12. (2007 National High School Mathematics League Competition Jiangsu Province Preliminary Question) Given the graph of the function $y=f(x)$ as shown in Figure $1-4$, then the range of $x$ that satisfies $f\left(\frac{2 x^{2}-x-1}{x^{2}-2 x+1}\right) \cdot f\left(\lg \left(x^{2}-6 x+20\right)\right) \leqslant 0$ is $\qquad$ .
12. $[-2,1)$ Since $\lg \left[x^{2}-6 x+20\right]=\lg \left[(x-3)^{2}+11\right] \geqslant \lg 11>1$, it follows that $f\left[\lg \left(x^{2}-6 x+20\right)\right]<0$. Therefore, from the graph, $\frac{2 x+1}{x-1} \leqslant 1$. That is, $\frac{x+2}{x-1} \leqslant 0$, solving this yields $-2 \leqslant x<1$. Hence, the range of $x$ is $x \in[-2,1)$.