The one-to-one function \( f \) is defined below. \[ f(x)=\frac{5 x}{6 x-5} \] Find \( f^{-1}(x) \), where \( f^{-1} \) is the inverse of \( f \). Also state the domain and range of \( f^{-1} \) in interval notation. \[ f^{-1}(x)= \]

To find the inverse of a function, we need to switch the roles of \( x \) and \( y \) and then solve for the new \( y \). Given function: \( f(x) = \frac{5x}{6x-5} \) Let's denote \( y = f(x) \), so we have: \[ y = \frac{5x}{6x-5} \] Now, let's switch \( x \) and \( y \): \[ x = \frac{5y}{6y-5} \] Next, we'll solve this equation for \( y \) to find the inverse function \( f^{-1}(x) \). \[ x = \frac{5y}{6y-5} \] \[ x(6y-5) = 5y \] \[ 6xy - 5x = 5y \] \[ 6xy - 5y = 5x \] \[ y(6x-5) = 5x \] \[ y = \frac{5x}{6x-5} \] Therefore, the inverse function is: \[ f^{-1}(x) = \frac{5x}{6x-5} \] Now, let's determine the domain and range of \( f^{-1} \) in interval notation. The domain of \( f^{-1} \) is the set of all real numbers except for the values that make the denominator zero. In this case, the denominator \( 6x-5 \) cannot be zero. So, the domain of \( f^{-1} \) is: \[ \text{Domain of } f^{-1} = (-\infty, \frac{5}{6}) \cup (\frac{5}{6}, \infty) \] The range of \( f^{-1} \) is the set of all real numbers. So, the range of \( f^{-1} \) is: \[ \text{Range of } f^{-1} = (-\infty, \infty) \] Therefore, the inverse function \( f^{-1}(x) \) is: \[ f^{-1}(x) = \frac{5x}{6x-5} \] The domain of \( f^{-1} \) is \( (-\infty, \frac{5}{6}) \cup (\frac{5}{6}, \infty) \), and the range of \( f^{-1} \) is \( (-\infty, \infty) \).