4. (2004 National College Entrance Examination, Zhejiang Paper) Given $f(x)=\left\{\begin{array}{ll}1 & x \geqslant 0 \\ -1 & x<0\end{array}\right.$, then the solution set of the inequality $x+(x+2) \cdot$ $f(x+2) \leqslant 5$ is
4. $\left(-\infty, \frac{3}{2}\right]$ Hint: Classify and discuss according to $x \geqslant-2$ and $x<-2$.

