The one-to-one function \( g \) is defined below. \( g(x)=\frac{3 x}{8 x-3} \) 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 \( g^{-1}(x)= \) Range of \( g^{-1} \)

To find the inverse function \( g^{-1}(x) \), we start with the equation: \[ y = g(x) = \frac{3x}{8x - 3} \] Now, we swap \( x \) and \( y \) and solve for \( y \): \[ x = \frac{3y}{8y - 3} \] Multiplying both sides by \( 8y - 3 \) gives us: \[ x(8y - 3) = 3y \] Distributing \( x \): \[ 8xy - 3x = 3y \] Now, isolate the terms involving \( y \): \[ 8xy - 3y = 3x \] Factoring out \( y \): \[ y(8x - 3) = 3x \] Now solve for \( y \): \[ y = \frac{3x}{8x - 3} \] Thus, we find: \[ g^{-1}(x) = \frac{3x}{8x - 3} \] Next, we need to determine the domain and range of \( g^{-1} \). For the domain of \( g^{-1}(x) \), we note that the denominator \( 8x - 3 \) cannot be zero: \[ 8x - 3 = 0 \implies x = \frac{3}{8} \] Thus, the domain of \( g^{-1}(x) \) is: \[ \text{Domain of } g^{-1}(x) = (-\infty, \frac{3}{8}) \cup (\frac{3}{8}, \infty) \] For the range of \( g^{-1}(x) \), since \( g \) is a one-to-one function that also has the horizontal asymptote as \( y = \frac{3}{8} \), we conclude that the range of \( g^{-1}(x) \) consists of all real numbers except \( \frac{3}{8} \): \[ \text{Range of } g^{-1} = (-\infty, \frac{3}{8}) \cup (\frac{3}{8}, \infty) \] So summarizing: \[ g^{-1}(x) = \frac{3x}{8x - 3} \] \[ \text{Domain of } g^{-1}(x) = (-\infty, \frac{3}{8}) \cup (\frac{3}{8}, \infty) \] \[ \text{Range of } g^{-1} = (-\infty, \frac{3}{8}) \cup (\frac{3}{8}, \infty) \]