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Please help check work

  1. Sep 20, 2005 #1
    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!
    Last edited: Sep 20, 2005
  2. jcsd
  3. Sep 21, 2005 #2


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    On the contrary, f is not even defined at -3(pi)/2! It is not enough that the limit exist there!

    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.

    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.

    ??? (-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.

    ??? Your definition only goes up to "0<x(less than or equal to)1".
    Last edited by a moderator: Sep 21, 2005
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