## anonymous 5 years ago Points of discontinuity for y = (x-1)^{1/3} ?

1. anonymous

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 $[1,\infty)$

2. anonymous

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

3. anonymous

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

4. anonymous

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