Functions and the Intermediate Value theorem

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Homework Statement


Let f : [0; 1] --> R be continuous on [0, 1], and assume that the range of f is contained
in [0; 1]. Prove that there exists an x in [0, 1] satisfying f(x) = x.


Homework Equations





The Attempt at a Solution



Well i am almost positive I need to use the intermediate value theorem.

First I could claim that either f(0)>x>f(1) or f(0)<x<f(1). where x is a value in (0,1)
Not too sure what to do, i think the key is something to do with the range of f being contained in [0,1], but any help would be great!
 
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If f(0)= 0, we are done. If f(1)= 1, we are done. So we can assume that f(0)\ne 0 and that f(1)\ne 1. But f(0) must be in [0, 1]. If it is not equal to 0, then f(0)> 0. Similarly, we must have f(1)< 1. Define H(x)= f(x)- x. What is H(0)? What is H(1)? Apply the intermediate value theorem to H(x).
 
Thanks so much for the help!
 
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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