1. **State the problem:** Find the inverse function $f^{-1}(x)$ of the function $f(x) = \sqrt{x} + 5$. 2. **Recall the definition of inverse function:** The inverse function $f^{-1}(x)$ satisfies $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. 3. **Set $y = f(x)$:** $$y = \sqrt{x} + 5$$ 4. **Swap $x$ and $y$ to find the inverse:** $$x = \sqrt{y} + 5$$ 5. **Isolate the square root term:** $$x - 5 = \sqrt{y}$$ 6. **Square both sides to remove the square root:** $$ (x - 5)^2 = y $$ 7. **Rewrite $y$ as $f^{-1}(x)$:** $$f^{-1}(x) = (x - 5)^2$$ 8. **Domain and range considerations:** - Since $f(x) = \sqrt{x} + 5$, the domain of $f$ is $x \geq 0$. - The range of $f$ is $y \geq 5$. - Therefore, the domain of $f^{-1}$ is $x \geq 5$. **Final answer:** $$f^{-1}(x) = (x - 5)^2, \quad x \geq 5$$