1. **Problem statement:** Given functions $f(x) = \frac{x+1}{x-3}$ and $g(x) = x + 3$, find the composition $(f \circ g)(x)$ and determine its domain. 2. **Formula and rules:** The composition $(f \circ g)(x)$ means $f(g(x))$. 3. **Calculate $f(g(x))$:** Substitute $g(x)$ into $f$: $$ (f \circ g)(x) = f(g(x)) = f(x+3) = \frac{(x+3)+1}{(x+3)-3} = \frac{x+4}{x}. $$ 4. **Domain of $(f \circ g)(x)$:** The domain consists of all $x$ values for which the expression is defined. - The denominator cannot be zero: $$ x \neq 0. $$ - Also, consider the domain of $g(x)$ which is all real numbers. - The original $f(x)$ is undefined at $x=3$, but since $g(x) = x+3$, we check when $g(x) = 3$: $$ x + 3 = 3 \implies x = 0. $$ - At $x=0$, $(f \circ g)(x)$ is undefined due to division by zero. 5. **Final answer:** $$(f \circ g)(x) = \frac{x+4}{x}, \quad \text{with domain } \{x \in \mathbb{R} : x \neq 0\}.$