The one-to-one function \( g \) is defined below. \( g(x)=\frac{5 x}{9 x-5} \) Find \( g^{-1}(x) \), where \( g^{-1} \) is the inverse of \( g \). Also state the domain and range of \( g^{-1} \) in interval notation. \( g^{-1}(x)=\square \) Domain of \( g^{-1}: \square \)

To find \( g^{-1}(x) \), we start with the equation \( y = g(x) = \frac{5x}{9x - 5} \). To find the inverse, we switch \( x \) and \( y \) and solve for \( y \): 1. Replace \( g(x) \) with \( y \): \[ x = \frac{5y}{9y - 5} \] 2. Cross-multiply to clear the fraction: \[ x(9y - 5) = 5y \] 3. Distribute \( x \): \[ 9xy - 5x = 5y \] 4. Rearranging gives: \[ 9xy - 5y = 5x \] Factor out \( y \): \[ y(9x - 5) = 5x \] 5. Solve for \( y \): \[ y = \frac{5x}{9x - 5} \] Thus, the inverse function is: \[ g^{-1}(x) = \frac{5x}{9x - 5} \] Next, we determine the domain and range of \( g^{-1} \). The function \( g(x) \) has a vertical asymptote at \( 9x - 5 = 0 \), or \( x = \frac{5}{9} \), which means the range of \( g(x) \) excludes this value. Thus, the domain of \( g^{-1}(x) \) is \( (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty) \). In summary: \[ g^{-1}(x) = \frac{5x}{9x - 5} \] Domain of \( g^{-1}: (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty) \)