1. **State the problem:** Find the inverse of the function $$f(x) = (x - 3)^2 + 3$$ for $$x \geq 3$$ and verify the result. 2. **Recall the definition of inverse function:** The inverse function $$f^{-1}(y)$$ satisfies $$f(f^{-1}(y)) = y$$ and $$f^{-1}(f(x)) = x$$. 3. **Find the inverse:** Start with $$y = (x - 3)^2 + 3$$. 4. Solve for $$x$$ in terms of $$y$$: $$y - 3 = (x - 3)^2$$ Take the square root on both sides: $$x - 3 = \pm \sqrt{y - 3}$$ 5. Since the domain is $$x \geq 3$$, the function is increasing and we take the positive root: $$x = 3 + \sqrt{y - 3}$$ 6. **Write the inverse function:** $$f^{-1}(y) = 3 + \sqrt{y - 3}$$ 7. **Check the inverse:** - Compute $$f(f^{-1}(y))$$: $$f(3 + \sqrt{y - 3}) = ((3 + \sqrt{y - 3}) - 3)^2 + 3 = (\sqrt{y - 3})^2 + 3 = y - 3 + 3 = y$$ - Compute $$f^{-1}(f(x))$$ for $$x \geq 3$$: $$f^{-1}((x - 3)^2 + 3) = 3 + \sqrt{(x - 3)^2 + 3 - 3} = 3 + |x - 3| = 3 + (x - 3) = x$$ Thus, the inverse function is correct. **Final answer:** $$f^{-1}(x) = 3 + \sqrt{x - 3}$$