Find \( f(g(x)) \) and \( g(f(x)) \) and determine whether the pair of functions \( f \) and \( g \) are inverses of each other. \( f(x)=9 x-8 \) and \( g(x)=\frac{x+9}{8} \) a. \( f(g(x))=\frac{9 x+17}{8} \) (Simplify your answer. Use integers or fractions for any numbers in the expression.) b. \( g(f(x))=\frac{9 x+1}{8} \) (Simplify your answer. Use integers or fractions for any numbersin the expression.) c. \( f \) and \( g \) are inverses of each other. cand \( g \) are not inverses of each other.

To find the compositions \( f(g(x)) \) and \( g(f(x)) \), we will substitute one function into the other. 1. For \( f(g(x)) \): \[ g(x) = \frac{x + 9}{8} \] Now, substitute \( g(x) \) into \( f \): \[ f(g(x)) = f\left(\frac{x+9}{8}\right) = 9\left(\frac{x+9}{8}\right) - 8 \] Simplifying gives: \[ f(g(x)) = \frac{9(x+9)}{8} - 8 = \frac{9x + 81}{8} - \frac{64}{8} = \frac{9x + 17}{8} \] 2. For \( g(f(x)) \): \[ f(x) = 9x - 8 \] Now, substitute \( f(x) \) into \( g \): \[ g(f(x)) = g(9x - 8) = \frac{(9x-8) + 9}{8} \] Simplifying gives: \[ g(f(x)) = \frac{9x + 1}{8} \] Now, let's analyze whether \( f \) and \( g \) are inverses. For two functions to be inverses, the compositions \( f(g(x)) \) and \( g(f(x)) \) must equal \( x \). From our results: - \( f(g(x)) = \frac{9x + 17}{8} \) (not equal to \( x \)) - \( g(f(x)) = \frac{9x + 1}{8} \) (not equal to \( x \)) Thus, \( f \) and \( g \) are **not** inverses of each other.