The domain of definition of `f(x) = sqrt(e^(cos-1)(log_(4) x^(2)))`is

Answer

To find the domain of the function \( f(x) = \sqrt{e^{\cos^{-1}(\log_{4}(x^2))}} \), we need to ensure that the expression inside the square root is defined and non-negative.

### Step-by-step Solution:

1. **Understanding the Square Root**:
The square root function is defined for non-negative values. Therefore, we need to ensure that the expression inside the square root, \( e^{\cos^{-1}(\log_{4}(x^2))} \), is non-negative. Since the exponential function \( e^x \) is always positive for any real number \( x \), we can conclude that this part does not impose any restrictions.

**Hint**: Remember that the exponential function is always positive.

2. **Analyzing the Cosine Inverse Function**:
The function \( \cos^{-1}(y) \) is defined for \( y \) in the interval \([-1, 1]\). Therefore, we need to find the values of \( x \) for which:
\[ -1 \leq \log_{4}(x^2) \leq 1 \]

**Hint**: The range of the cosine inverse function is limited, so we need to restrict \( \log_{4}(x^2) \).

3. **Splitting into Two Inequalities**:
We can split the inequality into two parts:
- \( \log_{4}(x^2) \geq -1 \)
- \( \log_{4}(x^2) \leq 1 \)

4. **Solving the First Inequality**:
For \( \log_{4}(x^2) \geq -1 \):
\[ x^2 \geq 4^{-1} = \frac{1}{4} \]
Taking the square root gives:
\[ |x| \geq \frac{1}{2} \quad \Rightarrow \quad x \leq -\frac{1}{2} \text{ or } x \geq \frac{1}{2} \]

**Hint**: Remember to consider both positive and negative roots when solving inequalities involving squares.

5. **Solving the Second Inequality**:
For \( \log_{4}(x^2) \leq 1 \):
\[ x^2 \leq 4^{1} = 4 \]
Taking the square root gives:
\[ |x| \leq 2 \quad \Rightarrow \quad -2 \leq x \leq 2 \]

**Hint**: Again, consider both the positive and negative roots when dealing with squares.

6. **Combining the Results**:
Now we combine the results from both inequalities:
- From the first inequality, we have \( x \leq -\frac{1}{2} \) or \( x \geq \frac{1}{2} \).
- From the second inequality, we have \( -2 \leq x \leq 2 \).

The valid intervals from both conditions are:
- From \( -2 \leq x \leq -\frac{1}{2} \)
- From \( \frac{1}{2} \leq x \leq 2 \)

Thus, the domain of \( f(x) \) is:
\[ x \in [-2, -\frac{1}{2}] \cup [\frac{1}{2}, 2] \]

### Final Domain:
The domain of the function \( f(x) \) is:
\[ [-2, -\frac{1}{2}] \cup [\frac{1}{2}, 2] \]