Hi there, I'm having a little trouble with working out the implied domain of this equation:
I thought that since the range of must be a subset or equal to the domain of , then . Making the implied domain
But apparently this is wrong...
Thanks for your help.
Just did a quick google and found this Complex Analysis/Elementary Functions/Inverse Trig Functions - Wikibooks, collection of open-content textbooks and http://docs.google.com/viewer?a=v&q=...jqEaNUEA&pli=1
I guess this is what my teacher was talking about...
Back to my question though; if the domain of was restricted to would my original answer be correct?
By the way, I had a look at your paintings Grandad. They are amazing
The use of the upper-case initial letter, then, restricts the domain of the functions to their principal values. This has the effect of making the functions one-to-one, and therefore makes it possible to define the inverse functions. In that case, the domain of
would indeed be
Since this upper- and lower-case convention appears not to be very widespread, I should use it with caution if I were you.
(Thanks for the comment about the paintings. If you just looked at the ones on my profile, you'll find some more in the Chat Room at http://www.mathhelpforum.com/math-he...-painting.html.)