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Why there is no max / min

  1. Nov 15, 2009 #1
    Given equation
    f(x) = 3x^2 / x - 3 on [2,8]

    I found the critical points were [3, 5/2, 0]
    f ' (x) = 3x^2 (x/-5) / (x-3)^2

    I want to know why there is no absolute min / max in this case?

    I think the reason is because this [a,b] is out of the bound of the critical point, thus we cannot compare [a,b] against the critical points?

    Also, is 0 the critical point?
    f ' (x) = 3x^2 (x/-5) / (x-3)^2
    3x^2 = 0
    so i got 0
    i always got confused with 1/2 =/ 0
     
  2. jcsd
  3. Nov 15, 2009 #2
    When you write an equation like this, use parentheses. I.e. f(x)=3x2/(x-3).

    Is this function continuous on the interval [2,8]? If it is, then it must have an absolute maximum and minimum in the interval [2,8], but if not, it may have only an absolute minimum, absolute maximum, or neither.
     
  4. Nov 15, 2009 #3
    Thanks for the reminder.
    Oh right... 3 is undefined.

    Oh I did not notice 3 was also my C.P (dumb)
    Thanks!!!
     
  5. Nov 15, 2009 #4
    Right! Although, 3 isn't a critical point because it isn't in the domain of the function. Just take the limits as x approaches 3 from the left and right hand sides to show that the function is unbounded in the interval [2,8].
     
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