问题描述: 已知函数f(x)=x+1分之2-X,证明f(x)在(-1.+∞)上为减函数,急,在线等. 1个回答 分类:数学 2014-11-05 问题解答: 我来补答 f(x)=(2-x)/(x+1)=(3-x-1)/(x+1)=3/(x+1) -1∵x+1在(-1.+∞)上单调增∴3/(x+1)在(-1.+∞)上单调减∴f(x)=在(-1.+∞)上单调减如果用定义证明:令-1<x1<x2f(x2)-f(x1)=3/(x2+1) -1-3/(x1+1) +1=3[1/(x2+1) -1/(x1+1) ]=3(x1+1-x2-1) / [(x2+1)(x1+1) ]=3(x1-x2) / [(x2+1)(x1+1) ]∵(x1-x2)<0,[(x2+1)(x1+1) >0∴3(x1-x2) / [(x2+1)(x1+1) ] <0即:f(x2)-f(x1)<0,f(x2)<f(x1),得证. 展开全文阅读