5. Given that $f(x)$ is an increasing function on $\mathbf{R}$, and $A(0,-1)$, $B(3,1)$ are two points on its graph. Then, the complement of the solution set of $|f(x+1)|<1$ is ( ).
(A) $(3,+\infty)$
(B) $[2,+\infty$ )
(C) $(-1,2)$
(D) $(-\infty,-1] \cup[2,+\infty)$
5.D.

Using $0<x+1<3$ we get the solution set of the original inequality as $(-1,2)$.