Fixed point theorem -GR9367

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Nameless
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Joined: Sun Aug 31, 2008 4:42 pm

Fixed point theorem -GR9367

Post by Nameless » Sun Aug 31, 2008 6:33 pm

Hi All
I am getting stuck with the following problem :
Let f be a real-valued function continuous on the closed interval [0,1] and differentiable on (0,1) with f(0)=1 and f(1)=0. which of the following must be true :
I . There exists x in (0,1) st. f(x)=x
II.There exists x in (0,1) st. f'(x)=-1
III. f(x)>0 for all x in [0,1)

(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, III

Whey said "on the closed interval [0,1]" does it mean that : f :[0,1] -->[0,1] ?

thanks so much
Last edited by Nameless on Mon Sep 01, 2008 12:30 am, edited 1 time in total.

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diogenes
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Post by diogenes » Sun Aug 31, 2008 9:36 pm

Well, here are my thoughts on this question.
For:
1) it looks like that we have f in C[0,1] and can apply the fixed point theorem to f and conclude 1 is true.
2) Using Mean Value Theorem, we have a c in (0,1) such that (f(b)-f(a))/(b-a) = f’(c) . So, we have (0-1)/1 = -1 = f’(c) and, so, 2 holds.

Nameless
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Joined: Sun Aug 31, 2008 4:42 pm

Post by Nameless » Sun Aug 31, 2008 11:07 pm

I knew that id f:[0,1]----->[0,1] then f has a fixed point but not for sute that if f: [0,1]--->R and f is differentiable on (0,1) then f has a fixed point :D
so if some one know , please explain ?

Ah, When said "on the closed interval [0,1]" does it mean that : f :[0,1] -->[0,1] ? ??

thanks

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lime
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Post by lime » Mon Sep 01, 2008 2:13 am

When said "on the closed interval [0,1]" does it mean that : f :[0,1] -->[0,1] ?
Of course not.
Not for sure that if f: [0,1]--->R and f is differentiable on (0,1) then f has a fixed point
Absolutely correct. According to the theorem, in order to have fixed point, differentiable function f(x) must be I-->I (it is also possible to show that instead of I=[0,1] it can be any closed interval [a,b]).
Nevertheless, in this concrete case f has fixed point!!! Because f(0)=1 and f(1)=0. You can prove it rigorously but just draw graph of function y=x
and put points (0,1) and (1,0). You can see, that there is no way to connect these points without crossing the line y=x.
Let's go back to questions.
I. Yes.
II. Yes. See diogenes's answer.
III. No. Consider function y=cos((3/2)*Pi*x).

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm

Post by Nameless » Mon Sep 01, 2008 8:56 am

Hi Lime,

Thanks for your awesome answer. The counterexample is great :D



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