Question

Find the domain of the function \( f(x) = \sin^{-1}(x^2 - 4) \). Also, find its range.

Hint:

To determine the domain and range of composite functions, solve inequalities for the input range of the inner function, and apply the output range of the outer function.

Answer 1

Step 1: Domain of the inverse sine function.
The inverse sine function \( \sin^{-1}(y) \) is defined for \( -1 \leq y \leq 1 \). For the function \( f(x) = \sin^{-1}(x^2 - 4) \), the argument \( x^2 - 4 \) must fall within this range: \[ -1 \leq x^2 - 4 \leq 1. \] Step 2: Solve the inequality. Rewrite the inequality: \[ -1 + 4 \leq x^2 \leq 1 + 4 \quad \Rightarrow \quad 3 \leq x^2 \leq 5. \] Now, take the square root on both sides: \[ \sqrt{3} \leq |x| \leq \sqrt{5}. \] Thus, we get the possible values for \( x \): \[ x \in [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]. \] Step 3: Domain of \( f(x) \).
The domain of \( f(x) = \sin^{-1}(x^2 - 4) \) is therefore: \[ \boxed{x \in [-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]}. \] Step 4: Range of \( f(x) \).
The range of the inverse sine function is \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). Since \( x^2 - 4 \) is always between \( -1 \) and \( 1 \) for the domain found above, the range of \( f(x) \) is: \[ \boxed{[-\frac{\pi}{2}, \frac{\pi}{2}]}. \]