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Math Help - Finding T.P using differentiation

  1. #1
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    Finding T.P using differentiation

    Hi
    I am having trouble finding the turning point and POI for this function:
    \frac{3x}{(x+8)^2}

    i know that the asymptote is x=-8 and y=0.

    x an y intercepts are 0.

    this is what i get for the first derivative: \frac{-3x^2+96x+192}{(x+8)^4}
    This is where i am stuck.

    P.S
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  2. #2
    Junior Member piglet's Avatar
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I am having trouble finding the turning point and POI for this function:
    \frac{3x}{(x+8)^2}

    i know that the asymptote is x=-8 and y=0.

    x an y intercepts are 0.

    this is what i get for the first derivative: \frac{-3x^2+96x+192}{(x+8)^4}
    This is where i am stuck.

    P.S
    Well to find the turning points, just solve for  \frac{dy}{dx} = 0

    and for the points of inflection just find the 2nd derivative of position i.e.
     \frac{d2y}{dx^{2}} and let that = 0 and solve for x. Then sub back in value(s) of x and find y

    These formula's should be in your class notes... and i would hazard a guess that this is pre-uni calculus...?
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  3. #3
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    i have found the turning point from the derivative \frac{-3x+24}{(x+8)^3} . values (8, \frac{3}{32})
    However i cannot seen to get a Point of Inflection value.
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