1. **State the problem:** Find the inverse function $f^{-1}(x)$ of the function $$f(x) = \frac{2 - 3x}{x + 1}.$$ 2. **Recall the formula and method for finding inverses:** To find the inverse function, we swap $x$ and $y$ in the equation $y = f(x)$ and then solve for $y$. The inverse function $f^{-1}(x)$ satisfies $$x = \frac{2 - 3y}{y + 1}.$$ 3. **Start with the equation:** $$x = \frac{2 - 3y}{y + 1}.$$ Multiply both sides by $y + 1$ to clear the denominator: $$x(y + 1) = 2 - 3y.$$ 4. **Distribute $x$ on the left:** $$xy + x = 2 - 3y.$$ 5. **Group terms with $y$ on one side:** $$xy + 3y = 2 - x.$$ 6. **Factor out $y$:** $$y(x + 3) = 2 - x.$$ 7. **Solve for $y$:** $$y = \frac{2 - x}{x + 3}.$$ 8. **Conclusion:** The inverse function is $$f^{-1}(x) = \frac{2 - x}{x + 3}.$$ 9. **Domain of $f^{-1}$:** The domain is all real numbers except where the denominator is zero, so $$x + 3 \neq 0 \implies x \neq -3.$$ 10. **Range of $f^{-1}$:** The range of $f^{-1}$ is the domain of $f$, which is all real numbers except where the denominator of $f$ is zero, i.e., $$x + 1 \neq 0 \implies x \neq -1.$$ **Final answers:** - Inverse function: $$f^{-1}(x) = \frac{2 - x}{x + 3}.$$ - Domain of $f^{-1}$: $$(-\infty, -3) \cup (-3, \infty).$$ - Range of $f^{-1}$: $$(-\infty, -1) \cup (-1, \infty).$$