1. **State the problem:** Find the domain and range of the composite function $f \circ g$, where $f(x) = \sqrt{x + 1}$ and $g(x) = \frac{1}{x - 3}$. 2. **Recall the definitions:** - The composite function $f \circ g$ is defined as $f(g(x))$. - The domain of $f \circ g$ consists of all $x$ values in the domain of $g$ such that $g(x)$ is in the domain of $f$. 3. **Find the domain of $g(x)$:** - $g(x) = \frac{1}{x - 3}$ is defined for all $x \neq 3$. - So, $D_g = \mathbb{R} \setminus \{3\}$. 4. **Find the domain of $f(x)$:** - $f(x) = \sqrt{x + 1}$ requires $x + 1 \geq 0$. - So, $D_f = [-1, \infty)$. 5. **Find the domain of $f \circ g$:** - We need $g(x) \in D_f$, i.e., $g(x) = \frac{1}{x - 3} \geq -1$. 6. **Solve the inequality:** $$\frac{1}{x - 3} \geq -1$$ Multiply both sides by $x - 3$, considering cases: - Case 1: $x - 3 > 0$ (i.e., $x > 3$): $$1 \geq -1(x - 3) \Rightarrow 1 \geq -x + 3 \Rightarrow x \geq 2$$ Since $x > 3$, this reduces to $x > 3$. - Case 2: $x - 3 < 0$ (i.e., $x < 3$): Multiply inequality by negative number reverses inequality: $$1 \leq -1(x - 3) \Rightarrow 1 \leq -x + 3 \Rightarrow x \leq 2$$ Since $x < 3$, this reduces to $x \leq 2$. 7. **Combine domain restrictions:** - From $g(x)$ domain: $x \neq 3$. - From inequality: $x \leq 2$ or $x > 3$. So, domain of $f \circ g$ is: $$(-\infty, 2] \cup (3, \infty)$$ 8. **Find the range of $f \circ g$:** - Since $f(x) = \sqrt{x + 1}$, range of $f$ is $[0, \infty)$. - The input to $f$ is $g(x) = \frac{1}{x - 3}$. - From step 5, $g(x) \geq -1$. - So, the range of $f \circ g$ is $[0, \infty)$. **Final answer:** - Domain: $(-\infty, 2] \cup (3, \infty)$ - Range: $[0, \infty)$