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Making a piecewise function continuous

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    find the values of b and c that make the function f continuous on (-[tex]\infty[/tex],[tex]\infty[/tex])

    f(x) = [tex]\frac{sin2x}{x}[/tex] if x< 0
    3-3c+b(x+1) if 0[tex]\leq[/tex]x<2
    5-cx+bx^2 if x[tex]\geq[/tex] 2
    2. Relevant equations

    lim as x [tex]\rightarrow[/tex] 0- of [tex]\frac{sin2x}{x}[/tex]
    works out to be 0
    3. The attempt at a solution
     
  2. jcsd
  3. Feb 17, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi meaganjulie! Welcome to PF! :wink:
    Nope :redface:

    what makes you think that? :smile:
     
  4. Feb 17, 2010 #3
    nevermind, that limit is actually 2. the limit for zero from the right also must equal 2, and the right and left limits at x=2 must also match.

    thats all ive got.
     
  5. Feb 17, 2010 #4

    Char. Limit

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    Just figure out what c and b values make the first and second equal at x=0.

    The other one follows a similar process.
     
  6. Feb 20, 2010 #5
    i dont understand how to find the values for c and b because i dont understand how i can use the point x=0 because the first part of the function has no c or b values. i have to use the last to equations at x=2, but how do i know what that limit is?
     
  7. Feb 21, 2010 #6

    tiny-tim

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    Hi meaganjulie! :smile:

    (just got up :zzz: …)
    The first equation tells you that the limit at x = 0 must be 2.

    The second equation tells you conditions on b and c which agree with that limit (at x = 0), and that gives you a formula (in b and c) for the limit at x = 2.

    And the third equation tells you conditions on b and c which agree with that limit at x = 2.
     
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