Which function has an inverse that is also a function?

A. \{(-4, 3), (-2, 7), (-1, 0), (4, -3), (11, -7)\}
B. \{(-4, 6), (-2, 2), (−1, 6), (4, 2), (11, 2)\}
C. \{(-4, 5), (-2, 9), (-1, 8), (4, 8), (11, 4)\}
D. \{(-4, 4), (-2, -1), (-1, 0), (4, 1), (11, 1)\}



Answer :

To determine which function has an inverse that is also a function, we need to check if, for each of the given sets, the inverse relation satisfies the criteria for being a function. In other words, we'll check if each value in the codomain (the second values in the pairs) maps uniquely to a value in the domain (the first values in the pairs). A function is only an inverse if each element in the codomain is a unique output, meaning it maps to one and only one element in the domain.

Let's analyze each set:

### Set 1
[tex]\[ \{(-4, 3), (-2, 7), (-1, 0), (4, -3), (11, -7)\} \][/tex]
- Inverse: [tex]\(\{(3, -4), (7, -2), (0, -1), (-3, 4), (-7, 11)\}\)[/tex]
- Check if each second element (3, 7, 0, -3, -7) is unique.
- Yes, each second element is unique.

### Set 2
[tex]\[ \{(-4, 6), (-2, 2), (-1, 6), (4, 2), (11, 2)\} \][/tex]
- Inverse: [tex]\(\{(6, -4), (2, -2), (6, -1), (2, 4), (2, 11)\}\)[/tex]
- The second element (codomain) has duplicates: 6 and 2 appear more than once.
- No, the second elements are not unique.

### Set 3
[tex]\[ \{(-4, 5), (-2, 9), (-1, 8), (4, 8), (11, 4) \][/tex] \}
- Inverse: [tex]\(\{(5, -4), (9, -2), (8, -1), (8, 4), (4, 11)\}\)[/tex]
- The second element (codomain) has duplicates: 8 appears more than once.
- No, the second elements are not unique.

### Set 4
[tex]\[ \{(-4, 4), (-2, -1), (-1, 0), (4, 1), (11, 1)\} \][/tex]
- Inverse: [tex]\(\{(4, -4), (-1, -2), (0, -1), (1, 4), (1, 11)\}\)[/tex]
- The second element (codomain) has duplicates: 1 appears more than once.
- No, the second elements are not unique.

### Conclusion
Only the first set, [tex]\(\{(-4, 3), (-2, 7), (-1, 0), (4, -3), (11, -7)\}\)[/tex], has an inverse that is also a function because each element in the codomain (the set of second values) is unique.

Thus, the function in set 1 has an inverse that is also a function.