How to Prove Cauchy’s and Lagrange’s Mean Value Theorems Without Rolle’s Theorem

You can prove both mean value theorems without citing Rolle’s theorem by using the extreme value theorem and Fermat’s theorem on stationary points. The key is to construct a function whose values at the two endpoints are equal, then find a point inside the interval where its derivative is zero.

This proof assumes the usual hypotheses: the functions are continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$. It does not divide by an endpoint change, so it also covers cases where one or both functions have equal endpoint values.

Cauchy’s mean value theorem

Let $f$ and $g$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Cauchy’s mean value theorem says there is some $c\in(a,b)$ such that

$$(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c).$$

To prove it, define the auxiliary function

$$H(x)=(f(b)-f(a))g(x)-(g(b)-g(a))f(x).$$

It is continuous on $[a,b]$ and differentiable on $(a,b)$. Write $\Delta f=f(b)-f(a)$ and $\Delta g=g(b)-g(a)$. The difference between its endpoint values is

$$H(b)-H(a)=\Delta f\,\Delta g-\Delta g\,\Delta f=0,$$

so $H(a)=H(b)$.

There are two cases:

1. $H$ is constant. Then $H'(x)=0$ at every $x\in(a,b)$, so the desired derivative equation holds at any such point.
2. $H$ is not constant. Since its endpoint values agree, some interior value must differ from that shared endpoint value. If an interior value is larger, the maximum on $[a,b]$ is attained at an interior point; if one is smaller, the minimum is attained there. This follows from the extreme value theorem. At that interior maximum or minimum, Fermat’s theorem gives $H'(c)=0$.

Differentiating the definition of $H$ gives

$$H'(x)=\Delta f\,g'(x)-\Delta g\,f'(x).$$

Thus $H'(c)=0$ is exactly

$$(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c),$$

which proves Cauchy’s theorem without using Rolle’s theorem.

Lagrange’s mean value theorem follows

Take $g(x)=x$ in Cauchy’s theorem. Then $g(b)-g(a)=b-a$ and $g'(c)=1$, so the equation becomes

$$f(b)-f(a)=(b-a)f'(c).$$

Equivalently,

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$

for some $c\in(a,b)$. This is Lagrange’s mean value theorem, also called the ordinary mean value theorem.

The same argument can be written directly: define $H(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)$. It has equal endpoint values, and an interior point with $H'(c)=0$ gives the Lagrange equation.

Worked examples

For Cauchy’s theorem, take $f(x)=x^2$ and $g(x)=x^3$ on $[1,2]$. Their endpoint changes are $\Delta f=3$ and $\Delta g=7$. The theorem asks for $c\in(1,2)$ such that

$$3(3c^2)=7(2c).$$

Since $c$ is positive, this gives $c=14/9$, which lies between 1 and 2. At that point the required derivative equation holds.

For Lagrange’s theorem, take $f(x)=x^2$ on $[1,3]$. The secant slope is $(9-1)/(3-1)=4$. Since $f'(x)=2x$, the point where the tangent slope equals 4 is $c=2$, which is in $(1,3)$.

Why the proof does not use Rolle’s theorem

The argument uses two standard results instead: a continuous function on a closed interval attains its maximum and minimum, and a differentiable function has derivative zero at an interior local extremum. The second fact, Fermat’s theorem, follows from the definition of derivative: at a local maximum, the difference quotients from the left and right have opposite weak signs, so if their common limit exists it must be zero; the minimum case is analogous.

This proof is sometimes viewed as a proof of Rolle’s theorem’s conclusion by a different route. That is fine: the point is that it establishes the needed zero-derivative fact directly from extrema, rather than assuming Rolle’s theorem as a prior result.

Common mistakes

• Dividing by $g(b)-g(a)$. That quantity can be zero. Cauchy’s theorem is stated as a cross-multiplied equation and does not require division.
• Assuming the auxiliary function has a strict extremum. It need not. Fermat’s theorem applies to a non-strict local maximum or minimum as well.
• Forgetting the constant case. If the auxiliary function is constant, its derivative is already zero throughout the open interval.
• Applying the theorem at an endpoint. The point $c$ is guaranteed to lie in $(a,b)$, where differentiability is assumed; it need not be an endpoint.