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anonymous
 5 years ago
Points of discontinuity for y = (x1)^{1/3} ?
anonymous
 5 years ago
Points of discontinuity for y = (x1)^{1/3} ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you just posting all your homework questions on here or what? \[(x1)^{1/3}=\sqrt[3]{x1}\] Cannot take the root of a negative, f(x) is continuous on \[[1,\infty)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hey. cot x [0, 2 pi] is continous everywhere?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cotx is 1/tanx so it has an infinite discontinuity where tanx=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
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