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Math Help - Continuity of a Piecewise Function

  1. #1
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    [SOLVED] Continuity of a Piecewise Function

    I need this answered as soon as possible please

    y={(x^2-4)/(x-2) if x is less than 2
    {ax^2-bx+3 if x is greater than and equal to 2 and less than 3
    {2x-a+b if x is greater than and equal to 3

    find a and b

    I have done some work and found that a=(-3/2) and b=(-7/2) but this does not work for the continuity at the point x=3
    please help

    Thank you
    Dylan Renke
    Last edited by DylanRenke; October 3rd 2012 at 10:36 PM.
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  2. #2
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    Re: Continuity of a Piecewise Function

    is the (a + b) in the third equation in brackets
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  3. #3
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    Re: Continuity of a Piecewise Function

    the answer for the equation is

    a = 0.5
    b = 0.5

    plug it in and all the piecewise functions will work because all the equations will continue at x = 2
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  4. #4
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    Re: Continuity of a Piecewise Function

    i tried that, but then when approaches 2 from the left, it equals 4, and from the right it equals 2
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  5. #5
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    Re: Continuity of a Piecewise Function

    the a and b are not in brackets
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  6. #6
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    Re: Continuity of a Piecewise Function

    (x^2 - 4) / (x - 2) = (x+2)(x-2) / (x-2) = (x +2)
    x= 2 then (2 + 2) = 4

    ax^2 - bx + 3
    a =0.5
    b = 0.5
    then (0.5 * 2^2 - 0.5 * 2) +3
    1 + 3 = 4

    2x - a + b
    2 * 2 - 0.5 + 0.5
    =4
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  7. #7
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    Re: Continuity of a Piecewise Function

    I am just too tired, i have tried everything and just keep messing up on stupid things, the answer is 0.5 and 0.5, thank you for the help (showing work took me like an hour cause of stupid mistakes)
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