The domain of the function f(x)=sin^-1(x-3)/√9-x^2 is a)[2,3] b)[1,2] c)[1,2) d)[2,3)

To determine the domain of the function \( f(x) = \frac{\sin^{-1}(x-3)}{\sqrt{9 - x^2}} \), we need to consider the restrictions imposed by both the inverse sine function and the square root function. Let's break this down step by step.

Analyzing the Inverse Sine Function

The expression \( \sin^{-1}(x-3) \) requires that \( x-3 \) must be within the valid range for the inverse sine function. The range for the \( \sin^{-1} \) function is between -1 and 1. Therefore, we can establish the inequality:

• -1 ≤ \( x - 3 \) ≤ 1

Solving these inequalities gives us:

• From \( x - 3 \geq -1 \):
• Adding 3 to both sides results in \( x \geq 2 \).
• From \( x - 3 \leq 1 \):
• Adding 3 to both sides gives \( x \leq 4 \).

Combining these results, we find that:

2 ≤ x ≤ 4

Considering the Square Root Function

Next, we need to analyze the denominator \( \sqrt{9 - x^2} \). For the square root to be defined, the expression inside it must be non-negative:

• 9 - \( x^2 \) ≥ 0

Rearranging this gives us:

• 9 ≥ \( x^2 \)
• This can be rewritten as \( -3 ≤ x ≤ 3 \).

Finding the Intersection of Both Domain Conditions

Now we have two conditions:

• From the inverse sine function: \( 2 ≤ x ≤ 4 \)
• From the square root function: \( -3 ≤ x ≤ 3 \)

The domain of the function \( f(x) \) will be the intersection of these two ranges. The overlapping interval is:

2 ≤ x ≤ 3

Final Domain Representation

In interval notation, this is expressed as:

[2, 3]

Thus, the correct option from the choices provided is a) [2, 3].