用单调性定义证明函数f(x)=x−2x+1在(-1,+∞)上是增函数.
用单调性定义证明函数f(x)=
在(-1,+∞)上是增函数.
x−2 |
x+1 |
数学人气:528 ℃时间:2019-10-23 07:21:36
优质解答
∵f(x)=x−2x+1=1-3x+1,设x1,x2∈(-1,+∞),且x1<x2,∴f(x1)-f(x2)=1-3x1+1-1+3x2+1=3(x1−x2)(x1+1)(x2+1),因为-1<x1<x2,所以x1-x2<0,x1+1>0,x2+1>0,所以f(x1)-f(x2)<0,即f(x1)<f...
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