Find the inflection point of the function y=x/(x^2+1)

The inflection points could be found by calculating the
roots of the second derivative (if there are any).

First of
all, we'll calculate the first derivative applying the quotient
rule:

f'(x)=[x'*(x^2+1)-x*(x^2+1)']/(x^2+1)^2

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

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

f"(x)
=
[f'(x)]'

f''(x)={(1-x^2)'(x^2+1)^2-(1-x^2)[(x^2+1)^2]'}/(x^2+1)^4

f''(x)=2x(x^2-3)/(x^2+1)^2

After
f"(x) calculus, we'll try to determine the roots of
f"(x).

For this,
f"(x)=0

Because (x^2+1)^2>0, only the
numerator could be
cancelled.

 2x(x^2-3)=0.

2x=0,
for x=0

x^2-3=0, for x=-sqrt3 or
x=sqrt3

So, the inflection points
are:

(0,f(0)),
(-sqrt3,f(-sqrt3)),(sqrt3,f(sqrt3))