Max/min question

1. Nov 6, 2006

I have a quick question,

I have t/f question that I'm not sure about,

"if a function is continuous on a closed domain then the function has a maximum?"

My thoughts on it are that yes it would have a maximum. I mean you are restricting the domain so it can't go to -infinity, +infinity. What worries me though is a straight line. Does this have a max to begin with?

it's been awhile since I've had this class,

2. Nov 6, 2006

AKG

How does your book/notes define "closed domain"?

3. Nov 6, 2006

Well this was actually for a friend, and he found a blurb in his notes that says "a function in a closed domain always has a max and min"

so he's putting yes.

He also doesn't have a book, just notes from the professor. I actually don't have calc book either.

4. Nov 7, 2006

quasar987

"Closed domain" probably means that the domain of the function is a closed set. Not that the domain of the function is a closed open connected set. That would be kind of awkward, especially if the course is an introductory one. So, closed domain = domain of the function which is a closet set.

5. Nov 7, 2006

quasar987

So a closed domain will be a finite reunion of closed intervals and points. A continuous fonction map those to a reunion of closed intervals and points. And the sup of this is a point of the set. So it's the max. So there is a max.

6. Nov 7, 2006

HallsofIvy

The critical question is: is a closed 'domain' necessarily bounded?

A closed and bounded set of real numbers is compact. The continuous image of a compact set is compact, therefore closed and bounded. Since it is bounded it has both upper and lower bounds, therefore both least upper and greatest lower bounds. Since the set is closed those are in the set and are the maximum and minimum values.

However, "bounded" is important there. If A is all of R, and f(x)= x, the A is closed and so is f(A). But since neither A nor f(A) is bounded, they do not contain maximum and minimum values.

7. Nov 7, 2006

quasar987

I confused closed and compact, nevermind.

8. Nov 7, 2006

quasar987

A simple counter-exemple then is f:$\mathbb{R}\rightarrow \mathbb{R}$ define by f(x)=x. The real line is closed but f is not bounded ==> there is no max.

9. Nov 8, 2006

I've tried to follow along with you guys. I somewhat understand what you are talking about, but a lot of the terminology is above me without further research. Where do the terms you have used come from? Is this real analysis?

The class my friend is in is brief calc. When he showed me the question, I said yes to begin with, then questioned myself.

Say we have a function f(x) = 5.

If we restrict the domain from 0 to 10. Then how can we determine where the max/min is? Actually even if we don't restrict the domain, I'm still unsure about this question.

So we take the derivative and set it to zero.

f'(x) = 0

So how do I handle this? Do I just say every point x0 is a critical point. Where x0 is the domain of f'(x), and x0 actually will always satisfy the condition. So the domain of f(x) consists of only critical points. Now how do we classify the points? There is no slope, so we cannot determine if it is a max or a min.

10. Nov 8, 2006

HallsofIvy

Okay, without using the word "compact", the theorem from calculus is that a continuous function, defined on a closed and bounded set, must take on both maximum and minimum values on that set.
We were wondering whether your definition of "domain" might include "bounded". Probably not so that the answer to your original question is "No".

Both max and min of the constant function f(x)= 5 is, of course, 5! Yes, everypoint is a "critical" point but you shouldn't worry about derivatives or critical points for a simple function like that- just use the definitions of "maximum" and "minimum". Since the function never takes on any value larger than 5, 5 is its maximum value and, since the function never takes on any value less than 5, 5 is its minimum value.

11. Nov 8, 2006