Proof of Cauchy's Mean Value Theorem

The Proof Strategy: Build Equal Endpoint Values

Cauchy’s Mean Value Theorem asserts that two functions have a common interior point where their derivatives satisfy a particular relation. The proof will use Rolle’s Theorem, which gives a point where a derivative is zero when a function has equal endpoint values. The key step is therefore to combine the two functions into one auxiliary function whose endpoint values match.

Write the endpoint increments as \(\Delta f=f(b)-f(a)\) and \(\Delta g=g(b)-g(a)\). A useful choice is \(H(x)=\Delta f\,g(x)-\Delta g\,f(x)\). Its change from \(a\) to \(b\) is \(\Delta f\,\Delta g-\Delta g\,\Delta f=0\). Thus \(H(a)=H(b)\), and Rolle’s Theorem can be applied. Differentiating \(H\) then produces exactly the terms in Cauchy’s identity.

This construction is designed around the conclusion. The coefficients \(\Delta f\) and \(-\Delta g\) are not arbitrary: they make the endpoint change cancel, while differentiation places the two derivatives in the required relation. The argument does not divide by either increment, so it also covers cases in which an increment is zero.

A Weighted Cancellation Principle

The auxiliary-function method can be stated in a form that makes its mechanism explicit. This principle will let us apply Rolle’s Theorem once, then obtain Cauchy’s theorem by choosing the coefficients appropriately.

Proof. Define \(H:[a,b]\to\mathbb{R}\) by \(H(x)=\alpha u(x)+\beta v(x)\). Since \(u\) and \(v\) are continuous on \([a,b]\), so is \(H\); since they are differentiable on \((a,b)\), so is \(H\) there. The assumed relation gives

Therefore \(H(a)=H(b)\). Rolle’s Theorem supplies a point \(c\in(a,b)\) such that \(H'(c)=0\). By the Sum Rule and the fact that \(\alpha,\beta\) are constants,

This is the claimed cancellation. The proof remains valid if one coefficient, or both coefficients, are zero: the hypotheses and the conclusion still make sense, and Rolle’s Theorem still applies. \(\square\)

The lemma gives a general way to locate a cancellation point. First choose weights that make the weighted endpoint changes sum to zero; then the weighted derivatives cancel at some interior point. For Cauchy’s theorem, the weights will be the other function’s endpoint increment, with one minus sign.

Proof of Cauchy’s Mean Value Theorem

Proof. Set \(\Delta f=f(b)-f(a)\) and \(\Delta g=g(b)-g(a)\). In the Weighted Derivative Cancellation Lemma, take \(u=f\), \(v=g\), \(\alpha=\Delta g\), and \(\beta=-\Delta f\). These choices satisfy the lemma’s endpoint condition because

The lemma therefore gives some \(c\in(a,b)\) such that

Rearranging this equality yields

as required. The lemma’s hypotheses hold because \(f\) and \(g\) are continuous on the closed interval and differentiable on its interior. No assumption that either endpoint increment is nonzero has been used. \(\square\)

Equivalently, the proof applies Rolle’s Theorem directly to the auxiliary function \(H(x)=\Delta f\,g(x)-\Delta g\,f(x)\). The weighted cancellation version simply packages the construction: its coefficients are \(-\Delta g\) for \(f\) and \(\Delta f\) for \(g\), which also make the endpoint changes cancel. Either choice of overall sign produces the same conclusion.

Worked Examples: Following the Construction

Why the Proof Avoids Division

The central proof step is an endpoint cancellation, not a ratio calculation. This matters because the hypotheses of Cauchy’s Mean Value Theorem do not require \(\Delta f\) or \(\Delta g\) to be nonzero. If either increment vanishes, the auxiliary function is still defined and its endpoint values still agree. Even when both increments vanish, Rolle’s Theorem applies.

After the theorem has been proved, a ratio form may be obtained when the relevant quantities are nonzero. For example, if \(\Delta g\ne0\) and \(g'(c)\ne0\), division gives

Those nonzero conditions belong to this rearrangement, not to the theorem or its proof. Keeping the identity as the primary conclusion prevents an invalid division from silently removing cases that the theorem covers.

The method is also a useful proof pattern beyond this theorem: form a linear combination whose endpoint change vanishes, apply Rolle’s Theorem, and interpret the resulting derivative equation. The coefficients must be chosen to cancel the endpoint changes exactly. In this proof, that cancellation is the algebraic equality \(\Delta f\,\Delta g-\Delta g\,\Delta f=0\).