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

Question

↳Redirected from "How do ionic compounds dissolve?"

2108 views around the world

You can reuse this answer
Creative Commons License

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#.