Question

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

Answer 1

分析 令f（x）=t，則f[f（x）]≤3即為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：函數(shù)f（x）=$\left\{\begin{array}{l}{{x}^{2}+2x（x＜0）}\\{-{x}^{2}（x≥0）}\end{array}\right.$，
令f（x）=t，則f[f（x）]≤3即為f（t）≤3，
即$\left\{\begin{array}{l}{t＜0}\\{{t}^{2}+2t≤3}\end{array}\right.$或$\left\{\begin{array}{l}{t≥0}\\{-{t}^{2}≤3}\end{array}\right.$，
則-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≤$\sqrt{3}$．
則解集為（-∞，$\sqrt{3}$]，
故答案為：（-∞，$\sqrt{3}$]．

點(diǎn)評(píng) 本題考查分段函數(shù)及運(yùn)用，考查解不等式時(shí)注意各段的自變量的范圍，同時(shí)考查換元法的運(yùn)用，考查運(yùn)算能力，屬于中檔題．