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Math Help - Calculus: Solving Piece wise function for constants a and b

  1. #1
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    Calculus: Solving Piece wise function for constants a and b

    Hi
    I have been working on this problem for days and i think im just missing a concept or something. the question is;
    Let f(x)={cos (pi*x/2) +a if x<-2
    {100 if x=-2
    {2x^2 +b if -2<x<0
    {2^x +1 if x>0 where a and b are constants. Find a and b for which both lim x-->0 f(x) and lim x-->-2f(x) exist.
    So far i work it out and i can get either a=11 and b=2 or a=101 and b=92 but with these answers f(x) isnt continuous.
    Help meh pleaaasseee!! (with working would be great )
    THANK YOU!!!
    Last edited by meggzie7; April 21st 2013 at 12:03 AM.
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  2. #2
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    Re: Calculus: Solving Piece wise function for constants a and b

    In order for \lim_{x\to-2}f(x) to exist, we require:

    \lim_{x\to-2^{-}}f(x)=\lim_{x\to-2^{+}}f(x)

    Now, using the definition of f(x), we find this means:

    \lim_{x\to-2^{-}}\left(\cos\left(\frac{\pi x}{2} \right)+a \right)=\lim_{x\to-2^{+}}\left(2x^2+b \right)

    \cos\left(\frac{\pi\cdot2}{2} \right)+a=2(2)^2+b

    -1+a=8+b

    a=9+b

    In order for \lim_{x\to0}f(x) to exist, we require:

    \lim_{x\to0^{-}}f(x)=\lim_{x\to0^{+}}f(x)

    Now, using the definition of f(x), we find this means:

    \lim_{x\to0^{-}}\left(2x^2+b \right)=\lim_{x\to0^{+}}\left(2^x+1 \right)

    2(0)^2+b=2^0+1

    b=2\,\therefore\,a=11

    This ensures the limits exist, and while there is a discontinuity at x=-2, this is allowed as the function need not have the value of the limits at that point.
    Thanks from agentmulder
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    Re: Calculus: Solving Piece wise function for constants a and b

    Thank you MarkFL that was the answer that i has but i didnt know that discontinuity would allow the limits to still exist thanks again
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