Three inflection points are on a line Given the function $f(x)=\frac{x+1}{x^2+1}$. I am asked to show that $f$ has three inflection points which are lie on a same line.
I know what to find those points by taking $f'$ and then $f''$, but how can I show that there is a line contaning these points?
 A: Unless, I miscalculated, the numerator of $f''$ becomes
$$2x^3+6x^2-6x-2.$$
Next we find the roots of $x^3+3x^2-3x-1$. Luckily $x_1:=1$ is a root and the polynomial factors as $(x-1)(x^2+4x+1)$. The other two roots hence are $x_{2,3}:=-2\pm\sqrt3$.
Their square: ${x_{2,3}}^2 = 7\mp4\sqrt3 $. And we get the three points:
$$\begin{align} x_1&=1 & y_1:=f(x_1)&=1 \\
x_{2,3}&=-2\pm\sqrt3 & y_{2,3}:=f(x_{2,3})&= \frac{-1\pm\sqrt3}{8\mp 4\sqrt3}=\ldots=\frac{1\pm\sqrt3}4.
 \end{align}$$
Then the slopes $\displaystyle\frac{y_2-y_1}{x_2-x_1}$ and $\displaystyle\frac{y_3-y_1}{x_3-x_1}$ become
$$\frac{y_{2,3}-y_1}{x_{2,3}-x_1} = \frac{\frac{1\pm\sqrt3}4-1}{-3\pm\sqrt3} =
\frac14\cdot\frac{-3\pm\sqrt3}{-3\pm\sqrt3} = \frac14. $$
$$ $$
A: Here is a partial answer: it shows that the inflection points (IP) are all on the same line. But it does not show that there are three of them.
The equation of the graph of $f$ is $y(x^2+1)=x+1$. Call it $(E)$. By Implicit Differentiation, we have 
$$(E') \hskip 5mm y'(x^2+1) +2xy=1 $$ 
and 
$$(E'') \hskip 5mm y''(x^2+1) +4xy'+2y = 0 $$
At an IP, we have $y''=0$, so $(E'')$ gives $2xy'=-y$. Multiplying $(E')$ by $2x$ and substituting, we have $-y(x^2+1)+4x^2y=2x$ which simplifies to 
$$(F) \hskip 5mm y(3x^2-1) = 2x $$
If we subtract $3$ times $(E)$ from $(F)$ we get $-4y=-x-3$. Therefore, the IP of $f$ are all on the line $x-4y+3=0$.
