# Find the points of inflection of the curve y=(1+x)/(1+x^2)?

Oct 17, 2014

$y = \frac{1 + x}{1 + {x}^{2}}$

By Quotient Rule,

y'={1cdot(1+x^2)-(1+x)cdot2x}/{(1+x^2)^2} ={1-2x-x^2}/{(1+x^2)^2}

By Quotient Rule,

$y ' ' = \frac{\left(- 2 - 2 x\right) \cdot {\left(1 + {x}^{2}\right)}^{2} - \left(1 - 2 x - {x}^{2}\right) \cdot 2 \left(1 + {x}^{2}\right) \left(2 x\right)}{{\left(1 + {x}^{2}\right)}^{4}}$

$= \frac{2 \left(1 + {x}^{2}\right) \left(- 1 - {x}^{2} - x - {x}^{3} - 2 x + 4 {x}^{2} + 2 {x}^{3}\right)}{{\left(1 + {x}^{2}\right)}^{4}}$

$= \frac{2 \left({x}^{3} + 3 {x}^{2} - 3 x - 1\right)}{{\left(1 + {x}^{2}\right)}^{3}}$

$= \frac{2 \left(x - 1\right) \left({x}^{2} + 4 x + 1\right)}{{\left(1 + {x}^{2}\right)}^{3}}$

By setting $y ' ' = 0$,

$\left\{\begin{matrix}x - 1 = 0 R i g h t a r r o w x = 1 \\ {x}^{2} + 4 x + 1 = 0 R i g h t a r r o w x = - 2 \pm \sqrt{3}\end{matrix}\right.$

Using the x-values found above to split $\left(- \infty , \infty\right)$ into intervals

$\left(- \infty , - 2 - \sqrt{3}\right) , \left(- 2 - \sqrt{3} , - 2 + \sqrt{3}\right) , \left(- 2 + \sqrt{3} , 1\right) , \text{ and } \left(1 , \infty\right)$.

Using sample points: $x = - 4 , - 2 , 0$, and $2$ for the intevals, respectively,

$y ' ' \left(- 4\right) < 0$, $y ' ' \left(- 2\right) > 0$, $y ' ' \left(0\right) < 0$, and $y ' ' \left(2\right) > 0$,

which indicate concavity changes at each point; therefore, there are inflection points at $x = 1 , - 2 \pm \sqrt{3}$.

The inflection points are

$\left(1 , y \left(1\right)\right) = \left(1 , 1\right)$,

(-2-sqrt{3},y(-2-sqrt{3}))=(-2-sqrt{3},{1-sqrt{3 }}/4),

and

$\left(- 2 + \sqrt{3} , y \left(- 2 + \sqrt{3}\right)\right) = \left(- 2 + \sqrt{3} , \frac{1 + \sqrt{3}}{4}\right)$.

I hope that this was helpful.