Find the domain of definitions of the functions
(Read the symbols [*] and {*} as greatestintegers and fractional part functions respectively.)
`f(x) = sqrt(x^(2) - |x|) + (1)/(sqrt( 9 - x^(2)))`

Answer

To find the domain of the function \( f(x) = \sqrt{x^2 - |x|} + \frac{1}{\sqrt{9 - x^2}} \), we need to consider the conditions under which the function is defined. This involves analyzing both parts of the function: the square root and the denominator.

### Step 1: Analyze the square root \( \sqrt{x^2 - |x|} \)

1. **Condition for the square root**: The expression inside the square root must be non-negative:
\[ x^2 - |x| \geq 0 \]
This can be rewritten based on the definition of absolute value:
- For \( x \geq 0 \): \( |x| = x \) so the inequality becomes:
\[ x^2 - x \geq 0 \implies x(x - 1) \geq 0 \]
The solutions to this inequality are \( x \leq 0 \) or \( x \geq 1 \).

- For \( x < 0 \): \( |x| = -x \) so the inequality becomes:
\[ x^2 + x \geq 0 \implies x(x + 1) \geq 0 \]
The solutions to this inequality are \( x \leq -1 \) or \( x \geq 0 \).

Combining both cases, we have:
\[ x \in (-\infty, -1] \cup [1, \infty) \]

### Step 2: Analyze the denominator \( \frac{1}{\sqrt{9 - x^2}} \)

2. **Condition for the denominator**: The expression inside the square root must be positive (since the denominator cannot be zero):
\[ 9 - x^2 > 0 \implies x^2 < 9 \]
This implies:
\[ -3 < x < 3 \]

### Step 3: Combine the conditions

Now we need to find the intersection of the two conditions:
1. From the square root: \( x \in (-\infty, -1] \cup [1, \infty) \)
2. From the denominator: \( x \in (-3, 3) \)

Now we find the intersection:
- For \( (-\infty, -1] \) and \( (-3, 3) \): The intersection is \( (-3, -1] \).
- For \( [1, \infty) \) and \( (-3, 3) \): The intersection is \( [1, 3) \).

### Final Domain

Thus, the domain of the function \( f(x) \) is:
\[ \text{Domain} = (-3, -1] \cup [1, 3) \]