Answer
$(-\infty,\infty)$
Work Step by Step
We are given the function:
$g(x)=\dfrac{3x^2-6x+7}{x^2+x+1}$
$g(x)$ is a rational function, therefore according to Theorem 2.10 b), the function is continuous for all $x$ for which the denominator is not zero.
We check the zeros of the denominator:
$x^2+x+1=0$
$\Delta=1^2-4(1)(1)=-3<0$
As the denominator has no zeros, the interval on which the function is continuous is:
$(-\infty,\infty)$
