The one-to-one function \( g \) is defined below. \( g(x)=\frac{7 x}{3-8 x} \) Find \( g^{-1}(x) \), where \( g^{-1} \) is the inverse of \( g \). Also state the domain and range of \( g^{-1} \) in interval notation \( g^{-1}(x)=\square \) Domain of \( g^{-1}: \)

To find the inverse of the function \( g(x) = \frac{7x}{3 - 8x} \), we need to switch the roles of \( x \) and \( y \) and solve for \( y \). Let \( y = g(x) \), so we have: \[ y = \frac{7x}{3 - 8x} \] Now, we solve for \( x \) in terms of \( y \): \[ y(3 - 8x) = 7x \] \[ 3y - 8xy = 7x \] \[ 3y = 7x + 8xy \] \[ 3y = x(7 + 8y) \] \[ x = \frac{3y}{7 + 8y} \] Now, we replace \( x \) with \( g^{-1}(x) \) and \( y \) with \( x \) to get the inverse function: \[ g^{-1}(x) = \frac{3x}{7 + 8x} \] The domain of \( g^{-1}(x) \) is the set of all \( x \) values for which \( g^{-1}(x) \) is defined. Since \( g(x) \) is a rational function, the domain of \( g^{-1}(x) \) is all real numbers except for the values of \( x \) that make the denominator zero. In this case, the denominator is \( 7 + 8x \), so we need to find the value of \( x \) that makes \( 7 + 8x = 0 \): \[ 7 + 8x = 0 \] \[ 8x = -7 \] \[ x = -\frac{7}{8} \] Therefore, the domain of \( g^{-1}(x) \) is all real numbers except \( -\frac{7}{8} \), which can be written in interval notation as: \[ (-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, \infty) \] The range of \( g^{-1}(x) \) is the set of all \( y \) values that \( g(x) \) can take. Since \( g(x) \) is a rational function, its range is all real numbers except for the value that makes the numerator zero. However, since the numerator is \( 7x \), it can take on any real value, and the denominator \( 3 - 8x \) is never zero for any real \( x \) (except at the point where \( x = -\frac{7}{8} \), which is not in the domain of \( g^{-1}(x) \)). Therefore, the range of \( g^{-1}(x) \) is all real numbers, which can be written in interval notation as: \[ (-\infty, \infty) \] So, the final answer is: \[ g^{-1}(x) = \frac{3x}{7 + 8x} \] Domain of \( g^{-1}: \) \( (-\infty, -\frac{7}{8}) \cup (-\frac{7}{8}, \infty) \)