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Concave up and concave down

  1. Oct 27, 2006 #1
    problem: Use the graphing strategy to sketch the graph of y=(4x)/(x^2+1). check the intervals where it is concave up and where it is concave down. Then graph it. please use sign charts.

    to find this we have to first find y''.
    so I used the quotient rule twice to get this
    y'' = (8x^5 - 16x^3 - 24x)/(x^2 + 1)^4
    to find the inflection points we set y'' = 0 and solve for x.
    I have a question on this
    while solving y'' = (8x^5 - 16x^3 - 24x)/(x^2 + 1)^4 = 0
    i come across this step
    (8x)(x^2 + 1)(x^2 - 3) = 0
    so that would mean

    8x = 0 this mean x = 0
    (x^2 + 1) = 0 this means x = sqrt(-1)
    (x^2 - 3) = 0 this means x = +-sqrt(3)

    but do we just ignore the x = sqrt(-1) and conclude that the inflection points are x = 0, x = sqrt(3) and x = -sqrt(3)

    So once we get the inflection points we use the sign charts to find concave up and concave down.

    intervals where graph is concave up: (-sqrt(3), 0) & (sqrt(3), infinity)
    intervals where graph is concave down: (-infinity, -sqrt(3)) & (0, sqrt(3))

    are these intervals i found correct?
  2. jcsd
  3. Oct 28, 2006 #2
    i did not get the imaginary answer you got

    i got this for the second derivative
    using mathematica
    [tex] -\frac{16x}{(x^2+1)^2} +4x\left(\frac{8x^3}{(1+x^2}^3} - \frac{2}{(1+x^2)^3} = 0 [/tex]

    it looks like you are assuming the denominator to be zero... taht cant be. that would not make this expression equal to zero.
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