1. ## Intermediate Value Theorem

a) Let f :[a,b] --> [a,b] be continuous on [a,b]. Prove that f has a fixed point, i.e., prove that (exist) c in [a,b] such that f(c)=c

b) let f,g : [a,b]->[a,b] be two continuous function on [a,b] such that f(a)>or equal g(a) and f(b)<or equal g(a).
Prove that exist c in [a,b] such that f(c) =g(c)
Plx help

2. I see that you have severval other postings.
You should understand that this is not a homework service nor is it a tutorial service.
PLease either post some of your own work on this problem or explain what you do not understand about the question.

3. Sorry,
I just want to get some idea for how to start the question since I have no idea how to start the question.

4. Here is a standard hint.

Define $\displaystyle g(x)=f(x)-x$, a continuous function.

Now show that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$.

5. For b)

Define new function H(x)=f(x)-g(x),

1. prove that H continuous...
2. Prove that exist x=t in [a,b] such that H(t)=0...What is the meaning of that?

6. Originally Posted by Also sprach Zarathustra
For b)

Define new function H(x)=f(x)-g(x),

1. prove that H continuous...
2. Prove that exist x=t in [a,b] such that H(t)=0...What is the meaning of that?
I get a problem when I am trying to prove that H(t)=0, how can I say that H(t)=0?
I think I have to show that f(t)is not > g(t) and also g(t) is not > f(t) then g(t)=f(t) but I have no idea how can I show that is true.

7. Try to look at H(a) and H(b) and then use Intermediate Value Theorem.

8. My favorite way to prove a)

Is to suppose that $\displaystyle f(x)\ne x$ for all $\displaystyle x\in[a,b]$ then $\displaystyle g:[a,b]\to\{-1,1\}:x\mapsto\frac{f(x)-x}{|f(x)-x|}$ is continuous and surjective, but this is impossible...why?

9. Originally Posted by Plato
Here is a standard hint.

Define $\displaystyle g(x)=f(x)-x$, a continuous function.

Now show that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$.
when I when that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$ f(c) will equal 0. how can I relat to c???

10. Originally Posted by mathbeginner
when I when that $\displaystyle g(a)\ge 0~\&~g(b)\le 0$ f(c) will equal 0. how can I relat to c???
Go to your textbook and class notes and review examples that apply the Intermediate Value Theorem. Spend more than just a few minutes doing this. Then come back if you still have this question. (You will find that the question you posted is a standard example in many textbooks that cover this material).

11. Originally Posted by mr fantastic
Go to your textbook and class notes and review examples that apply the Intermediate Value Theorem. Spend more than just a few minutes doing this. Then come back if you still have this question. (You will find that the question you posted is a standard example in many textbooks that cover this material).
I find out that what is c depend on how do u set up the question, ex. g(a)<c<g(b) c can be anything ex. 0 or 1 if I put 0 on it then f(c)=0 or if I put 1 on it then f(c)=1

Am I right? I just want to make sure that am I get it or not. Thanks.

12. Perhaps you are missing the point of this problem.
You are not asked to actually find the value of c.
You are asked to show that one actually exists.
Of course, it does depend upon the setup.

13. Originally Posted by Plato
Perhaps you are missing the point of this problem.
You are not asked to actually find the value of c.
You are asked to show that one actually exists.
Of course, it does depend upon the setup.
So you are saying that I should prove that f(c)is not > or < c is equal c ?
Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?

I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.

Thanks

14. Originally Posted by mathbeginner
So you are saying that I should prove that f(c)is not > or < c is equal c ?
Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?

I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.

Thanks
Your confusion is so profound that I have absolutely nothing more to say.

15. Originally Posted by mathbeginner
So you are saying that I should prove that f(c)is not > or < c is equal c ?
Or saying I did find out f(c)=0 since it equal 0 so that is a c exist?

I am so confused abt this because when I was doing that first I did prove that that f(c)is not > or < c is equal c. Then I thought I get it wrong because what I am doing is proving that IVT is true. That's why I use the other way, to use g(a)<0<g(b) find out that f(c)= 0.

Thanks
Please take the advice given in post #10. Then come back with fresh eyes on this thread.