# 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 $$g’\left( x \right) \ne 0$$ for all $$x \in \left( {a,b} \right).$$ Then there is a point $$x = c$$ in this interval such that

${\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}}} = {\frac{{f’\left( c \right)}}{{g’\left( c \right)}}.}$

#### 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 $$g’\left( {d} \right) = 0.$$ This, however, contradicts the hypothesis that $$g’\left( x \right) \ne 0$$ for all $$x \in \left( {a,b} \right).$$

We introduce the auxiliary function

$F\left( x \right) = f\left( x \right) + \lambda g\left( x \right)$

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

${f\left( a \right) + \lambda g\left( a \right) = f\left( b \right) + \lambda g\left( b \right),\;\;}\Rightarrow {f\left( b \right) – f\left( a \right) = \lambda \left[ {g\left( a \right) – g\left( b \right)} \right],\;\;}\Rightarrow {\lambda = – \frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}}.}$

and the function $$F\left( x \right)$$ takes the form

${F\left( x \right) } = {f\left( x \right) – \frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}}g\left( x \right).}$

This function is continuous on the closed interval $$\left[ {a,b} \right],$$ differentiable on the open interval $$\left( {a,b} \right)$$ and takes equal values at the boundaries of the interval at the chosen value of $$\lambda.$$ Then by Rolle’s theorem, there exists a point $$c$$ in the interval $$\left( {a,b} \right)$$ such that

$F’\left( c \right) = 0.$

Hence,

${f’\left( c \right) }- {\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}}g’\left( c \right) = 0}$

or

${\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}} } = {\frac{{f’\left( c \right)}}{{g’\left( c \right)}}.}$

By setting $$g\left( x \right) = x$$ in the Cauchy formula, we can obtain the Lagrange formula:

$\frac{{f\left( b \right) – f\left( a \right)}}{{b – a}} = f’\left( c \right).$

Cauchy’s mean value theorem has the following geometric meaning. Suppose that a curve $$\gamma$$ is described by the parametric equations $$x = f\left( t \right),$$ $$y = g\left( t \right),$$ where the parameter $$t$$ ranges in the interval $$\left[ {a,b} \right].$$ When changing the parameter $$t,$$ the point of the curve in Figure $$2$$ runs from $$A\left( {f\left( a \right), g\left( a \right)} \right)$$ to $$B\left( {f\left( b \right),g\left( b \right)} \right).$$ According to the theorem, there is a point $$\left( {f\left( {c} \right), g\left( {c} \right)} \right)$$ on the curve $$\gamma$$ where the tangent is parallel to the chord joining the ends $$A$$ and $$B$$ of the curve.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

The function $$f\left( x \right)$$ is differentiable on the interval $$\left[ {a,b} \right],$$ where $$ab \gt 0.$$ Show that the following equality ${\frac{1}{{a – b}}\left| {\begin{array}{*{20}{c}} a&b\\ {f\left( a \right)}&{f\left( b \right)} \end{array}} \right|} = {f\left( c \right) – c f’\left( c \right)}$ holds for this function, where $$c \in \left( {a,b} \right).$$

### Example 2

Check the validity of Cauchy’s mean value theorem for the functions $$f\left( x \right) = {x^4}$$ and $$g\left( x \right) = {x^2}$$ on the interval $$\left[ {1,2} \right].$$

### Example 3

Check the validity of Cauchy’s mean value theorem for the functions $$f\left( x \right) = {x^3}$$ and $$g\left( x \right) = \arctan x$$ on the interval $$\left[ {0,1} \right].$$

### Example 4

Check the validity of Cauchy’s mean value theorem for the functions $$f\left( x \right) = \cos x$$ and $$g\left( x \right) = \sin x$$ on the interval $$\left[ {a,b} \right].$$

### Example 5

Show that at $$x \ne 0$$ the following inequality holds:
$1 – \frac{{{x^2}}}{2} \lt \cos x.$

### Example 1.

The function $$f\left( x \right)$$ is differentiable on the interval $$\left[ {a,b} \right],$$ where $$ab \gt 0.$$ Show that the following equality ${\frac{1}{{a – b}}\left| {\begin{array}{*{20}{c}} a&b\\ {f\left( a \right)}&{f\left( b \right)} \end{array}} \right|} = {f\left( c \right) – c f’\left( c \right)}$ holds for this function, where $$c \in \left( {a,b} \right).$$

Solution.

