1. **State the problem:** Find the inverse function $g^{-1}(x)$ of the one-to-one function $g(x) = \frac{4x}{5x - 9}$ and determine the domain and range of $g^{-1}$. 2. **Find the inverse function:** Start with the equation: $$y = \frac{4x}{5x - 9}$$ To find the inverse, swap $x$ and $y$: $$x = \frac{4y}{5y - 9}$$ 3. **Solve for $y$:** Multiply both sides by the denominator to clear the fraction: $$x(5y - 9) = 4y$$ Distribute $x$: $$5xy - 9x = 4y$$ Bring all terms involving $y$ to one side: $$5xy - 4y = 9x$$ Factor out $y$: $$y(5x - 4) = 9x$$ Divide both sides by $(5x - 4)$: $$y = \frac{9x}{5x - 4}$$ Show cancellation step: $$y = \frac{9x}{\cancel{5x - 4}}$$ 4. **Write the inverse function:** $$g^{-1}(x) = \frac{9x}{5x - 4}$$ 5. **Determine the domain of $g^{-1}$:** The denominator cannot be zero: $$5x - 4 \neq 0 \implies x \neq \frac{4}{5}$$ So the domain is: $$(-\infty, \frac{4}{5}) \cup (\frac{4}{5}, \infty)$$ 6. **Determine the range of $g^{-1}$:** The range of $g^{-1}$ is the domain of $g$. For $g(x) = \frac{4x}{5x - 9}$, the denominator cannot be zero: $$5x - 9 \neq 0 \implies x \neq \frac{9}{5}$$ So the domain of $g$ (and range of $g^{-1}$) is: $$(-\infty, \frac{9}{5}) \cup (\frac{9}{5}, \infty)$$ --- **Final answers:** $$g^{-1}(x) = \frac{9x}{5x - 4}$$ Domain of $g^{-1}$: $(-\infty, \frac{4}{5}) \cup (\frac{4}{5}, \infty)$ Range of $g^{-1}$: $(-\infty, \frac{9}{5}) \cup (\frac{9}{5}, \infty)$