Proving the Existence of f(x)=x: A Continuos Function Problem | [0,1]

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Homework Statement



Suppose that f is ais continuos function defined on [0,1] with f(0)=1 and f(1)=0. show that there is a value of x that in [0,1] such that f(x)=x. Thank You.
 
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This is slightly simpler:
Let's say I have a function g:[0,1] \rightarrow \Re, which is continuous, with g(0)=1 and g(1)=-1 can you show that there is an x \in [0,1] so that g(x)=0?
 
Im Still Not Following. Can You Explain It A Little Bit Better To Me. Thank You For Your Post By The Way.
 
If a function that's continuous is negative at one point, and positive at another point, does it necesarily cross the x-axis (i.e. is zero somewhere in between)?

That's what he's driving at, but puts it in terms that are more obviously applicable to the problem at hand
 
Define the function G(x)=f(x)-x. Now four questions. i) is G continuous? ii) What are G(0) and G(1)? iii) What does it mean if G(x)=0 in terms of f? iv) Might this have something to do with the NateTG's and Office_Shredder's hints?
 
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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