Question

7．已知函数f（x）=$\left\{\begin{array}{l}{{x}^{2}+2x（x＜0）}\\{-{x}^{2}（x≥0）}\end{array}\right.$，则不等式f[f（x）]≤3的解集为（-∞，$\sqrt{3}$]．

Answer 1

分析 令f（x）=t，则f[f（x）]≤3即为f（t）≤3，解得t≥-3，得到$\left\{\begin{array}{l}{x＜0}\\{{x}^{2}+2x≥-3}\end{array}\right.$或$\left\{\begin{array}{l}{x≥0}\\{-{x}^{2}≥-3}\end{array}\right.$，解得即可．

解答 解：函数f（x）=$\left\{\begin{array}{l}{{x}^{2}+2x（x＜0）}\\{-{x}^{2}（x≥0）}\end{array}\right.$，
令f（x）=t，则f[f（x）]≤3即为f（t）≤3，
即$\left\{\begin{array}{l}{t＜0}\\{{t}^{2}+2t≤3}\end{array}\right.$或$\left\{\begin{array}{l}{t≥0}\\{-{t}^{2}≤3}\end{array}\right.$，
则-3≤t＜0或t≥0，
即有t≥-3．
即f（x）≥-3．
即有$\left\{\begin{array}{l}{x＜0}\\{{x}^{2}+2x≥-3}\end{array}\right.$或$\left\{\begin{array}{l}{x≥0}\\{-{x}^{2}≥-3}\end{array}\right.$，
解得x＜0或0≤x≤$\sqrt{3}$，即有x≤$\sqrt{3}$．
则解集为（-∞，$\sqrt{3}$]，
故答案为：（-∞，$\sqrt{3}$]．

点评 本题考查分段函数及运用，考查解不等式时注意各段的自变量的范围，同时考查换元法的运用，考查运算能力，属于中档题．