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Find all x such that |x-1|+|x-2|>1.

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Find all x such that ##|x-1|+|x-2|>1##.

    2. Relevant equations

    Definition of absolute value:
    |x| = x if x ≥ 0.
    |x| = -x if x ≤ 0.

    3. The attempt at a solution

    I figured the most straightforward way if to do this case-wise:

    Case 1: ##(x-1)>0 \wedge (x-2)>0## then

    ##(x-1) + (x-2) > 1 \implies x > 2.##


    Case 2: ##(x-1)<0 \wedge (x-2)<0## then

    ## (1-x) + (2-x) > 1 \implies x<2.##


    Case 3: ##(x-1)>0 \wedge (x-2)<0## then

    ## (x-1)+(1-x) > 1 \implies 0 >1. ##


    Case 4: ##(x-1)<0 \wedge (x-2)>0## then

    ## (1-x) + (x+2) > 1 \implies 3 > 1. ##


    Cases 3 and 4 are bothering me because x 'drops out.' Case 3 makes no sense and has me wondering if Case 4 even makes sense.
     
  2. jcsd
  3. Feb 12, 2012 #2
    Case 1 looks right to me. Plugging 2 in you get |2-1|+|2-2|>1 ==> 1+0>1 so x has to be at least bigger than 2.
     
  4. Feb 12, 2012 #3
    Yeah. I don't really understand what is going on with the other cases. Even case 2 fails if you plug in x = 1. I thought I needed to evaluate all cases and then take what is common between them?
     
  5. Feb 12, 2012 #4

    Dick

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    You need to intersect your assumptions with the solution. Take case 2. You've assumed x<1 AND x<2 AND you have concluded x<2. Of course, x=1 doesn't need to work. It violates your assumption that x<1.
     
    Last edited: Feb 12, 2012
  6. Feb 12, 2012 #5
    Oof. Ok, I see that one now. Now what about case 3 and 4? Do they actually give me any useful information? I guess I need to take a second look here and do what you said.

    Case 3 I assumed ##(x-1 > 0) \wedge (x-2 < 0)## which translates to ## x>1 \wedge x<2##. So this means that I am only looking at numbers on the interval (1,2). So does the fact that I got a 'nonsense' answer of 0>1 mean that there are no numbers on the interval (1,2) that satisfy the inequality?

    And as for case 4, it seems I have assumed that ##x < 1 \wedge x>2## which is nonsense from the start! So it means that I need not consider this case (because it is not even a case!). Correct?
     
  7. Feb 12, 2012 #6

    Dick

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    That's exactly what it means.
     
  8. Feb 12, 2012 #7
    Don't know if you saw the edit about case 4, but I think I got it right. Thanks again Dick! I know you're sick of seeing my posts, but I promise there are plenty more to come. :smile: High school drop-outs turned engineers need to relearn math right sometimes :biggrin:
     
  9. Feb 12, 2012 #8

    Dick

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    When I'm sick of hearing from you I'll stop answering your posts. Until then, you are fine. And yes, you are correct that some of the cases are nonsense. Just throw them away.
     
  10. Feb 13, 2012 #9

    Ray Vickson

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    You may gain more insight by plotting the function f(x) = |x-1| + |x-2|. Think of adding the graph of y = |x-1| to the graph of y = |x-2|. Look in detail: when x is to the left of 1, both graphs have slope -1, so the combined graph has slope -2. For x between 1 and 2, one graph has slope -1 and the other has slope +1, so the combined slope = 0 (that is, f(x) is constant for 1 < x < 2). To the right of x = 2, both graphs have slope +1, so the combined graph has slope +2.

    RGV
     
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