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Need to clarify Extreme value theorem

  1. Nov 24, 2009 #1
    From what i know if a graph has say one turning point, the relative global max(/min) is that point depending on the concavity correct? However as i was going through some notes, i notice that according to the mean value theorem that in a closed and bounded interval there exist a relative global max/min. Does that mean so long the interval is close and bounded, there will be a global max/min and it does not need to be the highest/lowest point in the graph? Lets say for a simple graph 1/x the global min will be x=0 and no global max? So if the interval is [2,5] is the global max and global min at x=5 and x=2 respectively? Thanks
  2. jcsd
  3. Nov 24, 2009 #2


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    No, a "global max" is, by definition, the highest point on a graph and the "global min" is, by definition, the lowest point on a graph. Any continuous function has a global max and min on a closed and bounded interval. The function 1/x, for all x, does not have either max or min. 0 is not the min because 1/x is never equal to 0. In fact, 1/x does not have a global max or min on any closed and bounded interval that contains 0. The global max of 1/x on the interval [2, 5] occurs at x= 2 and is 1/2. The global min occurs at x= 5 and is 1/5.

    But stricly speaking the answer to your question "So if the interval is [2,5] is the global max and global min at x=5 and x=2 respectively?" is "No" because the order inwhich they are given implies that you are asking if the global max is at 5 and the global min is at 2. That's not true, they are reversed.
  4. Nov 24, 2009 #3
    Sorry my bad i wanted to say 1/x^2. I see now so basically if a close and bounded interval is given there will be a global max/min but if the interval is not given then we have to look at the graph to decide? Thanks
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