by bard
Tags: check, work
 P: 65 Can someone please check my work, thank you Discuss any discontinuities. classify each discontinuity as removable or nonremovable. (this is a peicewise function) f(x)={sin(x), x<-3(pi)/2 {tan(x/2), -3(pi)/2
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PF Gold
P: 39,569
 Quote by bard Can someone please check my work, thank you Discuss any discontinuities. classify each discontinuity as removable or nonremovable. (this is a peicewise function) f(x)={sin(x), x<-3(pi)/2 {tan(x/2), -3(pi)/2
On the contrary, f is not even defined at -3(pi)/2! It is not enough that the limit exist there!

 (2) At x=-pi, f(x-) sin(x)= +infinity, while f(x+)tan(x/2)=- infinity, not equal, therefore f(x) is discontinuous, which is nonremovable.
What? sin(pi)= 0 not infinity! In any case, f(x)= sin(x) only for x<-3pi/2, not anywhere near -pi so that's irrelevant. You are correct that tan(pi/2) is not defined so there is a removable discontinuity there.

 (3) At x=0, f(x+)=x^3+x = 0, while f(x-)= tan(x/2)=0, so f(x) is continuous.
I'm sorry, where did you get f(x)= x^3+ 3? Did you intend to define f(x)= x^3+ 3 for 0< x<= 1? You missed writing that. If that was what you intended, then, yes, f is continuous at x= 0.

 (4) At x=1, f(x+)=(-3x+1)/(x-2)=1, f(x-)=x^3+x=2, so they are not equal, therefore f(x) is discontinuous, nonremovable
??? (-3(1)+1)/(1-2)= (-2)/(-1)= 2! Now, was f(1) defined to be 1^2+ 1= 2? If so then f is continuous at x= 1. If not then there is a removable discontinuity.

 (5) At x=3, f(x+)=-sqrt(x+6)=-3, f(x-)=(-3x+1)/(x-2)=-8, not equal, so f(x) is discontinuous, nonremovable. (6) at x=2 is not in the domain of the function, therefore there is a discontinuity, removable
??? Your definition only goes up to "0<x(less than or equal to)1".

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