# 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?

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Calculate the two slopes between P1 and P2, and between P1 and P3. If they're the same then P1,P2,P3 on a line. – coffeemath Jan 25 '13 at 10:01
Can you share us what is the nominator of $f''$? – Berci Jan 25 '13 at 10:45

Unless, I miscalculated, the nominator of $f''$ becomes $$2x^3+6x^2-6x-2$$ So to find the roots of $x^3+3x^2-3x-1$. Luckily $x_1:=1$ is a root, 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 got 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 $$\frac{y_{2,3}-y_1}{x_{2,3}-x_1} = \frac{\frac{1\pm\sqrt3}4-1}{-3\pm\sqrt3} = \frac14\cdot\frac{-3\pm\sqrt3}{-3\pm\sqrt3} = \frac14.$$