# Thread: Find minimum value

1. ## Find 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..

2. Hello,

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

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

3. Originally Posted by Raidan
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..
the min value would be $2-ln4$ at x=4
(using differentiation)

4. 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.

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

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

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

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

Min value: $\sqrt 4 - ln4$

x at which it occurs: $x = 4$

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

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.

5. thnx..

6. 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...

7. n/n^2= 1/n, doesn't it?

8. its actually ln(n)/n^2...a suitable comparison to show that this converges or diverges using the first question above...