To determine which of the given functions has an inverse that is also a function, we need to ensure that the function in question has unique [tex]\( y \)[/tex]-values for each [tex]\( x \)[/tex]-value. In other words, no [tex]\( y \)[/tex]-value should be repeated for different [tex]\( x \)[/tex]-values. This property ensures that the function is one-to-one and therefore has an inverse that is also a function.
Given sets:
1. [tex]\(\{(-1, -2), (0, 4), (1, 3), (5, 14), (7, 4)\}\)[/tex]
2. [tex]\(\{(-1, 2), (0, 4), (1, 5), (5, 4), (7, 2)\}\)[/tex]
3. [tex]\(\{(-1, 3), (0, 4), (1, 14), (5, 6), (7, 2)\}\)[/tex]
4. [tex]\(\{(-1, 4), (0, 4), (1, 2), (5, 3), (7, 1)\}\)[/tex]
We will check each set to see if [tex]\( y \)[/tex]-values are unique.
### 1. [tex]\(\{(-1, -2), (0, 4), (1, 3), (5, 14), (7, 4)\}\)[/tex]
[tex]\[
y\text{-values: } \{-2, 4, 3, 14, 4\}
\][/tex]
Here, the [tex]\( y \)[/tex]-value 4 is repeated (for [tex]\( x = 0 \)[/tex] and [tex]\( x = 7 \)[/tex]). Therefore, this function does not have a unique inverse.
### 2. [tex]\(\{(-1, 2), (0, 4), (1, 5), (5, 4), (7, 2)\}\)[/tex]
[tex]\[
y\text{-values: } \{2, 4, 5, 4, 2\}
\][/tex]
Here, the [tex]\( y \)[/tex]-values 2 and 4 are repeated. Thus, this function does not have a unique inverse.
### 3. [tex]\(\{(-1, 3), (0, 4), (1, 14), (5, 6), (7, 2)\}\)[/tex]
[tex]\[
y\text{-values: } \{3, 4, 14, 6, 2\}
\][/tex]
Here, all the [tex]\( y \)[/tex]-values are unique. Therefore, this function potentially has an inverse that is also a function, making it one-to-one.
### 4. [tex]\(\{(-1, 4), (0, 4), (1, 2), (5, 3), (7, 1)\}\)[/tex]
[tex]\[
y\text{-values: } \{4, 4, 2, 3, 1\}
\][/tex]
Here, the [tex]\( y \)[/tex]-value 4 is repeated (for [tex]\( x = -1 \)[/tex] and [tex]\( x = 0 \)[/tex]). Thus, this function does not have a unique inverse.
Based on the analysis, we can conclude that the function:
[tex]\[
\{(-1, 3), (0, 4), (1, 14), (5, 6), (7, 2)\}
\][/tex]
has an inverse that is also a function. Thus, the correct function is the third one.
