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http://i359.photobucket.com/albums/oo31/tanzl/IntermediateValueTheorem.jpg

I try to do this question but I am not really what the question want. I will appreciate if anyone can tell me my mistakes. Thanks.

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- #1

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http://i359.photobucket.com/albums/oo31/tanzl/IntermediateValueTheorem.jpg

I try to do this question but I am not really what the question want. I will appreciate if anyone can tell me my mistakes. Thanks.

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http://i359.photobucket.com/albums/oo31/tanzl/IntermediateValueTheorem2.jpg

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Start by showing lim f(x)=infinity as x goes to +/-infinity. Pick a definite value of x, say f(0). Can you define an interval [-a,a] such that f(x)>f(0) for x<-a or x>a? Then the global min f(y) is the minimum of f on [-a,a].

Can you explain more about this? Because I don't really got your idea.

I don understand why I need to show that lim f(x)=infinity as x goes to +/-infinity.

Pick a definite value of x, say f(0). Can you define an interval [-a,a] such that f(x)>f(0) for x<-a or x>a?

I think this is something like the idea in my proof but instead makes the minimum becomes an arbitrary point, M(a).

Then the global min f(y) is the minimum of f on [-a,a].

What implies that global min f(y) is the minimum of f on [-a,a]?

Sorry for asking so many questions but I don really understand.

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http://i359.photobucket.com/albums/oo31/tanzl/IntermediateValueTheorem3.jpg

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http://i359.photobucket.com/albums/oo31/tanzl/IntermediateValueTheorem3.jpg

It's still WAY too complicated. Just i) prove f(x)->infinity. ii) Pick a number N>f(0). Then since f(x)->infinity there is an interval [-a,a] such that f(x)>=N for x outside of [-a,a]. So the min over [-a,a] is the global min. That's it. I don't think you need a lot of that other stuff. And don't pick an 'arbitrary interval [-a,a]' first. An arbitrary interval won't work. The size of the interval has to be based on the values of f.

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You are right. The infinitive limit is sufficient to prove that f(x)>=f(0).

I modified my proof again. Thanks.

http://i359.photobucket.com/albums/oo31/tanzl/ExtremeValueTheorem.jpg

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But if the interval a is big enough, then it will includes the global minimum. Isnt it?

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But if the interval a is big enough, then it will includes the global minimum. Isnt it?

Sure, but how do you make sure that 'a' is big enough? Yes, later on you pick an 'a' based on the fact f(x)->infinity. But then what was the first 'a' for? I would start with a blank screen and rewrite the proof rather than tinkering with the wording on the existing proof. Parts of the proof are just leftovers from previous versions and don't really do anything. If you can get the proof to about half it's current size, it's probably pretty close to correct.

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I think I see what you mean. I try to explain what I understand by words first.

In my solution, I set the interval first. But the problem is I did not set a condition for choosing a. So, it is not true for any a but only for "big enough" a which I did not define properly.

In your solution, you let a number N larger than any value of f(x) first such as f(x0). It doesn't matter whether f(x0) is the minimum o not because N will still be larger than the minimum. Now you already have a bound N for f(x) so I guess the next step you do is to define the corresponding interval a for x by the condition f(x) larger than the bound N(since the limit for f(x) is infinity) for x outside of the interval a.

Correct?

http://i359.photobucket.com/albums/oo31/tanzl/ExtremeValueTheorem2.jpg

In my solution, I set the interval first. But the problem is I did not set a condition for choosing a. So, it is not true for any a but only for "big enough" a which I did not define properly.

In your solution, you let a number N larger than any value of f(x) first such as f(x0). It doesn't matter whether f(x0) is the minimum o not because N will still be larger than the minimum. Now you already have a bound N for f(x) so I guess the next step you do is to define the corresponding interval a for x by the condition f(x) larger than the bound N(since the limit for f(x) is infinity) for x outside of the interval a.

Correct?

http://i359.photobucket.com/albums/oo31/tanzl/ExtremeValueTheorem2.jpg

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for all x.

For this question, I can exactly the same thing as previous one.

But I have problem when I need to assume f(x) > N since I do not know anything about limit of f.

Is it true for any polynomial function? Let's say -(x

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http://i359.photobucket.com/albums/oo31/tanzl/ExtremeValueTheorem3.jpg

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I thought that is exactly the same with my solution. What is wrong with my solution?

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I think it is the same. Maybe I'm just getting proof fatigue from looking at so many of these :).

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http://i359.photobucket.com/albums/oo31/tanzl/Untitled-1.jpg

I think it should be correct. I am quite clear with every reasonings.

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It is almost the same but I use f(x)>0 instead of f(x)=0

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http://i359.photobucket.com/albums/oo31/tanzl/JavaPrinting.jpg

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I followed your instructions and wrote this proof.

http://i359.photobucket.com/albums/oo31/tanzl/JavaPrinting-1.jpg

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I think that pretty much says what you want.

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Thanks. New question...

http://i359.photobucket.com/albums/oo31/tanzl/JavaPrinting-2.jpg

http://i359.photobucket.com/albums/oo31/tanzl/JavaPrinting-2.jpg

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