Results 1 to 4 of 4

Thread: Concavity of a function

  1. #1
    Member
    Joined
    Dec 2007
    Posts
    137

    Question Concavity of a function

    I am to determine the inflection point and intervals where the graph is concave up and concave down, and I have gotten a wrong answer somehow. I'm guessing it's probably when I solved for y'', it was a very long derivative to simplify:

    $\displaystyle y = \frac{ln(x)}{6\sqrt{x}} $

    $\displaystyle y' = \frac{6\sqrt(x)-ln(x)3\sqrt{x}}{36x^2} $

    $\displaystyle y'' = \frac{3\sqrt(x)ln(x)-12\sqrt{x}}{36x^3} $

    $\displaystyle y''=0 $ when $\displaystyle x=e^4 $
    Follow Math Help Forum on Facebook and Google+

  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,849
    Quote Originally Posted by xxlvh View Post
    I am to determine the inflection point and intervals where the graph is concave up and concave down, and I have gotten a wrong answer somehow. I'm guessing it's probably when I solved for y'', it was a very long derivative to simplify:

    $\displaystyle y = \frac{ln(x)}{6\sqrt{x}} $

    $\displaystyle y' = \frac{6\sqrt(x)-ln(x)3\sqrt{x}}{36x^2} $

    $\displaystyle y'' = \frac{3\sqrt(x)ln(x)-12\sqrt{x}}{36x^3} $

    $\displaystyle y''=0 $ when $\displaystyle x=e^4 $
    $\displaystyle y'=\frac{2-\ln{x}}{12\sqrt[3]{x^2}}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    898
    Thanks
    200
    ...which is equivalent to what xxlvh has above. This gives solution to y'=0 as x=e^2.
    In von Nemo19's from it is much easier to find second deriv.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2007
    Posts
    137
    I was able to find the inflection point but still having some difficulties with the concavity.

    The simplified second derivative is $\displaystyle y'' = \frac{18\sqrt{x}ln(x)-48\sqrt{x}}{144x^3} $

    Here's the inequalities I determined:
    For it to be concave up,
    $\displaystyle 18\sqrt{x}ln(x)-48\sqrt{x}> 0 $ and $\displaystyle 144x^3 > 0 $
    or $\displaystyle 18\sqrt{x}ln(x)-48\sqrt{x} < 0 $ and $\displaystyle 144x^3 < 0 $

    When I solved these I had $\displaystyle x < 0, x > e^{8/3} $

    For it to be concave down,
    $\displaystyle 18\sqrt{x}ln(x)-48\sqrt{x}> 0 $ and $\displaystyle 144x^3 < 0 $
    or $\displaystyle
    18\sqrt{x}ln(x)-48\sqrt{x} < 0 $ and $\displaystyle 144x^3 > 0 $

    And the solution for this one didn't work out at all.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Oct 19th 2011, 04:49 AM
  2. Concavity of this function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 2nd 2011, 04:18 PM
  3. Proving concavity of a 2 variable function
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Feb 25th 2011, 03:42 PM
  4. Concavity/Convexity of a function
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Aug 6th 2010, 01:42 PM
  5. decreasing function and concavity
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Oct 5th 2008, 06:46 PM

Search Tags


/mathhelpforum @mathhelpforum