Question

Verify Cauchy's mean value theorem for f(x)=ln x and g(x) =1/x , where x belongs to [1,e]​

Answer 1

Answer:

$\boxed{\bf \: \: c =\dfrac{e}{e - 1} \: \in \: (1,e) \: \: }$

Step-by-step explanation:

Given that,

$\sf \: f(x) = ln(x) \: \: \: x \: \in \: [ 1,e]$

and

$\sf \: g(x) = \dfrac{1}{x} \: \: \: x \: \in \: [ 1,e]$

Now,

$\sf \: f(x) \: and \: g(x) \: are \: continuous \: functions \:on \: \: x \: \in \: [ 1,e]$

Now,

$\sf \: f'(x) = \dfrac{1}{x}$

and

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

So,

$\sf \: f(x) \: and \: g(x) \: are \: differentiable \: functions \:on \: \: x \: \in \: ( 1,e)$

So, Cauchy's mean value theorem is applicable.

Therefore, there exist atleast one real number c $\in$ (1, e) such that

$\sf \: \dfrac{f'(c)}{g'(c)} = \dfrac{f(e) - f(1)}{g(e) - g(1)}$

$\sf \: \dfrac{\dfrac{1}{c} }{ - \dfrac{1}{ {c}^{2} } } = \dfrac{ln(e) - ln(1)}{\dfrac{1}{e} - 1}$

$\sf \: - c = \dfrac{1 - 0}{\dfrac{1 - e}{e}}$

$\sf \: - c =\dfrac{e}{1 - e}$

$\implies\sf \: \boxed{\sf \: c =\dfrac{e}{e - 1} \: \in \: (1,e)\: }$