1. **State the problem:** Find the inverse function $h^{-1}(x)$ of the one-to-one function $h(x) = \frac{5x - 3}{6x + 1}$ and determine the domain and range of $h^{-1}$. 2. **Formula and rules:** To find the inverse, swap $x$ and $y$ in the equation $y = h(x)$ and solve for $y$. 3. **Set up the equation:** Let $y = \frac{5x - 3}{6x + 1}$. To find $h^{-1}(x)$, swap $x$ and $y$: $$x = \frac{5y - 3}{6y + 1}$$ 4. **Solve for $y$:** Multiply both sides by $6y + 1$: $$x(6y + 1) = 5y - 3$$ $$6xy + x = 5y - 3$$ 5. **Group $y$ terms on one side:** $$6xy - 5y = -x - 3$$ 6. **Factor out $y$:** $$y(6x - 5) = -x - 3$$ 7. **Divide both sides by $(6x - 5)$:** $$y = \frac{-x - 3}{6x - 5}$$ Intermediate step showing cancellation: $$y = \frac{\cancel{-1}(x + 3)}{6x - 5} = \frac{-(x + 3)}{6x - 5}$$ 8. **Final inverse function:** $$h^{-1}(x) = \frac{-(x + 3)}{6x - 5} = \frac{-x - 3}{6x - 5}$$ 9. **Domain of $h^{-1}$:** The denominator cannot be zero: $$6x - 5 \neq 0 \implies x \neq \frac{5}{6}$$ Domain is all real numbers except $\frac{5}{6}$: $$(-\infty, \frac{5}{6}) \cup (\frac{5}{6}, \infty)$$ 10. **Range of $h^{-1}$:** The range of $h^{-1}$ is the domain of $h$. Domain of $h$ is where denominator $6x + 1 \neq 0 \implies x \neq -\frac{1}{6}$. So range of $h^{-1}$ is: $$(-\infty, -\frac{1}{6}) \cup (-\frac{1}{6}, \infty)$$