Cauchy’s Mean Value Theorem

VSAT 2022

What is Cauchy’s Mean Value Theorem?

Cauchy’s Mean Value Theorem is a very important theorem used in the world of derivatives. The formula used in Cauchy's Theorem is

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


It is very important to note that the given function should be continuous and differentiable over its entire domain. We will see this theorem in more detail in this article.


History of the Mathematician Cauchy

Augustin Louis Cauchy, Image Credit: Wikimedia


Augustin Louis Cauchy

Name: Augustin Louis Cauchy

Born: 21 August 1789

Died: 23 May 1857

Field: Mathematics

Nationality: French


General Statement of Cauchy’s Theorem

Cauchy’s Mean Value Theorem states that for any two functions $f\left( x \right)$ and$g\left( x \right)$, which are continuous on the interval $\left[ a,b \right]$ and differentiable on the interval $\left( a,\text{ }b \right)$ and $g\left( x \right)\ne 0$ for all $x\in \left( a,b \right)$, then there will be a point $x=c$ in the given interval such that

$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$.


Cauchy’s Theorem Proof

Let $f\left( x \right)$ and $g(x)$ be continuous on the interval $\left[ a,b \right]$ and differentiable on the interval $(a,b)$ and $g\left( x \right)\ne 0 \forall x\in \left( a,b \right)$.

Now let $F\left( x \right)=f\left( x \right)+P\times g\left( x \right)$

Where P is taken such that $F\left( x \right)$ satisfies the Rolle’s Theorem in $\left[ a,b \right]$

So, from Rolle’s Theorem $F\left( a \right)=F\left( b \right)$

$\Rightarrow f\left( a \right)+P\times g\left( a \right)=f\left( b \right)+P\times g\left( b \right)$

$\Rightarrow f\left( a \right)-f\left( b \right)=P\times\left( g\left( b \right)-g\left( a \right) \right)$

$\Rightarrow P=\dfrac{f(b)-f(a)}{g(b)-g(a)}$

Now $F(x)=f(x)-\left(\dfrac{f(b)-f(a)}{g(b)-g(a)}\right)g(x)$

Since $F\left( x \right)$ satisfies Rolle’s Theorem, $F\left( c \right)=0$

$\Rightarrow f'(c)-\left(\dfrac{f(b)-f(a)}{g(b)-g(a)}\right)g'(c)=0$

$\Rightarrow \dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$

This proves the Cauchy Formula.

Hence proved.


Limitations of the Theorem

  • The first limitation of Cauchy's mean value theorem is that it is not applicable if the functions are not continuous in the given interval.

  • It is not applicable even if the given functions are not differentiable.

Applications of the Theorem

We know that from Cauchy’s Mean Value Theorem,

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

Here if$g\left( x \right)=x$, the formula will be reduced to Lagrange’s Mean Value Theorem.

That is, $f'(c)=\dfrac{f(b)-f(a)}{b-a}$


Solved Examples

1. Find C of Cauchy’s Mean Value Theorem for the functions $\dfrac{1}{x}$ and $\dfrac{1}{{{x}^{2}}} \in \left[ 4,6 \right]$.

Ans. Let $f(x)=\dfrac{1}{x}$ and $g(x)=\dfrac{1}{{{x}^{2}}}$

Both $f\left( x \right)$and $g(x)$ are continuous and differentiable in $\left[ 4,6 \right]$

$g'(x)=\dfrac{-2}{{{x}^{3}}}$

So $g\left( x \right)\ne 0 \in \left[ 4,6 \right]$

Now we know that Cauchy’s Equation is

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

$f'(x)=\dfrac{-1}{{{x}^{2}}}$

Now:

$f(a)=f(4)=\dfrac{1}{4}$

$f(b)=f(6)=\dfrac{1}{6}$

$g(a)=g(4)=\dfrac{1}{16}$

$g(b)=g(6)=\dfrac{1}{36}$

$f'(c)=-\dfrac{1}{{{c}^{2}}}$

$g'(c)=\dfrac{-2}{{{c}^{3}}}$

Substituting these values in the formula: $\dfrac{\dfrac{-1}{{{c}^{2}}}}{\dfrac{-2}{{{c}^{3}}}}=\dfrac{\dfrac{1}{6}-\dfrac{1}{4}}{\dfrac{1}{36}-\dfrac{1}{16}}$

By solving this, we get $c=4.8$.


2. If $f(x)$ is differentiable and $g\left( x \right)\ne 0$ such that $f(1)=4$,$f(2)=16$, $f\left( x \right)=8g\left( x \right)$ and $g(2)=4$ then what is the value of $g(1)$?

