anonymous
  • anonymous
How can I solve this inequation with absolute values? |x| + |x-2| > 3
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
First explicitate the formula |x| (consider both cases x>0 so |x|=x and |x-2|=x-2 and case 2 X<0 so x=-x and |x-2|=2-x) solve the ineq for x in both cases and the solution is the reunion of the solution of case 1 and 2
anonymous
  • anonymous
|dw:1332624153478:dw|
anonymous
  • anonymous
|dw:1332624250882:dw|

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anonymous
  • anonymous
|dw:1332624388157:dw|
anonymous
  • anonymous
|dw:1332624548770:dw|
anonymous
  • anonymous
These are your critical values on the number line. Check values in each part to see which is greater than.
anonymous
  • anonymous
|dw:1332624848769:dw|
anonymous
  • anonymous
|dw:1332624951514:dw|
anonymous
  • anonymous
Thank you very much
anonymous
  • anonymous
Take three cases 1. x <0 in this case ur eq will become -x-(x-2)>3 ie -2x+2>3 ie x<-1/2 Case 2 when 03 ie 2>3 which is not possible case 3 x>2 in this case ur eq will be x+x-2>3 ie 2x-2>3 ie x>5/2
anonymous
  • anonymous
See attached, so when x<0, if 2-2x>3, x<-1/2 when x>2, if 2x-2>3,x>5/2 so when x<-1/2 or x>5/2, y>3
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