What is the domain of the function #1/(x^2+x+1)+x# ?

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Answer 1 · 2017-04-02T10:21:27

#RR#

Notice that:

#x^2+x+1 = (x+1/2)^2+3/4#

So:

#x^2+x+1 >= 3/4 > 0# for all real values of #x#

Hence, for any real value of #x#, the given expression:

#1/(x^2+x+1)+x#

is well defined.

So the domain is the whole of the real numbers #RR#.