Tofind the inflexion point of y = x/(x^2+1).

At the infection point d^2/dx^2 = 0.

So we find d2y/dx^2 and equate it ti zero and solve for x and y .

dy/dx=y' = (x/x^2+1)'

y' = { (x)'(x^2+1)-x(x^2+1)}/(x^2+1

y' = {x^2+1-2x^2}/(x^2+1)

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

y' = -(x^2-1)/(x^2+1) = (-1) { 1 -2/(x^2+1)

y" = {-1+2/(x^2+1)}'

y" = -2*2x/(x^2+1)^2.

y" = -4x/(x^2+1)^2 .

Equating to zero, we get: -4x = 0 or x = 0 and corresponding y = 0/(0+1) = 0

So the inflection point is at (0,0)

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))**