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Math Help - Lots of confusion on factoring this...

  1. #1
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    Lots of confusion on factoring this...



    The function f is given by the first formula
    when x < -2 and




    by the second formula
    when -2 is less than or equal to x.

    I can't figure out what value I have to put in for a to make the function continuous at -2?
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  2. #2
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    Quote Originally Posted by frozenflames View Post


    The function f is given by the first formula
    when x < -2 and




    by the second formula
    when -2 is less than or equal to x.

    I can't figure out what value I have to put in for a to make the function continuous at -2?
    The only problem with finding the limit- and the only reason why you want to factor- is that both numerator and denominator are 0 when x= -12. And that tells you that (x- (-2))= x+ 2 is a factor. Divide 4x^3+ 8x^2+ 8x+ 16 by x+ 2 ("synthetic division" works nicely here) to find the other factor.
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  3. #3
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    Quote Originally Posted by frozenflames View Post


    The function f is given by the first formula
    when x < -2 and




    by the second formula
    when -2 is less than or equal to x. x > -2

    I can't figure out what value I have to put in for a to make the function continuous at -2?
    note ...

    \frac{4x^3+8x^2+8x+16}{x+2} = \frac{4x^2(x+2) + 8(x+2)}{x+2} = 4x^2+8


    to be continuous at x = -2, \lim_{x \to -2} f(x) must exist ...

    \lim_{x \to -2^+} x^2 - 6x + a = \lim_{x \to -2^-} 4x^2 + 8

    a = ?
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  4. #4
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    8!

    That makes sense. The reason why I was flustered was because I forgot the x^2 and converted the first equation to [(4x+8)(x+2)]/(x+2).

    Thanks!
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