The domain of definition of the real function` f(x) = sqrt""(log_(16)x^2)` of the real variable x is

Options

A. ` x gt 0`
B. ` |x| ge 1`
C. `|x| ge 4`
D. ` x ge 4`

Answer

To find the domain of the function \( f(x) = \sqrt{\log_{16}(x^2)} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ \log_{16}(x^2) \geq 0 \]

### Step 1: Understand the logarithmic condition
The logarithm \( \log_{16}(x^2) \) is non-negative when \( x^2 \) is greater than or equal to \( 16^0 \) (which is 1). Therefore, we can rewrite the inequality as:

\[ x^2 \geq 1 \]

### Step 2: Solve the inequality
To solve \( x^2 \geq 1 \), we can rearrange it:

\[ x^2 - 1 \geq 0 \]

### Step 3: Factor the expression
Next, we factor the left-hand side:

\[ (x - 1)(x + 1) \geq 0 \]

### Step 4: Determine the critical points
The critical points from the factors are \( x = -1 \) and \( x = 1 \).

### Step 5: Test intervals
Now we will test the intervals determined by these critical points:

1. \( (-\infty, -1) \)
2. \( (-1, 1) \)
3. \( (1, \infty) \)

- For \( x < -1 \) (e.g., \( x = -2 \)):
\((x - 1)(x + 1) = (-2 - 1)(-2 + 1) = (-3)(-1) = 3 \geq 0\) (True)

- For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\((x - 1)(x + 1) = (0 - 1)(0 + 1) = (-1)(1) = -1 < 0\) (False)

- For \( x > 1 \) (e.g., \( x = 2 \)):
\((x - 1)(x + 1) = (2 - 1)(2 + 1) = (1)(3) = 3 \geq 0\) (True)

### Step 6: Combine the results
From our tests, the inequality holds true for:

- \( x \leq -1 \)
- \( x \geq 1 \)

### Final Domain
Thus, we can express the domain in interval notation as:

\[ (-\infty, -1] \cup [1, \infty) \]