1. ## Mean value theorem

Hello,

The question is: use the mean value theorem to show that for all x > 0

$\displaystyle \frac{1}{\sqrt{1 + x}} - 1 > -\frac{1}{2}x$

and

$\displaystyle \ln(3 + x) - \ln(3) \leq \frac{x}{3}$

How do I do this? I thought you always need 2 x-values (a and b) and then use $\displaystyle \frac{f(b)-f(a)}{b-a}$ to find the mean slope, but this is an entire different context...
How do I do this? Thanks!

2. ## Re: Mean value theorem

For 1., rearranging gives $\displaystyle \frac{1}{\sqrt{x+1}} + \frac{1}{2}x > 1$.

The LHS is a strictly increasing function in terms of x, so the minimal value occurs when x is minimized. If x = 0, the LHS = 1, but if x > 0, the LHS is greater than 1.

For 2., divide both sides by x:

$\displaystyle \frac{\ln(x+3) - \ln(3)}{x} \le \frac{1}{3}$.

Here, we want to show that the slope between two points (3, ln 3) and (x+3, ln (x+3)) is at most 1/3. This is true, because the derivative of y = ln x is 1/x, and the derivative is strictly decreasing. The maximal value of the LHS occurs when x gets infinitely close to 3, which is 1/3.