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Math Help - concave...

  1. #1
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    concave...

    the question is

    given f'(x) = (2x)/(x^2 - 4)^2 and f"(x) = -2(3x^2 + 4)/(x^2 - 4)^2
    B) Use calculus to find all intervals on which f(x) is concave up or down and the find the x-coordinate of all inflection points, if they exist.


    i've tride f"(x) is 0 when -2(3x^2 + 4) = 0 x= ... ?

    f"(x) is undefined when (x^2 - 4)^2 = 0, x=2, -2

    what should i do with " -2(3x^2 + 4) = 0 x= ... ? " <-- here?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by haebinpark View Post
    the question is

    given f'(x) = (2x)/(x^2 - 4)^2 and f"(x) = -2(3x^2 + 4)/(x^2 - 4)^2
    B) Use calculus to find all intervals on which f(x) is concave up or down and the find the x-coordinate of all inflection points, if they exist.


    i've tride f"(x) is 0 when -2(3x^2 + 4) = 0 x= ... ?

    f"(x) is undefined when (x^2 - 4)^2 = 0, x=2, -2

    what should i do with " -2(3x^2 + 4) = 0 x= ... ? " <-- here?
    Concave up and down depends on the sign of the second derivative. Inflection points are when f''(x)=0.

    And f''(x)=\frac{-2(3x^2+4)}{(x^2-4)^{\color{red}3}}

    So find where f''(x)<0 (down) and where f''(x)>0 (up).
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