The one-to-one function \( f \) is defined below. \[ f(x)=\frac{-x+8}{3+2 x} \] Find \( f^{-1}(x) \), where \( f^{-1} \) is the inverse of \( f \). Also state the domain and range of \( f^{-1} \) in interval notation. Domain of \( f^{-1} \) Range of \( f^{-1} \)

To find the inverse function \( f^{-1}(x) \), we start by setting \( y = f(x) \): \[ y = \frac{-x + 8}{3 + 2x} \] Next, we switch \( x \) and \( y \): \[ x = \frac{-y + 8}{3 + 2y} \] Now, we solve for \( y \). First, multiply both sides by \( (3 + 2y) \): \[ x(3 + 2y) = -y + 8 \] Expanding and combining like terms gives us: \[ 3x + 2xy = -y + 8 \] Rearranging terms to isolate \( y \): \[ 2xy + y = 8 - 3x \] Factoring out \( y \): \[ y(2x + 1) = 8 - 3x \] Finally, we can solve for \( y \): \[ y = \frac{8 - 3x}{2x + 1} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{8 - 3x}{2x + 1} \] Now, let's determine the domain and range of \( f^{-1} \). Since \( f(x) \) is a one-to-one function and a rational function, its domain consists of all real numbers except where the denominator is zero. For \( f(x) \), the denominator \( 3 + 2x \) equals zero when \( x = -\frac{3}{2} \). Consequently, the range of \( f(x) \) excludes the corresponding value of \( f(-\frac{3}{2}) \), which can be calculated. Overall, the domain and range of \( f^{-1}(x) \) will be: Domain of \( f^{-1} \): \((- \infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)\) Range of \( f^{-1} \): \((- \infty, \frac{8}{3}) \cup (\frac{8}{3}, \infty)\)