证明函数f(x)=x+1⼀x在(0,+∞)的单调性

2025-06-22 00:38:54
推荐回答(1个)
回答1:

设任意x1,x2∈(0,+∞),且x1则f(x1)-f(x2)=x1+1/x1-x2-1/x2
=(x1-x2)+(x2-x1)/x1x2
=(x1-x2)+[(x2-x1)]/x1x2
=(x1-x2)[1-1/x1x2]
=(x1-x2)[(x1x2-1)/(x1x2)]
x1-x2<0,x1x2>0,
若x1 x2∈(0,1],则x1x2<1 所以f(x1)>f(x2)
所以,f(x)在(0,1]上是减函数
若x1x2∈(1,+∞),则x1x2>1 所以f(x1)所以,f(x)在[1,+∞)上是增函数