1. **State the problem:** We are given two functions: $$f(x) = 3x - 30$$ $$g(x) = x^2 - x - 90$$ We need to find the function $\frac{f}{g}(x)$ and determine its domain. 2. **Formula used:** The division of two functions is defined as: $$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$$ The domain of $\frac{f}{g}(x)$ includes all values of $x$ where $g(x) \neq 0$. 3. **Find $\frac{f}{g}(x)$:** $$\frac{f}{g}(x) = \frac{3x - 30}{x^2 - x - 90}$$ 4. **Factor numerator and denominator:** Numerator: $$3x - 30 = 3(x - 10)$$ Denominator: Factor $x^2 - x - 90$: find two numbers that multiply to $-90$ and add to $-1$. These are $-10$ and $9$, so: $$x^2 - x - 90 = (x - 10)(x + 9)$$ 5. **Simplify the fraction:** $$\frac{3(x - 10)}{(x - 10)(x + 9)}$$ Cancel the common factor $x - 10$ (noting $x \neq 10$ to avoid division by zero): $$\frac{3\cancel{(x - 10)}}{\cancel{(x - 10)}(x + 9)} = \frac{3}{x + 9}$$ 6. **Domain:** The original denominator $g(x) = (x - 10)(x + 9)$ cannot be zero, so: $$x - 10 \neq 0 \Rightarrow x \neq 10$$ $$x + 9 \neq 0 \Rightarrow x \neq -9$$ Therefore, the domain of $\frac{f}{g}(x)$ is all real numbers except $x = 10$ and $x = -9$. **Final answers:** $$\left(\frac{f}{g}\right)(x) = \frac{3}{x + 9}$$ Domain: $x \neq 10, -9$