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Math Help - Concavity of a function

  1. #1
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    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:

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

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

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

    y''=0 when  x=e^4
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  2. #2
    No one in Particular VonNemo19's Avatar
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    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:

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

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

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

    y''=0 when  x=e^4
    y'=\frac{2-\ln{x}}{12\sqrt[3]{x^2}}
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  3. #3
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    ...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.
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  4. #4
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    I was able to find the inflection point but still having some difficulties with the concavity.

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

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

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

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

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

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