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

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Answer 1 · 2016-07-16T07:25:43

The domain is the whole of the Real numbers, i.e. $(-oo, oo)$

$x^2+x+1$ is in the form $ax^2+bx+c$ with $a = b = c = 1$

This has discriminant:

$Delta = b^2-4ac = 1-4 = -3$

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

Let:

$f(x) = 1/(x^2+x+1)$

Then for all Real values of $x$ , $x^2+x+1 != 0$ , so $f(x)$ is defined.

So the domain of $f(x)$ is all of the Real numbers $RR$ , i.e. $(-oo, oo)$

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