Question

Find the domain of the function $f(x) = \sin^{-1}(-x^2)$.

Hint:

When determining the domain of inverse trigonometric functions, ensure that the argument lies within the valid range of the function (for $\sin^{-1}$, the range is $[-1, 1]$).

Answer 1

The domain of the inverse sine function, $\sin^{-1} x$, is restricted to $x \in [-1, 1]$. Consequently, for $f(x) = \sin^{-1}(-x^2)$, the argument $-x^2$ must satisfy: \[ -1 \leq -x^2 \leq 1. \] Multiplying the inequality by -1 reverses the signs, yielding: \[ 1 \geq x^2 \geq 0. \] This condition simplifies to $x^2 \leq 1$, which implies that $x$ must be in the interval: \[ -1 \leq x \leq 1. \] Therefore, the domain of $f(x) = \sin^{-1}(-x^2)$ is $x \in [-1, 1]$.