This problem is taken straight from a free textbook of Mooculus. The problem itself has nothing to do with calculus as it's part of introductory review section of the book. I hope it goes into right sub-forum. If not, I kindly ask one of the site admins to more it to right section, if possible.

This is not the original problem and the question presented in the book (page 17, exercise 5 to be exact) is easily solvable by just looking at graphs, so my interest here is purely out of curiosity and for learning.

Now that the preliminaries are out of the way, onwards to the actual problem.

Let us have a function

ie. f(x) = 18*sin((pi*x)/7)+20 and let us also restrict it to [3.5,10.5] in order to make it inversible. With my feeble algebraic skills I can come up with

ie. f^1-(x) = 7 * arcsin((t - 20)/18)/pi which is blatantly incorrect. From another source I got the right answer which is

ie. 7 * (-arcsin((t - 20)/18) + pi)/pi. What I do not understand is how to get to this. I know this has everything to do with the fact that arcsin is only defined in [-pi/2, pi/2], but I can't quite work this out on my own. Could someone explain the process in as simple steps as possible? After two days of occasional poking at this I feel like being on the verge of "finally getting it", yet I can't just quite grasp what on earth I'm actually pondering about.