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December 18th, 2012, 08:37 AM   #1
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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:

Prove that there exists an inverese of . In what set is differentiable? Calculate

I would really appreciate a thorough explanation. Thank you.
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December 18th, 2012, 08:57 AM   #2
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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 .
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