Thread: 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..

Hello,

Define .

There is a minimum when i.e .

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 at x=4
(using differentiation)

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.





@

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

Min value:

x at which it occurs:

Now how do we provide for all ?

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.

thnx..

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...

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

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