Question

13．若f（x）=$\left\{\begin{array}{l}{-{x}^{2}+2x（x≥0）}\\{-{x}^{2}-2x（x＜0）}\end{array}\right.$，则不等式f（x）+3＜0的解集为{x|x＞3或x＜-3}．

Answer 1

分析 由题意可得$\left\{\begin{array}{l}{-{x}^{2}+2x+3＜0}\\{x≥0}\end{array}\right.$或$\left\{\begin{array}{l}{-{x}^{2}-2x+3＜0}\\{x＜0}\end{array}\right.$，分别解不等式组即可．

解答 解：f（x）=$\left\{\begin{array}{l}{-{x}^{2}+2x（x≥0）}\\{-{x}^{2}-2x（x＜0）}\end{array}\right.$，则不等式f（x）+3＜0，
则$\left\{\begin{array}{l}{-{x}^{2}+2x+3＜0}\\{x≥0}\end{array}\right.$或$\left\{\begin{array}{l}{-{x}^{2}-2x+3＜0}\\{x＜0}\end{array}\right.$，
即$\left\{\begin{array}{l}{（x-3）（x+1）＞0}\\{x≥0}\end{array}\right.$或$\left\{\begin{array}{l}{（x+3）（x-1）＞0}\\{x＜0}\end{array}\right.$，
解得x＞3或x＜-3，
故不等式的解集为{x|x＞3或x＜-3}，
故答案为：{x|x＞3或x＜-3}．

点评 本题考查了分段函数和不等式组的解法，属于基础题．