Question

已知函数f(x)=|1/x-1|

Answer 1

（1）f(x)=|1/x-1|在x∈[1，+∞)上单调递增
证明：设1≤x₁＜x₂则
x₁-x₂＜0
x₂x₁＞1＞0
f(x₁)-f(x₂)
=|1/x₁-1|-|1/x₂-1|
=(1-1/x₁)-(1-1/x₂)
=1/x₂-1/x₁
=(x₁-x₂)/(x₂x₁)
＜0
∴ f(x₁)＜f(x₂)
∴ f(x)=|1/x-1|在x∈[1，+∞)上单调递增
（2）若1/2≤x₁＜x₂＜1则
f(x₁)-f(x₂)
=|1/x₁-1|-|1/x₂-1|
=(1/x₁-1)-(1/x₂-1)
=1/x₁-1/x₂
=(x₂-x₁)/(x₂x₁)
＞0
∴ f(x)=|1/x-1|在x∈[1/2，1)上单调递减
又 f(1/2)=1 f(2)=1/2
∴ f(x)=|1/x-1|在x∈[1/2，2]上有最小值且f(x)最小=f(1)=0
∴ f(x)=|1/x-1|在x∈[1/2，2]上的值域为[0,1]
∴A=B=[0,1]

追问

thanks =^3^=