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Homework Help: Extreme value theorem question?

  1. Jun 18, 2012 #1
    1. The problem statement, all variables and given/known data

    Given g(x) = 1/x-2 is continuous and defined on [3, infinity). g(x) has no minimum value on the interval [3, infinity ). Does this function contradict the EVT? Explain.

    3. The attempt at a solution

    was wondering what would be a better explanation for this one?

    No, because the the extreme value theorems says that a continuous function defined on a DEFINITE interval will attain a max/min in the interval. [3, ∞) is not a definite interval. [3, a] where a is some number greater than 3 is and would achieve its minimum at x = a.


    g has no minimum on this interval even though its continuous. It does not contradict how ever because the interval is not open.
  2. jcsd
  3. Jun 19, 2012 #2


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    I don't think the EVT says anything about DEFINITE intervals, nor does it says anything about open intervals. What kind of intervals does it talk about?
  4. Jun 19, 2012 #3
    The thought process behind this answer seems appropriate and correct. :smile:

    Though, there is no 'definite interval'. You need to rephrase that word looking back at the definition of EVT.
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