To determine which two functions are inverses of each other, we need to check if the composition of each function gives us the identity function \( f(g(x)) = x \) and \( g(f(x)) = x \).
Let's apply this principle to each pair of functions:
-
\( f(x) = 15x + 2 \) and \( g(x) = 2x - 15 \):
- \( f(g(x)) = f(2x - 15) = 15(2x - 15) + 2 = 30x - 225 + 2 = 30x - 223 \) (not equal to \( x \))
- \( g(f(x)) = g(15x + 2) = 2(15x + 2) - 15 = 30x + 4 - 15 = 30x - 11 \) (not equal to \( x \))
-
\( f(x) = -4x - 13 \) and \( g(x) = -3x - 14 \):
- \( f(g(x)) = f(-3x - 14) = -4(-3x - 14) - 13 = 12x + 56 - 13 = 12x + 43 \) (not equal to \( x \))
- \( g(f(x)) = g(-4x - 13) = -3(-4x - 13) - 14 = 12x + 39 - 14 = 12x + 25 \) (not equal to \( x \))
-
\( f(x) = \frac{1}{11}x + 2 \) and \( g(x) = 11x - 2 \):
- \( f(g(x)) = f(11x - 2) = \frac{1}{11}(11x - 2) + 2 = x - \frac{2}{11} + 2 = x + \frac{20}{11} \) (not equal to \( x \))
- \( g(f(x)) = g\left(\frac{1}{11}x + 2\right) = 11\left(\frac{1}{11}x + 2\right) - 2 = x + 22 - 2 = x + 20 \) (not equal to \( x \))
-
\( f(x) = 3x + 75 \) and \( g(x) = \frac{3}{5}x - 7 \):
- First, we need to check \( g(x) \) for correctness since it seems to be incorrectly represented in the prompt. Assuming it's \( g(x) = \frac{3}{5}x - 7 \) (not \( 35x-7 \)):
- \( f(g(x)) = f(\frac{3}{5}x - 7) = 3(\frac{3}{5}x - 7) + 75 = \frac{9}{5}x - 21 + 75 = \frac{9}{5}x + 54 \) (not equal to \( x \))
- \( g(f(x)) = g(3x + 75) = \frac{3}{5}(3x + 75) - 7 = \frac{9}{5}x + 45 - 7 = \frac{9}{5}x + 38 \) (not equal to \( x \))
After checking all pairs, none of the provided pairs are functions that serve as inverses of each other. Therefore, we might want to double-check for errors in function details or look for an option that meets the criteria.
If you can provide additional context or corrected function pairs, I can assist further!