The domain of the function` f (x) = sqrt""(2-2x-x^2)` is

Options

A. `-3 le x le sqrt""3`
B. ` -1-sqrt"" 3 le x le -1 + sqrt""3`
C. `-2 le x lt 2 `
D. None of these

Answer

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

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

### Step 1: Rearranging the Inequality
First, we can rearrange the inequality:

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

Multiplying the entire inequality by -1 (which flips the inequality sign) gives:

\[ x^2 + 2x - 2 \leq 0 \]

### Step 2: Finding the Roots
Next, we need to find the roots of the quadratic equation \( x^2 + 2x - 2 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = 2, c = -2 \):

\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]
\[ x = \frac{-2 \pm \sqrt{4 + 8}}{2} \]
\[ x = \frac{-2 \pm \sqrt{12}}{2} \]
\[ x = \frac{-2 \pm 2\sqrt{3}}{2} \]
\[ x = -1 \pm \sqrt{3} \]

Thus, the roots are:

\[ x_1 = -1 - \sqrt{3}, \quad x_2 = -1 + \sqrt{3} \]

### Step 3: Testing Intervals
Now we need to test the intervals determined by the roots \( x_1 \) and \( x_2 \):

1. **Interval 1:** \( (-\infty, -1 - \sqrt{3}) \)
2. **Interval 2:** \( (-1 - \sqrt{3}, -1 + \sqrt{3}) \)
3. **Interval 3:** \( (-1 + \sqrt{3}, \infty) \)

We will test a point from each interval to determine where the inequality \( x^2 + 2x - 2 \leq 0 \) holds.

- **For Interval 1:** Choose \( x = -5 \)
\[ (-5)^2 + 2(-5) - 2 = 25 - 10 - 2 = 13 \quad (\text{positive}) \]

- **For Interval 2:** Choose \( x = 0 \)
\[ (0)^2 + 2(0) - 2 = -2 \quad (\text{negative}) \]

- **For Interval 3:** Choose \( x = 2 \)
\[ (2)^2 + 2(2) - 2 = 4 + 4 - 2 = 6 \quad (\text{positive}) \]

### Step 4: Conclusion
The inequality \( x^2 + 2x - 2 \leq 0 \) holds true in the interval:

\[ [-1 - \sqrt{3}, -1 + \sqrt{3}] \]

Thus, the domain of the function \( f(x) = \sqrt{2 - 2x - x^2} \) is:

\[ \text{Domain: } [-1 - \sqrt{3}, -1 + \sqrt{3}] \]