1. **State the problem:** Find the inverse function $f^{-1}(x)$ for the function $f(x) = \sqrt{x - 1}$ where $x \geq 1$. 2. **Recall the definition of inverse functions:** The inverse function $f^{-1}(x)$ reverses the effect of $f(x)$, so $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. 3. **Start with the equation:** $$y = \sqrt{x - 1}$$ 4. **Swap $x$ and $y$ to find the inverse:** $$x = \sqrt{y - 1}$$ 5. **Solve for $y$:** Square both sides: $$x^2 = y - 1$$ Add 1 to both sides: $$y = x^2 + 1$$ 6. **Determine the domain of the inverse:** Since $f(x)$ has domain $x \geq 1$, the range of $f(x)$ is $y \geq 0$ (because square root outputs non-negative values). Therefore, the domain of $f^{-1}(x)$ is $x \geq 0$. 7. **Write the inverse function with domain:** $$f^{-1}(x) = x^2 + 1, \quad x \geq 0$$ 8. **Match with the options:** Option C matches exactly. **Final answer:** C. $f^{-1}(x) = x^2 + 1, x \geq 0$