6. (The 13th "Hope Cup" Grade 1 High School 2nd Test) Given that the function $f(x)$ is an increasing function on $\mathbf{R}$, and its graph passes through the points $A(0,-2), B(3,2)$, then the solution set of the inequality $|f(x+1)| \geqslant 2$ is
A. $(-\infty,-1) \cup[2,+\infty)$
B. $[2,+\infty)$
C. $(-\infty,-1]$
D. $[3,+\infty)$
6. The original inequality is equivalent to $f(x+1) \geqslant 2$ or $f(x+1) \leqslant-2$.

Since the graph of the function $f(x)$ passes through the points $A(0,-2), B(3,2)$,
we have $f(x+1) \geqslant f(3), f(x+1) \leqslant f(0)$.
Also, the function $f(x)$ is an increasing function on $\mathbf{R}$.
Therefore, $x+1 \geqslant 3$ or $x+1 \leqslant 0$,
which means $x \geqslant 2$ or $x \leqslant-1$, so the answer is $\mathrm{A}$.
Another solution: This problem only specifies that the function $f(x)$ is a decreasing function on $\mathbf{R}$ passing through the points $A(0,-2)$ and $B(3,2)$. It does not specify the type of curve. We can assume the curve is a straight line passing through points $A$ and $B$. Using the two-point form, we get $f(x)=\frac{4}{3} x-2$, so $f(x+1)=$ $\frac{4}{3}(x+1)-2=\frac{4}{3} x-\frac{2}{3}$. Solving $\left|\frac{4}{3} x-\frac{2}{3}\right| \geqslant 2$ yields $x \geqslant 2$ or $x \leqslant-11$, so the answer is A.

This is a specialization strategy, which involves finding a special function that meets the requirements of the problem based on the given information, and using this special function to solve the problem.