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Help, Concaved downward?

  1. Feb 13, 2005 #1
    Help, Concaved downward??

    Let f(x) = 12x^(2/3) Find all the intervals on which f(x) is concaved downward.

    I know I have to take the second derivative to find the inflection point to find the interval. I figured the first derivative to be (8-4x^(1/3)/x^(1/3)

    I can't seem to get the second derivative to work out. Please Help!
     
  2. jcsd
  3. Feb 13, 2005 #2

    dextercioby

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    Is this your function:[tex] f(x)=12x^{\frac{2}{3}} [/tex] ??

    If so,then your first derivative is incorrect.

    Daniel.
     
  4. Feb 14, 2005 #3
    The first derivative of [tex] f(x) = 12x^{\frac{2}{3}}[/tex]

    is [tex](\frac{2}{3})12x^\frac{-1}{3}[/tex]

    which simplified gives you

    [tex] 8x^\frac{-1}{3}[/tex]

    ------------------------------
    Sorry for messing up the Latex thing... I'm still new to it
     
    Last edited: Feb 14, 2005
  5. Feb 14, 2005 #4

    dextercioby

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    Would you care to correct your latex graphics in your post...?

    Daniel.
     
  6. Feb 14, 2005 #5
    sorry the orginial function is wrong...it is actually f(x)= 12x^(2/3)-4x
    I still got the derivative of (8-4x^(1/3)/x^(1/3) but I am still having trouble getting through the second derivative...any help would be great...
     
  7. Feb 14, 2005 #6

    dextercioby

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    Okay,why didn't you leave it in the original handy form...?
    [tex]f'(x)=8x^{-\frac{1}{3}}-4 [/tex]

    Now differentiate once more...

    Daniel.
     
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