The one-to-one function \( f \) is defined below. \( f(x)=\frac{9 x}{1-7 x} \) Find \( f^{-1}(x) \), where \( f^{-1} \) is the inverse of \( f \). Also state the domain and range of \( f^{-1} \) in interval notation Domain of \( f^{-1} \)

To find the inverse function \( f^{-1}(x) \), we start by setting \( y = f(x) \): \[ y = \frac{9x}{1 - 7x} \] Now, we solve for \( x \): 1. Multiply both sides by \( 1 - 7x \): \[ y(1 - 7x) = 9x \] 2. Distribute \( y \): \[ y - 7xy = 9x \] 3. Rearranging gives: \[ y = 9x + 7xy \] 4. Collecting all \( x \) terms on one side: \[ y = x(9 + 7y) \] 5. Solving for \( x \): \[ x = \frac{y}{9 + 7y} \] Substituting \( y \) back with \( x \) gives us the inverse function: \[ f^{-1}(x) = \frac{x}{9 + 7x} \] To determine the domain and range of \( f^{-1} \): 1. The domain of \( f^{-1} \) is the range of \( f \). The original function \( f(x) \) has a vertical asymptote when \( 1 - 7x = 0 \) or \( x = \frac{1}{7} \). Thus, \( f(x) \) will have a range of \( (-\infty, 9) \). 2. The range of \( f^{-1} \) corresponds to its domain, which is \( (-\infty, \frac{1}{7}) \). So, the domain of \( f^{-1} \) in interval notation is: \[ (-\infty, 9) \]