The inverse of f = {(1,2), (2,3), (3,4), (4,1), (5,2)} is a function if the domain of f is limited to a) {1,3,5} b) (1,2,3,4} c){1,5}, d) {1,2,4,5} e) {1,2,3,4,

Question

This question is from a SAT 2 book. B is the correct answer, and the explanation given is that for the inverse function, two of the points would be (2,1) and (2,5). Since no two points in a function can have the same x-value but different y-values, 1 and 5 can't both be in the domain. B is the only option that abides to this. However, after drawing the original function on a graph I saw that it does not pass the horizontal line test, no matter which of the 5 domain options are used. Am I doing something wrong? Sorry for the lengthiness of this question.

Answer 1

Notice that it never says that "f" is a function, it says that the inverse of "f" is a function if the domain of f is limited.

 

So if you eliminate the last point, (5,2), then the inverse will be a function because it will contain only the points

 

{(2,1), (3,2), (4,3), (1,4)}

 

If you put in the inverse of the point (5,2), you will be including the point (2,5) in the inverse which would cause the inverse to NOT be a function, as you have pointed out because the inverse would no longer pass the vertical line test.  

 

So the domain of "f" is the set of all the x-coordinates of the points that can be included in the inverse.  {2, 3, 4, 1}, or put in correct order, {1, 2, 3, 4}