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When is a function defined and discontinuous at the same point?

  1. Apr 5, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to create a function and pick a point, I call "c" , where;
    (1)the limits from the right and left are equal,
    (2)the function is defined at "c", but...
    (3)the function is not continuous at "c"

    I don't get how this works.

    2. Relevant equations



    3. The attempt at a solution
    I used a removable discontinuity function: f(x) = x^2 - 4/x - 2 but then it is not defined at x = 2. So I know I could redefine it and say "at x = 2, let f(x) have the value 0f 4".

    Can it be that simple? I factored the original equation and canceled out all but x + 2, but I am not sure why I was trying that. Except that I keep trying all these things I have learned in math to try and make sense of this. I can almost see the answer but then I get to "lim (as x approaches 2) of f(x) is equal to lim ( as x approaches 2) (x+2) = 4.

    I have also used x^2-9/x-3 (which is basically the same)
    I have used x^2 - 9/x+3 but then I get lost again after factoring and get x-3.
     
  2. jcsd
  3. Apr 5, 2009 #2

    Dick

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    There is no rule that tells you how to define the value of a function at 'c'. You can 'define' it any way you want. You can define f(x)=0 at all values of x not equal to c and f(c)=1. Now you have a limit from both sides but it's not equal to f(c). It's probably a lot simpler than you think.
     
  4. Apr 5, 2009 #3
    You are actually making this a little more complicated than you need to be. Try looking at one of the most simple type of functions you know of. See if it satisfies the continuity conditions in your textbook. What can you do to a continuous function so that it satisfies condition (3) you have listed above? If a function satisfies the first two conditions that you've listed, then what is the third condition that will give you that f is continuous at c?
     
  5. Apr 5, 2009 #4

    symbolipoint

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    Might a linear function be a source to develop such a function as cinda wants? Would this need to be a combination of two functions? You could introduce a simple linear factor to numerator AND denominator so that for factor of (x-c), the function would be discontinuous at x=c; but the student could simply make a statement that the function has some maybe arbitrary value when x=c. Would any of this help?
     
  6. Apr 5, 2009 #5
    Edit: I see that this is what Dick might have been egging you toward, and I don't want to give the answer right away. Think about what a function is: it isn't just a formula, it can take many different forms. By definition, a function serves to map numbers from one set to another, which any number of formulas do fine, such as f(x)=x^3 or otherwise, but is there some way which you could redefine the function at c so that it is discontinuous? You have probably seen this type of function many times.
     
    Last edited: Apr 5, 2009
  7. Apr 5, 2009 #6
    This could be potentially confusing. Dick has already given a solution, and I gave some hints trying to get cinda to re-read the section on continuity. The original poster needs to post back showing their progress or that they understand, as at this point there is enough to go off of.
     
  8. Apr 17, 2009 #7
    I think I may be confusing the notation as much as the math itself. When I pick a point "c", does that become the same number as x? As in the number I use to do the limit notation? Does c need to be the same number as the number that x is approaching in the limit? Say I used this function...x^3 + 1/x and I pick a point c that is equal to 1. I will write it out like : lim as x approaches 1 = 2. I got that by just plugging in 1. When I look at the graph I see that I can't use 0 because then I am undefined at x = 0, so I chose 1. At x = 1 however I am at the bottom of a parabola with a discontinuity when
    -1< x < 1.

    Am I getting the idea?
     
  9. Apr 17, 2009 #8
    Unfortunately, no. The function [itex]f(x)=x^3+1/x[/itex] has a discontinuity at x=0, but the limits from the left and right are not equal. You are picking your point c=1, but the limits from the left and right at 1 are equal, the function is defined at 1, AND the function is continuous at 1. So you know this doesn't work either.

    I will give you an example of a solution to your problem since you aren't understanding, but I encourage you to come up with your own example.

    Let the function f(x) be defined by
    [tex]
    f(x) =
    \begin{cases}
    2 &, x=1 \\
    x &, x\not=1
    \end{cases}
    [/tex]
    Let c=1. Does this function and the point c satisfy the statements given in your problem?

    Also, this problem was posted a while ago. Are you in a calculus course? If you are still having trouble with this idea, I strongly recommend going to speak to your professor.
     
  10. Apr 18, 2009 #9
    Okay, so in the function I used, all the criteria is met for the problem I am working; limits are equal on the left and the right, it is defined at point c also (when c = 1) BUT it also continuous at 1 (which is what I am trying to get away from) So I should be able to define my function, with the proper notation, using a piecewise definition like you showed???
    I am in a calculus course at an on-line university, they provide web lectures that I can watch over and over, and I have a mentor to talk with, but I don't have anyone to sit and work through this with me directly. My mentor has worked on the phone with me but is restricted from giving specific details. They have provided me with tons of resources to look at and study.
    I have not ever been very good with math. I have not had much luck finding anyone to answer my specific questions and trying to learn calculus on line has been hard for me because I am not as confident as I should be probably.
     
  11. Apr 18, 2009 #10
    Yes, that is what you need to do. Also, you were trying to start off with a complicated function, which was throwing you off.
    That's okay, and yes it is difficult to learn by yourself and possibly even more difficult over the internet. Although, your mentor should provide details when you're really lost, and the only reason I asked is because this topic is usually covered very early in the semester.
     
  12. Apr 18, 2009 #11
    OOPs... so I could define my function like this: f(x) = { 2, x = 1
    {x, x = 0

    that is the part I don't get, if my function is defined at a point won't it always be continuous at that point also? I thought that was one of the fundamental rules. And a continuous at c function however, is not always defined at c.
     
  13. Apr 18, 2009 #12
    I have just finished a pre-calculus review and this is the beginning of my calculus (actually March 1). Trouble with the review was I have been out of high school 25 years.

    I am very much enjoying the challenges in this course, and have wanted to study physics since I was a very small child...but I never was confident enough at the math. Now I have decided I will learn calculus and physics both.
     
  14. Apr 18, 2009 #13
    No. That doesn't satisfy any of the requirements in your problem statements as it is just a function defined at only two points, one at x=0 and another at x=1. I recommend sketching the function that I posted to get a feel for what's going on here. I think you're missing the picture in your head.

    For example, look at this figure:
    http://upload.wikimedia.org/wikipedia/en/2/24/Removable-discontinuity.svg
    This function's definition is
    [tex]
    f(x) =
    \begin{cases}
    x^2 &, x<1 \\
    0 &, x=1 \\
    2-x &, x>1
    \end{cases}
    [/tex]
    The point c=1 satisfies your problem statements. Go the link http://en.wikipedia.org/wiki/Classification_of_discontinuities" [Broken] at Wikipedia for help as well.

    Just because your function is defined at a point, doesn't mean it will be continuous there. This is what this problem is trying to get at.
     
    Last edited by a moderator: May 4, 2017
  15. Apr 18, 2009 #14
    Thanks so much, I will follow the link and also sketch out the function, pictures are what I need. I have a graphing calculator but, the calculator as we know, is only as good as the input.

    Thanks again for suggestions.
     
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