Question 280940
{{{graph(600,600,-1,2,-1000,1000,1/(x^3-x^2))}}}


The graph of your equation is shown above.


It looks like you have points of discontinuity at x = 0 and x = 1


The points of discontinuity would be when the denominator in the equation equals 0.


Your equation is y = 1/(x^3 - x^2)


The denominator is 0 when x^3 - x^2 = 0


Solve the equation of x^3 - x^2 = 0 to find the points of discontinuity.


x^3 - x^2 factors out to be x^2 * (x-1) = 0


This equation will be true when either x^2 = 0 or when x-1 = 0 or when both are equal to 0.


x^2 = 0 when x = 0.


x-1 = 0 when x = 1.


Your points of discontinuity are when x = 0 and x = 1.


The graph confirms that.