Discussion
The domain of \(f(x) = \frac{x^{2} + 1}{x^{2} + x + 1}\) is
Answer: A
The domain is \(\mathbb{R}\), all real numbers.
A rational function is undefined only where its denominator is zero. The denominator \(x^{2} + x + 1\) has discriminant \(1 - 4 = -3 \lt 0\), so it has no real roots; in fact
\[ \begin{aligned} x^{2} + x + 1 &= \left(x + \tfrac{1}{2}\right)^{2} + \tfrac{3}{4} \\ &\gt 0 \end{aligned} \]
for every real \(x\). Since the denominator never vanishes, no real number is excluded, and the options removing 2, 1, 0 or \(\pm 1\) are wrong; none of those values makes the denominator zero.
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