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Continuity with piecewise functions.

  1. Sep 11, 2006 #1
    I have two problems.. I'll put them both here, and show my work on both of them.

    1. Is this function continuous on [-1, 1]?
    f(x) =
    x / |x| , x does not equal 0
    0 , x = 0

    After graphing this function, the first statement gives y = -1 for all x < 0, and y = 1 for all x > 0. The second statement makes y = 0 when x = 0. Putting everything together, y values are defined at -1, 0, 1, but nothing in-between. Does this make the function not continuous because of the jumping at x = 0?
    2.
    Discuss the continuity of the function g:
    g(x) =
    sin x / x , for x does not equal 0
    1/2 , for x = 0

    There is a hole in the graph at x = 0 (for the first statement), but the second statement defines x = 0 to be 1/2, it's just lower than the curve in the first statement. Would this be a continuous function or not?
     
  2. jcsd
  3. Sep 12, 2006 #2
    There are two types of discontinuities but I'll get to that later. A function is discontinuous if you can't draw the entire graph without picking your pencil off the paper. In question 2, you say there's a point just below the rest of the graph. Clearly you can't draw the entire graph unless you pick up your pencil off the paper so it's discontinuous. The same goes for question 1.

    The two questions have different types of discontinuities. In question 2, only 1 point is off of the rest of the graph. The point at x=0. In question one, you have an infinite number of points that are off the rest of the graph (in this case they just don't exist). There is an infinite number of points between -1 and 0 and between 0 and 1 (for example .01, .001, .009, etc are all undefined). Question 2 has a finite discontinuity because you can count how many points are off the rest of the graph. Question 1 has inifinite discontinuity.

    The "picking up the pencil" explanation is just a simple way to think about it. You should really get used to thinking about it in terms of left and right sided limits but I assume you already learned that stuff.


    EDIT: Hey sorry I misread something in question 1. There is only 1 point that is discontinuous so it has a finite discontinuity, not infinite as I first said.
     
    Last edited: Sep 12, 2006
  4. Sep 12, 2006 #3

    I think most people would consider at least 3 types of discontinuity, the first being a hole where a single point is missing from teh curve, nex being a jump discontinuity where the function seems to jump to a different function at some point (this is not the best way to explain this) and the third being an infinite discontinuity where the function is undefined at a point, typ[ically because of division by zero. Your analysis of question one seems correct, but at question 2 there are not an infinite number of discontinuities but only one at x equals 0, the function is continuous on either side of 0, but has a jump discontinuity at x=0. And the points corresponding to x= .01, .001, .009 all do exist so I'm not sure what you meant by that.
     
  5. Sep 12, 2006 #4
    I think you misread what I typed :) . I wrote:
    Question 2 has a finite discontinuity because you can count how many points are off the rest of the graph. Question 1 has inifinite discontinuity. Also the x=.01 and stuff was still referring to question 1. Sorry for the messiness. I should use more paragraphs :-p
     
  6. Sep 12, 2006 #5
    What I said still stands but I said question one when I meant two and vice versa.

    The function defined in question one IS defined for all values of x on [-1,1], but there is a jump discontinuity at x=0, it is not an infinite discontinuity.
     
  7. Sep 12, 2006 #6
    Oh, really sorry. The original poster wrote: "y values are defined at -1, 0, 1, but nothing in-between." I wasn't paying attention and just noticed the words "nothing in-between" and ignored the [-1,1] domain info. Yes you're right and I'll edit my first post. It is a finite discontinuity

    It's also 2AM. good night
     
  8. Sep 12, 2006 #7

    HallsofIvy

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    The definition of "continuous at x= a" is
    [tex]\lim_{x\rightarrow a}f(x)= f(a)[/tex]

    What are the limits of the functions in 1 and 2 as x goes to 0?
     
  9. Sep 12, 2006 #8
    thanks everyone... I was trying to do it without using limits because we haven't gotten there yet... (we're still in the introductory review chapter of the calculus book..).. I've done it with limits before in my high school calculus class, but we haven't gotten to it yet in this college calculus... But, I understand all of the explanations, thanks.
     
  10. Sep 12, 2006 #9

    HallsofIvy

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    Then what definition of "continuous" were you using?
     
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