The one-to-one function \( g \) is defined below. \( g(x)=\frac{4 x}{5 x-9} \) Find \( g^{-1}(x) \), where \( g^{-1} \) is the inverse of \( g \). Also state the domain and range of \( g^{-1} \) in interval notation. Domain of \( g^{-1} \) Range of \( g^{-1}(x)= \)

To find the inverse function \( g^{-1}(x) \) of \( g(x) = \frac{4x}{5x - 9} \), we first set \( g(x) = y \): \[ y = \frac{4x}{5x - 9} \] Next, we swap \( x \) and \( y \) to find the inverse: \[ x = \frac{4y}{5y - 9} \] Now we solve for \( y \): 1. Multiply both sides by \( 5y - 9 \): \[ x(5y - 9) = 4y \] 2. Expand: \[ 5xy - 9x = 4y \] 3. Rearrange to isolate \( y \): \[ 5xy - 4y = 9x \] 4. Factor out \( y \): \[ y(5x - 4) = 9x \] 5. Solve for \( y \): \[ y = \frac{9x}{5x - 4} \] Thus, the inverse function is: \[ g^{-1}(x) = \frac{9x}{5x - 4} \] Now, to find the domain and range of \( g^{-1} \): - The domain of \( g^{-1} \) is the set of all possible outputs from \( g \). Since \( g \) is a rational function and is one-to-one, its range can be calculated. In \( g(x) \), the term \( 5x - 9 \) cannot equal zero, hence \( 5x \neq 9 \) or \( x \neq \frac{9}{5} \). The output \( g(x) \) approaches but never reaches \( \frac{4}{5} \) as \( x \to +\infty \) or \( x \to -\infty \). Thus: Domain of \( g^{-1} = (-\infty, \frac{4}{5}) \cup (\frac{4}{5}, +\infty) \] - The range of \( g^{-1} \) corresponds to the original function's domain, which must also exclude the point that makes \( 5x - 9 = 0 \). Hence: Range of \( g^{-1}(x) = (-\infty, \frac{9}{5}) \cup (\frac{9}{5}, +\infty) \]