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

Let \( y = g(x) \), so \[ y = \frac{5x}{9x-5}. \] **Step 1.** Multiply both sides by the denominator to eliminate the fraction: \[ y(9x-5) = 5x. \] **Step 2.** Distribute \( y \) on the left-hand side: \[ 9xy - 5y = 5x. \] **Step 3.** Collect all \( x \)-terms on one side: \[ 9xy - 5x = 5y. \] **Step 4.** Factor out \( x \) on the left-hand side: \[ x(9y-5) = 5y. \] **Step 5.** Solve for \( x \): \[ x = \frac{5y}{9y-5}. \] Since \( x = g^{-1}(y) \), replacing \( y \) by \( x \) we obtain the inverse function: \[ g^{-1}(x) = \frac{5x}{9x-5}. \] **Step 6.** Determine the domain and range of \( g^{-1} \). - The domain of \( g^{-1} \) is given by the range of \( g \). For the original function, - The domain of \( g(x) \) is all \( x \) such that \( 9x-5 \neq 0 \), i.e., \( x \neq \frac{5}{9} \). - To find the range of \( g \), note that if we solve \( y = \frac{5x}{9x-5} \) for \( x \), the result imposes a restriction \( 9y-5 \neq 0 \) (since dividing by \( 9y-5 \) is required). This gives \( y \neq \frac{5}{9} \). Therefore, the range of \( g \) is \[ (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty). \] This is the domain of \( g^{-1} \). - The range of \( g^{-1} \) is the domain of \( g \), which is also \[ (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty). \] **Final Answer:** \[ g^{-1}(x) = \frac{5x}{9x-5} \] Domain of \( g^{-1} \): \( (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty) \) Range of \( g^{-1} \): \( (-\infty, \frac{5}{9}) \cup (\frac{5}{9}, \infty) \)