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)#