[tex]y = \dfrac3{(x+1)^2+2}[/tex]
is continuous over its entire domain (all real numbers). As an interval, you could write this as (-∞, ∞).
A rational expression like this one would be discontinuous wherever the denominator vanishes. But this does not happen in this case, since (x + 1)² is always non-negative (the square of any real number is zero or positive), so
(x + 1)² ≥ 0 ===> (x + 1)² + 2 ≥ 2
