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

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(A) The function is defined as $f(x) = \sin^{-1}[\log_2(x/2)]$. Since the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$,we must have: $-1 \le \log_2(x/2

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$[1, 4]$

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(A) The function is defined as $f(x) = \sin^{-1}[\log_2(x/2)]$. Since the domain of $\sin^{-1}(u)$ is $u \in [-1, 1]$,we must have: $-1 \le \log_2(x/2) \le 1$. Applying the definition of the logarithm,we exponentiate with base $2$: $2^{-1} \le x/2 \le 2^1$. This simplifies to: $1/2 \le x/2 \le 2$. Multiplying the entire inequality by $2$,we get: $1 \le x \le 4$. Therefore,the domain of the function is $x \in [1, 4]$.

The domain of the function $f(x) = \sin^{-1}[\log_2(x/2)]$ is

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