The domain of the inverse sine function, \( \sin^{-1}x \), is \( -1 \le x \le 1 \). In this problem, we are given that \( -1 \le x \le 0 \).
We need to find the values of \( \sin^{-1}x \) at \( x = -1 \) and \( x = 0 \). $$ \sin^{-1}(-1) = -\frac{\pi}{2} $$ $$ \sin^{-1}(0) = 0 $$
Since \( \sin^{-1}x \) is a continuous and increasing function on the interval \( [-1, 0] \), the range of \( y = \sin^{-1}x \) for \( -1 \le x \le 0 \) is \( [-\frac{\pi}{2}, 0] \).
Final Answer: \([-\frac{\pi}{2}, 0]\)<\/span>
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