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Math Help - One interval where the graphs concave upwards, and one where it is concave downwards

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    One interval where the graphs concave upwards, and one where it is concave downwards

    Where is the function Concave Up and Down?

    Hi im trying to find one interval where the function is concave up and one interval where it is concave down. The function is: f(x)=(x+5)/(x^2-16). Heres my information. What points do i use from my notes?

    would it be between the inflection point and -4 asymptote (theres a inflextion)?
    Asymptotoes are x=4 x=-4


    Local minimum/maximum:

    f'(x) = 0
    −(xイ+10x+16)/(xイ−16)イ = 0
    xイ+10x+16 = 0
    (x+8) (x+2) = 0
    x = −8, x = −2

    f(−2) = (−2+5)/(4−16) = −1/4


    f''(x) = 2(xウ+15xイ+48x+80)/(xイ−16)ウ
    f''(−8) = 2(−512+960−384+80)/(64−16)ウ = 1/384 > 0 -----> local minimum
    f''(−2) = 2(−8+60−96+80)/(4−16)ウ = −1/24 < 0 ------------> local maximum

    Local minimum at point (−8, −1/16)
    Local maximum at point (−2, −1/4)

    覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧

    Points of inflection:

    f''(x) = 0
    2(xウ+15xイ+48x+80)/(xイ−16)ウ = 0
    xウ + 15xイ + 48x + 80 = 0

    There is no easy way to solve this. Using graphing calculator, I found:
    x = −5 − 3^(4/3) − 3^(2/3) ≈ −11.406832534

    f(−5 − 3^(4/3) − 3^(2/3)) ≈ −0.056143242

    Point of inflection: (−11.406832534, −0.056143242)
    Last edited by iFuuZe; March 26th 2013 at 06:57 AM.
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