12. Let $f(x)$ be continuous on $\mathbf{R}$, $f(0)=1$, and $f(x+y) \geqslant f(x) f(y)$, find $f(x)$.
12. $f(x+y)-f(x) \geqslant f(x)[f(y)-1]$, that is,
$$
f(x+y)-f(x) \geqslant f(x)[f(y)-f(0)],
$$
(1) Dividing both sides by $y$,
when $y>0$, we get
$$
\frac{1}{(x+y)-x}[f(x+y)-f(x)] \geqslant f(x) \frac{f(y)-f(0)}{y-0}
$$

Let $y \rightarrow 0$. We get $f^{\prime}(x) \geqslant f(x) \cdot f^{\prime}(0)$
head $y>0)$.