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

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Answer 1 · 2017-01-15T06:13:19

POI : O(0, 0)

Resolving into partial fractions,

$y=-x-1/2(1/(x-1)+1/(x+1))$

$y''=1/(x-1)^3+1/(x+1)^3=0$ , when

$(x+1)^3=-(x-1)^3$ , giving the cubic $x(x^2+1/2)=0$

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]}

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