The one-to-one function \( f \) is defined below. \( f(x)=\frac{-5 x+2}{5+6 x} \) 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 \( f^{-1}(x)= \) Range of \( f \)

To find the inverse function \( f^{-1}(x) \), we start by setting \( y = f(x) \): \[ y = \frac{-5x + 2}{5 + 6x} \] Next, we swap \( x \) and \( y \) and solve for \( y \): \[ x = \frac{-5y + 2}{5 + 6y} \] Cross-multiplying gives: \[ x(5 + 6y) = -5y + 2 \] Expanding this, we find: \[ 5x + 6xy = -5y + 2 \] Rearranging terms: \[ 5x - 2 = -5y - 6xy \] Factoring out \( y \): \[ 5x - 2 = y(-5 - 6x) \] Now, solving for \( y \): \[ y = \frac{5x - 2}{-5 - 6x} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{5x - 2}{-5 - 6x} \] To find the domain of \( f^{-1}(x) \), we need to determine when the denominator is not zero: \[ -5 - 6x \neq 0 \implies x \neq -\frac{5}{6} \] So, the domain of \( f^{-1}(x) \) is: \[ \text{Domain of } f^{-1}(x) = (-\infty, -\frac{5}{6}) \cup (-\frac{5}{6}, \infty) \] The range of \( f \) is the same as the domain of \( f^{-1}(x) \): \[ \text{Range of } f = (-\infty, -\frac{5}{6}) \cup (-\frac{5}{6}, \infty) \]