The one-to-one function \( g \) is defined below. \( g(x)=\frac{5 x-2}{9 x+8} \) Find \( g^{-1}(x) \), where \( g^{-1} \) is the inverse of \( g \). Also state the domain and range of \( g^{-1} \) in interval notation. Domain of \( g^{-1}(x)=\square \) Range of \( g^{-1} \)

To find the inverse function \( g^{-1}(x) \), we start by setting \( y = g(x) \): \[ y = \frac{5x - 2}{9x + 8} \] Next, we swap \( x \) and \( y \) to solve for \( y \): \[ x = \frac{5y - 2}{9y + 8} \] Now, we can cross-multiply to eliminate the fraction: \[ x(9y + 8) = 5y - 2 \] Expanding and rearranging gives us: \[ 9xy + 8x = 5y - 2 \] Bringing terms involving \( y \) to one side, we have: \[ 9xy - 5y = -8x - 2 \] Factoring out \( y \): \[ y(9x - 5) = -8x - 2 \] Now, solving for \( y \) yields: \[ y = \frac{-8x - 2}{9x - 5} \] Thus, the inverse function is: \[ g^{-1}(x) = \frac{-8x - 2}{9x - 5} \] Now, let's analyze the domain and range of \( g^{-1}(x) \). The domain of the inverse function corresponds to the range of the original function \( g(x) \). The range of \( g(x) \) is found by evaluating the limits. As \( x \) approaches values causing the denominator of \( g(x) \) to vanish, we find that \( g(x) \) approaches certain asymptotic values. Specifically, the transformation shows: 1. As \( x \to \infty \), \( g(x) \to \frac{5}{9} \) 2. As \( x \to -\frac{8}{9} \), \( g(x) \to \infty \) Thus, the range of \( g(x) \) is \( (-\infty, \frac{5}{9}) \) and therefore the domain of \( g^{-1}(x) \) is: \[ \text{Domain of } g^{-1}(x) = (-\infty, \frac{5}{9}) \] The range of \( g^{-1}(x) \) corresponds to the domain of \( g(x) \). Since \( g(x) \) can take all real numbers apart from the point where the denominator is zero, the range of \( g^{-1}(x) \) is: \[ \text{Range of } g^{-1}(x) = (-\frac{8}{9}, \infty) \] So, the final answers are: Domain of \( g^{-1}(x) = (-\infty, \frac{5}{9}) \) Range of \( g^{-1}(x) = (-\frac{8}{9}, \infty) \)