Answer to Question #247396 in Real Analysis for Sree

Question

Verify Cauchy's mean value theorem for log x and 1/x in [1,e]

Answer 1

logx and 1/x in [1,e]

Let

$f(x)=logx\\g(x)=\dfrac{1}{x}\\$

Taking differentiation of f(x) and g(x), we get

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

1.) f(x) and g(x) both are continuous in [1,e]

2.) f(x) and g(x) both are differentiable in (1,e)

So, there must be a value 'c' between [1,e] where

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

Here, a = 1 and b=e

So,

$$\dfrac{f(e)-f(1)}{g(e)-g(1)}=\dfrac{f'(c)}{g'(c)}\\\ \\\dfrac{loge-log1}{1/e-1}=\dfrac{1/c}{-1/c^2}\\\ \\\implies \dfrac{1}{1/e-1}=-c\\\ \\\implies -c=\dfrac{e}{1-e}\\\ \\\implies \boxed{c=\dfrac{e}{e-1}} \in [1,e]$$

Hence, Cauchy's mean value theorem is verified.