What are the points of inflection, if any, of f(x) =x/(x+8)^2?

1 Answer
Apr 4, 2018

Below

Explanation:

f(x)=x/(x+8)^2
f'(x)=((x+8)^2(1)-(x)2(x+8))/(x+8)^4
f'(x)=((x+8)-2x)/(x+8)^3
f'(x)=(8-x)/(x+8)^3
f''(x)=((x+8)^3(-1)-(8-x)(3)(x+8)^2)/(x+8)^6
f''(x)=((-1)(x+8)-3(8-x))/(x+8)^4
f''(x)=(-x-8-24+3x)/(x+8)^4
f''(x)=(2x-32)/(x+8)^4

For points of inflexion, f''(x)=0

ie

(2x-32)/(x+8)^4=0
2x-32=0
x=16

Test x=16

When:
x=15, f''(x)=-2/279841
x=16, f''(x)=0
x=17, f''(x)=2/390625

Since there is a change in concavity, then there is a point of inflexion at (16,1/36)