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.