To determine which function assigns the domain value [tex]\( x = 12 \)[/tex] to the range value [tex]\( y = 7 \)[/tex], we'll evaluate each of the given functions at [tex]\( x = 12 \)[/tex] and see which one yields [tex]\( y = 7 \)[/tex].
Given the functions:
A. [tex]\( f(x) = 2x - 2 \)[/tex]
B. [tex]\( f(x) = 3x - 4 \)[/tex]
C. [tex]\( f(x) = \frac{1}{2} x + 1 \)[/tex]
D. [tex]\( f(x) = \frac{3}{7} x + 12 \)[/tex]
Let's evaluate each function at [tex]\( x = 12 \)[/tex]:
For function A:
[tex]\[
f_A(12) = 2(12) - 2 = 24 - 2 = 22
\][/tex]
Since [tex]\( f_A(12) = 22 \)[/tex], it does not equal 7.
For function B:
[tex]\[
f_B(12) = 3(12) - 4 = 36 - 4 = 32
\][/tex]
Since [tex]\( f_B(12) = 32 \)[/tex], it does not equal 7.
For function C:
[tex]\[
f_C(12) = \frac{1}{2}(12) + 1 = 6 + 1 = 7
\][/tex]
Since [tex]\( f_C(12) = 7 \)[/tex], this function assigns the domain value 12 to the range value 7.
For function D:
[tex]\[
f_D(12) = \frac{3}{7}(12) + 12 = \frac{36}{7} + 12 = \frac{36}{7} + \frac{84}{7} = \frac{120}{7} \approx 17.14
\][/tex]
Since [tex]\( f_D(12) \neq 7 \)[/tex], it does not equal 7.
After evaluating all the functions, we find that only function C, [tex]\( f(x) = \frac{1}{2} x + 1 \)[/tex], assigns the domain value 12 to the range value 7.
Thus, the correct answer is:
C. [tex]\( f(x) = \frac{1}{2} x + 1 \)[/tex]
