1. The problem is to find the inverse function $f^{-1}(x)$ for the function $f(x) = \sqrt{x - 1}$ with domain $x \geq 1$. 2. To find the inverse, start by setting $y = f(x) = \sqrt{x - 1}$. 3. Swap $x$ and $y$ to find the inverse: $x = \sqrt{y - 1}$. 4. Solve for $y$: square both sides to get $x^2 = y - 1$. 5. Add 1 to both sides: $y = x^2 + 1$. 6. The domain of $f$ is $x \geq 1$, so the range of $f$ (and domain of $f^{-1}$) is $y \geq 0$ because $f(x) = \sqrt{x-1} \geq 0$. 7. Therefore, the inverse function is $f^{-1}(x) = x^2 + 1$ with domain $x \geq 0$. 8. Comparing with the options, the correct answer is C: $f^{-1}(x) = x^2 + 1$, $x \geq 0$. Final answer: $f^{-1}(x) = x^2 + 1$, $x \geq 0$.