Question

7．已知函數(shù)f（x）是奇函數(shù)，且當(dāng)x＞0時(shí)，$f（x）=-\sqrt{x+1}$，則當(dāng)x∈R時(shí)，f（x）的解析式為f（x）=$\left\{\begin{array}{l}{-\sqrt{x+1}，x＞0}\\{0，x=0}\\{\sqrt{-x+1}，x＜0}\end{array}\right.$．

Answer 1

分析 要求函數(shù)的解析式，已知已有x＞0時(shí)的函數(shù)解析式，只要根據(jù)題意求出x＜0及x=0時(shí)的即可，根據(jù)奇函數(shù)的性質(zhì)容易得f（0）=0，而x＜0時(shí)，由-x＞0及f（-x）=-f（x）可求．

解答 解：設(shè)x＜0，則-x＞0
∵當(dāng)x＞0時(shí)，$f（x）=-\sqrt{x+1}$，∴f（-x）=-$\sqrt{-x+1}$
由函數(shù)f（x）為奇函數(shù)可得f（-x）=-f（x）
∴f（x）=-f（-x）=$\sqrt{-x+1}$，x＜0
∵f（0）=0
∴f（x）=$\left\{\begin{array}{l}{-\sqrt{x+1}，x＞0}\\{0，x=0}\\{\sqrt{-x+1}，x＜0}\end{array}\right.$．
故答案為：f（x）=$\left\{\begin{array}{l}{-\sqrt{x+1}，x＞0}\\{0，x=0}\\{\sqrt{-x+1}，x＜0}\end{array}\right.$．

點(diǎn)評(píng) 本題主要考查了利用函數(shù)的奇偶性求解函數(shù)的解析式，解題中要注意函數(shù)的定義域是R，不用漏掉對(duì)x=0時(shí)的考慮．