We are asked to find the inverse function of $y = \sqrt{x-2} + 1$ for $x \ge 2$.

1. Problem Description

We are asked to find the inverse function of $y = \sqrt{x-2} + 1$ for $x \ge 2$.

2. Solution Steps

To find the inverse function, we need to swap $x$ and $y$ and then solve for $y$.

Starting with the given function:

$y = \sqrt{x-2} + 1$, $x \ge 2$

Swap $x$ and $y$:

$x = \sqrt{y-2} + 1$

Now, solve for $y$:

$x - 1 = \sqrt{y-2}$

Square both sides:

$(x-1)^2 = y - 2$

$y = (x-1)^2 + 2$

Now we need to find the domain of the inverse function, which is the range of the original function. Since $x \ge 2$, $\sqrt{x-2} \ge 0$.

Therefore, $y = \sqrt{x-2} + 1 \ge 0 + 1 = 1$.

So the range of the original function is $y \ge 1$.

Therefore, the domain of the inverse function is $x \ge 1$.

So the inverse function is $y = (x-1)^2 + 2$ for $x \ge 1$.

This matches option D: $y = 2 + (x-1)^2$ $(x \ge 1)$.

3. Final Answer

D. $y = 2 + (x-1)^2$ $(x \ge 1)$