Note that due to the condition $$ab \gt 0,$$ the segment $$\left[ {a,b} \right]$$ does not contain the point $$x = 0.$$ Consider the two functions $$F\left( x \right)$$ and $$G\left( x \right)$$ having the form:

${F\left( x \right) = \frac{{f\left( x \right)}}{x},}\;\;\;\kern-0.3pt {G\left( x \right) = \frac{1}{x}.}$

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

${\frac{{F\left( b \right) – F\left( a \right)}}{{G\left( b \right) – G\left( a \right)}} } = {\frac{{F’\left( c \right)}}{{G’\left( c \right)}},}$

where the point $$x = c$$ lies in the interval $$\left( {a,b} \right).$$

Find the derivatives:

${F’\left( x \right) = {\left( {\frac{{f\left( x \right)}}{x}} \right)^\prime } = \frac{{f’\left( x \right)x – f\left( x \right)}}{{{x^2}}},}\;\;\;\kern-0.3pt {G’\left( x \right) = {\left( {\frac{1}{x}} \right)^\prime } = – \frac{1}{{{x^2}}}.}$

Substituting this in the Cauchy formula, we get

${\frac{{\frac{{f\left( b \right)}}{b} – \frac{{f\left( a \right)}}{a}}}{{\frac{1}{b} – \frac{1}{a}}} } = {\frac{{\frac{{c f’\left( c \right) – f\left( c \right)}}{{{c^2}}}}}{{ – \frac{1}{{{c^2}}}}},\;\;}\Rightarrow {\frac{{\frac{{af\left( b \right) – bf\left( a \right)}}{{ab}}}}{{\frac{{a – b}}{{ab}}}} } = { – \frac{{\frac{{c f’\left( c \right) – f\left( c \right)}}{{{c^2}}}}}{{\frac{1}{{{c^2}}}}},\;\;}\Rightarrow {\frac{{af\left( b \right) – bf\left( a \right)}}{{a – b}} = f\left( c \right) – c f’\left( c \right)}$

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

${\frac{1}{{a – b}}\left| {\begin{array}{*{20}{c}} a&b\\ {f\left( a \right)}&{f\left( b \right)} \end{array}} \right|} = {f\left( c \right) – c f’\left( c \right).}$

### Example 2.

Check the validity of Cauchy’s mean value theorem for the functions $$f\left( x \right) = {x^4}$$ and $$g\left( x \right) = {x^2}$$ on the interval $$\left[ {1,2} \right].$$

Solution.

The derivatives of these functions are

${f’\left( x \right) = \left( {{x^4}} \right) = 4{x^3},}\;\;\;\kern-0.3pt {g’\left( x \right) = \left( {{x^2}} \right) = 2x.}$

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

$\require{cancel} {\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}} = \frac{{f’\left( c \right)}}{{g’\left( c \right)}},\;\;}\Rightarrow {\frac{{{b^4} – {a^4}}}{{{b^2} – {a^2}}} = \frac{{4{c^3}}}{{2c }},\;\;}\Rightarrow {\frac{{\cancel{\left( {{b^2} – {a^2}} \right)}\left( {{b^2} + {a^2}} \right)}}{\cancel{{b^2} – {a^2}}} = 2{c^2},\;\;}\Rightarrow {{c^2} = \frac{{{a^2} + {b^2}}}{2},\;\;}\Rightarrow {c = \pm \sqrt {\frac{{{a^2} + {b^2}}}{2}}.}$

We take into account that the boundaries of the segment are $$a = 1$$ and $$b = 2.$$ Consequently,

${c = \pm \sqrt {\frac{{{1^2} + {2^2}}}{2}} } = { \pm \sqrt {\frac{5}{2}} \approx \pm 1,58.}$

In this case, the positive value of the square root $$c = \sqrt {\large\frac{5}{2}\normalsize} \approx 1,58$$ is relevant. It is evident that this number lies in the interval $$\left( {1,2} \right),$$ i.e. satisfies the Cauchy theorem.

### Example 3.

Check the validity of Cauchy’s mean value theorem for the functions $$f\left( x \right) = {x^3}$$ and $$g\left( x \right) = \arctan x$$ on the interval $$\left[ {0,1} \right].$$

Solution.

Calculate the derivatives of these functions:

${f’\left( x \right) = {\left( {{x^3}} \right)^\prime } = 3{x^2},}\;\;\;\kern-0.3pt {g’\left( x \right) = {\left( {\arctan x} \right)^\prime } = \frac{1}{{1 + {x^2}}}.}$

Substitute the functions $$f\left( x \right)$$, $$g\left( x \right)$$ and their derivatives in the Cauchy formula:

${\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}} = \frac{{f’\left( c \right)}}{{g’\left( c \right)}},\;\;}\Rightarrow {\frac{{{b^3} – {a^3}}}{{\arctan b – \arctan a}} = \frac{{3{c^2}}}{{\frac{1}{{1 + {c^2}}}}},\;\;}\Rightarrow {\frac{{{b^3} – {a^3}}}{{\arctan b – \arctan a}} = \frac{{1 + {c^2}}}{{3{c^2}}}.}$

For the values of $$a = 0$$, $$b = 1,$$ we obtain:

