Write the domain of the following real functions
`sqrt(1-2x-3x^(2))`

Answer

To find the domain of the function \( f(x) = \sqrt{1 - 2x - 3x^2} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ 1 - 2x - 3x^2 \geq 0 \]

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

\[ -3x^2 - 2x + 1 \geq 0 \]

Multiplying through by -1 (and reversing the inequality sign) gives us:

\[ 3x^2 + 2x - 1 \leq 0 \]

### Step 2: Finding the Roots
Next, we need to find the roots of the quadratic equation \( 3x^2 + 2x - 1 = 0 \) using the quadratic formula:

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

where \( a = 3 \), \( b = 2 \), and \( c = -1 \).

Calculating the discriminant:

\[ b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot (-1) = 4 + 12 = 16 \]

Now substituting into the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{16}}{2 \cdot 3} = \frac{-2 \pm 4}{6} \]

This gives us two solutions:

1. \( x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3} \)
2. \( x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1 \)

### Step 3: Testing Intervals
Now we have the roots \( x = -1 \) and \( x = \frac{1}{3} \). We will test the intervals determined by these roots to find where the quadratic is less than or equal to zero.

The intervals to test are:
- \( (-\infty, -1) \)
- \( (-1, \frac{1}{3}) \)
- \( (\frac{1}{3}, \infty) \)

1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[ 3(-2)^2 + 2(-2) - 1 = 12 - 4 - 1 = 7 \quad (\text{positive}) \]

2. **Interval \( (-1, \frac{1}{3}) \)**: Choose \( x = 0 \):
\[ 3(0)^2 + 2(0) - 1 = -1 \quad (\text{negative}) \]

3. **Interval \( (\frac{1}{3}, \infty) \)**: Choose \( x = 1 \):
\[ 3(1)^2 + 2(1) - 1 = 3 + 2 - 1 = 4 \quad (\text{positive}) \]

### Step 4: Conclusion on the Domain
From our testing, we find that \( 3x^2 + 2x - 1 \leq 0 \) in the interval \( [-1, \frac{1}{3}] \). Therefore, the domain of the function \( f(x) \) is:

\[ \text{Domain of } f(x) = [-1, \frac{1}{3}] \]