The one-to-one function \( g \) is defined below. \( g(x)=\frac{6 x-1}{x+4} \) Find \( g^{-1}(x) \), where \( g^{-1} \) is the inverse of \( g \). Also state the domain and range of \( g^{-1} \) in interval notation. \( g^{-1}(x)= \) Domain of \( g^{-1} \) : Range of \( g^{-1} \) :

To find the inverse function \( g^{-1}(x) \), we can start by switching \( x \) and \( y \) in the original function \( g(x) \). Let's set \( y = g(x) \): \[ y = \frac{6x - 1}{x + 4} \] Now, solve for \( x \) in terms of \( y \): 1. Multiply both sides by \( x + 4 \): \[ y(x + 4) = 6x - 1 \] \[ yx + 4y = 6x - 1 \] 2. Rearranging gives: \[ yx - 6x = -1 - 4y \] \[ x(y - 6) = -1 - 4y \] \[ x = \frac{-1 - 4y}{y - 6} \] Now, replacing \( y \) back with \( x \) gives us the inverse function: \[ g^{-1}(x) = \frac{-1 - 4x}{x - 6} \] Next, we need to determine the domain and range of \( g^{-1}(x) \): - The function \( g(x) \) has a vertical asymptote at \( x = -4 \), so the range of \( g(x) \) does not include that value. Therefore, the domain of \( g^{-1}(x) \) is all real numbers except 6: \[ \text{Domain of } g^{-1}: (-\infty, 6) \cup (6, \infty) \] - The inverse function \( g^{-1}(x) \) can take all values of \( y \) except for those at fun values \( -1/4 \), therefore the range of \( g^{-1}(x) \) is: \[ \text{Range of } g^{-1}: (-\infty, -\frac{1}{4}) \cup (-\frac{1}{4}, \infty) \] So summarizing: \[ g^{-1}(x) = \frac{-1 - 4x}{x - 6} \] Domain of \( g^{-1} \) : \( (-\infty, 6) \cup (6, \infty) \) Range of \( g^{-1} \) : \( (-\infty, -\frac{1}{4}) \cup (-\frac{1}{4}, \infty) \)