19. The function $f$ is defined on $\mathbf{R}$ and satisfies the following conditions:
(1) $f(x)$ is strictly increasing:
(2) For any $x \in \mathbf{R}^{+}$, $f(\vec{c})>-\frac{1}{x}$;
(3) For any $x \in \mathbf{R}$, $f(x) f\left[f(x)+\frac{1}{x}\right]=1$. Find $f(1)$ and provide a function that satisfies the above conditions.
19. Let $f(1)=a$, from condition (3) we know
$$
\text { af }(a+1)=1 \text {, }
$$

so
$$
f(a+1)=\frac{1}{a} .
$$

Also,
$$
f(a+1) f\left[f(a+1)+\frac{1}{a+1}\right]=1,
$$

so
$$
f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a=f(1) .
$$

Since $f$ is strictly increasing, we have
$$
\frac{1}{a}+\frac{1}{a+1}=1
$$

Solving for $a$ gives $a=\frac{1 \pm \sqrt{5}}{2}$.
If $a=\frac{1+\sqrt{5}}{2}$, then $1<a=f(1)<f(a+1)=\frac{1}{a}<1$, which is a contradiction.
Therefore, $f(1)=a=\frac{1-\sqrt{5}}{2}$.
Additionally, constructing the function $f(x)=\frac{1-\sqrt{5}}{2 x}$, it is easy to verify that this function satisfies the above three conditions.