y=x^3/(1-x^2)y=x31−x2
Find the first derivative of y,
dy/dx=−(x^2(x^2-3))/(x^2-1)^2dydx=−x2(x2−3)(x2−1)2 ( using quotient rule )
Let the first derivative be 00 to find the stationary points,
−(x^2(x^2-3))/(x^2-1)^2=0−x2(x2−3)(x2−1)2=0
color(white)(^^^)x^2(x^2-3)=0⋀x2(x2−3)=0
color(white)(^^xxxxx/....)x=0,+-sqrt3
Subsititute in x-values of stationary points to find their y-values,
When x=0,
y=(0)^3/(1-(0)^2)
color(white)(y)=0
When x=sqrt3,
y=(sqrt3)^3/(1-(sqrt3)^2)
color(white)(y)=-2.598
When x=-sqrt3,
y=(-sqrt3)^3/(1-(-sqrt3)^2)
color(white)(y)=2.598
Find the second derivative of y,
(d^2y)/dx^2=-(2x(x^2+3))/(x^2-1)^3 ( using quotient rule )
Substitute in x-values of stationary points to find their respective nature,
When x=0,
(d^2y)/dx^2=-(2(0)((0)^2+3))/((0)^2-1)^3
color(white)(xxx)=0 ( usually inflexion point )
When x=sqrt3,
(d^2y)/dx^2=-(2(sqrt3)((sqrt3)^2+3))/((sqrt3)^2-1)^3
color(white)(xxx)=2.598 > 0 ( minimum )
When x=-sqrt3,
(d^2y)/dx^2=-(2(-sqrt3)((-sqrt3)^2+3))/((-sqrt3)^2-1)^3
color(white)(xxx)=-2.598 > 0 ( maximum )
Hence, the stationary points are maximum(-sqrt3,2.598), inflexion(0,0), and minimum(sqrt3,-2.598)
Check:graph{x^3/(1-x^2) [-10, 10, -5, 5]}
