Question

11．定義在R上的函數(shù)f（x）是奇函數(shù)，且當(dāng)x＞0時(shí)，f（x）=e^x+1，則x∈R時(shí)，f（x）=$\left\{\begin{array}{l}{{e}^{x}+1，}&{x＞0}\\{0，}&{x=0}\\{-{e}^{-x}-1，}&{x＜0}\end{array}\right.$，．

Answer 1

分析 根據(jù)函數(shù)奇偶性的性質(zhì)，利用對(duì)稱性進(jìn)行求解即可．

解答 解：若x＜0，則-x＞0，則f（-x）=e^-x+1，
∵函數(shù)f（x）是R上的奇函數(shù)，
∴f（0）=0，f（-x）=e^-x+1=-f（x），
即f（x）=-e^-x-1，
故f（x）=$\left\{\begin{array}{l}{{e}^{x}+1，}&{x＞0}\\{0，}&{x=0}\\{-{e}^{-x}-1，}&{x＜0}\end{array}\right.$，
故答案為：$\left\{\begin{array}{l}{{e}^{x}+1，}&{x＞0}\\{0，}&{x=0}\\{-{e}^{-x}-1，}&{x＜0}\end{array}\right.$．

點(diǎn)評(píng) 本題主要考查函數(shù)解析式的求解，利用函數(shù)奇偶性的性質(zhì)是解決本題的關(guān)鍵．