(7) Given that the function $f(x)$ is an increasing function on $\mathbf{R}$, and $A(0,-1)$, $B(3,1)$ are two points on its graph, then the solution set of $|f(x+1)|<1$ is ( ).
(A) $(1,4)$
(B) $(-1,2)$
(C) $(-\infty, 1) \cup[4,+\infty)$
(D) $(-\infty,-1) \cup[2,+\infty)$
7 Because $f(x)$ is an increasing function on $\mathbf{R}$, and $f(0)=-1, f(3)=1$, when $|f(x+1)|<1$, we have $-1<f(x+1)<1$, which means $0<x+1<3$, thus $-1<x<2$. Therefore, the answer is B.