Key Concepts and Formulas

• Domain of $\sin^{-1}(y)$: The inverse sine function $\sin^{-1}(y)$ is defined for $-1 \le y \le 1$.
• Domain of $\log_b(y)$: The logarithmic function $\log_b(y)$ is defined for $y > 0$, $b > 0$, and $b \neq 1$.
• Logarithm Inequality Property: For a base $b > 1$, $\log_b(A) < \log_b(B) \iff A < B$. For a base $0 < b < 1$, $\log_b(A) < \log_b(B) \iff A > B$.

Step-by-Step Solution

The function given is $f\left( x \right) = {\sin ^{ - 1}}\left[ {{{\log }_3}\left( {{x \over 3}} \right)} \right]$. To find the domain of this function, we need to consider the restrictions imposed by each part of the composite function.

Step 1: Apply the domain restriction for the inverse sine function.

The argument of the $\sin^{-1}$ function must lie in the interval $[-1, 1]$. In this case, the argument is ${{\log }_3}\left( {{x \over 3}} \right)$. Therefore, we must have: $-1 \le {{\log }_3}\left( {{x \over 3}} \right) \le 1$ To solve this compound inequality, we convert the logarithmic form to an exponential form. Since the base of the logarithm is $3$ (which is greater than $1$), the inequality signs remain unchanged. We raise $3$ to the power of each part of the inequality: ${3^{ - 1}} \le {x \over 3} \le {3^1}$ This simplifies to: ${1 \over 3} \le {x \over 3} \le 3$ Now, we multiply all parts of the inequality by $3$ to isolate $x$: $3 \times {1 \over 3} \le x \le 3 \times 3$ $1 \le x \le 9$ This condition implies that $x$ must be in the interval $[1, 9]$.

Step 2: Apply the domain restriction for the logarithmic function.

The argument of the logarithmic function must be strictly positive. In this case, the argument is ${x \over 3}$. Therefore, we must have: ${x \over 3} > 0$ Multiplying both sides by $3$ to solve for $x$: $x > 0$ This condition implies that $x$ must be in the interval $(0, \infty)$.

Step 3: Combine the domain restrictions.

For the function $f(x)$ to be defined, both conditions derived in Step 1 and Step 2 must be satisfied simultaneously. We need to find the intersection of the intervals obtained:

• From the $\sin^{-1}$ condition: $x \in [1, 9]$
• From the $\log_3$ condition: $x \in (0, \infty)$

The intersection of these two intervals is: $[1, 9] \cap (0, \infty)$ The interval $[1, 9]$ is a subset of $(0, \infty)$ because all numbers between $1$ and $9$ (inclusive) are greater than $0$. Therefore, the intersection is simply $[1, 9]$.

Common Mistakes & Tips

• Argument of Logarithm: Always remember that the argument of a logarithm must be strictly positive ($> 0$). Forgetting this can lead to incorrect lower bounds for the domain.
• Base of Logarithm: Pay close attention to the base of the logarithm when dealing with inequalities. If the base is between $0$ and $1$, the direction of the inequality reverses when converting to exponential form. In this problem, the base is $3$, so the inequality direction remains the same.
• Composite Functions: For composite functions, ensure that the domain restrictions of all constituent functions are considered. The final domain is the intersection of all these individual restrictions.

Summary

To find the domain of $f\left( x \right) = {\sin ^{ - 1}}\left[ {{{\log }_3}\left( {{x \over 3}} \right)} \right]$, we first applied the condition that the argument of $\sin^{-1}$ must be between $-1$ and $1$, which yielded $1 \le x \le 9$. Next, we applied the condition that the argument of $\log_3$ must be strictly positive, which yielded $x > 0$. The domain of the function is the intersection of these two conditions, which is $[1, 9]$.

The final answer is $\boxed{[1, 9]}$.