The domain of the function `f(x)=sin^(-1)log_(3)(x/3))` is

Options

A. [1,9]
B. [-1,9]
C. [-9,1]
D. [-9,-1]

Answer

To find the domain of the function \( f(x) = \sin^{-1}(\log_3(\frac{x}{3})) \), we need to ensure that the expression inside the sine inverse function is valid. The sine inverse function, \( \sin^{-1}(y) \), is defined for \( -1 \leq y \leq 1 \). Therefore, we need to solve the inequalities:

1. \( -1 \leq \log_3\left(\frac{x}{3}\right) \leq 1 \)
2. Additionally, the argument of the logarithm must be positive: \( \frac{x}{3} > 0 \).

### Step 1: Solve the left inequality \( -1 \leq \log_3\left(\frac{x}{3}\right) \)

To solve this inequality, we can rewrite it in exponential form:

\[ \log_3\left(\frac{x}{3}\right) \geq -1 \implies \frac{x}{3} \geq 3^{-1} = \frac{1}{3} \]

Multiplying both sides by 3 gives:

\[ x \geq 1 \]

### Step 2: Solve the right inequality \( \log_3\left(\frac{x}{3}\right) \leq 1 \)

Similarly, we rewrite this in exponential form:

\[ \log_3\left(\frac{x}{3}\right) \leq 1 \implies \frac{x}{3} \leq 3^1 = 3 \]

Multiplying both sides by 3 gives:

\[ x \leq 9 \]

### Step 3: Combine the results from Step 1 and Step 2

From the two inequalities, we have:

\[ 1 \leq x \leq 9 \]

This means:

\[ x \in [1, 9] \]

### Step 4: Ensure the argument of the logarithm is positive

We also need to ensure that the argument of the logarithm is positive:

\[ \frac{x}{3} > 0 \implies x > 0 \]

### Step 5: Combine all conditions

The condition \( x > 0 \) is already satisfied by the interval \( [1, 9] \). Therefore, the domain of the function \( f(x) \) is:

\[ \text{Domain of } f(x) = [1, 9] \]

### Final Answer:

The domain of the function \( f(x) = \sin^{-1}(\log_3(\frac{x}{3})) \) is \( [1, 9] \).

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