To solve the problem, we need to analyze the function given by:
\[ f(x) = (x^3 + x + 1) \tan(\pi [x]) \]
where \([x]\) represents the greatest integer part of \(x\).
### Step 1: Understand the greatest integer function
The greatest integer function, \([x]\), gives the largest integer less than or equal to \(x\). For example:
- If \(x = 2.5\), then \([x] = 2\).
- If \(x = 3\), then \([x] = 3\).
- If \(x = -1.2\), then \([x] = -2\).
### Step 2: Analyze the tangent function
The tangent function \(\tan(\pi n)\) where \(n\) is an integer is equal to 0. This is because:
\[ \tan(n \pi) = \frac{\sin(n \pi)}{\cos(n \pi)} = \frac{0}{(-1)^n} = 0 \]
### Step 3: Substitute into the function
Since \([x]\) is an integer, we can conclude:
\[ \tan(\pi [x]) = 0 \]
Thus, substituting back into the function:
\[ f(x) = (x^3 + x + 1) \cdot 0 = 0 \]
### Step 4: Determine the domain and range
- **Domain**: The function \(f(x)\) is defined for all real numbers \(x\) since there are no restrictions on \(x\) in the expression.
- **Range**: Since \(f(x) = 0\) for all \(x\), the range is simply \{0\}.
### Step 5: Check if \(f(x)\) is even
To check if the function is even, we need to see if \(f(-x) = f(x)\):
\[ f(-x) = ((-x)^3 + (-x) + 1) \tan(\pi [-x]) \]
Since \([-x] = -[x] - 1\) if \(x\) is not an integer, and \([-x] = -[x]\) if \(x\) is an integer, we can see that:
- For any real \(x\), \(\tan(\pi [-x]) = 0\) as well.
Thus, \(f(-x) = 0 = f(x)\), confirming that \(f(x)\) is even.
### Step 6: Check if \(f(x)\) is periodic
A constant function (in this case, \(f(x) = 0\)) is considered periodic because it repeats its value. However, it does not have a fundamental period.
### Summary of Results
1. **Domain**: All real numbers \(\mathbb{R}\)
2. **Range**: \{0\}
3. **Even Function**: Yes, \(f(x) = f(-x)\)
4. **Periodic**: Yes, but with no fundamental period.
### Conclusion
The correct statements about the function \(f(x)\) are:
- The domain is \(\mathbb{R}\).
- The range is \{0\}.
- The function is even.
- The function is periodic.
