Question:
Find the domain of:
\[ f(x) = \sin^{-1}\!\left(\sqrt{x^2 – 1}\right) \]
Concept:
For \( \sin^{-1}(t) \) to be defined:
\[ -1 \leq t \leq 1 \]
Also, the square root requires:
\[ x^2 – 1 \geq 0 \]
—Solution:
Step 1: Square root condition
\[ x^2 – 1 \geq 0 \Rightarrow x^2 \geq 1 \Rightarrow x \leq -1 \text{ or } x \geq 1 \]
Step 2: Inverse sine condition
\[ 0 \leq \sqrt{x^2 – 1} \leq 1 \]
Squaring:
\[ 0 \leq x^2 – 1 \leq 1 \Rightarrow 1 \leq x^2 \leq 2 \]
\[ 1 \leq |x| \leq \sqrt{2} \]
Step 3: Final domain
\[ x \in [-\sqrt{2}, -1] \cup [1, \sqrt{2}] \]
—Final Answer:
\[ \boxed{[-\sqrt{2}, -1] \cup [1, \sqrt{2}]} \]