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The domain of the inverse cosine function, $\cos^{-1}(u)$, is $-1 \le u \le 1$. This means that the input to the $\cos^{-1}$ function must be between -1 and 1, inclusive.
For the function $f(x) = \cos^{-1}(x^2 - 4)$, we must have $-1 \le x^2 - 4 \le 1$.
We can split this compound inequality into two separate inequalities:\r\n\r\n1. $x^2 - 4 \le 1$ which implies $x^2 \le 5$, so $-\sqrt{5} \le x \le \sqrt{5}$.\r\n2. $x^2 - 4 \ge -1$ which implies $x^2 \ge 3$, so $x \le -\sqrt{3}$ or $x \ge \sqrt{3}$.
We need to find the intersection of the two solution sets:\r\n\r\n$-\sqrt{5} \le x \le \sqrt{5}$ and ($x \le -\sqrt{3}$ or $x \ge \sqrt{3}$).\r\n\r\nThis gives us the intervals $[-\sqrt{5}, -\sqrt{3}]$ and $[\sqrt{3}, \sqrt{5}]$.
Therefore, the domain of the function $f(x) = \cos^{-1}(x^2 - 4)$ is $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$.
Final Answer: $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$