${\frac{{{1^3} – {0^3}}}{{\arctan 1 – \arctan 0}} = \frac{{1 + {c^2}}}{{3{c^2}}},\;\;}\Rightarrow {\frac{{1 – 0}}{{\frac{\pi }{4} – 0}} = \frac{{1 + {c^2}}}{{3{c^2}}},\;\;}\Rightarrow {\frac{4}{\pi } = \frac{{1 + {c^2}}}{{3{c^2}}},\;\;}\Rightarrow {12{c^2} = \pi + \pi {c^2},\;\;}\Rightarrow {\left( {12 – \pi } \right){c^2} = \pi ,\;\;}\Rightarrow {{c^2} = \frac{\pi }{{12 – \pi }},\;\;}\Rightarrow {c = \pm \sqrt {\frac{\pi }{{12 – \pi }}}.}$

Given that we consider the segment $$\left[ {0,1} \right],$$ we choose the positive value of $$c.$$ Make sure that the point $$c$$ lies in the interval $$\left( {0,1} \right):$$

${c = \sqrt {\frac{\pi }{{12 – \pi }}} } {\approx \sqrt {\frac{{3,14}}{{8,86}}} \approx 0,60.}$

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

### Example 4.

Check the validity of Cauchy’s mean value theorem for the functions $$f\left( x \right) = \cos x$$ and $$g\left( x \right) = \sin x$$ on the interval $$\left[ {a,b} \right].$$

Solution.

For these functions the Cauchy formula is written as

${\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}} = \frac{{f’\left( c \right)}}{{g’\left( c \right)}},\;\;}\Rightarrow {\frac{{\cos b – \cos a}}{{\sin b – \sin a}} = \frac{{{{\left( {\cos c } \right)}^\prime }}}{{{{\left( {\sin c } \right)}^\prime }}},\;\;}\Rightarrow {\frac{{\cos b – \cos a}}{{\sin b – \sin a}} = – \frac{{\sin c }}{{\cos c }}} = {- \tan c ,}$

where the point $$c$$ lies in the interval $$\left( {a,b} \right).$$

Using the sum-to-product identities, we have

$\require{cancel} {\frac{{ – \cancel{2}\sin \frac{{b + a}}{2}\cancel{\sin \frac{{b – a}}{2}}}}{{\cancel{2}\cos \frac{{b + a}}{2}\cancel{\sin \frac{{b – a}}{2}}}} = – \tan c ,\;\;}\Rightarrow {- \tan \frac{{a + b}}{2} = – \tan c ,\;\;}\Rightarrow {c = \frac{{a + b}}{2} + \pi n,\;n \in Z.}$

In the context of the problem, we are interested in the solution at $$n = 0,$$ that is

$c = \frac{{a + b}}{2}.$

As you can see, the point $$c$$ is the middle of the interval $$\left( {a,b} \right)$$ and, hence, the Cauchy theorem holds.

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

${\left\{ \begin{array}{l} \cos \frac{{b + a}}{2} \ne 0\\ \sin\frac{{b – a}}{2} \ne 0 \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} \frac{{b + a}}{2} \ne \frac{\pi }{2} + \pi n\\ \frac{{b – a}}{2} \ne \pi k \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} a + b \ne \pi + 2\pi n\\ b – a \ne 2\pi k \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} a \ne \frac{\pi }{2} + \pi n\\ b \ne \frac{\pi }{2} + \pi k \end{array} \right.,}$

where $$n, k \in \mathbb{Z}.$$

### Example 5.

CShow that at $$x \ne 0$$ the following inequality holds:
$1 – \frac{{{x^2}}}{2} \lt \cos x.$

Solution.

We introduce the functions

${f\left( x \right) = 1 – \cos x,}\;\;\;\kern-0.3pt {g\left( x \right) = \frac{{{x^2}}}{2}}$

and apply the Cauchy formula on the interval $$\left[ {0,x} \right].$$ As a result, we get

${\frac{{f\left( x \right) – f\left( 0 \right)}}{{g\left( x \right) – g\left( 0 \right)}} = \frac{{f’\left( \xi \right)}}{{g’\left( \xi \right)}},\;\;}\Rightarrow {\frac{{1 – \cos x – \left( {1 – \cos 0} \right)}}{{\frac{{{x^2}}}{2} – 0}} = \frac{{\sin \xi }}{\xi },\;\;}\Rightarrow {\frac{{1 – \cos x}}{{\frac{{{x^2}}}{2}}} = \frac{{\sin \xi }}{\xi },}$

where the point $$\xi$$ is in the interval $$\left( {0,x} \right).$$

The expression $${\large\frac{{\sin \xi }}{\xi }\normalsize}\;\left( {\xi \ne 0} \right)$$ in the right-hand side of the equation is always less than one. Indeed, this follows from Figure $$3,$$ where $$\xi$$ is the length of the arc subtending the angle $$\xi$$ in the unit circle, and $$\sin \xi$$ is the projection of the radius-vector $$OM$$ onto the $$y$$-axis. In this case we can write

${\frac{{1 – \cos x}}{{\frac{{{x^2}}}{2}}} = \frac{{\sin \xi }}{\xi } \lt 1,\;\;}\Rightarrow {1 – \cos x \lt \frac{{{x^2}}}{2}\;\;\text{or}}\;\; {1 – \frac{{{x^2}}}{2} \lt \cos x.}$