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

To find the inverse function \( f^{-1}(x) \) of the function \( f(x) = \frac{-5x + 7}{8 + x} \), we will follow these steps: 1. **Replace \( f(x) \) with \( y \)**: \[ y = \frac{-5x + 7}{8 + x} \] 2. **Swap \( x \) and \( y \)**: \[ x = \frac{-5y + 7}{8 + y} \] 3. **Solve for \( y \)**: - Multiply both sides by \( 8 + y \): \[ x(8 + y) = -5y + 7 \] - Distribute \( x \): \[ 8x + xy = -5y + 7 \] - Rearrange to isolate terms involving \( y \): \[ xy + 5y = 7 - 8x \] - Factor out \( y \): \[ y(x + 5) = 7 - 8x \] - Solve for \( y \): \[ y = \frac{7 - 8x}{x + 5} \] 4. **Thus, the inverse function is**: \[ f^{-1}(x) = \frac{7 - 8x}{x + 5} \] Next, we need to determine the domain and range of \( f^{-1}(x) \). ### Domain of \( f^{-1}(x) \): The domain of \( f^{-1}(x) \) is determined by the values of \( x \) for which the function is defined. The function \( f^{-1}(x) = \frac{7 - 8x}{x + 5} \) is undefined when the denominator is zero: \[ x + 5 = 0 \implies x = -5 \] Thus, the domain of \( f^{-1}(x) \) is all real numbers except \( -5 \): \[ \text{Domain of } f^{-1}: (-\infty, -5) \cup (-5, \infty) \] ### Range of \( f^{-1}(x) \): The range of \( f^{-1}(x) \) corresponds to the domain of \( f(x) \). To find the domain of \( f(x) \), we note that \( f(x) \) is defined for all \( x \) except when the denominator is zero: \[ 8 + x = 0 \implies x = -8 \] Thus, the domain of \( f(x) \) is all real numbers except \( -8 \): \[ \text{Domain of } f: (-\infty, -8) \cup (-8, \infty) \] Since \( f \) is a one-to-one function, the range of \( f \) is the same as the domain of \( f^{-1} \). Therefore, the range of \( f^{-1}(x) \) is: \[ \text{Range of } f^{-1}: (-\infty, -8) \cup (-8, \infty) \] ### Final Answers: \[ f^{-1}(x) = \frac{7 - 8x}{x + 5} \] \[ \text{Domain of } f^{-1}: (-\infty, -5) \cup (-5, \infty) \]