x\) in Cauchy's theorem: \(g'(x) = 1\) and \(g(b) - g(a) = b - a\), which gives exactly Lagrange's theorem.
The geometric picture. Think of \(x = g(t)\), \(y = f(t)\) as a curve traced out as \(t\) runs from \(a\) to \(b\). The slope of the curve at time \(t\) is \(\dfrac{dy}{dx} = \dfrac{f'(t)}{g'(t)}\). Cauchy's theorem says that at some moment \(c\), the tangent to this curve is parallel to the chord joining its start and end points: the same picture as Lagrange's theorem, for a curve given in parametric form.
Cauchy's theorem is also the key step in proving L'Hôpital's rule.
Proof (using Rolle's theorem)
- As explained above, \(g(b) \ne g(a)\). Let \(k = \dfrac{f(b) - f(a)}{g(b) - g(a)}\) and define \[\phi(x) = f(x) - f(a) - k\,\big[g(x) - g(a)\big]\]
- \(\phi\) is continuous on \([a, b]\) and differentiable on \((a, b)\), because \(f\) and \(g\) are.
- \(\phi(a) = 0\), and \(\phi(b) = f(b) - f(a) - k\,[g(b) - g(a)] = 0\).
- By Rolle's theorem, \(\phi'(c) = f'(c) - k\,g'(c) = 0\) for some \(c \in (a, b)\). Since \(g'(c) \ne 0\), divide to get \(\dfrac{f'(c)}{g'(c)} = k\). \(\blacksquare\)
How to verify the theorem
- Continuity of both \(f\) and \(g\) on \([a, b]\).
- Differentiability of both on \((a, b)\).
- Check \(g'(x) \ne 0\) on \((a, b)\).
- Calculate \(\dfrac{f(b) - f(a)}{g(b) - g(a)}\), simplify, set it equal to \(\dfrac{f'(c)}{g'(c)}\), and keep the values of \(c\) inside \((a, b)\).
Solved examples
Examples 2, 3 and 4 give a lovely pattern worth remembering: for three standard pairs of functions, \(c\) turns out to be the arithmetic, geometric and harmonic mean of \(a\) and \(b\).
Example 1: Two powers
EasyVerify Cauchy's mean value theorem for \(f(x) = x^2\) and \(g(x) = x^3\) on \([1, 2]\).
- Both are polynomials, so continuity and differentiability hold. \(g'(x) = 3x^2 \ne 0\) on \((1, 2)\).
- \(\dfrac{f(2) - f(1)}{g(2) - g(1)} = \dfrac{4 - 1}{8 - 1} = \dfrac{3}{7}\).
- \(\dfrac{f'(c)}{g'(c)} = \dfrac{2c}{3c^2} = \dfrac{2}{3c}\). Setting \(\dfrac{2}{3c} = \dfrac{3}{7}\) gives \(c = \dfrac{14}{9} \approx 1.56\), which lies in \((1, 2)\).
Answer\(c = \dfrac{14}{9}\)
Example 2: The arithmetic mean
MediumVerify the theorem for \(f(x) = e^x\) and \(g(x) = e^{-x}\) on \([a, b]\).
- Both are continuous and differentiable everywhere, and \(g'(x) = -e^{-x} \ne 0\).
- \(g(b) - g(a) = e^{-b} - e^{-a} = \dfrac{e^a - e^b}{e^{a+b}}\), so \(\dfrac{f(b) - f(a)}{g(b) - g(a)} = \dfrac{(e^b - e^a)\,e^{a+b}}{e^a - e^b} = -e^{a+b}\).
- \(\dfrac{f'(c)}{g'(c)} = \dfrac{e^c}{-e^{-c}} = -e^{2c}\). Setting \(-e^{2c} = -e^{a+b}\) gives \(2c = a + b\).
Answer\(c = \dfrac{a + b}{2}\), the arithmetic mean of \(a\) and \(b\)
Example 3: The geometric mean
MediumVerify the theorem for \(f(x) = \sqrt{x}\) and \(g(x) = \dfrac{1}{\sqrt{x}}\) on \([a, b]\), where \(0 < a < b\).
- Both are continuous and differentiable for \(x > 0\), and \(g'(x) = -\dfrac{1}{2x^{3/2}} \ne 0\).
- \(g(b) - g(a) = \dfrac{1}{\sqrt b} - \dfrac{1}{\sqrt a} = \dfrac{\sqrt a - \sqrt b}{\sqrt{ab}}\), so the ratio is \(\dfrac{(\sqrt b - \sqrt a)\sqrt{ab}}{\sqrt a - \sqrt b} = -\sqrt{ab}\).
- \(\dfrac{f'(c)}{g'(c)} = \dfrac{1/(2\sqrt c)}{-1/(2c^{3/2})} = -c\). So \(c = \sqrt{ab}\), which lies between \(a\) and \(b\).
Answer\(c = \sqrt{ab}\), the geometric mean
Example 4: The harmonic mean
MediumVerify the theorem for \(f(x) = \dfrac{1}{x^2}\) and \(g(x) = \dfrac{1}{x}\) on \([a, b]\), where \(0 < a < b\).
- Both are continuous and differentiable for \(x > 0\), and \(g'(x) = -\dfrac{1}{x^2} \ne 0\).
- \(f(b) - f(a) = \dfrac{a^2 - b^2}{a^2 b^2}\) and \(g(b) - g(a) = \dfrac{a - b}{ab}\), so the ratio is \(\dfrac{a^2 - b^2}{a^2b^2}\cdot\dfrac{ab}{a - b} = \dfrac{a + b}{ab}\).
- \(\dfrac{f'(c)}{g'(c)} = \dfrac{-2/c^3}{-1/c^2} = \dfrac{2}{c}\). Setting \(\dfrac{2}{c} = \dfrac{a + b}{ab}\) gives \(c = \dfrac{2ab}{a + b}\).
