The domain of the function `f(x)=sin^(-1)("log"_(2)(x^(2))/(2))` is given by…

Explanation

To find the domain of the function \( f(x) = \sin^{-1}\left(\frac{\log_2(x^2)}{2}\right) \), we need to ensure that the argument of the inverse sine function is valid. The inverse sine function, \( \sin^{-1}(y) \), is defined for \( -1 \leq y \leq 1 \). Therefore, we need to solve the inequality:

\[ -1 \leq \frac{\log_2(x^2)}{2} \leq 1 \]

### Step 1: Solve the left inequality
Starting with the left side of the inequality:

\[ -1 \leq \frac{\log_2(x^2)}{2} \]

Multiplying both sides by 2 (which does not change the direction of the inequality):

\[ -2 \leq \log_2(x^2) \]

Now, converting the logarithmic inequality to its exponential form:

\[ x^2 \geq 2^{-2} = \frac{1}{4} \]

Taking the square root of both sides gives:

\[ |x| \geq \frac{1}{2} \]

### Step 2: Solve the right inequality
Now, we solve the right side of the inequality:

\[ \frac{\log_2(x^2)}{2} \leq 1 \]

Multiplying both sides by 2:

\[ \log_2(x^2) \leq 2 \]

Converting this to exponential form:

\[ x^2 \leq 2^2 = 4 \]

Taking the square root of both sides gives:

\[ |x| \leq 2 \]

### Step 3: Combine the results
Now we combine the results from both inequalities:

1. From the left inequality, we have \( |x| \geq \frac{1}{2} \).
2. From the right inequality, we have \( |x| \leq 2 \).

This can be expressed as:

\[ \frac{1}{2} \leq |x| \leq 2 \]

### Step 4: Express the domain in interval notation
The absolute value condition implies two intervals:

1. \( x \) can be in the interval \( [-2, -\frac{1}{2}] \)
2. \( x \) can be in the interval \( [\frac{1}{2}, 2] \)

Thus, the domain of the function \( f(x) \) is:

\[ [-2, -\frac{1}{2}] \cup [\frac{1}{2}, 2] \]

### Final Answer
The domain of the function \( f(x) = \sin^{-1}\left(\frac{\log_2(x^2)}{2}\right) \) is:

\[ [-2, -\frac{1}{2}] \cup [\frac{1}{2}, 2] \]

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