INVERSE TRIGONOMETRIC FUNCTIONS 3

Note that the domain of the function f is the set D, which is the set { x , y} and the
range of the function f is the set {c} . Also, note that the function f is an onto
function. The set E is the range of the function f . The function f maps x in the
set D to c in the set E. The function f also maps y in the set D to c in the set E. So,
the inverse function would map c back to either x or y. Which one do you use?
The problem here is that the function f is not a one-to-one function. In order for a
function to be a one-to-one function, you may only use each element in the set E
once. In order for a function to have an inverse, it must be a one-to-one
function. If a function is not a one-to-one function, then the lack of this needed
condition is not as easy to fix as the lack of the onto condition. In order to fix the
lack of the one-to-one condition, you must put a restriction on the domain of the
function. In other words, you must eliminate elements from the set D. What
elements in the set D are you going to chose to eliminate? This is the reason that
fixing the lack of the one-to-one condition is harder. For the function f , the
domain is the set D = { x , y}. Thus, we will either eliminate x or y. Each restricted
domain will produce an inverse function. Thus, these two choices for the restricted
domain will produce two inverse functions.
If we eliminate y, then we get the following inverse function: