# How do you find the domain of 1/(x^2+x+1)?

Jul 16, 2016

The domain is the whole of the Real numbers, i.e. $\left(- \infty , \infty\right)$

#### Explanation:

${x}^{2} + x + 1$ is in the form $a {x}^{2} + b x + c$ with $a = b = c = 1$

This has discriminant:

$\Delta = {b}^{2} - 4 a c = 1 - 4 = - 3$

Since $\Delta < 0$ this quadratic has no Real zeros.

Let:

$f \left(x\right) = \frac{1}{{x}^{2} + x + 1}$

Then for all Real values of $x$, ${x}^{2} + x + 1 \ne 0$, so $f \left(x\right)$ is defined.

So the domain of $f \left(x\right)$ is all of the Real numbers $\mathbb{R}$, i.e. $\left(- \infty , \infty\right)$