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)#