Question

4．已知直线y₁=-3x与双曲线y₂=-$\frac{9}{x}$，满足y₁＜y₂的x的取值范围为x＜-$\sqrt{3}$或x＞$\sqrt{3}$．

Answer 1

分析 联立直线y₁=-3x与双曲线y₂=-$\frac{9}{x}$得到交点坐标，即可得到结论．

解答 解：联立直线y₁=-3x与双曲线y₂=-$\frac{9}{x}$，
依题意有：$\left\{\begin{array}{l}{{y}_{1}=-3x}\\{{y}_{2}=-\frac{9}{x}}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=-\sqrt{3}}\\{y=3\sqrt{3}}\end{array}\right.$，$\left\{\begin{array}{l}{x=\sqrt{3}}\\{y=-3\sqrt{3}}\end{array}\right.$，
故满足y₁＜y₂的x的取值范围为x＜-$\sqrt{3}$或x＞$\sqrt{3}$．
故答案为：x＜-$\sqrt{3}$或x＞$\sqrt{3}$．

点评 本题考查的是反比例函数与一次函数的交点问题，解答此题的关键是求出直线y₁=-3x与双曲线y₂=-$\frac{9}{x}$的交点坐标．