1. The problem is to find the inverse function $f^{-1}(x)$ of the function $f(x) = x^2 - 8$ with the domain restriction $x \geq 0$. 2. To find the inverse, start by setting $y = f(x) = x^2 - 8$. 3. Swap $x$ and $y$ to get $x = y^2 - 8$. 4. Solve for $y$: $$x = y^2 - 8 \implies y^2 = x + 8 \implies y = \pm \sqrt{x + 8}$$ 5. Since the original function has domain $x \geq 0$, the inverse must be the positive root: $$f^{-1}(x) = \sqrt{x + 8}$$ 6. The other options given, such as $\sqrt{x - 8} + 3$, $\sqrt{x + 8} + 6$, and $\sqrt{x - 8} - 3$, do not match the inverse derived from the function and domain. 7. Therefore, the correct inverse function is: $$f^{-1}(x) = \sqrt{x + 8}$$