Question

14．函數(shù)y=$\left\{\begin{array}{l}{2x，x≥0}\\{-{x}^{2}，x＜0}\\{\;}\end{array}\right.$的反函數(shù)是（　�。�

• A． y=$\left\{\begin{array}{l}{\frac{x}{2}，x≥0}\\{\sqrt{-x}，x＜0}\\{\;}\end{array}\right.$
• B． y=$\left\{\begin{array}{l}{\frac{x}{2}，x≥0}\\{-\sqrt{-x}，x＜0}\\{\;}\end{array}\right.$
• C． y=$\left\{\begin{array}{l}{2x，x≥0}\\{\sqrt{-x}，x＜0}\end{array}\right.$
• D． y=$\left\{\begin{array}{l}{2x，x≥0}\\{-\sqrt{-x}，x＜0}\\{\;}\end{array}\right.$

Answer 1

分析 利用反函數(shù)的求法、分段函數(shù)的性質(zhì)即可得出．

解答 解：∵y=$\left\{\begin{array}{l}{2x，x≥0}\\{-{x}^{2}，x＜0}\\{\;}\end{array}\right.$，x≥0時(shí)，由y=2x，解得x=$\frac{1}{2}y$，把x與y互換可得：y=$\frac{1}{2}$x；
x＜0，由y=-x²，解得x=-$\sqrt{-y}$，把x與y互換可得：y=$-\sqrt{-x}$．
∴函數(shù)y=$\left\{\begin{array}{l}{2x，x≥0}\\{-{x}^{2}，x＜0}\\{\;}\end{array}\right.$的反函數(shù)是y=$\left\{\begin{array}{l}{\frac{x}{2}，x≥0}\\{-\sqrt{-x}，x＜0}\end{array}\right.$．
故選：B．

點(diǎn)評(píng) 本題考查了反函數(shù)的求法、分段函數(shù)的性質(zhì)，考查了推理能力與計(jì)算能力，屬于中檔題．