14. Given the quadratic function $f(x)$ satisfies: (1) $f(-1)=0$; (2) for all values of $x$, $x \leqslant f(x) \leqslant \frac{1+x^{2}}{2}$, find the analytical expression of $f(x)$.
14. From (2), let $x=1, f(1)=1$, assume $f(x)=a x^{2}+b x+c$, then $a+b+c=1, a-b+c=0$. Therefore, $b=$ $a+c=\frac{1}{2}$, and $a x^{2}+b x+c \geqslant x$, for all real numbers $x$, hence $a>0$, and $\Delta=(b-1)^{2}-4 a c \leqslant 0, a c \geqslant \frac{1}{16}$. So $a=c=\frac{1}{4}, b=\frac{1}{2}, f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.