Answer to Question #310126 in Real Analysis for Kaygold

Question #310126

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

1
Expert's answer
2022-03-14T18:50:59-0400

Solution


Cauchy's Mean Value Theorem is defined as:


f(c)g(c)=f(b)f(a)g(b)g(a)\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)=x2f(x)=x^2 and g(x)=xg(x)=x and the interval is [a,b][a, b]


Then,


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


Therefore, as per Cauchy's Mean Value Theorem


f(c)g(c)=f(b)f(a)g(b)g(a)\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)}}


2c1=b2a2ba\frac{2c}{1} = \frac{b^2-a^2}{b-a}


2c1=(ba)(b+a)ba\frac{2c}{1} = \frac{(b-a)(b+a)}{b-a}


2c=a+b2c=a+b


Hence the Cauchy's Mean Value Theorem gives,


a+b=2ca+b=2c




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