Question

Points of discontinuity for y = (x-1)^{1/3} ?

Answer 1

Are you just posting all your homework questions on here or what?
\[(x-1)^{1/3}=\sqrt[3]{x-1}\]
Cannot take the root of a negative, f(x) is continuous on \[[1,\infty)\]

Answer 2

Hey. cot x [0, 2 pi] is continous everywhere?

Answer 3

cotx is 1/tanx so it has an infinite discontinuity where tanx=0

Answer 4

No, because
\[\cot(x)=\cos(x) \div \sin(x)\]
Cannot divide by zero. on \[[0,2\pi]\] sin(x) is 0 at x=0, x=pi, and x=2pi. So cot(x) is a periodic function that has vertical asymptotes at the beforementioned points. Discontinuous at x=0, x=pi, x=2pi