18. If $b, c \in \mathbf{R}$, the quadratic function $f(x)=x^{2}+b x+c$ has two distinct intersections with the $x$-axis in the interval $(0,1)$. Find the range of $c^{2}+(1+b) c$.

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$$
\begin{array}{l}
f(x)=(x-r)(x-s) . \\
\text { Then } c=f(0)=r s, \\
1+b+c=f(1)=(1-r)(1-s) . \\
\text { Therefore } 0<c^{2}+(1+b) c=f(0) f(1) \\
=r s(1-r)(1-s)<\left(\frac{r+1-r}{2}\right)^{2}\left(\frac{s+1-s}{2}\right)^{2} \\
=\frac{1}{16} .
\end{array}
$$