18. Let the variable $x$ satisfy the inequality: $x^{2}+b x \leqslant-x(b<-1)$, and the minimum value of $f(x)=x^{2}+b x$ is $-\frac{1}{2}$, find the value of the real number $b$.
18. From $x^{2}+(b+1) x \leqslant 0$ and $b<-1$, we can solve to get $0 \leqslant x \leqslant-(b+1)$. Also, $f(x)=\left(x+\frac{b}{2}\right)^{2}-\frac{b^{2}}{4}$.
(1) When $-(b+1)<-\frac{b}{2}$, i.e., $-2<b<-1$, $f(x)$ is monotonically decreasing on $[0,-(b+1)]$, then $f[-(b+1)]$ is the minimum value, i.e., $f(-b-1)=\left(\frac{b}{2}+1\right)^{2}-\frac{b^{2}}{4}=b+1=-\frac{1}{2}$.
From this, we can solve to get $b=-\frac{3}{2} \in(-2,-1)$.
(2) When $-\frac{b}{2} \leqslant-(b+1)$, i.e., $b \leqslant-2$, $f\left(-\frac{b}{2}\right)$ is the minimum value, i.e., $f\left(-\frac{b}{2}\right)=-\frac{b^{2}}{4}=-\frac{1}{2}$, solving gives $b=$ $\pm \sqrt{2} \notin(-\infty,-2]$.
Therefore, the required $b=-\frac{3}{2}$.