Answer\(c = \dfrac{2ab}{a + b}\), the harmonic mean
Example 5: A trigonometric pair
MediumVerify the theorem for \(f(x) = \sin x\) and \(g(x) = \cos x\) on \(\left[0, \tfrac{\pi}{2}\right]\).
- Both are continuous and differentiable, and \(g'(x) = -\sin x \ne 0\) on \(\left(0, \tfrac{\pi}{2}\right)\).
- \(\dfrac{\sin(\pi/2) - \sin 0}{\cos(\pi/2) - \cos 0} = \dfrac{1}{-1} = -1\).
- \(\dfrac{f'(c)}{g'(c)} = \dfrac{\cos c}{-\sin c} = -\cot c = -1\) gives \(\cot c = 1\), so \(c = \dfrac{\pi}{4}\).
Answer\(c = \dfrac{\pi}{4}\)
Example 6: When the theorem cannot be applied
Exam levelCan Cauchy's mean value theorem be applied to \(f(x) = x^3\) and \(g(x) = x^2\) on \([-1, 1]\)?
- Both functions are continuous and differentiable.
- But \(g'(x) = 2x = 0\) at \(x = 0\), which lies inside \((-1, 1)\). The third condition fails.
- Indeed \(g(1) - g(-1) = 1 - 1 = 0\), so the left-hand side is not even defined.
AnswerNo, because \(g'(0) = 0\).
Common mistakes
1. Applying Lagrange's theorem to \(f\) and \(g\) separately, then dividing
Lagrange's theorem gives \(f(b) - f(a) = (b - a) f'(c_1)\) and \(g(b) - g(a) = (b - a) g'(c_2)\), but \(c_1\) and \(c_2\) are usually different points. Dividing gives \(\dfrac{f'(c_1)}{g'(c_2)}\), not \(\dfrac{f'(c)}{g'(c)}\). The whole point of Cauchy's theorem is that one single \(c\) works for both.
2. Forgetting to check \(g'(x) \ne 0\)
This is the condition students most often skip, and it's exactly the one that fails in Example 6.
3. Using the quotient rule
\(\dfrac{f'(c)}{g'(c)}\) is the ratio of two separate derivatives. It is not the derivative of \(\dfrac{f}{g}\).
How it's asked in exams
- Verify the theorem for a given pair of functions and find \(c\). The AM, GM and HM pairs in Examples 2 to 4 are favourites.
- State and prove the theorem, and show that Lagrange's theorem is a special case.
- Show that \(c\) is a particular mean of \(a\) and \(b\) for a given pair of functions.
Practice questions
Try each one on paper before you open the answer.
Q1Verify the theorem for \(f(x) = x^2\) and \(g(x) = x^3\) on \([0, 1]\).
The ratio is \(\dfrac{1 - 0}{1 - 0} = 1\), and \(\dfrac{2c}{3c^2} = \dfrac{2}{3c} = 1\) gives \(c = \dfrac{2}{3}\). (Note \(g'(0) = 0\) is fine: only the open interval matters.)
Q2Verify the theorem for \(f(x) = \log x\) and \(g(x) = \dfrac{1}{x}\) on \([1, e]\).
The ratio is \(\dfrac{1 - 0}{\frac{1}{e} - 1} = \dfrac{e}{1 - e}\). \(\dfrac{1/c}{-1/c^2} = -c\), so \(c = \dfrac{e}{e - 1} \approx 1.58\), which lies in \((1, e)\).
Q3Verify the theorem for \(f(x) = e^x\) and \(g(x) = e^{-x}\) on \([0, 1]\).
By Example 2, \(c = \dfrac{0 + 1}{2} = \dfrac{1}{2}\).
Q4Show that for \(f(x) = \sin x\), \(g(x) = \cos x\) on \([a, b]\) with \(0 < a < b < \tfrac{\pi}{2}\), the value of \(c\) is \(\dfrac{a + b}{2}\).
Using sum-to-product formulas, \(\dfrac{\sin b - \sin a}{\cos b - \cos a} = \dfrac{2\cos\frac{a+b}{2}\sin\frac{b-a}{2}}{-2\sin\frac{a+b}{2}\sin\frac{b-a}{2}} = -\cot\dfrac{a + b}{2}\).
Since \(\dfrac{f'(c)}{g'(c)} = -\cot c\), we get \(c = \dfrac{a + b}{2}\).
Q5Show that Lagrange's mean value theorem follows from Cauchy's.
Take \(g(x) = x\). Then \(g'(x) = 1 \ne 0\), and Cauchy's conclusion becomes \(\dfrac{f(b) - f(a)}{b - a} = \dfrac{f'(c)}{1}\).
Q6Challenge. Can the theorem be applied to \(f(x) = x\) and \(g(x) = |x|\) on \([-1, 2]\)?
No. \(g\) is not differentiable at \(x = 0\), which lies inside \((-1, 2)\).
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Frequently asked questions
What is Cauchy's mean value theorem in simple words?
For two smooth functions, there is a point where the ratio of their rates of change equals the ratio of their total changes over the interval.
What is the difference between Lagrange's and Cauchy's mean value theorems?
Lagrange's theorem is about one function. Cauchy's theorem is about two functions at once, and reduces to Lagrange's theorem when \(g(x) = x\).
Why do we need g′(x) ≠ 0?
So that \(\dfrac{f'(c)}{g'(c)}\) is defined, and so that \(g(b) \ne g(a)\), which keeps the left-hand side defined.
Where is Cauchy's mean value theorem used?
Mainly to prove L'Hôpital's rule, and Taylor's theorem with the remainder term.
Where Cauchy's theorem leads next
← All 33 Calculus topics
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.