Question

Find the domain of $\sin^{-1} \sqrt{x - 1}$.

Hint:

When determining the domain of an inverse trigonometric function, ensure the expression inside the inverse satisfies the required range. For $\sin^{-1}(y)$, $y$ must be in the range $[-1, 1]$.

Answer 1

The domain of $\sin^{-1} \sqrt{x - 1}$ is sought. The $\sin^{-1}(y)$ function is defined for $-1 \leq y \leq 1$. Consequently, we require: \[-1 \leq \sqrt{x - 1} \leq 1.\] Given that $\sqrt{x - 1} \geq 0$, the inequality simplifies to: \[0 \leq \sqrt{x - 1} \leq 1.\] Squaring both sides yields: \[0 \leq x - 1 \leq 1.\] This implies: \[1 \leq x \leq 2.\] Therefore, the domain of $\sin^{-1} \sqrt{x - 1}$ is the interval $[1, 2]$.