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

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(B) To define the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$,we must satisfy two conditions:$1$. The argument of the square root in the

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$[2, 3)$

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(B) To define the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$,we must satisfy two conditions:$1$. The argument of the square root in the denominator must be strictly positive:$9 - x^2 > 0$$x^2 $-3 $2$. The argument of the inverse sine function must lie in the interval $[-1, 1]$:$-1 \le x - 3 \le 1$Adding $3$ to all parts:$2 \le x \le 4$  --- $(ii)$Taking the intersection of conditions $(i)$ and $(ii)$:$x \in (-3, 3) \cap [2, 4]$$x \in [2, 3)$Thus,the domain of the function is $[2, 3)$.

The domain of the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$ is

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