Results 1 to 8 of 8

Math Help - Find minimum value

  1. #1
    Junior Member
    Joined
    May 2009
    Posts
    34

    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..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member Infophile's Avatar
    Joined
    May 2009
    Posts
    50
    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.

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    96
    Quote Originally Posted by Raidan View Post
    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)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Apr 2009
    Posts
    166
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2009
    Posts
    34
    thnx..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2009
    Posts
    34
    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...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,583
    Thanks
    1418
    n/n^2= 1/n, doesn't it?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2009
    Posts
    34
    its actually ln(n)/n^2...a suitable comparison to show that this converges or diverges using the first question above...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find minimum possible value..
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: August 12th 2009, 01:02 AM
  2. Find the possible minimum
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 9th 2009, 12:43 PM
  3. Find minimum value of...?
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: August 29th 2008, 01:33 AM
  4. Find the minimum value
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 16th 2008, 03:55 AM
  5. find the minimum
    Posted in the Calculus Forum
    Replies: 7
    Last Post: May 6th 2007, 07:20 PM

Search Tags


/mathhelpforum @mathhelpforum