Why is there no absolute min / max for the given equation?

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Discussion Overview

The discussion centers around the function f(x) = 3x^2 / (x - 3) on the interval [2, 8] and the question of why there is no absolute minimum or maximum for this function within that range. Participants explore the implications of critical points and the continuity of the function.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests that the absence of an absolute minimum or maximum may be due to the critical points being outside the bounds of the interval [2, 8].
  • Another participant questions the continuity of the function on the interval, noting that if the function is continuous, it should have an absolute maximum and minimum, but if not, it may not.
  • A later reply clarifies that the point x = 3 is undefined for the function, indicating it is not a critical point within the domain.
  • Another participant proposes taking limits as x approaches 3 from both sides to demonstrate that the function is unbounded in the interval [2, 8].

Areas of Agreement / Disagreement

Participants generally agree that x = 3 is not a valid critical point due to it being undefined in the function's domain. However, there remains uncertainty regarding the implications of continuity and the behavior of the function within the specified interval.

Contextual Notes

There is a lack of consensus on the continuity of the function and its implications for the existence of absolute extrema. The discussion also highlights the need for careful consideration of critical points and their relevance to the function's behavior.

jwxie
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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
 
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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.
 
Thanks for the reminder.
Oh right... 3 is undefined.

Oh I did not notice 3 was also my C.P (dumb)
Thanks!
 
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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