My Math Forum prove there exists an inverse function

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 December 18th, 2012, 08:37 AM #1 Newbie   Joined: Dec 2012 Posts: 23 Thanks: 0 prove there exists an inverse function I'm quite new to calculus and frankly speaking I'm studying it on my own. Could you tell me how to solve this: $f: [\frac{\pi}{2}, \pi) \ni x \rightarrow \frac{1}{\sin x} \in \mathbb{R}$ Prove that there exists an inverese of $f$. In what set is $f^{-1}$ differentiable? Calculate $f^{-1'}$ I would really appreciate a thorough explanation. Thank you.
 December 18th, 2012, 08:57 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: prove there exists an inverse function Okay, do you remember the requirements that a function be invertible? In order that for any given y we be able to find the x such that y= f(x), there must be at least one such x (f is "onto") and there must be just one such x (f is "one-to-one"). Neither of those is true for sin(x) itself- there is NO x such that sin(x)= 2 and there exist infinitely many x such that sin(x)= 0. I would suggest starting by looking at graph of $y= \frac{1}{sin(x)}$.

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# prove a function exists

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