Identify the type of the functions: `f(x)=log((x^4+x^2+1)/(x^2+x+1))`

Options

A. Odd
B. Even
C. Neither
D. Both

Explanation

To identify the type of the function \( f(x) = \log\left(\frac{x^4 + x^2 + 1}{x^2 + x + 1}\right) \), we need to determine whether it is an odd function, an even function, or neither.

### Step-by-Step Solution:

1. **Definition of Even and Odd Functions**:
- A function \( f(x) \) is **even** if \( f(-x) = f(x) \) for all \( x \).
- A function \( f(x) \) is **odd** if \( f(-x) = -f(x) \) for all \( x \).

2. **Calculate \( f(-x) \)**:
- Substitute \(-x\) into the function:
\[ f(-x) = \log\left(\frac{(-x)^4 + (-x)^2 + 1}{(-x)^2 + (-x) + 1}\right) \]
- Simplifying the terms:
- \((-x)^4 = x^4\)
- \((-x)^2 = x^2\)
- Therefore, the numerator becomes \( x^4 + x^2 + 1 \).
- The denominator becomes \( x^2 - x + 1 \).
- Thus, we have:
\[ f(-x) = \log\left(\frac{x^4 + x^2 + 1}{x^2 - x + 1}\right) \]

3. **Compare \( f(-x) \) with \( f(x) \)**:
- We already have:
\[ f(x) = \log\left(\frac{x^4 + x^2 + 1}{x^2 + x + 1}\right) \]
- Now, we need to check if \( f(-x) \) is equal to \( f(x) \) or \(-f(x)\):
- \( f(-x) \) is:
\[ \log\left(\frac{x^4 + x^2 + 1}{x^2 - x + 1}\right) \]
- Clearly, \( f(-x) \neq f(x) \) because the denominators are different.
- Now, check if \( f(-x) = -f(x) \):
- For \( f(-x) = -f(x) \), we would need:
\[ \log\left(\frac{x^4 + x^2 + 1}{x^2 - x + 1}\right) = -\log\left(\frac{x^4 + x^2 + 1}{x^2 + x + 1}\right) \]
- This implies:
\[ \frac{x^4 + x^2 + 1}{x^2 - x + 1} = \frac{1}{\frac{x^4 + x^2 + 1}{x^2 + x + 1}} \]
- This equation does not hold true for all \( x \).

4. **Conclusion**:
- Since \( f(-x) \neq f(x) \) and \( f(-x) \neq -f(x) \), we conclude that the function \( f(x) \) is neither even nor odd.

### Final Answer:
The function \( f(x) = \log\left(\frac{x^4 + x^2 + 1}{x^2 + x + 1}\right) \) is neither even nor odd.