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 \( h^{-1}(x) \) of the function \( h(x)=\frac{2 x}{7 x-6} \), we first set \( y = h(x) \): \[ y = \frac{2 x}{7 x - 6} \] Next, we solve for \( x \) in terms of \( y \): \[ y(7x - 6) = 2x \\ 7xy - 6y = 2x \\ 7xy - 2x = 6y \\ x(7y - 2) = 6y \\ x = \frac{6y}{7y - 2} \] Thus, the inverse function is: \[ h^{-1}(x) = \frac{6x}{7x - 2} \] Now, to find the domain and range of \( h^{-1}(x) \), we examine \( h(x) \): 1. The original function \( h(x) \) has a vertical asymptote when the denominator \( 7x - 6 = 0 \), which gives \( x = \frac{6}{7} \). Thus, the range of \( h(x) \) excludes the value \( \frac{2}{7} \). 2. Therefore, since the range of \( h(x) \) is \( \mathbb{R} \setminus \{\frac{2}{7}\} \), the domain of \( h^{-1}(x) \) is also \( \mathbb{R} \setminus \{\frac{2}{7}\} \). Domain of \( h^{-1}(x) = (-\infty, \frac{2}{7}) \cup (\frac{2}{7}, \infty) \) Range of \( h^{-1}(x) = (-\infty, \frac{6}{7}) \cup (\frac{6}{7}, \infty) \)