1. **State the problem:** Find the inverse of the function $$f(x) = x^2 - 8$$ with the domain restriction $$x \geq 0$$. 2. **Recall the definition of inverse function:** The inverse function $$f^{-1}(x)$$ reverses the roles of $$x$$ and $$y$$. If $$y = f(x)$$, then $$x = f^{-1}(y)$$. 3. **Write the function with $$y$$:** $$y = x^2 - 8$$. 4. **Swap $$x$$ and $$y$$ to find the inverse:** $$x = y^2 - 8$$. 5. **Solve for $$y$$:** $$x = y^2 - 8$$ Add 8 to both sides: $$x + 8 = y^2$$ 6. **Take the square root of both sides:** $$y = \pm \sqrt{x + 8}$$ 7. **Apply the domain restriction:** Since the original function has $$x \geq 0$$, the inverse must have $$y \geq 0$$ to be a function. Therefore, we take the positive root: $$f^{-1}(x) = \sqrt{x + 8}$$ 8. **State the domain of the inverse:** The expression under the square root must be non-negative: $$x + 8 \geq 0 \implies x \geq -8$$ So the domain of $$f^{-1}$$ is $$[-8, \infty)$$. **Final answer:** $$f^{-1}(x) = \sqrt{x + 8}, \quad x \geq -8$$