Show that arctan(x) exists and is continuous

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The discussion focuses on demonstrating the existence and continuity of the inverse function arctan(x) given the continuous and strictly increasing nature of tan(x) on the interval (-pi/2, pi/2). Participants suggest that showing tan(x) is bijective is a valid approach, but there is uncertainty about proving the continuity of arctan(x). One proposed method is to assume arctan(x) is not continuous and derive a contradiction using the relationship tan(arctan(x))=x. Additionally, the role of continuous functions and their inverses is highlighted, emphasizing that continuous functions can have discontinuous inverses under certain conditions. The conversation encourages exploring the properties of tan(x) further, particularly its behavior with open sets.
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Homework Statement



Show tan(x): (-pi/2, pi/2) -> R has a continuous inverse arctan(x) : R -> (-pi/2, pi/2).

You may assume that tan(x) is continuous and strictly increasing on the given domain, and tends to +/- \infty at +/- pi/2

Homework Equations





The Attempt at a Solution



I think I have shown that tan(x) has an inverse on this domain by showing it is bijective.

However, I am unsure how to go about showing that arctan(x) is continuous.
 
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You could try considering tan(arctan(x))=x and show that if arctan(x) is not continuous, it leads to a contradiction.

(I'm just offering a suggestion. I think it'll work, but I didn't actually try it.)
 
Continuous functions can have discontinuous inverses.

I'm not sure if this is possible when the domain and codomain are intervals, though. I suspect not.
 
What characterizations of continuity have you learnt? Try showing that tan takes every open set to an open set.
 
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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