1. The problem asks for the domain of the inverse of the function $f(x) = 2 + \sqrt{x - 1}$.\n\n2. First, recall that the domain of the inverse function is the range of the original function $f(x)$. So we need to find the range of $f(x)$.\n\n3. The function $f(x) = 2 + \sqrt{x - 1}$ is defined when the expression inside the square root is non-negative: $$x - 1 \geq 0 \implies x \geq 1.$$ So the domain of $f$ is $[1, \infty)$.\n\n4. Next, find the range of $f(x)$. Since $\sqrt{x - 1} \geq 0$ for $x \geq 1$, the smallest value of $f(x)$ is when $x=1$: $$f(1) = 2 + \sqrt{1 - 1} = 2 + 0 = 2.$$\n\n5. As $x$ increases, $\sqrt{x - 1}$ increases without bound, so $f(x)$ increases without bound. Therefore, the range of $f$ is $$[2, \infty).$$\n\n6. Since the domain of the inverse function is the range of the original function, the domain of $f^{-1}$ is $$[2, \infty).$$\n\nFinal answer: The domain of the inverse function is $[2, \infty)$, which corresponds to option (b).