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Homework Help: Basic definition of discontinuity

  1. Mar 17, 2010 #1
    i am looking for a simple definition of discontinuity using the example 3/(x^2+x-6)
     
  2. jcsd
  3. Mar 17, 2010 #2

    rock.freak667

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    f(x)=3/(x2+x-6)

    A simple definition would be that a point of discontinuity on the graph of f(x) occurs at x=a such that f(a) is undefined.

    So where would your function be undefined for?
     
  4. Mar 17, 2010 #3
    at x=2,-3. i understand that but when you say x=a what do u mean by that. can you show me a variable formula (using a, b and c) that would better explain this
     
    Last edited: Mar 17, 2010
  5. Mar 17, 2010 #4

    rock.freak667

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    Say your function was

    f(x)=1/(x-a)

    Wouldn't you agree that [itex]\lim_{x \rightarrow a} = \infty[/itex] and so f(a) is undefined?
     
  6. Mar 17, 2010 #5
    so you are saying that a would be the constant. like in my example if we factor x^2+x-6 we get (x-2)(x+3), so how would that be undefined
     
  7. Mar 17, 2010 #6

    rock.freak667

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    your f(x) is 3/(x-2)(x+3), as x→2, f(x)→∞, thus f(2) is undefined, similarly for f(-3)
     
  8. Mar 17, 2010 #7
    sorry but i still dont get it, its not like we have a denominator of zero
     
  9. Mar 17, 2010 #8

    rock.freak667

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    :confused: but if x=2, don't you have one?

    [tex]f(x)=\frac{3}{(x-2)(x+3)}[/tex]
     
  10. Mar 17, 2010 #9
    wat i ment was it dosent cancel out or anything
     
  11. Mar 17, 2010 #10
    ok then how would you compare that with a function that is not discontinuous like 1/(x+3)
     
  12. Mar 17, 2010 #11

    rock.freak667

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    cancel out with what?

    f(x)=1/(x+3) is discontinuous at x=-3.
     
    Last edited: Mar 17, 2010
  13. Mar 17, 2010 #12

    jav

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    A function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c).

    In this case, consider your c=2. f(c) is not defined. The limit of f(x) as x->2 from the positive side is +infinity. The limit of f(x) as x->2 from the negative side is -infinity.

    The first way you can know this function is discontinuous at 2 is that it is not defined at 2. Then no matter what the limit of f(x) is as x->2, it is not equal to f(2) (even if the limit does not exist).

    The second way you can know this function is discontinuous at 2 is that the limit of f(x) as x->2 from the positive side does not equal the limit of f(x) as x->2 from the negative side. Therefore the limit does not exist. If the limit does not exist, then it does not equal f(2) (even if f(2) is not defined).

    Remember to keep in mind that a function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c). f(c) could exist, it just happens to not exist in this example.
     
  14. Mar 17, 2010 #13

    jav

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    Actually the limit of f(x) as x->a from the left is negative infinity, while the limit of f(x) as x->a from the right is positive infinity.

    Therefore the limit of f(x) as x->a doesn't exist.

    I think you mean x = -3.
     
  15. Mar 17, 2010 #14
    so does that mean all rational functions are discontiuous.
     
  16. Mar 17, 2010 #15

    jav

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    Not necessarily, a rational function just means that it can be written as the ratio of two polynomials. But if, for example, the "denominator polynomial" is a constant, then the function is continuous over its domain.
     
  17. Mar 17, 2010 #16
    do you mean it would be contiuous if the rational function was a polynomial divied by a constant like for ex. (x+1)/3. but wouldnt that give you a linear function
     
  18. Mar 17, 2010 #17

    jav

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    If you choose your numerator to be a first order polynomial, then yes. But what about f(x) = [(x+1)^2]/3
     
  19. Mar 18, 2010 #18
    would (x)/(x^2-4) be a discontinuity
     
  20. Mar 18, 2010 #19

    jav

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    That function would be discontinuous at x=2 and x=-2
     
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