How to Prove a Function Has a Fixed Point on a Closed Interval?

SomeRandomGuy
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1.) Suppose f:[a,b]->R and g:[a,b]->R are continuous such that f(a)<=g(a) and f(b)=>g(b). Prove that f(c)=g(c) for some c in [a,b].

I started out by using the Intermediate Value Property for some c1 and c2 with f(c1)=L1 and g(c2)=L2. I am trying to conclude that L1=L2. This was one approach. Then, I also tried setting up a ratio where f(a)/g(a) > 1 and f(b)/g(b) < 1 by assumption. These are continuous if g !=0, so there must exist a point where f/g = 1. But, what if g(x)=0. This method seems to fall apart as well, and now I'm stuck.

2.) Let f:[0,1]->[0,1] be a continuous function. Prove that f must have a fixed point; that is, show f(x)=x for some x in [0,1]. I'm trying to show that f must intersect the graph g(x)=x by using the IVP again. Don't quite know how, though. Seems that if I figure out the first problem, this one would be very similiar.

Thanks for any help, I appreciate it.
 
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Hint for 1. Consider the function h(x)=f(x)-g(x) and apply the intermediate value theorem.

2 is easy once you know 1.
 
Wow... I tell ya, sometimes I just feel like a moron with how obvious some of these proofs are.

Thanks for your help, both of these problems were cake.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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