Prove (x-1)/x < lnx < (x-1) using 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:

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

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?

 

[thepatientmental]

To prove the inequality (x-1)/x < lnx < (x-1), we can use the Mean Value Theorem.

The Mean Value Theorem states that for a function f(x) that is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), there exists a point c in (a,b) such that f'(c) = (f(b)-f(a))/(b-a).

In this case, we can apply the Mean Value Theorem to the function f(x) = lnx on the interval [1,x].

Since lnx is continuous on [1,x] and differentiable on (1,x), there exists a point c in (1,x) such that f'(c) = (f(x)-f(1))/(x-1).

From the definition of the derivative, we know that f'(c) = 1/c. Therefore, we can rewrite the equation as 1/c = (lnx - ln1)/(x-1).

Simplifying this further, we get 1/c = lnx/(x-1).

Rearranging this inequality, we get (x-1)/x < lnx.

Similarly, we can apply the Mean Value Theorem to the function f(x) = (x-1) on the interval [1,x].

Since (x-1) is continuous on [1,x] and differentiable on (1,x), there exists a point c in (1,x) such that f'(c) = (f(x)-f(1))/(x-1).

From the definition of the derivative, we know that f'(c) = 1. Therefore, we can rewrite the equation as 1 = ((x-1)-1)/(x-1).

Simplifying this further, we get 1 = (x-2)/(x-1).

Rearranging this inequality, we get (x-1) < x.

Combining these two inequalities, we get (x-1)/x < lnx < (x-1).

Therefore, we have proved the inequality (x-1)/x < lnx < (x-1) using the Mean Value Theorem.