Discussion
Let the functions f and g be defined by \(f(x) = 3\sin x,\ -\frac{\pi}{2} \le x \le \frac{\pi}{2}\) and \(g(x) = 6 - 3x^{2},\ x \in \mathbb{R}\). Then \(f^{-1}(g(x))\) =
Answer: A
\(f^{-1}(g(x)) = \sin^{-1}(2 - x^{2})\).
Invert f: if \(y = 3\sin x\) with \(x \in [-\frac{\pi}{2}, \frac{\pi}{2}]\), then \(x = \sin^{-1}\frac{y}{3}\). So \(f^{-1}(y) = \sin^{-1}\frac{y}{3}\).
\[ \begin{aligned} f^{-1}(g(x)) &= \sin^{-1}\left(\frac{6 - 3x^{2}}{3}\right) \\ &= \sin^{-1}(2 - x^{2}) \end{aligned} \]
The options with a factor 3 or \(\frac{1}{3}\) outside the inverse sine put the 3 in the wrong place: it divides the argument. \(\sin^{-1}(6 - 3x^{2})\) forgets the division altogether.
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