# Cauchy’s Theorem

Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/cauchy-theorem/
Fetched from algebrica.org post 6206; source modified 2026-02-28T20:15:41.

## Introduction

**Cauchy’s Theorem** establishes a relationship between the changes of two functions over a given interval. Specifically, if \\( f(x) \\) and \\( g(x) \\) are [continuous](../continuous-functions.md) on a closed interval \\(\[a, b\]\\) and differentiable in its interior, with \\( g’(x) \\neq 0 \\), then there exists at least one point \\( c \\) in \\( (a, b) \\) where the ratio of their [derivatives](../derivatives.md) matches the ratio of their overall change across the interval:

## Statement

The Cauchy’s theorem states the following. Let \\( f(x) \\) and \\( g(x) \\) be two functions such that:

-   \\( f(x) \\) and \\( g(x) \\) are continuous on the interval \\(\[a, b\]\\).
-   \\( f(x) \\) and \\( g(x) \\) are differentiable at every point in the interior of the interval.
-   \\( g’(x) \\neq 0 \\) for every \\( x \\) in the interior of \\(\[a, b\]\\).

Then, there exists at least one point \\( c \\) in the interior of the interval \\(\[a, b\]\\) such that:

\\[
\\frac{f’ \\left(c \\right)}{g’ \\left(c \\right)} = \\frac{f(b)-f(a)}{g(b)-g(a)}
\\]

that is, the ratio of the increments of the functions \\( f(x) \\) and \\( g(x) \\) over the interval \\(\[a, b\]\\) is equal to the ratio of their respective derivatives evaluated at a point \\( c \\) in the interior of the interval.

Setting \\(g(x) = x\\) reduces Cauchy’s theorem directly to [Lagrange’s theorem](../lagrange-theorem.md). With \\(g’(x) = 1\\) and \\(g(b) - g(a) = b - a\\), the conclusion takes the form:

\\[
\\frac{f’(c)}{1} = \\frac{f(b) - f(a)}{b - a}
\\]

 Lagrange’s theorem is therefore a special case of Cauchy’s theorem, obtained when one of the two functions is the identity.

##### Cauchy’s theorem provides the theoretical basis for the proof of [L’Hôpital’s Rule](../hopital-rule.md).

## Proof of Cauchy’s theorem

To prove the theorem, define a new function \\( \\varphi(x) \\) where \\( \\lambda \\) is a constant to be determined.:

\\[
\\varphi(x) = f(x)-\\lambda g(x)
\\]

We want to choose \\( \\lambda \\) such that \\( \\varphi(a) = \\varphi(b) \\). From this condition, we obtain:

\\[
\\lambda = \\frac{f(b)-f(a)}{g(b)-g(a)}
\\]

* * *

Now the function \\( \\varphi(x) \\) is continuous on \\(\[a, b\]\\) and differentiable on \\((a, b)\\), with \\( \\varphi(a) = \\varphi(b) \\). We can apply the [Rolle’s Theorem](../rolles-theorem.md), which guarantees that there exists at least one point \\( c \\in (a, b) \\) such that \\( \\varphi’ \\left(c \\right) = 0 \\). By calculating the derivative of \\( \\varphi(x) \\) and setting \\( \\varphi’ \\left(c \\right) = 0 \\), we obtain:

\\[
\\varphi’(x) = f’(x) - \\lambda g’(x)
\\]



\\[
f’ \\left(c \\right) = \\lambda g’ \\left(c \\right)
\\]

* * *

Substituting the value of \\( \\lambda \\), we get:

\\[
f’ \\left(c \\right) = \\frac{f(b) - f(a)}{g(b) - g(a)} g’ \\left(c \\right)
\\]

By dividing both sides by \\( g’ \\left(c \\right ) \\), we obtain:

\\[
\\frac{f’ \\left(c \\right)}{g’ \\left(c \\right)} = \\frac{f(b) - f(a)}{g(b) - g(a)}
\\]

Thus, we have proven the conclusion of the theorem.

## Example 1

Let’s verify, for example, that the theorem is applicable to the functions \\(f(x)\\) and \\(g(x) \\) in the interval \[1,3\]:

\\[
f(x) = 2x^2-4x+2
\\]



\\[
g(x) = x^2
\\]

The two functions are [polynomials](../polynomials.md), therefore they are continuous and differentiable for every \\( x \\in \\mathbb{R} \\). Moreover, \\( g’(x) = 2x \\neq 0 \\). This satisfies the hypotheses of the theorem.

* * *

Let’s now verify the existence of a point \\( c \\in (1,3) \\) such that:

\\[
\\frac{f’ \\left(c \\right)}{g’ \\left(c \\right)} = \\frac{f(3)-f(1)}{g(3)-g(1)}
\\]

We obtain:

\\[
\\frac{f(3)-f(1)}{g(3)-g(1)} = \\frac{18-12+2-2+4-2}{9-1} = \\frac{8}{8} = 1
\\]

* * *

Now let’s calculate:

\\[
\\frac{f’ \\left(c \\right)}{g’ \\left(c \\right)} = \\frac{4c-4}{2c}
\\]

From \\(1\\), the equality becomes:

\\[
\\begin{align} \\frac{4c - 4}{2c} &= 1 \\[0.5em] \\frac{4c - 4}{2c} &= \\frac{2c}{2c} \\[0.5em] 4c - 4 &= 2c \\[0.5em] 2c &= 4 \\[0.5em] c &= 2\\\\ \\end{align}
\\]

Since \\( c = 2 \\in \[1,3\] \\), the theorem is verified.

## A note on the hypothesis \\(g’(x) \\neq 0\\)

Consider the particular case in which \\(g(b) = g(a)\\). In this situation, the denominator of the following ratio is zero, and the expression is undefined:

\\[
\\frac{f(b)-f(a)}{g(b)-g(a)}
\\]

 For this reason, the constant cannot be introduced:

\\[
\\lambda = \\frac{f(b)-f(a)}{g(b)-g(a)}
\\]

 As a result, the proof utilising the following auxiliary function is no longer valid:

\\[
\\varphi(x) = f(x) - \\lambda g(x)
\\]

 Under the hypotheses of Cauchy’s theorem, this situation does not arise. The condition \\(g’(x) \\neq 0\\) within the interior of the interval ensures that g is strictly monotonic, which in turn guarantees that \\(g(b) \\neq g(a)\\). Therefore, the case \\(g(b) = g(a)\\) is precluded by the theorem’s assumptions.

## Selected references

-   **University of Florida J. Keesling**. [The Cauchy Mean Value Theorem](https://people.clas.ufl.edu/kees/files/CauchyMeanValue.pdf)

-   **University of Maryland**. [The Cauchy Mean Value Theorem and Consequences](https://math.umd.edu/~immortal/MATH410/lecturenotes/ch4-4.pdf)

-   **UC Berkeley A. Vizeff**. [Mean Value Theorem and L’Hôpital’s Rule](https://math.berkeley.edu/~avizeff/calculus-I-F23/lecture-16.pdf)