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

Let the given function be:

$$ f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}} $$

For the function to be defined, the conditions for both the numerator and the denominator must be satisfied simultaneously.

Condition 1: Domain of the inverse sine function in the numerator.

The argument of the inverse sine function must lie within the closed interval from -1 to 1:

$$ -1 \leq x - 3 \leq 1 $$

Add 3 to all parts of the inequality to isolate x:

$$ -1 + 3 \leq x \leq 1 + 3 $$

$$ 2 \leq x \leq 4 $$

This gives the first valid interval for x:

$$ x \in [2, 4] $$

Condition 2: Domain of the square root function in the denominator.

The expression inside the square root in the denominator must be strictly greater than 0 because division by 0 is undefined:

$$ 9 - x^2 > 0 $$

Rearrange the inequality by multiplying by -1, which flips the inequality sign:

$$ x^2 - 9 < 0 $$

Factor the difference of squares:

$$ (x - 3)(x + 3) < 0 $$

This quadratic inequality is satisfied when x lies strictly between the two roots:

$$ -3 < x < 3 $$

This gives the second valid interval for x:

$$ x \in (-3, 3) $$

To find the domain of the complete function, take the intersection of the two individual intervals:

$$ \text{Domain} = [2, 4] \cap (-3, 3) $$

Comparing the boundaries, the lower bound must be at least 2, and the upper bound must be strictly less than 3. Combining these restrictions yields:

$$ 2 \leq x < 3 $$

Written in interval notation, this represents a half-open interval.

Final Answer:

The domain of the function is $$ [2, 3) $$.