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Absolute values

  1. Sep 4, 2008 #1
    How could I go about expressing |x|-|x-6| without using absolute value signs?
     
  2. jcsd
  3. Sep 4, 2008 #2
    I would just look at when the expression inside each absolute value is negative or positive. For instance, if x is negative, then you easily get x - (x+6) = -6 so if x is any negative value, you'll get the constant function f(x) = -6.

    You'll ultimately express it as a piecewise function but there shouldn't be many "pieces".
     
  4. Sep 5, 2008 #3

    CRGreathouse

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    Here are two thoughts, entirely unsimplified (so you have something to do!).

    1. |x| - |x - 6| =
    (x) - (x - 6), x >= 0 and x - 6 >= 0
    (x) - -(x - 6), x >= 0 and x - 6 < 0
    -(x) - (x - 6), x < 0 and x - 6 >= 0
    -(x) - -(x - 6), x < 0 and x - 6 < 0

    2. sqrt(x^2) - sqrt((x - 6)^2)

    Note that both assume that x is real.
     
  5. Sep 5, 2008 #4

    HallsofIvy

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    If x< 0, both x and x-6 are negative so |x|- |x-6|= -x-(-(x-6)= -x+ x- 6= -6.

    If [itex]0\le x< 6[/itex], x- 6 is negative so |x|- |x-6|= x- (-(x-6))= x+x- 6= 2x- 6
    If [itex]6\le x[/itex], both x and x- 6 are positive so |x|- |x-6|= x- (x-6)= 6.

    We can write
    [tex]|x|-|x-6|= \left\{\begin{array}{cc}-6 & if x< 0\\2x-6 & if 0\le x< 6\\6 & if x>6\end{array}\right[/tex]

    We could also use the Heaviside step function. H(x), which is 0 for x< 0 and 1 for [itex]x\ge 0[/itex]. We want to start with -6 for x<0 and for x> 0 we have to add 2x: -6+ 2xH(x).
    Now, if x is greater than 6, we need to change that -6 to 6 and eleminate the 2x. We can do that by adding 12- 2x.

    |x|- |x-6|= -6+ 2xH(x)+ (12- 2x)H(x- 6).
     
  6. Sep 5, 2008 #5
    Thanks to all who responded, your help is appreciated!
     
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