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 inverse sine function \( \sin^{-1}(z) \) is defined for \( z \in [-1, 1] \). Therefore, for \( \sin^{-1}(x^2 - 3) \), the expression \( x^2 - 3 \) must satisfy \( -1 \leq x^2 - 3 \leq 1 \). Adding 3 to all parts of the inequality yields \( 2 \leq x^2 \leq 4 \). Taking the square root of all parts gives \( \sqrt{2} \leq |x| \leq 2 \). This inequality implies that \( x \) must be in the intervals \( [-\sqrt{2}, -2] \) or \( [\sqrt{2}, 2] \). However, \( \sqrt{2} \approx 1.41 \). Thus, the inequality \( \sqrt{2} \leq |x| \leq 2 \) translates to \( -2 \leq x \leq -\sqrt{2} \) or \( \sqrt{2} \leq x \leq 2 \). The domain of \( \sin^{-1}(x^2 - 3) \) is therefore \( x \in [-2, -\sqrt{2}] \cup [\sqrt{2}, 2] \).