1. Let $f(x)$ be a function defined on $R$, for any real number $x$ we have $f(x+3) \cdot f(x-4)=-1$. Also, when $0 \leq x<7$, $f(x)=\log _{2}(9-x)$, then the value of $f(-100)$ is . $\qquad$
1.

Answer: $-\frac{1}{2}$.
Solution: From the given condition, $f(x+14)=-\frac{1}{f(x+7)}=f(x)$, so $f(-100)=f(-100+14 \times 7)=f(-2)=-\frac{1}{f(5)}=-\frac{1}{\log _{2} 4}=-\frac{1}{2}$.