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Math Help - Urgent help on concavity

  1. #1
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    Urgent help on concavity

    Find any points of inflection of the graph of the function and determine whether the graph is concave upward or concave downward:
    G(x) = \frac{x}{x^2+4}
    Am needing any help... Am having trouble finding the 2nd derivative
    Here's what ive graphed
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find any points of inflection of the graph of the function and determine whether the graph is concave upward or concave downward:
    G(x) = \frac{x}{x^2+4}
    Use the quotient rule,
    G'(x)=\frac{(x)'(x^2+4)-(x)(x^2+4)'}{(x^2+4)^2}
    Thus,
    G'(x)=\frac{x^2+4-x(2x)}{(x^2+4)^2}
    Thus,
    G'(x)=\frac{-x^2+4}{(x^2+4)^2}
    Now the second derivative,
    G''(x)=\frac{(-x^2+4)'(x^2+4)^2-(-x^2+4)[(x^2+4)^2]'}{(x^2+4)^4}
    Thus,
    G''(x)=\frac{-2x(x^2+4)^2-4x(-x^2+4)(x^2+4)}{(x^2+4)^4}
    Factor,
    G''(x)=\frac{-2x(x^2+4)[(x^2+4)+2(-x^2+4)}{(x^2+4)^4}
    Thus,
    G''(x)=\frac{-2x[x^2+4-2x^2+8]}{(x^2+4)^4}
    Thus,
    G''(x)=\frac{-2x(-x^2+12)}{(x^2+4)^3}

    The inflection points is when G''(x)=0
    That is when the numerator is zero.
    Thus,
    -2x(-x^2+12)=0
    x=0
    x=+/-sqrt(12)
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