The one-to-one function \( g \) is defined below. \( g(x)=\frac{7 x}{6 x-1} \) 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 \( g^{-1} \) Range of \( g^{-1} \)

To find the inverse of the function \( g(x) = \frac{7x}{6x - 1} \), we start by replacing \( g(x) \) with \( y \): 1. \( y = \frac{7x}{6x - 1} \) Next, we solve for \( x \) in terms of \( y \): 2. Multiply both sides by \( 6x - 1 \) to eliminate the fraction: \( y(6x - 1) = 7x \) 3. Distributing \( y \): \( 6xy - y = 7x \) 4. Rearranging: \( 6xy - 7x = y \) 5. Factoring out \( x \): \( x(6y - 7) = y \) 6. Solving for \( x \): \( x = \frac{y}{6y - 7} \) Now we replace \( y \) with \( x \) to express the inverse function: \[ g^{-1}(x) = \frac{x}{6x - 7} \] Next, we identify the domain and range of \( g^{-1}(x) \): **Domain of \( g^{-1} \)**: The inverse function is defined for all \( x \) except where the denominator is zero: \[ 6x - 7 \neq 0 \Rightarrow x \neq \frac{7}{6} \] So, the domain of \( g^{-1} \) in interval notation is: \[ (-\infty, \frac{7}{6}) \cup (\frac{7}{6}, \infty) \] **Range of \( g^{-1} \)**: The range corresponds to the domain of \( g(x) \). The function \( g \) has a vertical asymptote at \( x = \frac{1}{6} \) (where the denominator is zero) and approaches \( 7/6 \) as \( x \) goes to infinity. Thus, the range is: \[ (-\infty, \frac{7}{6}) \cup (\frac{7}{6}, \infty) \] Combining all of this, we have: \[ g^{-1}(x) = \frac{x}{6x - 7} \] Domain of \( g^{-1} \): \( (-\infty, \frac{7}{6}) \cup (\frac{7}{6}, \infty) \) Range of \( g^{-1} \): \( (-\infty, \frac{7}{6}) \cup (\frac{7}{6}, \infty) \)