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Absolute Max/Min

  1. Nov 6, 2009 #1
    fuction is y=x sqr.root (3-x)

    Domain (-infinity, 3]
    i got my critical numbers to be x= 2, 3

    Now i know the absolute max is x=2.

    but for the absolute min, would it be at x=3? or none since the function continues to go to negative infinity as x approaches negative infinity?
     
  2. jcsd
  3. Nov 6, 2009 #2
    Well, assuming that the function goes to negative infinity, then there is no lower bound. It contradicts the definition of an absolute minimum value.
     
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