The one-to-one function \( g \) is defined below. \[ g(x)=\frac{x-5}{8 x+7} \] Find \( g^{-1}(x) \), where \( g^{-1} \) is the inverse of \( g \). Also state the domain and range of \( g^{-1} \) in interval notation. \( g^{-1}(x)=\square \) Romain of \( g^{-1}: \square \)

To find the inverse function \( g^{-1}(x) \) of the function \( g(x) = \frac{x-5}{8x+7} \), we will follow these steps: 1. **Replace \( g(x) \) with \( y \)**: \[ y = \frac{x-5}{8x+7} \] 2. **Solve for \( x \) in terms of \( y \)**: Multiply both sides by \( 8x + 7 \): \[ y(8x + 7) = x - 5 \] Expanding the left side: \[ 8xy + 7y = x - 5 \] Rearranging the equation to isolate \( x \): \[ 8xy - x = -5 - 7y \] Factor out \( x \) on the left side: \[ x(8y - 1) = -5 - 7y \] Now, solve for \( x \): \[ x = \frac{-5 - 7y}{8y - 1} \] 3. **Replace \( y \) with \( x \) to express the inverse**: \[ g^{-1}(x) = \frac{-5 - 7x}{8x - 1} \] Now we have the inverse function: \[ g^{-1}(x) = \frac{-5 - 7x}{8x - 1} \] Next, we need to determine the domain and range of \( g^{-1}(x) \). ### Domain of \( g^{-1}(x) \) The domain of \( g^{-1}(x) \) is determined by the values of \( x \) for which the denominator is not zero: \[ 8x - 1 \neq 0 \implies x \neq \frac{1}{8} \] Thus, the domain of \( g^{-1}(x) \) in interval notation is: \[ \text{Domain of } g^{-1}: (-\infty, \frac{1}{8}) \cup (\frac{1}{8}, \infty) \] ### Range of \( g^{-1}(x) \) The range of \( g^{-1}(x) \) corresponds to the domain of \( g(x) \). To find the domain of \( g(x) \), we need to check when the denominator is not zero: \[ 8x + 7 \neq 0 \implies x \neq -\frac{7}{8} \] Thus, the domain of \( g(x) \) is: \[ \text{Domain of } g: (-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, \infty) \] Since \( g \) is a one-to-one function, the range of \( g^{-1}(x) \) is the same as the domain of \( g(x) \): \[ \text{Range of } g^{-1}: (-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, \infty) \] ### Final Answer Thus, we have: \[ g^{-1}(x) = \frac{-5 - 7x}{8x - 1} \] \[ \text{Domain of } g^{-1}: (-\infty, \frac{1}{8}) \cup (\frac{1}{8}, \infty) \] \[ \text{Range of } g^{-1}: (-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, \infty) \]