Three Collinear Points of Inflection

[achilleas]

Show that the curve \(y=\frac{x+1}{x^{2}+1}\) has three points of inflection which lie on one straight line.

[Grigorios Kostakos]

The function \(f(x)=\frac{x+1}{x^2+1}\) has second derivative \(f''(x)=2\,\frac{x^3-3x+3x^2-1}{({x^2+1})^3}\) of which roots are \(x_1=1\), \(x_2=-2+\sqrt{3}\) and \(x_3=-2-\sqrt{3}\).

It's easy to check that \((x_1,f(x_1))\), \((x_2,f(x_2))\) and \((x_3,f(x_3))\) are the three points of inflection of curve \(y=\frac{x+1}{x^2+1}\).

Because the difference quotients \(\frac{f(1)-f(-2+\sqrt{3}\,)}{1-(-2+\sqrt{3}\,)}=\frac{1}{4}\frac {-9+5\,\sqrt{3}}{(-2+\sqrt {3}\,)(3-\sqrt{3}\,)}\)and \(\frac{f(1)-f(-2-\sqrt{3}\,)}{1-(-2-\sqrt{3}\,)}=\frac{1}{4}\frac {9+5\,\sqrt{3}}{(2+\sqrt {3}\,)(3+\sqrt{3}\,)}\) are both equal to \(\frac{1}{4}\), we have that the three points of inflection lie on one straight line.