5. Given that the function $f(x)$ is defined and increasing on $(0,+\infty)$, and satisfies $f(x) \cdot f\left(f(x)+\frac{1}{x}\right)=1$, then $f(1)=(\quad)$.
A. I
B. 0
C. $\frac{1+\sqrt{5}}{2}$
D. $\frac{1-\sqrt{5}}{2}$
5. I)

Let $f(1)=a, x=1$, from the known function equation we get $f(1) f(f(1)+1)=1, a f(a+1)$ $=1, f(a+1)=\frac{1}{a}$. Let $x=a+1$, then $f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1, \frac{1}{a} f$ $\left(\frac{1}{a}+\frac{1}{a+1}\right)=1, f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a=f(1)$. Since $f(x)$ is an increasing function, we have $\frac{1}{a}+$ $\frac{1}{a+1}=1$, solving for $a$ gives $a=\frac{1 \pm \sqrt{5}}{2}$. When $f(1)=\frac{1+\sqrt{5}}{2}$, we have $1<a=f(1)<f(a+1)=$ $\frac{1}{a}<1$, which is a contradiction. Therefore, $f(1)=\frac{1-\sqrt{5}}{2}$.