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#1
Sep2005, 06:20 PM

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<x(less than or equal to)0 {(3x+1)/(x2), 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)/(x2)=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)/(x2)=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! 


#2
Sep2105, 07:36 AM

Math
Emeritus
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P: 39,338




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