Fixed point theorem -GR9367

[Nameless]

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

[diogenes]

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]

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
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

[lime]

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]

Hi Lime,

Thanks for your awesome answer. The counterexample is great