How do I determine where a function has a removable and a jump discontinuity?

  1. 1. The problem statement, all variables and given/known data

    I am given the following function, piecewise:

    f(x) = (-x+b) (x<1)
    3 if x=1
    (-12/(x-b))-1 (x>1, x=/b)

    I am asked:

    1) For what value(s) of 'b' does 'f' have a removable discontinuity at 1?
    2) For what value(s) of 'b' does 'f' have a (finite) jump discontinuity at 1? Write your answer in interval notation.

    2. Relevant equations

    X

    3. The attempt at a solution

    Honestly, I have to clue where to start; especially in regards to the removable discontinuity. I tried making the functions equal each other for the jump discontinuity but, that was way out as far as I can tell.

    --------------------------

    If I could at least be given a starting point to go from, it would be appreciated. I'm not asking for the answers but, I'm so lost and just staring at this thing isn't helping. Thanks,

    RK
     
  2. jcsd
  3. statdad

    statdad 1,479
    Homework Helper

    If a function has a removable discontinuity at [tex] x = a [/tex], that means these things.

    [tex]
    \begin{align*}
    \lim_{x \to a^+} f(x) & = \lim_{x \to a^-} f(x)\\
    \text{ so } & \lim_{x \to a} f(x) \text{ exists}\\
    f(a) & \text{ is defined} \\
    f(a) & \ne \lim_{x \to a^+} f(x)
    \end{align*}
    [/tex]

    All the limits above are finite. Basically, if there is a removable discontinuity at [tex] x = a [/tex], the function has almost everything needed to be continuous there, but the function value is "wrong". Here is an example.

    [tex]
    f(x) = \begin{cases}
    3x+5 & \text{ if } x \ne 10\\
    20 & \text{ if } x = 10
    \end{cases}
    [/tex]

    Here the limit at 10 is equal to 35 (because both one-sided limits equal 35), but [tex] f(10) = 20 [/tex], so the function is not continuous there. We say there is a removable discontinuity at 10 because we can do this: make [tex] f [/tex] continuous merely by doing this:

    [tex]
    F(x) = \begin{cases}
    3x + 5 & \text{ if } x \ne 10\\
    35 & \text{ if } x = 10
    \end{cases}
    [/tex]

    i.e. - we remove the discontinuity by redefining the function at the problem point, in order to make the function continuous there.

    A function has a jump discontinuity at a spot if the two one-sided limits both exist but are not equal. Here is an example - the jump discontinuity is at [tex] x = 2 [/tex].

    [tex]
    w(x) = \begin{cases}
    2x - 1 & \text{ if } x <=2\\
    10x + 1 & \text{ if } x > 2
    \end{cases}
    [/tex]

    The limit from the left is 3, the limit from the right is 21. If you were to view the graph of [tex] w [/tex] you would see a 'jump' in the two portions of the graph at [tex] x = 2 [/tex]. What you need to do is find the values of [tex] a [/tex] and [tex] b [/tex] to make your function have the types of behavior discussed here. Good luck.
     
  4. If you could please relate that back to my question, it would be appreciated. Again, no answers but, I'm still a little confused and don't know where to start.
     
  5. It is also the case where f(a) is not defined but the overall limit as x-->a exists. The way one can remove this is by defining the function at x=a.

    P.S. I am sure you know this, but this is adressed to the OP.
     
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