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Graphing inequality with mod function

  1. Sep 27, 2010 #1
    How can we graph this inequality - |y|+1/2>=e-|x| ?
    I drew the function(actually a combination of functions) for equality. It would be symmetric in all quadrants and intersect the axes at +ln2 and -ln2(x-axis) and 1/2 and -1/2(y-axis).However since the various graphs are mixed up it is hard pinpoint what region to take.

    I have attached one half of the actual graph.
    The other half is it's reflection on the x-axis.

    Attached Files:

    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 27, 2010 #2


    Staff: Mentor

    Your inequality is equivalent to |y| >= e-|x| - 1/2. This can be rewritten as
    y >= e-|x| - 1/2 or -y >= e-|x| - 1/2

    So y >= e-|x| - 1/2 or y <= -e-|x| + 1/2

    Does that help?
  4. Sep 28, 2010 #3
    Yes I should have considered the intervals for the different inequalities.Thanks a lot.
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