# What are the points of inflection of f(x)=x/(1+x^2)?

Oct 22, 2016

There are 3 points of inflexion $\left(0.0\right)$ and $\left(\sqrt{3} , \frac{\sqrt{3}}{4}\right)$ and $\left(- \sqrt{3} , - \frac{\sqrt{3}}{4}\right)$

#### Explanation:

The domain of $f \left(x\right)$ is $\mathbb{R}$

To determine the points of inflexion, we calculte $f ' ' \left(x\right)$ and find out when it is $= 0$

We start by calculating the first derivative
$u \left(x\right) = x$ $\implies$ $u ' \left(x\right) = 1$
$v \left(x\right) = 1 + {x}^{2}$ $\implies$ $v ' \left(x\right) = 2 x$

$f ' \left(x\right) = \frac{u ' v - u v '}{v} ^ 2 = \frac{1 \cdot \left(1 + {x}^{2}\right) - x \cdot 2 x}{1 + {x}^{2}} ^ 2$
$= \frac{1 + {x}^{2} - 2 {x}^{2}}{1 + {x}^{2}} ^ 2 = \frac{1 - {x}^{2}}{1 + {x}^{2}} ^ 2$

$f \left(x\right) = 0$ when $1 - {x}^{2} = 0$ $\implies$ $x = \pm 1$
from $x = \infty$ to $x = - 1$
$f ' \left(x\right)$ is negative

from $x = - 1$ to $x = 1$
$f ' \left(x\right)$ is positive

from $x = 1$ to $x = \infty$
$f ' \left(x\right)$ is negative

So we have a minimum at $\left(- 1 , - \frac{1}{2}\right)$
and a maximum at $\left(1 , \frac{1}{2}\right)$

Let's calculate $f ' ' \left(x\right)$
$u = 1 - {x}^{2}$ , $u ' = - 2 x$
$v = {\left(1 + {x}^{2}\right)}^{2}$ so $v ' = 4 x \left(1 + {x}^{2}\right)$

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

$f ' ' \left(x\right) = 0$ when $x = 0$ and tis correspond to the point $\left(0 , 0\right)$

$f ' ' \left(x\right) = \frac{- 2 x \left(1 + {x}^{2}\right) \left(1 + {x}^{2} + 2 - 2 {x}^{2}\right)}{1 + {x}^{2}} ^ 4$
This is $= 0$ for $x = 0$

and $3 - {x}^{2} = 0$ $\implies$ $x = \pm \sqrt{3}$

graph{x/(1+x^2) [-5, 5, -2.5, 2.5]}