Answer
The three inflection points are: $(1,1), (-2-\sqrt{3}, \frac{1-\sqrt{3}}{4}),$ and $(-2+\sqrt{3}, \frac{1+\sqrt{3}}{4})$
The straight line connecting the three points has a slope of $\frac{1}{4}$
The three inflection points lie on one straight line.
Work Step by Step
$y = \frac{1+x}{1+x^2}$
$y' = \frac{(1+x^2)-(1+x)(2x)}{(1+x^2)^2}$
$y' = \frac{1+x^2-2x-2x^2}{(1+x^2)^2}$
$y' = \frac{1-2x-x^2}{(1+x^2)^2}$
$y'' = \frac{(-2-2x)(1+x^2)^2-(1-2x-x^2)(2)(1+x^2)(2x)}{(1+x^2)^4}$
$y'' = \frac{(-2-2x)(1+x^2)-(1-2x-x^2)(2)(2x)}{(1+x^2)^3}$
$y'' = \frac{-2-2x-2x^2-2x^3-4x+8x^2+4x^3}{(1+x^2)^3}$
$y'' = \frac{-2-6x+6x^2+2x^3}{(1+x^2)^3}$
We can find the values of $x$ such that $y''=0$:
$y'' = \frac{-2-6x+6x^2+2x^3}{(1+x^2)^3} = 0$
$-2-6x+6x^2+2x^3 = 0$
$x^3+3x^2-3x-1 = 0$
$(x-1)(x^2+4x+1) = 0$
$x=1$ or $x^2+4x+1 = 0$
We can use the quadratic formula:
$x = \frac{-4\pm \sqrt{(4)^2-(4)(1)(1)}}{2(1)}$
$x = \frac{-4\pm \sqrt{12}}{2}$
$x = -2 \pm \sqrt{3}$
When $x = 1$:
$y = \frac{1+1}{1+1^2} = 1$
When $x = -2-\sqrt{3}$:
$y = \frac{1+(-2-\sqrt{3})}{1+(-2-\sqrt{3})^2}$
$y = \frac{-1-\sqrt{3}}{1+4+4\sqrt{3}+3}$
$y = \frac{-1-\sqrt{3}}{8+4\sqrt{3}}\cdot \frac{8-4\sqrt{3}}{8-4\sqrt{3}}$
$y = \frac{4-4\sqrt{3}}{16}$
$y = \frac{1-\sqrt{3}}{4}$
When $x = -2+\sqrt{3}$:
$y = \frac{1+(-2+\sqrt{3})}{1+(-2+\sqrt{3})^2}$
$y = \frac{-1+\sqrt{3}}{1+4-4\sqrt{3}+3}$
$y = \frac{-1+\sqrt{3}}{8-4\sqrt{3}}\cdot \frac{8+4\sqrt{3}}{8+4\sqrt{3}}$
$y = \frac{4+4\sqrt{3}}{16}$
$y = \frac{1+\sqrt{3}}{4}$
The three inflection points are: $(1,1), (-2-\sqrt{3}, \frac{1-\sqrt{3}}{4}),$ and $(-2+\sqrt{3}, \frac{1+\sqrt{3}}{4})$
We can find the slope $m$ of the line connecting the points $(1,1)$ and $(-2-\sqrt{3}, \frac{1-\sqrt{3}}{4})$
$m = \frac{(\frac{1-\sqrt{3}}{4})-1}{(-2-\sqrt{3})-1}$
$m = \frac{\frac{-3-\sqrt{3}}{4}}{-3-\sqrt{3}}$
$m = \frac{-3-\sqrt{3}}{-12-4\sqrt{3}}$
$m = \frac{3+\sqrt{3}}{12+4\sqrt{3}}\cdot \frac{12-4\sqrt{3}}{12-4\sqrt{3}}$
$m = \frac{24}{96}$
$m = \frac{1}{4}$
We can find the slope $m$ of the line connecting the points $(1,1)$ and $(-2+\sqrt{3}, \frac{1+\sqrt{3}}{4})$
$m = \frac{(\frac{1+\sqrt{3}}{4})-1}{(-2+\sqrt{3})-1}$
$m = \frac{\frac{-3+\sqrt{3}}{4}}{-3+\sqrt{3}}$
$m = \frac{-3+\sqrt{3}}{-12+4\sqrt{3}}$
$m = \frac{-3+\sqrt{3}}{-12+4\sqrt{3}}\cdot \frac{-12-4\sqrt{3}}{-12-4\sqrt{3}}$
$m = \frac{24}{96}$
$m = \frac{1}{4}$
The straight line connecting the three points has a slope of $\frac{1}{4}$
The three inflection points lie on one straight line.
