1. **State the problem:** We have two functions: $f(x) = \sqrt{x + 3}$ and $g(x) = \frac{1}{x}$. We need to find the composition $g(f(x))$ and determine its domain. 2. **Find the composition $g(f(x))$:** By definition, $g(f(x)) = g\bigl(\sqrt{x + 3}\bigr) = \frac{1}{\sqrt{x + 3}}$. 3. **Determine the domain of $g(f(x))$:** - First, the expression inside the square root must be non-negative: $$x + 3 \geq 0 \implies x \geq -3.$$ - Second, the denominator cannot be zero: $$\sqrt{x + 3} \neq 0 \implies x + 3 \neq 0 \implies x \neq -3.$$ 4. **Combine domain restrictions:** - From the square root: $x \geq -3$ - From the denominator: $x \neq -3$ Therefore, the domain is $$x > -3.$$ 5. **Final answer:** $$g(f(x)) = \frac{1}{\sqrt{x + 3}}$$ with domain $$\boxed{x > -3}.$$