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To solve for \(\left(\frac{f}{g}\right)(x)\) given \(f(x) = x^2 + x + 1\) and \(g(x) = -x^2 - x - 1\), let's proceed with the following steps:

1. **Define the function** \(\frac{f}{g}(x)\):
   \[
   \frac{f}{g}(x) = \frac{f(x)}{g(x)}
   \]
   Substitute \(f(x)\) and \(g(x)\):
   \[
   \frac{f(x)}{g(x)} = \frac{x^2 + x + 1}{-x^2 - x - 1}
   \]

2. **Simplify the expression**:
   Notice that \(f(x) = -g(x)\) because \(g(x) = - (x^2 + x + 1)\). Therefore, we can rewrite the fraction as:
   \[
   \frac{f(x)}{g(x)} = \frac{x^2 + x + 1}{- (x^2 + x + 1)} = -1
   \]

3. **Conclusion**:
   Since \(\frac{f}{g}(x) = -1\) for all \(x\), the correct answer is:
   \[
   \frac{f}{g}(x) = -1
   \]

Thus, the answer is:
- **Option C: \(\left(\frac{f}{g}\right)(x) = -1\)**

Would you like further details or have any questions on this?

Here are 5 related questions for further practice:
1. How would you find \(\frac{f}{g}(x)\) if \(f(x) = x^2 + 2x + 1\) and \(g(x) = -(x^2 + 2x + 1)\)?
2. What is the result of \(\frac{f}{g}(x)\) if \(f(x) = x^2 + x + 2\) and \(g(x) = -x^2 - x - 2\)?
3. Can you generalize a rule for \(\frac{f}{g}(x)\) when \(f(x) = h(x)\) and \(g(x) = -h(x)\)?
4. How would you simplify \(\frac{f}{g}(x)\) if \(f(x)\) and \(g(x)\) were polynomials that differed only by a constant factor?
5. If \(f(x) = 2x^2 + 3x + 4\) and \(g(x) = -2x^2 - 3x - 4\), what would \(\frac{f}{g}(x)\) simplify to?

**Tip**: Always look for a possible relationship or factor between functions in the numerator and denominator before proceeding with division, as it may simplify your work.

Find \(\left(\frac{f}{g}\right)(x)\) if \(f(x) = x^2 + x + 1\) and \(g(x) = -x^2 - x - 1\).

Learn how to solve for \(\left(\frac{f}{g}\right)(x)\) when \(f(x) = x^2 + x + 1\) and \(g(x) = -x^2 - x - 1\). This problem explores function division and simplifying expressions, suitable for high school students in Grades 10-12.

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