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

• Mar 26th 2013, 05:41 AM
iFuuZe
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)

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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)