point of inflection of $f(x)={2x\over 1+x^2}$ could any one tell me how to determine the point of inflection of $f(x)={2x\over 1+x^2}$
Answer says It Has exactly $3$ point of inflection separated by a point of Maxima and a point of Minima.
I found $f''(x)=0$ at two point.  

The derivative, via the quotient rule, is
$$f'(x) = \frac{(1 + x^2) (2) - (2x)(2x)}{(1 + x^2)^2} = 2 \frac{1 - x^{2}}{(1 + x^2)^2}$$
Ignoring the leading $2$, the numerator of the second derivative is thus
$$(1 + x^2)^2 (-2x) - (1 - x^2) \cdot 2 (1 + x^2) 2x$$
Simplifying, this is
\begin{align}
(1 + x^2) \left((1 + x^2)(-2x) - (1 - x^2)(4x)\right) &= -2x (1 + x^2) \left(1 + x^2 + 2(1 - x^2) \right) \\
&= -4x (1 + x^2)(3 - x^2)
\end{align}
This has three real roots at $0, \pm \sqrt{3}$, and two imaginary roots at $\pm i$.
