Competency-Based
Assessment Made Simple.

Step 1: Determine the domain.

For the function \(f(x) = \sin^{-1}(x^2 - 4)\) to be defined, the argument of the inverse sine function must be between -1 and 1, inclusive. Therefore, we must have:

\(-1 \le x^2 - 4 \le 1\)

Step 2: Solve the inequality.

We can split this into two inequalities:

1. \(x^2 - 4 \le 1 \Rightarrow x^2 \le 5 \Rightarrow -\sqrt{5} \le x \le \sqrt{5}\)

2. \(x^2 - 4 \ge -1 \Rightarrow x^2 \ge 3 \Rightarrow x \le -\sqrt{3}\) or \(x \ge \sqrt{3}\)

Step 3: Combine the inequalities.

Combining these two conditions, we get the domain as:

\(-\sqrt{5} \le x \le -\sqrt{3}\) or \(\sqrt{3} \le x \le \sqrt{5}\)

In interval notation: \([-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]\)

Step 4: Determine the range.

To find the range, we need to find the minimum and maximum values of \(x^2 - 4\) for \(x\) in the domain \([-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]\).

When \(x = \pm \sqrt{3}\), \(x^2 - 4 = 3 - 4 = -1\). \(\sin^{-1}(-1) = -\frac{\pi}{2}\)

When \(x = \pm \sqrt{5}\), \(x^2 - 4 = 5 - 4 = 1\). \(\sin^{-1}(1) = \frac{\pi}{2}\)

Since the inverse sine function is continuous and increasing on the interval [-1, 1], the range of \(f(x)\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).