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Math Help - [SOLVED] Little help with "continuity"

  1. #1
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    [SOLVED] Little help with "continuity"

    I have trouble grasping the concept of "continuity". I understand it when the teacher explains it to me but when faced with problems I don't seem to be able to do them. Will someone help me with this?
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  2. #2
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    Loosely, a continuous function is one that has no "holes" (i.e., the function is undefined at a point), no "jumps" (i.e., the limit as x approaches a value c from the left is different from the limit as x approaches the same value c from the right), and no vertical asymptotes.

    If you say that a function is continuous on an interval, then it has all of these properties on that interval. However, if the interval is open, then the function may have a vertical asymptote at one of the open endpoints and it will still be continuous.
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  3. #3
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    Okay so how exactly must I approach this type of problem?

    Find the numbers a and b so that the function f (x) is continuous.

    -x^2 + 2 if x < 0
    ax + b if 0 less than or equal to x < 1
    (x - 1)^2 if 1 less than or equal to x
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  4. #4
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    The function is defined piecewise. At the x-values of 0 and 1, the limit of f(x) from the left must be the same as the limit of f(x) from the right.

    In terms of your question, this means that

    -x^2 + 2 = ax + b when x = 0 and

    ax + b = (x-1)^2 when x = 1.

    Thus,

    -0^2 + 2 = a(0) + b

    b = 2

    a(1) + 2 = (1-1)^2

    a + 2 = 0

    a = -2
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  5. #5
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    Wow it seems a lot simpler than I thought. My thanks to you sir.
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