#y=x^3/(1-x^2)#
Find the first derivative of y,
#dy/dx=−(x^2(x^2-3))/(x^2-1)^2# ( using quotient rule )
Let the first derivative be #0# to find the stationary points,
#−(x^2(x^2-3))/(x^2-1)^2=0#
#color(white)(^^^)x^2(x^2-3)=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]}
