# How do you find all points of inflection given y=-x^3/(x^2-1)?

Jan 15, 2017

POI : O(0, 0)

#### Explanation:

Resolving into partial fractions,

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

$y ' ' = \frac{1}{x - 1} ^ 3 + \frac{1}{x + 1} ^ 3 = 0$, when

${\left(x + 1\right)}^{3} = - {\left(x - 1\right)}^{3}$, giving the cubic $x \left({x}^{2} + \frac{1}{2}\right) = 0$

that has just one real zero x = 0. at which $y ' ' ' \ne 0$.

So, the origin is the poiit of inflexion . See the illustrative graph, for

tangent-crossing-curve, at O.

graph{(x^3/(1-x^2)-y)y=0 [-5, 5, -2.5, 2.5]}