Answer to Question #310126 in Real Analysis for Kaygold

Question

Applying Cauchy’s mean value theorem to the function f and g defined as f(x)=x2 and g(x)=x for all x∈[a,b], gives

Answer 1

Solution

Cauchy's Mean Value Theorem is defined as:

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

Here, given that

$f(x)=x^2$ and $g(x)=x$ and the interval is $[a, b]$

Then,

$f'(x)=2x\\ f'(c)=2c\\$ and $g'(x)=1\\ g'(c)=1\\$

Therefore, as per Cauchy's Mean Value Theorem

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

$\frac{2c}{1} = \frac{b^2-a^2}{b-a}$

$\frac{2c}{1} = \frac{(b-a)(b+a)}{b-a}$

$2c=a+b$

Hence the Cauchy's Mean Value Theorem gives,

$a+b=2c$