Question

3．若函數(shù)f（x）=$\left\{\begin{array}{l}{-2x+1，x＜0}\\{g（x），x＞0}\end{array}\right.$是奇函數(shù)，則f^-1（x）=$\left\{\begin{array}{l}{\frac{1-x}{2}，x＞1}\\{-\frac{x+1}{2}，x＜-1}\end{array}\right.$．

Answer 1

分析 由奇函數(shù)的性質(zhì)求出f（x）=$\left\{\begin{array}{l}{-2x+1，x＜0}\\{-2x-1，x＞0}\end{array}\right.$，由此能求出f^-1（x）．

解答 解：∵函數(shù)f（x）=$\left\{\begin{array}{l}{-2x+1，x＜0}\\{g（x），x＞0}\end{array}\right.$是奇函數(shù)，
∴x＞0時(shí)，g（x）=-f（-x）=-[-2（-x）+1]=-2x-1，
∴f（x）=$\left\{\begin{array}{l}{-2x+1，x＜0}\\{-2x-1，x＞0}\end{array}\right.$，
當(dāng)x＜0時(shí)，y=f（x）=-2x+1，x=$\frac{1-y}{2}$，∴f^-1（x）=$\frac{1-x}{2}$，x＞1，
當(dāng)x＞0時(shí)，y=f（x）=-2x-1，x=-$\frac{y+1}{2}$，∴f^-1（x）=-$\frac{x+1}{2}$，x＜-1．
∴f^-1（x）=$\left\{\begin{array}{l}{\frac{1-x}{2}，x＞1}\\{-\frac{x+1}{2}，x＜-1}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{\frac{1-x}{2}，x＞1}\\{-\frac{x+1}{2}，x＜-1}\end{array}\right.$．

點(diǎn)評 本題考查分段函數(shù)的反函數(shù)的求法，是基礎(chǔ)題，解題時(shí)要認(rèn)真審題，注意反函數(shù)的性質(zhì)的合理運(yùn)用．