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Piecewise Defined Function

  1. Mar 7, 2007 #1
    1. The problem statement, all variables and given/known data
    The equation [tex]\left| {x + a} \right| = \left| {x - b} \right|[/tex] has exactly one solution; at x=2. Find the value(s) of a and b

    2. The attempt at a solution
    Here is the way in which I have approached the situation:
    [tex]
    \begin{array}{l}
    \left| {x + a} \right| = \left| {x - b} \right| \\
    x + a = x + b \\
    \end{array}
    [/tex]
    this solution is not possible as the x's on either side of the equation cancel each other out, and a is not equal to b, so;
    [tex]
    \begin{array}{l}
    x + a = - x + b \\
    2x + a = b \\
    4 + a = b \\
    a = k \\
    a = k + 4 \\
    \end{array}
    [/tex]

    However I'm not sure on my my final answer, as I have not evaluated the values for a or b, as i have left them in the term of k. However what I have interpreted from the question, I have not been given enough information. Thank you to those who reply, and correct me if anything I have stated is wrong

    unique_pavadrin
     
  2. jcsd
  3. Mar 7, 2007 #2
    Not sure how you got this step:
    [tex]
    \begin{array}{l}
    \left| {x + a} \right| = \left| {x - b} \right| \\
    x + a = x + b \\
    \end{array}
    [/tex]
    If you are taking both to be positive, it should be [tex]x + a = x - b [/tex]
    Then you can kill the x's. so a=-b.
     
  4. Mar 12, 2007 #3
    thanks for the reply theperthvan

    ive used those steps as a simply does not equal -b as the x can be anywhere on the number line. by stating that a=-b, x is a fixed point at zero, which is incorrect as z is given the value of 2

    or is that completely wrong?
    thanks
     
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