The domain of the function f(x) = sin^{-1}lef | vedclass

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(C) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.For the given function,we must have $-1 \leq \log_2\left(\frac{x^2}{2}\right) \leq 1$.Applying the

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$[-2, -1] \cup [1, 2]$

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(C) The domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$.For the given function,we must have $-1 \leq \log_2\left(\frac{x^2}{2}\right) \leq 1$.Applying the base $2$ exponentiation to all parts,we get $2^{-1} \leq \frac{x^2}{2} \leq 2^1$.This simplifies to $\frac{1}{2} \leq \frac{x^2}{2} \leq 2$.Multiplying by $2$,we get $1 \leq x^2 \leq 4$.Taking the square root,we find $x \in [-2, -1] \cup [1, 2]$.

The domain of the function $f(x) = \sin^{-1}\left(\log_2\left(\frac{x^2}{2}\right)\right)$ is

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