Find the domain of each of the following functions: `f(x)=ln[x^(2)+x+1]`, where [.] GIF

Explanation

To find the domain of the function \( f(x) = \ln[\lfloor x^2 + x + 1 \rfloor] \), we need to ensure that the argument of the logarithm is positive, since the logarithm is only defined for positive values.

### Step-by-Step Solution:

1. **Identify the Argument of the Logarithm**:
The function is defined as \( f(x) = \ln[\lfloor x^2 + x + 1 \rfloor] \). The expression inside the logarithm is \( \lfloor x^2 + x + 1 \rfloor \).

2. **Set Up the Inequality**:
For the logarithm to be defined, we need:
\[ \lfloor x^2 + x + 1 \rfloor > 0 \]
This means that \( x^2 + x + 1 \) must be greater than or equal to 1, since the floor function \( \lfloor y \rfloor \) is greater than 0 when \( y \) is at least 1.

3. **Solve the Inequality**:
We need to solve:
\[ x^2 + x + 1 \geq 1 \]
Simplifying this gives:
\[ x^2 + x \geq 0 \]

4. **Factor the Expression**:
We can factor the left-hand side:
\[ x(x + 1) \geq 0 \]

5. **Determine the Critical Points**:
The critical points are \( x = 0 \) and \( x = -1 \). We will test the intervals determined by these points: \( (-\infty, -1) \), \( (-1, 0) \), and \( (0, \infty) \).

6. **Test the Intervals**:
- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2)(-2 + 1) = (-2)(-1) = 2 \quad (\text{positive}) \]
- For \( -1 < x < 0 \) (e.g., \( x = -0.5 \)):
\[ (-0.5)(-0.5 + 1) = (-0.5)(0.5) = -0.25 \quad (\text{negative}) \]
- For \( x > 0 \) (e.g., \( x = 1 \)):
\[ (1)(1 + 1) = (1)(2) = 2 \quad (\text{positive}) \]

7. **Combine the Results**:
The expression \( x(x + 1) \geq 0 \) is satisfied for:
- \( x \in (-\infty, -1] \)
- \( x \in [0, \infty) \)

8. **Write the Domain**:
Therefore, the domain of the function \( f(x) \) is:
\[ (-\infty, -1] \cup [0, \infty) \]

### Final Result:
The domain of the function \( f(x) = \ln[\lfloor x^2 + x + 1 \rfloor] \) is:
\[ (-\infty, -1] \cup [0, \infty) \]