Find \( f(g(x)) \) and \( g(f(x)) \) and determine whether the pair of functions \( f \) and \( g \) are inverses of each \( f(x)=9 x-7 \) and \( g(x)=\frac{x+9}{7} \) a. \( f(g(x))=\square \) (Simplify your answer. Use integers or fractions for any numbers in the expression.)

To find \( f(g(x)) \), we start by substituting \( g(x) \) into \( f(x) \). Given: \[ f(x) = 9x - 7 \] \[ g(x) = \frac{x + 9}{7} \] Now, substituting \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{x + 9}{7}\right) = 9\left(\frac{x + 9}{7}\right) - 7 \] Calculating: \[ = \frac{9(x + 9)}{7} - 7 = \frac{9x + 81}{7} - 7 \] To combine the terms, we need a common denominator: \[ = \frac{9x + 81}{7} - \frac{49}{7} = \frac{9x + 81 - 49}{7} = \frac{9x + 32}{7} \] Thus: \[ f(g(x)) = \frac{9x + 32}{7} \] Next, we will find \( g(f(x)) \): Substituting \( f(x) \) into \( g(x) \): \[ g(f(x)) = g(9x - 7) = \frac{(9x - 7) + 9}{7} = \frac{9x + 2}{7} \] So, our results are: \[ f(g(x)) = \frac{9x + 32}{7} \] \[ g(f(x)) = \frac{9x + 2}{7} \] To determine if \( f \) and \( g \) are inverses of each other, we check if \( f(g(x)) = x \) and \( g(f(x)) = x \). 1. For \( f(g(x)) = \frac{9x + 32}{7} \): This does not equal \( x \). 2. For \( g(f(x)) = \frac{9x + 2}{7} \): This also does not equal \( x \). Therefore, \( f \) and \( g \) are not inverses of each other. Final answer: a. \( f(g(x)) = \frac{9x + 32}{7} \)