Question

Find the domain of \( \sin^{-1}(x^2 - 3) \).

Hint:

For inverse trigonometric functions, the argument must lie within the valid range of the function. For \( \sin^{-1}(x) \), the argument must be between \([-1, 1]\).

Answer 1

The domain of the inverse sine function \( \sin^{-1}(z) \) is restricted to values of \( z \in [-1, 1] \), meaning for \( \sin^{-1}(x^2 - 3) \), the argument \( x^2 - 3 \) must lie within the interval \([-1, 1]\).
We start by setting up the inequality for the argument: \[ -1 \leq x^2 - 3 \leq 1 \] To solve for \( x \), first add 3 to all parts of the inequality: \[ -1 + 3 \leq x^2 - 3 + 3 \leq 1 + 3 \] \[ 2 \leq x^2 \leq 4 \] Now, solve for \( x \) by taking square roots: \[ \sqrt{2} \leq |x| \leq 2 \] This means \( x \) can take values in two ranges: \[ -\sqrt{2} \leq x \leq -2 \quad \text{or} \quad \sqrt{2} \leq x \leq 2 \] Since \( \sqrt{2} \approx 1.41 \), the simplified domain of \( x \) is: \[ x \in [-2, -1] \cup [1, 2] \] Thus, the domain of \( \sin^{-1}(x^2 - 3) \) is \( x \in [-2, -1] \cup [1, 2] \).