- #1
indigo1
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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.