The one-to-one function \( h \) is defined below. \( h(x)=\frac{2 x}{3 x-7} \) Find \( h^{-1}(x) \), where \( h^{-1} \) is the inverse of \( h \). Also state the domain and range of \( h^{-1} \) in interval notation. \( h^{-1}(x)=\square \) Domain of \( h^{-1}: \square \)

To find the inverse of the function \( h(x) = \frac{2x}{3x - 7} \), we need to switch the roles of \( x \) and \( y \) and solve for \( y \). Let \( y = h(x) \), so we have: \[ y = \frac{2x}{3x - 7} \] Now, we solve for \( x \) in terms of \( y \): \[ y(3x - 7) = 2x \] \[ 3xy - 7y = 2x \] \[ 3xy - 2x = 7y \] \[ x(3y - 2) = 7y \] \[ x = \frac{7y}{3y - 2} \] Now we have \( x \) in terms of \( y \), which means we have found the inverse function \( h^{-1}(y) \): \[ h^{-1}(y) = \frac{7y}{3y - 2} \] To express this in terms of \( x \), we replace \( y \) with \( x \): \[ h^{-1}(x) = \frac{7x}{3x - 2} \] The domain of \( h^{-1}(x) \) is the set of all \( x \) values for which \( h^{-1}(x) \) is defined. Since the denominator cannot be zero, we must exclude the value that makes \( 3x - 2 = 0 \): \[ 3x - 2 = 0 \] \[ 3x = 2 \] \[ x = \frac{2}{3} \] So, the domain of \( h^{-1}(x) \) is all real numbers except \( x = \frac{2}{3} \): \[ \text{Domain of } h^{-1}: (-\infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty) \] The range of \( h^{-1}(x) \) is the set of all \( y \) values that \( h(x) \) can take. Since \( h(x) \) is a rational function, its range is all real numbers except for the value that makes the denominator zero. We already found that this value is \( x = \frac{2}{3} \), so the range of \( h^{-1}(x) \) is: \[ \text{Range of } h^{-1}: (-\infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty) \] However, since we are looking for the range of \( h^{-1}(x) \) and not \( h(x) \), we need to consider the range of \( h(x) \) first. The range of \( h(x) \) is all real numbers except for the value that makes the denominator zero, which is \( x = \frac{2}{3} \). Therefore, the range of \( h^{-1}(x) \) is the same as the range of \( h(x) \), which is: \[ \text{Range of } h^{-1}: (-\infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty) \] In summary: \[ h^{-1}(x) = \frac{7x}{3x - 2} \] Domain of \( h^{-1}: (-\infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty) \] Range of \( h^{-1}: (-\infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty) \]