The one-to-one function \( g \) is defined below. \( g(x)=\frac{4 x}{5 x-9} \) 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 \( g^{-1} \) Range of \( g^{-1}(x)= \)

Let \( g(x)=\frac{4x}{5x-9} \). To find \( g^{-1}(x) \), we start by writing \[ y=\frac{4x}{5x-9}. \] Then, we proceed with the following steps: 1. Multiply both sides by the denominator \((5x-9)\): \[ y(5x-9)=4x. \] 2. Distribute \( y \) on the left-hand side: \[ 5xy-9y=4x. \] 3. Move the term containing \( x \) on the right-hand side: \[ 5xy-4x=9y. \] 4. Factor \( x \) on the left-hand side: \[ x(5y-4)=9y. \] 5. Solve for \( x \) by dividing both sides by \( (5y-4) \): \[ x=\frac{9y}{5y-4}. \] Now, swap \( x \) and \( y \) to obtain the inverse function: \[ g^{-1}(x)=\frac{9x}{5x-4}. \] Next, we determine the domain and range of \( g^{-1}(x) \). **Domain of \( g^{-1} \):** The domain of \( g^{-1}(x) \) consists of all \( x \) for which the expression is defined. The denominator \( 5x-4 \) cannot be zero: \[ 5x-4\neq 0 \quad \Longrightarrow \quad x\neq \frac{4}{5}. \] Thus, the domain of \( g^{-1}(x) \) in interval notation is: \[ (-\infty, \frac{4}{5})\cup(\frac{4}{5}, \infty). \] **Range of \( g^{-1} \):** The range of \( g^{-1} \) is the domain of the original function \( g(x) \). Since \[ g(x)=\frac{4x}{5x-9} \] has a denominator \( 5x-9 \) that cannot be zero: \[ 5x-9\neq 0 \quad \Longrightarrow \quad x\neq \frac{9}{5}, \] the domain of \( g(x) \) (and hence the range of \( g^{-1} \)) is: \[ (-\infty, \frac{9}{5})\cup(\frac{9}{5}, \infty). \] Thus, the final answers are: \[ g^{-1}(x)=\frac{9x}{5x-4}, \] Domain of \( g^{-1} \): \( (-\infty, \frac{4}{5})\cup(\frac{4}{5}, \infty) \), \\ Range of \( g^{-1} \): \( (-\infty, \frac{9}{5})\cup(\frac{9}{5}, \infty) \).