1. **State the problem:** We are given two functions: $$f(x) = \frac{x}{x - 9}$$ $$g(x) = \frac{x + 7}{x - 9}$$ We need to find the function $$\left(\frac{f}{g}\right)(x)$$ and determine its domain. 2. **Formula and rules:** The division of two functions $$\frac{f}{g}$$ is defined as: $$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$$ Important: The domain excludes values where the denominator is zero. 3. **Find $$\left(\frac{f}{g}\right)(x)$$:** $$\left(\frac{f}{g}\right)(x) = \frac{\frac{x}{x - 9}}{\frac{x + 7}{x - 9}}$$ 4. **Simplify the complex fraction:** $$= \frac{x}{x - 9} \times \frac{x - 9}{x + 7}$$ Since $$x - 9 \neq 0$$, we can cancel: $$= \frac{x}{x + 7}$$ 5. **Determine the domain:** - From $$f(x)$$ and $$g(x)$$, $$x - 9 \neq 0 \Rightarrow x \neq 9$$ - From the simplified function $$\frac{x}{x + 7}$$, denominator $$x + 7 \neq 0 \Rightarrow x \neq -7$$ 6. **Final domain:** All real numbers except $$x = 9$$ and $$x = -7$$. In interval notation: $$(-\infty, -7) \cup (-7, 9) \cup (9, \infty)$$ **Answer:** $$\left(\frac{f}{g}\right)(x) = \frac{x}{x + 7}$$ Domain: $$(-\infty, -7) \cup (-7, 9) \cup (9, \infty)$$