Question

What is Cauchy Mean Value Theorem?

Answer 1

Hint: Cauchy's Mean Value Theorem sums up Lagrange's Mean Value Theorem. This hypothesis is likewise called the Extended or Second Mean Value Theorem. It builds up the connection between the derivatives of two functions and changes in these functions on a limited interval.

Complete step by step solution:
Cauchy's Mean Value Theorem states that,
Let there be two functions,$f\left( x \right)$ and $g\left( x \right)$. These two functions shall be continuous on the interval, $\left[ a,b \right]$, and these functions are differentiable on the range $\left( a,b \right)$ , and ${g}'\left( x \right)\ne 0$ for all $x\in \left( a,b \right)$ .
Then there will be a point where $x=c$ in the given range or the interval such that,
$\dfrac{f\left( b \right)-f\left( a \right)}{g\left( b \right)-g\left( a \right)}=\dfrac{{f}'\left( c \right)}{{g}'\left( c \right)}$
Cauchy's Mean Value Theorem sums up the Lagrange's Mean Value Theorem.
We achieve this by setting up $g\left( x \right)=x$ in the formula.
On substituting the function, we get,
$\dfrac{f\left( b \right)-f\left( a \right)}{b-a}={f}'\left( c \right)$
Let us take an example on The Cauchy's Mean Value Theorem to understand better.
Let us now check the validity of Cauchy's Mean Value Theorem for the two given functions. The given functions are $f\left( x \right)={{x}^{4}}$ and $g\left( x \right)={{x}^{2}}$. We have to find the validity of them in the interval $\left[ 1,2 \right]$
For this let us first find the derivatives of the functions.
${f}'\left( x \right)=4{{x}^{3}}$ and ${g}'\left( x \right)=2x$
Now let us substitute these in Cauchy's Mean Value formula.
Cauchy's Mean Value formula is, $\dfrac{f\left( b \right)-f\left( a \right)}{g\left( b \right)-g\left( a \right)}=\dfrac{{f}'\left( c \right)}{{g}'\left( c \right)}$
On substituting we get,
$\Rightarrow \dfrac{{{b}^{4}}-{{a}^{4}}}{{{b}^{2}}-{{a}^{2}}}=\dfrac{4{{c}^{3}}}{2c}$
On further evaluation we get,
$\Rightarrow \dfrac{\left( {{b}^{2}}-{{a}^{2}} \right)\left( {{b}^{2}}+{{a}^{2}} \right)}{{{b}^{2}}-{{a}^{2}}}=\dfrac{4{{c}^{3}}}{2c}$
$\Rightarrow \left( {{b}^{2}}+{{a}^{2}} \right)=2{{c}^{2}}$
Hence on rearranging the terms we get,
$\Rightarrow \dfrac{\left( {{b}^{2}}+{{a}^{2}} \right)}{2}={{c}^{2}}$
On squaring on both sides, we get,
$\Rightarrow c=\pm \sqrt{\dfrac{{{b}^{2}}+{{a}^{2}}}{2}}$
Given that the boundaries of the function are $a=1;b=2$ .
Hence on placing the values we get,
$\Rightarrow c=\pm \sqrt{\dfrac{{{2}^{2}}+{{1}^{2}}}{2}}$
$\Rightarrow c=\pm \sqrt{\dfrac{5}{2}}$
Which is roughly equal to $c=\pm 1.581$
If we consider the positive value of this, the number lies in the interval $\left( 1,2 \right)$.
Hence it satisfies Cauchy's Mean Value Theorem.

Note: Cauchy's Mean Value Theorem is said to be the extended mean value theorem. The original Mean Value Theorem states that if there is a function $f\left( x \right)$ which is defined over $\left[ a,b \right]$ and is also a continuous function, and that is differential on the range $\left( a,b \right)$ , where $a