Key Concepts and Formulas

• Domain of $\sin^{-1}(u)$: The inverse sine function, $\sin^{-1}(u)$, is defined for arguments $u$ such that $-1 \le u \le 1$.
• Domain of $\sqrt{v}$: The square root function, $\sqrt{v}$, is defined for real numbers when the radicand $v$ is non-negative, i.e., $v \ge 0$.
• Domain of a Rational Function: For a function of the form $\frac{A}{B}$, the denominator $B$ must be non-zero, i.e., $B \ne 0$.
• Intersection of Intervals: To find the overall domain of a function with multiple restrictions, we must find the intersection of all individual domain intervals.

Step-by-Step Solution

The given function is $f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}$. To determine the domain of this function, we need to ensure that all its components are well-defined.

Step 1: Determine the restriction from the inverse sine function. The numerator of the function involves $\sin^{-1}(x-3)$. For this term to be defined, its argument, $(x-3)$, must lie within the interval $[-1, 1]$. $-1 \le x - 3 \le 1$ To isolate $x$, we add 3 to all parts of the inequality: $-1 + 3 \le x - 3 + 3 \le 1 + 3$ $2 \le x \le 4$ This gives us the first condition for the domain: $x \in [2, 4]$.

Step 2: Determine the restrictions from the square root in the denominator. The denominator of the function is $\sqrt{9-x^2}$. For this expression to be defined in real numbers, two conditions must be met: a) The radicand must be non-negative: $9 - x^2 \ge 0$. b) The denominator cannot be zero: $\sqrt{9-x^2} \ne 0$, which implies $9-x^2 \ne 0$.

Combining these two conditions, we require the radicand to be strictly positive: $9 - x^2 > 0$ To solve this inequality, we can rewrite it as: $x^2 < 9$ Taking the square root of both sides, we get: $\sqrt{x^2} < \sqrt{9}$ $|x| < 3$ This inequality means that $x$ must be between -3 and 3, excluding the endpoints: $-3 < x < 3$ This gives us the second condition for the domain: $x \in (-3, 3)$.

Step 3: Combine all the restrictions to find the overall domain. For the function $f(x)$ to be defined, $x$ must satisfy both the restriction from the inverse sine function (Step 1) and the restriction from the square root in the denominator (Step 2). We need to find the intersection of the two intervals: Interval 1: $x \in [2, 4]$ Interval 2: $x \in (-3, 3)$

The intersection of these two intervals is the set of all $x$ values that are common to both. We are looking for $x$ such that $2 \le x \le 4$ AND $-3 < x < 3$. The values of $x$ that satisfy both conditions are $x$ such that $2 \le x < 3$.

Therefore, the domain of the function $f(x)$ is the interval $[2, 3)$.

Common Mistakes & Tips

• Ignoring the Denominator Condition: A common error is to only consider $9-x^2 \ge 0$ for the square root, which would lead to $x \in [-3, 3]$. However, since $\sqrt{9-x^2}$ is in the denominator, it cannot be zero. Always remember to exclude values that make the denominator zero.
• Solving $x^2 < a^2$: Remember that $x^2 < a^2$ is equivalent to $|x| < a$, which means $-a < x < a$. Incorrectly solving this can lead to an incorrect interval.
• Intersection of Intervals: The domain of a function is the set of all input values for which all conditions are simultaneously met. Therefore, always find the intersection of the intervals derived from each condition.

Summary

The domain of the function $f(x)$ is determined by ensuring that the argument of the inverse sine function is within $[-1, 1]$ and that the expression under the square root in the denominator is strictly positive. The condition $-1 \le x-3 \le 1$ yields $2 \le x \le 4$. The condition $9-x^2 > 0$ yields $-3 < x < 3$. The intersection of these two intervals, $[2, 4]$ and $(-3, 3)$, is $[2, 3)$, which is the domain of the function.

The final answer is $\boxed{[2, 3)}$.