# Analysis- continuity and differentiability

Hi, could somebody please help me with the following question, I have been stuck on it for ages.

1. let f[0,1] -> R be continuous with f(0)=0, f(1)=1. Prove the following:

a.(i) If for c in (0,1) f is differentiable at c with f'(c)<0 then there are exists points y such that f(x)=y has more than one solution.

(ii). If f(x)>0 for x>0 show there is $$\delta$$ in (0,1) with $$\delta$$ not equal to f(x) for all x in [0.5,1].

(iii). If f is also differentiable on (0,1) prove there is an a in (0,1) such that

f'(a)/cos(a$$\pi/2$$) = $$2/\pi$$

b. If f:[-1,1]->R is such that f(sin(x)) is continuous on R then is f continuous. (give a proof or counterexample).

3. I am really stuck on this question and so have not got very far at all. I think part 1 must involve the application of the intermediate value theorem twice though I can't see how to do so. I don't even know how to start part 2. For part 3 I think I will need to use the mean value theorem (and probably a previous question) but can't see where they have got the cos function from.For part b I suspect that it is true but don't know how to write a proof.

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Some rough hints:

1a) i) You should be able to find one solution just by noting that f has to take on every value between 0 and 1 by IVT
ii) I think this should be handled by epsilon-delta proof. If f(x) > 0, you can take f(x) to be epsilon
iii) Rearrange the formula to isolate f'(a) on one side and see if you can make the equality into something that looks like it would be true by the MVT.

b. Have you seen the proof that if g is continuous at a and f is continuous at g(a), then the composition f of g is continuous at a? If you haven't, see if you can prove this on your own. Part b) is based on this theorem.

Ok, so for part i we know there is one solution (by IVT), what other things could I try in order to establish there is more than one solution?

for 2, I am aware of the theorem you stated but, is it an if and only if statement? If not I can't see how I could use it.