Problem 4. Solve the inequality

$$
\frac{7}{|x-1|-3} \geqslant|x+2|
$$
Solution. We construct the graphs of the functions $f(x)=|x+2|$, $\varphi(x)=|x-1|-3$, $\psi(x)=\frac{7}{\varphi(x)}$ (Fig. 3). The graphs of these functions can be easily constructed without using derivatives, as they are monotonic on the corresponding intervals.

The lines $x=-2$ and $x=4$ are vertical asymptotes of the graph of the function $\psi(x)$.

To determine the abscissas of points $B$ and $C$, we solve the equations

$$
\frac{7}{(x-1)-3}=x+2 \text{ and } \frac{7}{-(x-1)-3}=-x-2.
$$

We obtain $B(5 ; 7)$ and $C(-\sqrt{7}-2 ; \sqrt{7})$. Using Fig. 3, we get the answer: $[-\sqrt{7}-2 ;-2),(4 ; 5]$.