T or F. If a function is continuous on a closed interval, it must have an absolute maximum on that interval.?

Question

What do you think?

Answer 1

I'd have to go with T.

Answer 2

True. If it does not have an absolute maximum it either has an infinite value or a limit.

If it has a limit maximum (a value V such that there is an x in the interval such that V - f(x) < epsilon for any epsilon > 0) then since it is closed there is an X with f(X) = V.

If it has an infinite value (actually, if it has arbitrarily large values is a better way to say it) then you can choose a series of points xi where xi > 2^i. The limit point of the xi is a member of the interval since it is closed, and the function is not continuous at that point.

Therefore it has a finite maximum value because it is continuous, and the point that maps to that value is in the interval because it is closed. qed.

Answer 3

Closed periods are textbf{compact} and the Heine-Borel theorem tells us that if f:D->R is non-provide up with D compact, then f(D) is compact. Compact instruments are closed and bounded so contain all their accumulation factors. for this reason, a function it quite is non-provide up on a closed and bounded era attains its minimum and optimal values. In differential calculus, we added learn that a serious huge type, c, is one for which f'(c)=0 or for which the spinoff of f(x) at x=c would not exist. a super theorem tells us that the minimum and optimal values of a non-provide up function on a closed era the two happen on the endpoints of the era or on the intense numbers on the interior the era. So, you need to shop taking math and take calculus and learn it heavily, too! :)

Answer 4

let's be careful here. is 1/x continuous on the closed interval [0,1]?

Answer 5

I say False.

How about y=1/x on [0,1] ?

Answer 6

I would say False

Answer 7

T