The one-to-one function \( g \) is defined below. \( g(x)=\frac{6 x}{4-5 x} \) Find \( g^{-1}(x) \), where \( g^{-1} \) is the inverse of \( g \). Also state the domain and range of \( g^{-1} \) in interval notation. \( g^{-1}(x)=\square \) Domain of \( g^{-1}: \square \)

To find the inverse of the function \( g(x) = \frac{6x}{4-5x} \), we start by replacing \( g(x) \) with \( y \): \[ y = \frac{6x}{4-5x} \] Next, we swap \( x \) and \( y \): \[ x = \frac{6y}{4-5y} \] Now, we solve for \( y \). Cross multiplying gives: \[ x(4 - 5y) = 6y \] Expanding this, we have: \[ 4x - 5xy = 6y \] Rearranging the equation, we get: \[ 4x = 6y + 5xy \] Factoring \( y \) from the right side, we have: \[ 4x = y(6 + 5x) \] Dividing both sides by \( 6 + 5x \) yields: \[ y = \frac{4x}{6 + 5x} \] Thus, the inverse function is: \[ g^{-1}(x) = \frac{4x}{6 + 5x} \] Now, we'll determine the domain and range of \( g^{-1} \). The domain of \( g^{-1}(x) \) is found by identifying values of \( x \) that would make the denominator zero: Set \( 6 + 5x = 0 \) which gives \( x = -\frac{6}{5} \). Therefore, the domain of \( g^{-1}(x) \) is: Domain of \( g^{-1}: \ (-\infty, -\frac{6}{5}) \cup (-\frac{6}{5}, \infty) \) And since \( y = g(x) \) approaches \( \frac{6}{5} \) (the horizontal asymptote) as \( x \) approaches infinity and can take any other value, the range of \( g^{-1} \) is: Range of \( g^{-1}: \ (-\infty, \frac{6}{5}) \cup (\frac{6}{5}, \infty) \) Putting it all together: \( g^{-1}(x) = \frac{4x}{6 + 5x} \) Domain of \( g^{-1}: \ (-\infty, -\frac{6}{5}) \cup (-\frac{6}{5}, \infty) \)