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The domain of the inverse sine function, $\sin^{-1}(u)$, is $-1 \le u \le 1$. This means that the input to the $\sin^{-1}$ function must be between -1 and 1, inclusive.
In our case, the input to the $\sin^{-1}$ function is $\sqrt{x-1}$. Therefore, we must have $-1 \le \sqrt{x-1} \le 1$.
Since the square root function always returns a non-negative value, $\sqrt{x-1}$ is always greater than or equal to 0. Thus, we only need to consider the upper bound: $\sqrt{x-1} \le 1$.
To solve the inequality $\sqrt{x-1} \le 1$, we can square both sides (since both sides are non-negative): $ (\sqrt{x-1})^2 \le 1^2 $ $ x-1 \le 1 $ $ x \le 2 $
We also need to consider the domain of the square root function itself. The expression inside the square root must be non-negative: $ x-1 \ge 0 $ $ x \ge 1 $
Combining the inequalities $x \le 2$ and $x \ge 1$, we get the domain of the function: $ 1 \le x \le 2 $
The domain of the function $\sin^{-1}\sqrt{x-1}$ is the interval $[1, 2]$.
Final Answer: [1, 2]
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