Determine the domain of the function:
$$
y = \sqrt{\frac{x - 3}{x + 3}} + \sqrt{\frac{1 - x}{1 + x}}.
$$

Question

Determine the domain of the function:  
$$
y = \sqrt{\frac{x - 3}{x + 3}} + \sqrt{\frac{1 - x}{1 + x}}.
$$

Accepted Answer

To find the domain of the given function, we need to ensure that both square root terms are defined. This requires the expressions inside the square roots to be non-negative.

### Step 1: Analyze the first term
For $$ \sqrt{\frac{x - 3}{x + 3}} $$, the condition is:
$$
\frac{x - 3}{x + 3} \geq 0.
$$
This inequality holds when:
- $$ x - 3 $$ and $$ x + 3 $$ have the same sign (both positive or both negative).

The critical points are $$ x = -3 $$ and $$ x = 3 $$. Using a sign chart:
- For $$ x  0 $$, so the fraction is negative.
- For $$ x > 3 $$, both $$ x - 3 > 0 $$ and $$ x + 3 > 0 $$, so the fraction is positive.

Thus, the solution for this inequality is:
$$
x \in (-\infty, -3] \cup [3, \infty).
$$

### Step 2: Analyze the second term
For $$ \sqrt{\frac{1 - x}{1 + x}} $$, the condition is:
$$
\frac{1 - x}{1 + x} \geq 0.
$$
This inequality holds when:
- $$ 1 - x $$ and $$ 1 + x $$ have the same sign (both positive or both negative).

The critical points are $$ x = -1 $$ and $$ x = 1 $$. Using a sign chart:
- For $$ x  0 $$ and $$ 1 + x  0 $$ and $$ 1 + x > 0 $$, so the fraction is positive.
- For $$ x > 1 $$, $$ 1 - x  0 $$, so the fraction is negative.

Thus, the solution for this inequality is:
$$
x \in [-1, 1].
$$

### Step 3: Combine conditions
The domain of the function is where both conditions hold simultaneously. Combining:
- From Step 1: $$ (-\infty, -3] \cup [3, \infty) $$,
- From Step 2: $$ [-1, 1] $$.

The intersection of these intervals is:
$$
x \in (-1, 1) \cup (3, \infty).
$$

Thus, the domain of the function is:
$$
x \in (-1, 1) \cup (3, \infty).
$$

Answer

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