Graphing inequality with mod function

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SUMMARY

The discussion focuses on graphing the inequality |y| + 1/2 ≥ e - |x|. The contributors established that the equality intersects the x-axis at +ln2 and -ln2, and the y-axis at +1/2 and -1/2, indicating symmetry across all quadrants. The inequality can be rewritten as two separate conditions: y ≥ e - |x| - 1/2 and y ≤ -e + |x| + 1/2. This clarification aids in identifying the correct regions for the inequality.

PREREQUISITES
  • Understanding of absolute value functions
  • Familiarity with exponential functions, specifically e
  • Knowledge of graphing techniques in Cartesian coordinates
  • Ability to manipulate and solve inequalities
NEXT STEPS
  • Study the properties of absolute value functions in graphing
  • Learn about the behavior of exponential functions, particularly e - |x|
  • Explore techniques for graphing inequalities in two dimensions
  • Investigate the concept of symmetry in mathematical graphs
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Mathematics students, educators, and anyone interested in advanced graphing techniques, particularly those dealing with inequalities and absolute value functions.

aim1732
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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.
 

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aim1732 said:
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.
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?
 
Yes I should have considered the intervals for the different inequalities.Thanks a lot.
 

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