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. \( f(g(x))=\square \) (Simplify your answer. Use integers or fractions for any numbers in the expression.)

To find \( f(g(x)) \), we first need to substitute \( g(x) \) into \( f(x) \). We have: \[ g(x) = \frac{x + 9}{8} \] Now, substituting \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{x + 9}{8}\right) = 9\left(\frac{x + 9}{8}\right) - 8 \] Now we can simplify it: \[ = \frac{9(x + 9)}{8} - 8 = \frac{9x + 81}{8} - \frac{64}{8} = \frac{9x + 81 - 64}{8} = \frac{9x + 17}{8} \] So, \[ f(g(x)) = \frac{9x + 17}{8}. \] Now, let's find \( g(f(x)) \) by substituting \( f(x) \) into \( g(x) \): \[ f(x) = 9x - 8 \] Substituting this into \( g(x) \): \[ g(f(x)) = g(9x - 8) = \frac{(9x - 8) + 9}{8} = \frac{9x + 1}{8}. \] So, \[ g(f(x)) = \frac{9x + 1}{8}. \] Now, let's determine if \( f \) and \( g \) are inverses of each other. For \( f \) and \( g \) to be inverses, we need: 1. \( f(g(x)) = x \) 2. \( g(f(x)) = x \) From our calculations: - \( f(g(x)) = \frac{9x + 17}{8} \) (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.