# Intermediate Value Theorem

• tanzl

#### tanzl

I typed this in maple and I could not convert it into latex. I included line numbering in the pdf. If u need to quote just quote the line numbers. Sorry for any inconvenience caused.

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.

You must have some conditions on the function phi, right? I'm guessing it's continuous and bounded? Sorry, but the proof so far isn't making a whole lot of sense. Don't you have a theorem that tells you that a continuous function on a closed bounded interval has a minimum? Can you show you only need to consider a bounded interval and not the whole real line since n is even?

That's better, but I still don't think it really works. And thanks for including the conditions on phi this time. Your original M is really a function of your original a, M(a) and you sort of have to nail down a and M(a) at the same time. Let's write f(x)=x^4+phi(x). I'd do something like this. 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].

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].

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.

The idea is that I can find a min of f(x) on any closed bounded interval. To make sure that's the global min I ensure that f(x) for ALL x outside the interval is greater than SOME value of f inside the interval. Showing f(x)->infinity isn't really a problem since x^n goes to infinity and your condition psi(x)/x^n shows psi isn't very important. BTW, sorry, I've been writing x^4 rather than x^n once in a while. Change it for me, ok?

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.

But the minimum on the interval [-a,a] is not necessarily the global minimum until you say what 'a' is. My suggestion was to first pick any value f(x0) of the function and THEN choose a so you are sure that f(x) is greater then f(x0) using the fact f->infinity. THEN you have an interval to choose the global min on (which may not be f(x0)).

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

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.

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

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You now have the idea exactly. But you don't have to 'assume there exists a point x0 such that f(x0)<N'. Just say 'pick a value of N>f(x0) (for any x0)' or you could just say 'pick a value of N>f(0)' (or f(1), f(2) ...). As you said it doesn't matter what x0 is because it doesn't have to be the minimum. Finally, I think you could be a little more explicit in the proof f(x)->infinity. For example, since phi(x)/x^4 -> 0 there is an n such that for |x|>n, |phi(x)/x^4|<1/2. So x^4-phi(x)>x^4-(1/2)*x^4=x^4*(1/2). Now it's clear x^4->infinity. And you probably don't need two separate cases for +/-infinity, right? They are really the same thing.

Thanks again for your help. One more question. Can N takes any value (1,2,100,1000000)? From my understanding so far the value is not important in this question. I ask this because most of the questions I have done need to limit the bound to (0,1).

You can't estimate the bound N without knowing something more about phi(x). If all you know is phi(x)/x^4 -> 0, N (and the minimum) could be anything.

Let f be any polynomial function. Prove that there is some number y such that |f(y)|<=|f(x)|
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 -(x2), the maximum is obvious, but can we consider its limit of -$$\infty$$ as the minimum?

You want to think about |f(x)|. f(x) may go to plus or minus infinity as x->+/-infinity but |f(x)| always goes to +infinity, right? Yes, it's an awful lot like the previous problem.

No, now you want to pick a number 0<N<f(x0) for any value of x0. Define the interval [-a,a] such that f(x)<N for |x|>a. That guarantees that the max is on [-a,a], doesn't it?

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

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

If you are clear then it must be ok, right? It's too complicated for me to read easily. The concept is simple. Let c=least upper bound of {x: f(x)=0 and a<=x<x0}. Let d=greatest lower bound of {x:f(x)=0 and b>=x>x0}. Isn't it easy to prove f(c)=f(d)=0 since f is continuous? That may be the idea behind your proof also but I can't see the forest for the trees. What are p0, p1, q0 and q1 for? How can you define A and B without saying what they are?

Yea, but I am not good in proof so just to make sure that it is correct : )
It is almost the same but I use f(x)>0 instead of f(x)=0

I don't know if it's correct because I can't understand it. Because I don't know what p0, p1, q0 and q1 are when you introduce them in the proof. And I'm not sure anyone else will either. Can't you state it without them? You are right, you are not good at stating proofs. I do think you understand it, but I don't think the written proof conveys your clarity of understanding. :).

"A={x:a<=x<=p; f(x)>0 on [x,p]}". What is that set? It looks to me like the set of included x's depends on p. As such it should be written A(p). Furthermore if a=x then f(x)=0, so there isn't any p that works anyway. Look at my outline in post #24 and try to follow it.

I think that pretty much says what you want.

I think you should start a new thread with new questions. People will often just pass over an old thread with lots of replies, and I can't answer ALL your questions. Let's give some other people a chance.