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Homework Help: Discontinuity Help please!

  1. Oct 1, 2009 #1

    So i have a few questions and i cant seem to wrap my head around it.


    Decide which of the following functions has a removable discontinuity at a, and then if it is so..remove the discontinuity.

    f(x)=3[x]; a=-1.

    (I dont think there is a removable discontinuity because eventhough X is in a set bracket any values of x would not be undefined. Thats what i think anyway...if i am wrong can you please explain why?)


    Find the number of discontinuity points of f(x)=[sinx] within the open interval (-2pi, 2pi).

    (Isnt sinx always continous? therefore how can there be any discontinuity points?)


    Find the set of all points at which the function f(x)=secx is continous?

    (My answer is this: continous at all points except x=pi/2 +kpi, where k is a constant. (is this right?))

    Thank you so much too anyone that responds to this thread!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 1, 2009 #2


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    Does [x] mean the greatest integer less than or equal x? If so, that might affect your thinking...
  4. Oct 1, 2009 #3
    yeah i do believe it does mean that. However I dont know how to go about solving it then...

    if it were to mean that then for the first question:

    f(x)=3[x], a=-1, it would not exist because x E(is not) in the Z(intergers)?

    then for the second question

    where sinx=an interger that is when its discontinous?

    If that is the case how would you go about proving that?
  5. Oct 1, 2009 #4


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    I don't know what the a = -1 is for in your first problem. The function [x] is defined for all real x. Draw its graph. For each x ask yourself what is the greatest integer less than or equal x and plot that for the y value.

    Similarly for [sin(x)]. Draw their graphs. Then you may see what to do.
  6. Oct 1, 2009 #5
    Q1: [x] <---- brackets usually imply the greatest integer function, sometimes called the step function for the pattern that it produces. At -1, you should see a jump in the step function, which would make it an unremovable discontinuity.

    Q2: This is a tricky one because it makes you think outside the box a little bit. From -2pi to 2pi, the sine wave starts at the x-axis, and reaches relative extremas (max/min) 4 times, twice for each period. If you consider the graph beginning at x=0, the y value will be 0. As you increase x, the y value continues to be 0 until you hit pi/2, in which case the y-value jumps to 1 for just one point, then continues along the x-axis. When you hit pi, the value will still be 0 (pi, 0), but the point immediately afterwards will jump down to a y-value of -1. This line segment with an open circle at (pi,-1) will travel horizontally to the right from an x-value of pi all the way to 2pi, in which case the value jumps back to 0. But since 2pi is not included on the interval, you don't count that as one of your discontinuities.

    So from -2pi to 2pi, I count discontinuities occuring at -3pi/2, -pi, 0, pi/2, and pi.

    Looks like five?

    Q3: Your analysis looks appropriate. At every pi/2 +kpi for cosx, the value is 0. Secant, its reciprocal function, should then be an asymptote at those values.
  7. Oct 1, 2009 #6


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    Mrkuo, you aren't supposed to just work the problem for them in this forum.
  8. Oct 1, 2009 #7
    Really? I just found this forum a few days ago. (maybe i should've read the faq...)

    Sorry about that.
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