Extreme value theorem question?

[indigo1]

Homework Statement

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.

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.

or

g has no minimum on this interval even though its continuous. It does not contradict how ever because the interval is not open.

 

[Dick]

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?

 

[Infinitum]

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.

The thought process behind this answer seems appropriate and correct.

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