1. The problem is to find the domain and range of the inverse function of $$s(x) = \frac{6x - 1}{x + 5}$$. 2. To find the inverse, swap $$x$$ and $$y$$ and solve for $$y$$: $$x = \frac{6y - 1}{y + 5}$$ 3. Multiply both sides by $$y + 5$$: $$x(y + 5) = 6y - 1$$ 4. Distribute $$x$$: $$xy + 5x = 6y - 1$$ 5. Rearrange terms to isolate $$y$$: $$xy - 6y = -1 - 5x$$ 6. Factor out $$y$$: $$y(x - 6) = -1 - 5x$$ 7. Solve for $$y$$: $$y = \frac{-1 - 5x}{x - 6}$$ 8. The inverse function is: $$s^{-1}(x) = \frac{-1 - 5x}{x - 6}$$ 9. The domain of the inverse function is all real numbers except where the denominator is zero: $$x - 6 \neq 0 \Rightarrow x \neq 6$$ 10. The range of the inverse function is the domain of the original function, which is all real numbers except where the denominator of $$s(x)$$ is zero: $$x + 5 \neq 0 \Rightarrow x \neq -5$$ 11. Therefore, the domain of $$s^{-1}(x)$$ is $$\{x \in \mathbb{R} \mid x \neq 6\}$$ and the range is $$\{y \in \mathbb{R} \mid y \neq -5\}$$. 12. Comparing with the options, the correct answer is option a.