Proving the Intermediate Value Theorem and Range of 1:1 Continuous Functions

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To prove that f(x) = f(x+1) for some x in [0, 1], consider the function h(x) = f(x) - f(x+1), which is continuous since it is the difference of two continuous functions. By applying the Intermediate Value Theorem, if h(0) and h(1) have opposite signs, there exists an x in [0, 1] such that h(x) = 0, thus proving f(x) = f(x+1). For part b, since f is one-to-one and continuous on [a, b], the range of f must be the interval [f(a), f(b)], as a continuous function on a closed interval achieves all values between its endpoints. Both problems highlight the application of continuity and the Intermediate Value Theorem in proving function properties. Understanding these concepts is essential for solving similar problems in calculus.
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Homework Statement


a)Let f(x) be continuous on [0, 2], with f(0) = f(2). Show that f(x) = f(x+1) for some x ε [0, 1].
b)Let f(x) be 1:1 and continuous on the interval [a, b] with f(a) < f(b). Show that the range of f is the interval [f(a), f(b)].


Homework Equations





The Attempt at a Solution


I'm not really where to start for either of them. In a), I find it obvious that there exists an f(x) = f(x+1) for some x in that interval, but find it difficult to prove without any specific function. I find using the I.V.T. difficult in general without being applied to a specific function. Any help/hints appreciated. Thanks!
 
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Hello fellow Waterloo student! I would love to help, but alas, I am having the same problems as you! :S
 
Hello busterkomo try taking a function h such that:

h(x)=f(x)-f(x+1)

And h is obviously a continuous function as a difference of two continuous functions.
Now what can you do from that??
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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