The one-to-one function \( h \) is defined below. \[ h(x)=\frac{2 x}{7 x-6} \] Find \( h^{-1}(x) \), where \( h^{-1} \) is the inverse of \( h \). Also state the domain and range of \( h^{-1} \) in interval notation. Domain of \( h^{-1}(x)=\square \) Range of \( h^{-1}= \)

To find the inverse of the function \( h(x) = \frac{2x}{7x - 6} \), we will follow these steps: 1. **Replace \( h(x) \) with \( y \)**: \[ y = \frac{2x}{7x - 6} \] 2. **Swap \( x \) and \( y \)** to find the inverse: \[ x = \frac{2y}{7y - 6} \] 3. **Solve for \( y \)**: - Multiply both sides by \( 7y - 6 \): \[ x(7y - 6) = 2y \] - Distribute \( x \): \[ 7xy - 6x = 2y \] - Rearrange the equation to isolate \( y \): \[ 7xy - 2y = 6x \] - Factor out \( y \): \[ y(7x - 2) = 6x \] - Divide by \( 7x - 2 \): \[ y = \frac{6x}{7x - 2} \] 4. **Thus, the inverse function is**: \[ h^{-1}(x) = \frac{6x}{7x - 2} \] Next, we will determine the domain and range of \( h^{-1}(x) \). ### Domain of \( h^{-1}(x) \): The domain of \( h^{-1}(x) \) is determined by the values of \( x \) for which the function is defined. The function \( h^{-1}(x) = \frac{6x}{7x - 2} \) is undefined when the denominator is zero: \[ 7x - 2 = 0 \implies x = \frac{2}{7} \] Thus, the domain of \( h^{-1}(x) \) in interval notation is: \[ (-\infty, \frac{2}{7}) \cup (\frac{2}{7}, \infty) \] ### Range of \( h^{-1}(x) \): The range of \( h^{-1}(x) \) corresponds to the domain of the original function \( h(x) \). To find the domain of \( h(x) \), we set the denominator to zero: \[ 7x - 6 = 0 \implies x = \frac{6}{7} \] Thus, the domain of \( h(x) \) is: \[ (-\infty, \frac{6}{7}) \cup (\frac{6}{7}, \infty) \] Therefore, the range of \( h^{-1}(x) \) is: \[ (-\infty, \frac{6}{7}) \cup (\frac{6}{7}, \infty) \] ### Final Answer: - \( h^{-1}(x) = \frac{6x}{7x - 2} \) - Domain of \( h^{-1}(x) = (-\infty, \frac{2}{7}) \cup (\frac{2}{7}, \infty) \) - Range of \( h^{-1}(x) = (-\infty, \frac{6}{7}) \cup (\frac{6}{7}, \infty) \)