Question

f(x)= x/(1+x)^2


c) find the intervals of concavity and the inflection points.

My answers:

f is concave downward on (- infinity, 2) and concave upward on (2, infinity).

Inflection point: ( 2, 2/9)

Answer 1

f(x)=x*(1+x)^-2
f'= (1+x)^-2 -2x(1+x)^-3
f"=-2(1+x)^-3-2(1+x)^-3+6x(1+x)^-4

so where is f"=0? Setting it to zero, multiplying through by (1+x)^4 then
0=-2(1+x)-2(1+x)+6x
0=-2x-2x+6x -4
x= 2
for the inflection point, I didn't do y.

now for x<2, f"=negative, so concave downward, for x>2 upward.