To determine which function rule models the function over the specified domain, we need to test each given function rule against the table values.
First, let's list the given [tex]\( x \)[/tex] values and their corresponding [tex]\( f(x) \)[/tex] values, as per the table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-7 & -11 \\
\hline
-1 & 1 \\
\hline
3 & 9 \\
\hline
4 & 11 \\
\hline
7 & 17 \\
\hline
\end{array}
\][/tex]
Next, we will evaluate each of the given function rules:
1. [tex]\( f(x) = 3x + 10 \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 3x + 10 \\
\hline
-7 & 3(-7) + 10 = -21 + 10 = -11 \\
-1 & 3(-1) + 10 = -3 + 10 = 7 \\
3 & 3(3) + 10 = 9 + 10 = 19 \\
4 & 3(4) + 10 = 12 + 10 = 22 \\
7 & 3(7) + 10 = 21 + 10 = 31 \\
\end{array}
\][/tex]
Clearly, this doesn't match the given table values.
2. [tex]\( f(x) = 2x + 3 \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 2x + 3 \\
\hline
-7 & 2(-7) + 3 = -14 + 3 = -11 \\
-1 & 2(-1) + 3 = -2 + 3 = 1 \\
3 & 2(3) + 3 = 6 + 3 = 9 \\
4 & 2(4) + 3 = 8 + 3 = 11 \\
7 & 2(7) + 3 = 14 + 3 = 17 \\
\end{array}
\][/tex]
This matches perfectly with the given table values:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-7 & -11 \\
\hline
-1 & 1 \\
\hline
3 & 9 \\
\hline
4 & 11 \\
\hline
7 & 17 \\
\hline
\end{array}
\][/tex]
3. [tex]\( f(x) = 4x + 5 \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 4x + 5 \\
\hline
-7 & 4(-7) + 5 = -28 + 5 = -23 \\
-1 & 4(-1) + 5 = -4 + 5 = 1 \\
3 & 4(3) + 5 = 12 + 5 = 17 \\
4 & 4(4) + 5 = 16 + 5 = 21 \\
7 & 4(7) + 5 = 28 + 5 = 33 \\
\end{array}
\][/tex]
This does not match the table values.
4. [tex]\( f(x) = 3x - 10 \)[/tex]
[tex]\[
\begin{array}{c|c}
x & f(x) = 3x - 10 \\
\hline
-7 & 3(-7) - 10 = -21 - 10 = -31 \\
-1 & 3(-1) - 10 = -3 - 10 = -13 \\
3 & 3(3) - 10 = 9 - 10 = -1 \\
4 & 3(4) - 10 = 12 - 10 = 2 \\
7 & 3(7) - 10 = 21 - 10 = 11 \\
\end{array}
\][/tex]
This does not match the table values.
Thus, the function rule that models the function over the given domain is:
[tex]\[ \boxed{f(x) = 2x + 3} \][/tex]
