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The domain of the inverse sine function, $sin^{-1}(u)$, is $-1 \le u \le 1$. Therefore, for $f(x) = sin^{-1}(x^2 - 4)$ to be defined, we must have $-1 \le x^2 - 4 \le 1$.
We have two inequalities to solve: 1. $x^2 - 4 \ge -1$ 2. $x^2 - 4 \le 1$
$x^2 - 4 \ge -1$ $x^2 \ge 3$ This implies $x \le -\sqrt{3}$ or $x \ge \sqrt{3}$.
$x^2 - 4 \le 1$ $x^2 \le 5$ This implies $-\sqrt{5} \le x \le \sqrt{5}$.
We need to find the intersection of the two solution sets: $x \le -\sqrt{3}$ or $x \ge \sqrt{3}$ and $-\sqrt{5} \le x \le \sqrt{5}$. This gives us the domain: $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$.
To find the range, we need to determine the possible values of $f(x) = sin^{-1}(x^2 - 4)$ for $x$ in the domain $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$. When $x = \pm \sqrt{3}$, $x^2 - 4 = 3 - 4 = -1$. So, $sin^{-1}(-1) = -\frac{\pi}{2}$. When $x = \pm \sqrt{5}$, $x^2 - 4 = 5 - 4 = 1$. So, $sin^{-1}(1) = \frac{\pi}{2}$. Since $x^2 - 4$ varies continuously from -1 to 1 as $x$ varies from $\sqrt{3}$ to $\sqrt{5}$ (and from $-\sqrt{5}$ to $-\sqrt{3}$), the range of $f(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Final Answer: Domain: $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$; Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$
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