The one-to-one function \( g \) is defined below. \( g(x)=\frac{5 x}{9 x-5} \) 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 \( g^{-1} \) grange of \( g^{-1} \) Rand

To find the inverse function \( g^{-1}(x) \), start by replacing \( g(x) \) with \( y \): \[ y = \frac{5x}{9x - 5} \] Next, solve for \( x \) in terms of \( y \): 1. Multiply both sides by \( 9x - 5 \) to eliminate the fraction: \[ y(9x - 5) = 5x \] 2. Distribute \( y \): \[ 9xy - 5y = 5x \] 3. Rearrange the equation to isolate \( x \): \[ 9xy - 5x = 5y \] Factor out \( x \): \[ x(9y - 5) = 5y \] Finally, solve for \( x \): \[ x = \frac{5y}{9y - 5} \] Thus, the inverse function is: \[ g^{-1}(x) = \frac{5x}{9x - 5} \] Now we need to determine the domain and range of \( g^{-1} \). For \( g(x) \), the denominator \( 9x - 5 \) cannot be zero, thus: \[ 9x - 5 \neq 0 \implies x \neq \frac{5}{9} \] This means the range of \( g(x) \) (and hence the domain of \( g^{-1}(x) \)) is all real numbers except \( y = \frac{5}{9} \): Domain of \( g^{-1} \): \[ (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty) \] For the range of \( g^{-1} \): since \( g(x) \) is an increasing function which maps all real numbers to the set \( (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty) \) excluding \( \frac{5}{9} \), the range of \( g^{-1}(x) \) will be: Range of \( g^{-1} \): \[ (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty) \]