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Absolute value inequality

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data
    |1/(x-1)|<1
    3. The attempt at a solution
    is that the same as this -1<1/(x-1)<1

    can i do each side by it self then take the values at which they intersect

    so i subtracted the 1 then got (-x+2)/(x-1) then made a sign chart with 2 and 1 on it
    then took the less than terms so i got (1,2) for the first suoltions
    then for the second i got x/(x-1) then made a sign chart with 0 and 1 on it
    then i took the greater than terms geting (-inf,0) U (1,inf)
    is this right
     
  2. jcsd
  3. Apr 10, 2009 #2
    Well, let's check. Suppose x = 0. Does this inequality hold? No because -1 = -1.

    Maybe this will help:

    |x| < y
    => x < y or -x > -y
     
  4. Apr 10, 2009 #3
    ok i see so you took x<y then divided it by a -1 and then flipped the sign
    so then would our soultions to my original inequality be
    (-inf,1) U (2,inf)
     
  5. Apr 10, 2009 #4

    statdad

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    A shorter idea (for future reference)
    [tex]
    |x| < y
    [/tex]

    is the same as
    [tex]
    -y < x < y
    [/tex]

    If your problem begins with [tex] \le [/tex] then replace [tex] < [/tex] with [tex] \le [/tex] in the simplification.
     
  6. Apr 10, 2009 #5
    would we then have to do a sign chart
     
    Last edited: Apr 11, 2009
  7. Apr 10, 2009 #6

    symbolipoint

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    crager,
    Look again at Dunkle's first response. Graph the relation on a number line and you may more clearly find a path to a solution of his example and to your exercise problem.

    In your original expression on the left, the expression inside of the absolute value is either positive, or negative; examine each of these conditions separately. Do you yet need more detailed descriptions?
     
  8. Apr 11, 2009 #7
    (-inf,-2) U (2,inf) is this the answer
     
  9. Apr 11, 2009 #8

    symbolipoint

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    Reread post #6 and #2. Check your answer.
     
  10. Apr 11, 2009 #9

    symbolipoint

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    Start with |y|<1, where y = 1/(x-1).

    Use a number line to help understand this:
    Either y<1 OR -y<1.

    Replace y with the original expression and solve, and check or first solve for y and then replace y with the original expression and finish solving.
    ...but be careful. I'm not absolutely sure about my own answer, so I might have made a conceptual error. Still, not that for the particular exercise, x canNOT be 1.
     
  11. Apr 11, 2009 #10

    HallsofIvy

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    In my opinion, the simplest way to solve most inequalities is to first solve the associated equation. |1/(x-1)|= 1 reduces to 1/(x-1)= 1 or 1/(x-1)= -1. Multiplying both sides of each gives 1= x-1 and 1= 1- x. In the first case, x= 2 and in the second x= 0. The point of that is that continuous functions can change from "<" to ">" on where they are "=". Of course, |1/(x-1)| is NOT continuous at x= 1 so we must add that possibility: the inequality can change at x= 0, x= 1, and x= 2. If we take x= -1< 0, |1/(-1-1)|= 1/2< 1 so the given inequality is true for all x< 0. Check a value of x between 0 and 1, a value of x between 1 and 2, and a value of x larger than 2 to determine which of those intervals also satisfy the inequality.
     
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