Ans: We know that from Cauchy’s Mean Value Theorem,

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

Here $a=1$ and $b=2$

$\dfrac{f'(c)}{g'(c)}=\dfrac{f(2)-f(1)}{g(2)-g(1)}$

Now substituting all the given values in the equation:

$\dfrac{8g'(c)}{g'(c)}=\dfrac{16-4}{4-g(1)}$

$32-8g(1)=12$

$g(1)=\dfrac{5}{2}$


3. Using Cauchy’s Mean Value Theorem, find the mean value c for the function in $\left[ 2,5 \right]$ if $f\left( x \right)=\left| x-1 \right|$ and $g\left( x \right)=lnx$

Ans: $f\left( x \right)$ and $g\left( x \right)$are continuous and differentiable in the given interval.

We know that from Cauchy’s Mean Value Theorem,

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

So, $f'(x)=\dfrac{\left| x-1 \right|}{x-1}$ and $g'(x)=\dfrac{1}{x}$

Now $f'(c)=1$ and $g'(c)=\dfrac{1}{c}$

$f\left( a \right)=f(2)=\left| 2-1 \right|=1$

$f\left( b \right)=f(5)=\left| 5-1 \right|=4$

Now substituting these values in the formula:

$\dfrac{1}{\dfrac{1}{c}}=\dfrac{4-1}{\ln 5-\ln 2}$

$c=\dfrac{3}{\ln \left(\dfrac{5}{2}\right)}=3.27$


Conclusion

In conclusion, we learn that Cauchy’s Mean Value Theorem is derived with the help of Rolle’s Theorem. Lagrange’s mean value theorem can be deduced from Cauchy’s Mean Value Theorem. Cauchy’s Mean Value Theorem is the relationship between the derivatives of two functions and changes in these functions on a finite interval. The continuity and differentiability of the given functions must be adequately checked.


Important Points to Rememeber

  • The continuity of the given functions must be checked.

  • The differentiability of the given functions must also be checked.

  • The formula must be applied properly.

Important Formulae:

  • The final formula of Cauchy’s Mean value theorem is $\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$

  • The final formula of Rolle’s theorem is $F\left( a \right)=F\left( b \right)$

Competitive Exams after 12th Science

FAQs on Cauchy’s Mean Value Theorem

1. Briefly explain the geometrical interpretation of Cauchy’s Mean Value Theorem.

If we try to give the brief explanation of the Cauchy Mean Value Theorem it is as follow if $f\left( x \right)$and $g\left( x \right)$ are the given functions, geometrically the theorem tells us that there is a value $c$ in $\left( a,\text{ }b \right)$ for which the tangent line to the curve at $(f\left( c),g\left( c \right) \right)$ is parallel to the line connecting the points $(f\left( a),g\left( a \right) \right)$ and $(f\left( b),g\left( b \right) \right)$.

2. What is the difference between Cauchy’s Theorem and Lagrange’s Mean Value Theorem?

The Lagrange’s Mean Value Theorem states that “let $f(x)$ be differentiable in $(a,b)$ and continuous in $\left[ a,b \right]$ , then there exists at least one value $x=c$ in $(a,b)$ such that

$\mathbf{f}\left( \mathbf{c} \right)=\dfrac{\left[ \mathbf{f}\left( b \right)-\mathbf{f}\left( a \right) \right]}{b-a}$——— (1)

Cauchy’s Mean Value Theorem is just an extension of the mean value theorem. Let us say we have one more similar function $g\left( x \right)$ which is also differentiable in $(a,b)$ and continuous in $\left[ a,b \right]$, then there exists at least one value $x=c$in $(a,b)$ such that


$\mathbf{g}\left( \mathbf{c} \right)=\dfrac{\left[ \mathbf{g}\left( b \right)-\mathbf{g}\left( a \right) \right]}{b-a}$———-(2)


Now as per Cauchy’s Theorem, 


$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$_______(3)


Equation 3 is called Cauchy’s mean value theorem which can be seen as the generalisation of the mean value theorem.

3. Why does the function have to be differentiable in an open interval $(a,b)$ and not $\left[ a,b \right]$?

For a function to be differentiable at a point, the function has to be differentiable at both the left side and right side of the point. Still, at the end points of a function, this is not possible because for the rightmost point at the extreme there is a left side, but there is no right side. Similarly, for the point at the leftmost extreme, there is no left side, but it has a right side; due to this reason, a function is never differentiable at its endpoints.

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