The one-to-one function \( g \) is defined below. \( g(x)=\frac{6 x}{4-5 x} \) 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 the inverse of the function \( g(x) = \frac{6x}{4 - 5x} \), we will follow these steps: 1. **Replace \( g(x) \) with \( y \)**: \[ y = \frac{6x}{4 - 5x} \] 2. **Swap \( x \) and \( y \)** to find the inverse: \[ x = \frac{6y}{4 - 5y} \] 3. **Solve for \( y \)**: - Multiply both sides by \( 4 - 5y \): \[ x(4 - 5y) = 6y \] - Distribute \( x \): \[ 4x - 5xy = 6y \] - Rearrange to isolate terms involving \( y \): \[ 4x = 6y + 5xy \] - Factor out \( y \) on the right side: \[ 4x = y(6 + 5x) \] - Solve for \( y \): \[ y = \frac{4x}{6 + 5x} \] 4. **Thus, the inverse function is**: \[ g^{-1}(x) = \frac{4x}{6 + 5x} \] Next, we will determine the domain and range of \( g^{-1} \). ### Domain of \( g^{-1} \): The domain of \( g^{-1} \) is determined by the values of \( x \) for which the function is defined. The function \( g^{-1}(x) = \frac{4x}{6 + 5x} \) is undefined when the denominator is zero: \[ 6 + 5x = 0 \implies 5x = -6 \implies x = -\frac{6}{5} \] Thus, the domain of \( g^{-1} \) is all real numbers except \( -\frac{6}{5} \): \[ \text{Domain of } g^{-1}: (-\infty, -\frac{6}{5}) \cup (-\frac{6}{5}, \infty) \] ### Range of \( g^{-1} \): The range of \( g^{-1} \) corresponds to the domain of \( g \). To find the domain of \( g \), we need to determine when the denominator is not zero: \[ 4 - 5x \neq 0 \implies 5x \neq 4 \implies x \neq \frac{4}{5} \] Thus, the domain of \( g \) is all real numbers except \( \frac{4}{5} \): \[ \text{Domain of } g: (-\infty, \frac{4}{5}) \cup (\frac{4}{5}, \infty) \] This means the range of \( g^{-1} \) is also: \[ \text{Range of } g^{-1}: (-\infty, \frac{4}{5}) \cup (\frac{4}{5}, \infty) \] ### Final Answers: - \( g^{-1}(x) = \frac{4x}{6 + 5x} \) - Domain of \( g^{-1}: (-\infty, -\frac{6}{5}) \cup (-\frac{6}{5}, \infty) \) - Range of \( g^{-1}: (-\infty, \frac{4}{5}) \cup (\frac{4}{5}, \infty) \)