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## Main Question or Discussion Point

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<x(less than or equal to)0

{(-3x+1)/(x-2), 1<x<3

{-sqrt(x+6), x(greater than or equal to)3

{x^3+x, 0<x(less than or equal to)1

Note: all these functions are part of f(x), thus they are peicewise functions

All the possible discontinuity points are -3(pi)/2, -pi, 0, 1, 2,3

(1) Since at x=-3(pi)/2, f(x-)=sin(-3(pi)/2)=1, whilef(x+)= tan(-3(pi)/4)=1, which are equal, therefore f(x) is continuous at -3(pi)/2.

(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.

(3) At x=0, f(x+)=x^3+x = 0, while f(x-)= tan(x/2)=0, so f(x) is continuous.

(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

(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, thereore there is a discontinuity, removable

therefore the discontinuities are at x=-pi, x=1, x=3, and x=2

am i missing any, thanks!

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<x(less than or equal to)0

{(-3x+1)/(x-2), 1<x<3

{-sqrt(x+6), x(greater than or equal to)3

{x^3+x, 0<x(less than or equal to)1

Note: all these functions are part of f(x), thus they are peicewise functions

All the possible discontinuity points are -3(pi)/2, -pi, 0, 1, 2,3

(1) Since at x=-3(pi)/2, f(x-)=sin(-3(pi)/2)=1, whilef(x+)= tan(-3(pi)/4)=1, which are equal, therefore f(x) is continuous at -3(pi)/2.

(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.

(3) At x=0, f(x+)=x^3+x = 0, while f(x-)= tan(x/2)=0, so f(x) is continuous.

(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

(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, thereore there is a discontinuity, removable

therefore the discontinuities are at x=-pi, x=1, x=3, and x=2

am i missing any, thanks!

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