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Math Help - Curve Sketching - Concavity - Problem 2

  1. #1
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    Question Curve Sketching - Concavity - Problem 2

    I'm having a really hard time with this. Could anyone help me figuring out the solution for this? Thank you very much in advance!

    Last edited by wdma; July 6th 2009 at 09:39 AM. Reason: I accidently hit edit and erased the post content.
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  2. #2
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    Could anyone here possibly help me? I still don't get it. Thanks.
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  3. #3
    Super Member malaygoel's Avatar
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    g(x)=-2x^3-(-24)x^2+(-90)x-2

    g'(x)=-6x^2-(-48)x+(-90)

    Can you find x_1,x_2 now?

    Spoiler:

    [tex]g'(x)=0[tex] gives x_1=-5 and x_2=-3.

    g''(x)=-12x-(-24)

    g''(x_1)>0

    there is relative minimum at x_2,x_1



    g''(x_2)>0
    Last edited by malaygoel; July 6th 2009 at 10:00 AM.
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  4. #4
    Member McScruffy's Avatar
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    Take the your derivative function, set that equal to zero and solve for x. Those values are your critical numbers. You can then find the second derivative and put your critical values into the second derivative. If that results in a value greater than zero the function at that given point is a relative minimum, and if the value is less than zero then you have a relative maximum.
    Hope that helps.
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