use Calculus to find the minimum value of Sqrt(x)-lnx and the value of x for which it occurs.Hence show Sqrt(x)>lnx for all x>0. Thanx..(Nod)

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- May 1st 2009, 09:41 AMRaidanFind minimum value
**use Calculus to find the minimum value of Sqrt(x)-lnx and the value of x for which it occurs.Hence show Sqrt(x)>lnx for all x>0. Thanx..(Nod)** - May 1st 2009, 09:59 AMInfophile
Hello,

Define $\displaystyle f:x\mapsto \sqrt{x}-\ln(x)$.

There is a minimum when $\displaystyle f'(x)=0$ i.e $\displaystyle \frac{\sqrt{x}-2}{2x}=0\Rightarrow x=4$.

:) - May 1st 2009, 10:01 AMadhyeta
- May 1st 2009, 10:07 AMderfleurer
Your minimum value is going to occur at all x such that f'(x)=0 when f'(x - a) < 0 and f'(x + a) > 0.

$\displaystyle f(x) = \sqrt x - lnx$

$\displaystyle f'(x) = \frac{1}{2\sqrt x} - \frac{1}{x} = \frac{\sqrt x - 2}{x}$

$\displaystyle \sqrt x - 2 = 0$ @ $\displaystyle x = 4$

Go ahead and test it, but i'll just guaruntee you it's a min, so:

Min value: $\displaystyle \sqrt 4 - ln4$

x at which it occurs: $\displaystyle x = 4$

Now how do we provide $\displaystyle \sqrt x > lnx$ for all $\displaystyle x > 0$?

Well, our minimum is giving us a point when x^(1/2) and lnx are closest (being the lowest y on the graph, we must have the smallest difference between the functions). So if we test x=4 for both functions individually and find that one is greater than the other, we've proved this. - May 1st 2009, 06:45 PMRaidan
thnx..

- May 1st 2009, 06:49 PMRaidan
**Ahh I see...,I got another question relating to this one,How would I use this question to find a suitable comparison to show that**

**∞**

∑ ln n/n^2

**n=1**

converges or diverges...I believe it converges..but Iam not sure...(Shake)

- May 2nd 2009, 03:49 AMHallsofIvy
n/n^2= 1/n, doesn't it?

- May 2nd 2009, 09:40 PMRaidan
**its actually ln(n)/n^2...a suitable comparison to show that this converges or diverges using the first question above...**