Using the Intermediate Value Theorem to Solve for x in Continuous Functions

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Homework Statement



a) Suppose that f(x) is a continuous function on [0, 1] and 0 <= f(x) <= 1 for all x in [0, 1]. Show that there is an x in [0, 1] where f(x) = x.

b) Suppose that f(x) is a continuous function on [0, 2] with f(0) = f(2). Show that there is an x in [0, 1] such that f(x) = f(x + 1).

Homework Equations


The Attempt at a Solution



I assume I have to use the Intermediate Value Theorem, but I have no idea how to use it! For (a) I thought 0 <= f(x) <= 1 means since f(x) is between f(a) and f(b) then there exists a c or x in this question so that f(x) = x, but I have no idea! And for (b)... not a clue! Please help?!
 
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Try to construct a new function g from f, in such a way that answering the question means comparing g to a constant (so that it's easier to apply the intermediate value theorem).
 
Oh! Does that mean for (a) g(x) = f(x) - x? Or am I completely off track?!
 
It's a good thought -- run with it and see where you get to.
 
I am still very confused with how to form g(x)... it is still not making sense how to prove this question!
 
LilTaru said:
I am still very confused with how to form g(x)... it is still not making sense how to prove this question!

You already formed a good g(x) by setting g(x)=f(x)-x. If g(0)=0 then f(0)=0 and you are done. Otherwise g(0) is positive, right? What happens if g(1)=0? Suppose g(1) is not zero. What sign is it?
 
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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