Cauchy’s Mean Value Theorem

Cauchy's Mean Value Theorem generalizes Lagrange's Mean Value Theorem. This theorem is also called the Extended or Second Mean Value Theorem. It establishes the relationship between the derivatives of two functions and changes in these functions on a finite interval.

Let the functions $f\left(x\right)$ and $g\left(x\right)$ be continuous on an interval $\left[a,b\right],$ differentiable on $\left(a,b\right),$ and $g^{\prime }\left(x\right)\ne 0$ for all $x\in \left(a,b\right).$ Then there is a point $x=c$ in this interval such that

Proof.

First of all, we note that the denominator in the left side of the Cauchy formula is not zero: $g\left(b\right)-g\left(a\right)\ne 0.$ Indeed, if $g\left(b\right)=g\left(a\right),$ then by Rolle's theorem, there is a point $d\in \left(a,b\right),$ in which $g^{\prime }\left(d\right)=0.$ This, however, contradicts the hypothesis that $g^{\prime }\left(x\right)\ne 0$ for all $x\in \left(a,b\right).$

We introduce the auxiliary function

and choose $\lambda$ in such a way to satisfy the condition $F\left(a\right)=F\left(b\right).$ In this case we get

and the function $$ takes the form

This function is continuous on the closed interval $$ differentiable on the open interval $$ and takes equal values at the boundaries of the interval at the chosen value of $$ Then by Rolle's theorem, there exists a point $$ in the interval $$ such that

Hence,

or

By setting $$ in the Cauchy formula, we can obtain the Lagrange formula:

Cauchy's mean value theorem has the following geometric meaning. Suppose that a curve $$ is described by the parametric equations $$ $$ where the parameter $$ ranges in the interval $$ When changing the parameter $$ the point of the curve in Figure $$ runs from $$ to $$ According to the theorem, there is a point $$ on the curve $$ where the tangent is parallel to the chord joining the ends $$ and $$ of the curve.

Solved Problems

Solution.

Note that due to the condition $$ the segment $$ does not contain the point $$ Consider the two functions $$ and $$ having the form:

For these functions, the Cauchy formula is written in the form:

where the point $$ lies in the interval $$

Find the derivatives:

Substituting this in the Cauchy formula, we get

The left side of this equation can be written in terms of the determinant. Then

Solution.

The derivatives of these functions are

Substituting the functions and their derivatives in the Cauchy formula, we get

We take into account that the boundaries of the segment are $$ and $$ Consequently,

In this case, the positive value of the square root $$ is relevant. It is evident that this number lies in the interval $$ i.e. satisfies the Cauchy theorem.

Solution.

Calculate the derivatives of these functions:

Substitute the functions $$, $$ and their derivatives in the Cauchy formula:

For the values of $$, $$ we obtain:

Solving this equation, we find:

Given that we consider the segment $$ we choose the positive value of $$ Make sure that the point $$ lies in the interval $$

Thus, Cauchy's mean value theorem holds for the given functions and interval.

Solution.

For these functions the Cauchy formula is written as

where the point $$ lies in the interval $$

Using the sum-to-product identities, we have

In the context of the problem, we are interested in the solution at $$ that is

As you can see, the point $$ is the middle of the interval $$ and, hence, the Cauchy theorem holds.

Note that the above solution is correct if only the numbers $$ and $$ satisfy the following conditions:

where $$

Solution.

We introduce the functions

and apply the Cauchy formula on the interval $$ As a result, we get

where the point $$ is in the interval $$

The expression $$ in the right-hand side of the equation is always less than one. Indeed, this follows from Figure $$ where $$ is the length of the arc subtending the angle $$ in the unit circle, and $$ is the projection of the radius-vector $$ onto the $$-axis. In this case we can write