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Homework Help: Continuity of the function

  1. Nov 13, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x) = x^2 - 4x + a[/tex]
    [tex]g(x) = \lim_{n\rightarrow\infty} \frac {2|x-b|^n + 1}{|x-b|^n + 1}[/tex]

    let [tex]h(x) = f(x)g(x)[/tex]

    Find the sum of a+b that makes h(x) continuous for all x.

    2. Relevant equations
    Power Series??? Derivation to test continuity???

    3. The attempt at a solution

    Well I know a function is continuous if

    1)f(c) exists
    2)Limit as x goes to c of f(x) exists
    3)#1 is equal to #2

    But I don't know where to go from here. Please help.
  2. jcsd
  3. Nov 13, 2008 #2


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    How does it fail to be continuous? f(x) is continuous for all a, but g(x) has a jump from when |x-b| < 1 (in which case g(x)=1) to when |x-b|>1 (in which case g(x)=2). So you need to figure out how to take care of it.
  4. Nov 13, 2008 #3
    I understood why the discontinuity occurs, since |x-b|^n of n to the infinity could take you to 0 or a huge value (I think it was related to geometrical sum) depending on whether it's smaller than 1 or bigger than 1. I still do not have a clue how to get the specific values of a and b from it...
    Last edited: Nov 13, 2008
  5. Nov 13, 2008 #4


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    So, for f(x)g(x) to be continuous, you know that when |x-b| isn't near 1, the function's continuous. You should realize this gives us that the only two points of concern are x=b+1 and x=b-1. You have that as x approaches b+1 and b-1 from different sides, g is either 2 or 1. So you need f(b+1) and f(b-1) to be values such that f(b-1)*2=f(b-1)*1 and f(b+1)*2=f(b+1)
  6. Nov 13, 2008 #5
    Then that simply means

    f(b-1) = 1
    f(b+1) = 1

    So we have

    [tex]1 = (b-1)^2 - 4(b-1) + a[/tex]
    [tex]1 = (b+1)^2 - 4(b+1) + a[/tex]

    [tex]0 = (b-1)^2 - 4(b-1) - (b+1)^2 + 4(b+1)[/tex]
    [tex]0 = (b-1)^2 - (b+1)^2 + 8[/tex]

    [tex]b = 2[/tex]


    [tex]1 = (2+1)^2 - 4(2+1) + a[/tex]
    [tex]1 = 9 - 12+ a[/tex]
    [tex]a = 4[/tex]

    a + b = 6 is what I get, but supposedly this isn't the answer. I don't see what went wrong here, everything seems right.


    Actually, I thought about it, but for the two distinct points of x = b-1 and x = b+1 to be continuous, shouldn't that mean that
    f(b-1) = f(b+1) ?
  7. Nov 13, 2008 #6


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    f(b-1)=1. Then 2*f(b-1)=f(b-1) gives us 2=1. Are you sure that's the right value?

    f(b-1) and f(b+1) have nothing to do with each other really (although it is correct that f(b-1)=f(b+1) is necessary, this is more by coincidence than any deep connection).
  8. Nov 13, 2008 #7

    Oh, it's actually

    f(b-1) = 0
    f(b+1) = 0

    Doh! Thx
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