1. **State the problem:** Given the function $f(x) = \frac{x}{x+5}$, find its inverse function $f^{-1}(x)$ and verify that $f^{-1}(f(x)) = f(f^{-1}(x)) = x$. 2. **Find the inverse function:** Start by setting $y = f(x) = \frac{x}{x+5}$. 3. Swap $x$ and $y$ to find the inverse: $x = \frac{y}{y+5}$. 4. Solve for $y$ in terms of $x$: $$x = \frac{y}{y+5}$$ Multiply both sides by $y+5$: $$x(y+5) = y$$ $$xy + 5x = y$$ 5. Rearrange to isolate $y$: $$xy - y = -5x$$ $$y(x - 1) = -5x$$ 6. Divide both sides by $(x-1)$: $$y = \frac{-5x}{x - 1}$$ Show the cancellation step: $$y = \frac{-5x}{\cancel{x - 1}}$$ 7. So the inverse function is: $$f^{-1}(x) = \frac{-5x}{x - 1}$$ 8. **Verify $f^{-1}(f(x)) = x$:** Substitute $f(x)$ into $f^{-1}$: $$f^{-1}(f(x)) = f^{-1}\left(\frac{x}{x+5}\right) = \frac{-5 \cdot \frac{x}{x+5}}{\frac{x}{x+5} - 1}$$ Simplify denominator: $$\frac{x}{x+5} - 1 = \frac{x}{x+5} - \frac{x+5}{x+5} = \frac{x - (x+5)}{x+5} = \frac{-5}{x+5}$$ 9. Substitute back: $$f^{-1}(f(x)) = \frac{-5 \cdot \frac{x}{x+5}}{\frac{-5}{x+5}} = \frac{-5x/(x+5)}{-5/(x+5)}$$ 10. Simplify the complex fraction: $$= \frac{-5x}{x+5} \times \frac{x+5}{-5} = x$$ 11. **Verify $f(f^{-1}(x)) = x$:** Substitute $f^{-1}(x)$ into $f$: $$f(f^{-1}(x)) = f\left(\frac{-5x}{x-1}\right) = \frac{\frac{-5x}{x-1}}{\frac{-5x}{x-1} + 5}$$ 12. Simplify denominator: $$\frac{-5x}{x-1} + 5 = \frac{-5x}{x-1} + \frac{5(x-1)}{x-1} = \frac{-5x + 5(x-1)}{x-1} = \frac{-5x + 5x - 5}{x-1} = \frac{-5}{x-1}$$ 13. Substitute back: $$f(f^{-1}(x)) = \frac{-5x/(x-1)}{-5/(x-1)} = \frac{-5x}{x-1} \times \frac{x-1}{-5} = x$$ 14. **Conclusion:** The inverse function is $f^{-1}(x) = \frac{-5x}{x-1}$ and it satisfies $f^{-1}(f(x)) = f(f^{-1}(x)) = x$. 15. **Note on domain:** $f(x)$ is undefined at $x = -5$ (division by zero), and $f^{-1}(x)$ is undefined at $x = 1$. 16. **Table values check:** - For $x=0$, $f(0) = 0$. - For $x=10$, $f(10) = \frac{10}{15} = \frac{2}{3}$. - At $x=-5$, $f(x)$ is undefined. These match the given table.