2. Let the function $f(x)=x^{3}+3 x^{2}+6 x+14$, and $f(a)=1, f(b)=19$, then $a+b=$
2. -2 Explanation: Given $f(x)=x^{3}+3 x^{2}+6 x+14=(x+1)^{3}+3(x+1)+10$, let $g(y)=y^{3}+3 y$, then $g(y)$ is an odd function and monotonically increasing.
Since $f(a)=(a+1)^{3}+3(a+1)+10=1, f(b)=(b+1)^{3}+3(b+1)+10=19$, we have $g(a+1)=-9, g(b+1)=9, g(-b-1)=-9$, thus $g(a+1)=g(-b-1)$, which means $a+1=-b-1$, hence $a+b=-2$.