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Writing an absolute value function as a piecewise function

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Write f(x) = |x2-x-12| as a piecewise function.


    2. Relevant equations



    3. The attempt at a solution
    -x2+x+12 where x>1/2
    x2-x-12 where x[itex]\geq[/itex]0

    According to the answer book the answers are
    -x2+x+12 where -3<x<4
    x2-x-12 where x[itex]\geq[/itex]4
    2x2-5x-3, x[itex]\leq[/itex]1/2

    I am really bad at functions when it comes to domain and range.. :cry:

    Help please!
     
  2. jcsd
  3. Dec 30, 2011 #2

    SammyS

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    Start by writing |u| as a piecewise function:
    [itex]\displaystyle \left|u\right|=\left\{ \matrix{\ u\,,\ \text{ if }\ u\ge 0\\-u\,,\ \text{ if }\ u<0}\right.[/itex]​

    Doing that for |x2-x-12|, gives:
    [itex]\displaystyle \left|x^2-x-12\right|=\left\{ \matrix{\ x^2-x-12\,,\ \text{ if }\ x^2-x-12\ge 0\\-(x^2-x-12)\,,\ \text{ if }\ x^2-x-12<0}\right.[/itex]​

    That leaves you to solve:
    x2-x-12 ≥ 0​
    and
    x2-x-12 < 0​
     
  4. Jan 8, 2012 #3

    phinds

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    Let me ask you a question.

    In the quoted solutions, the first is for the range from -3 to +4. I choose to pick the point -1, which is in this range, and so I know that at the point x=-1, the equation to use is the first of the two.

    In the second answer, the range is anything where x is less than +1/2. I choose the point -1, which is certainly less than 1/2, and so I know that for the point x=-1, the second answer is the correct equation.

    Now the two equations are not equivalent but the stated answers give them both as the solution for the point x=-1.

    What do you think about that?
     
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