1. **Find the domain and range of the function** $h(x) = \sqrt{x^2 - 4}$. Step 1: The expression under the square root, called the radicand, must be non-negative for $h(x)$ to be real. $$x^2 - 4 \geq 0$$ Step 2: Solve the inequality. $$x^2 \geq 4$$ This means: $$x \leq -2 \quad \text{or} \quad x \geq 2$$ So, the domain of $h$ is: $$(-\infty, -2] \cup [2, \infty)$$ Step 3: Find the range. Since $h(x) = \sqrt{x^2 - 4} \geq 0$ for all $x$ in the domain. Step 4: Find minimum value. At $x=\pm 2$, $$h(2) = \sqrt{2^2 - 4} = \sqrt{0} = 0$$ As $|x|$ increases, $x^2 - 4$ increases, so $h(x)$ becomes larger without bound. Hence range is: $$[0, \infty)$$ 2. **Determine $(g \circ h)(x)$ given:** $$g(x) = x^2 + 2, \quad h(x) = \sqrt{x^2 - 4}$$ Step 1: Compose the functions: $$(g \circ h)(x) = g(h(x)) = \left(\sqrt{x^2 - 4}\right)^2 + 2$$ Step 2: Simplify: $$ (g \circ h)(x) = x^2 - 4 + 2 = x^2 - 2$$ Step 3: Remember that the domain of $(g \circ h)(x)$ comes from the domain of $h(x)$ since $h(x)$ must be defined. So, domain is: $$(-\infty, -2] \cup [2, \infty)$$ **Final answers:** - Domain of $h(x)$: $(-\infty, -2] \cup [2, \infty)$ - Range of $h(x)$: $[0, \infty)$ - $(g \circ h)(x) = x^2 - 2$ with domain $(-\infty, -2] \cup [2, \infty)$