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Find the values of a and b that make f continuous everywhere

  1. Sep 11, 2012 #1
    1. The problem statement, all variables and given/known data

    This is a piece wise function of course. f(x) =

    (x2-4) / (x-2) if x is less than two.

    ax2 - bx + 1 if x is greater than or equal to 2, or less than three.

    4x - a + b if x is greater than or equal to three.

    2. Relevant equations



    3. The attempt at a solution

    Alright, I know enough to factor the top of the fist equation and get x+2. That means when x is two, f(x) is four. We can use f(x) in this case because we are making the function continuous. I've gotten as far as plugging in this value in the second equation and getting

    4 = a4 - b2 + 1

    but I don't know what to do from here, or how to get the values of a and b. I think I subtract one from the right and get

    3 = a4 - 2b

    Now I am definitely stuck.
     
  2. jcsd
  3. Sep 11, 2012 #2

    Mark44

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    Written in a more useful way, your equation is

    4a - 2b = 3


    Now, what about at x = 3? You want the function to be continuous there, as well, right? What needs to happen for f to be continuous at x = 3?

    That should give you another equation so that you have a system of two equations in the unknowns a and b.
     
  4. Sep 11, 2012 #3

    SammyS

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    Hello SmittenWCalc. Welcome to PF!

    So, you have the following:
    [itex]\displaystyle \lim_{x\to\,2-}f(x)=\lim_{x\to\,2-}(x+2)=4\ .[/itex]

    [itex]\displaystyle \lim_{x\to\,2+}f(x)=\lim_{x\to\,2+} (ax^2+bx+1)=4a+2b+1\ .[/itex]

    Do something similar at x=3 .
     
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