Question

13．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：由y=2x，x≥0，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，x＜0，解得x=-$\sqrt{-y}$，把x與y互換可得：y=-$\sqrt{-x}$，（x＜0）．
∴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 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，屬于基礎(chǔ)題．