Mean Value Teorem

[iNCREDiBLE]

How do I prove that (x-1)/x < lnx < (x–1), for x > 1 by using the mean value teorem?

 

[TD]

Well, since $\ln \left( 1 \right) = 0$, we can write:

$$\begin{array}{l} \frac{{x - 1}}{x} < \ln \left( x \right) - \ln \left( 1 \right) < x - 1 \\ \\ \frac{{x - 1}}{{x\left( {x - 1} \right)}} < \frac{{\ln \left( x \right) - \ln \left( 1 \right)}}{{x - 1}} < \frac{{x - 1}}{{x - 1}} \\ \\ \frac{1}{x} < \frac{{\ln \left( x \right) - \ln \left( 1 \right)}}{{x - 1}} < 1 \\ \end{array}$$

Now you can apply the Mean Value Theorem to the middle expression, which states that, if $f\left( x \right)$ is differentiable over (a,b) and continuous over [a,b], there exists a $c \in \left( {a,b} \right)$ so that:
$$f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$$

Does that help?