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Graph of double absolute values

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data
    |x| + |y| ≤ 1
    What is the region in the plane that solves this inequality?


    2. Relevant equations

    3. The attempt at a solution
    I first tried graphing it by isolating the y variable
    |y| ≤ -|x| + 1
    Then I looked at the hint we were given, which was to assume that x and y are both positive, so I treated it as if it were written
    y ≤ -x+1
    And then constructed a graph of a line that was basically y = -x+1 with a domain of [0,3]
    But then what threw me off is that the y variable does have the absolute value bars, so there can't be negative y values in the graph, right? So now I'm thinking I just have to flip the part of the graph so that it more closely resembles y=|x-1| in the domain of [0,3]. Is that correct or am I completely missing this?
     
  2. jcsd
  3. Sep 5, 2013 #2

    eumyang

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    Homework Helper

    I'm not sure how can the domain be [0, 3]. Can you put any value between and including 0 & 3 in for x in |x| + |y| ≤ 1?

    What I would do is graph the four inequalities. (Yes, there are four; can you figure out what they would be? You have one of them: y ≤ -x+1.) The intersection of the four graphs would serve as the graph of |x| + |y| ≤ 1.
     
  4. Sep 5, 2013 #3
    I don't think I actually meant domain, that's my bad. I meant my line went from x=0 to x=3 just because of the size of my graph.

    Are the inequalities:
    x + y ≤ 1
    x - y ≤ 1
    -x + y ≤ 1
    -x - y ≤ 1
     
  5. Sep 6, 2013 #4

    eumyang

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    Yes.
     
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