Question

证明函数f(x)=x+1/x在区间(0,1]上是减函数

Answer 1

设渣局0<x1<x2≤1
f(x2)-f(x1)=x2-x1+1/x2 -1/御氏x1
=(x2-x1)+(x1-x2)/(x1x2)
=(x2-x1)[1-1/如拆让(x1x2)]
因为 0<x1<x2≤1
所以x1x2<1，从而
(x2-x1)[1-1/(x1x2)]<0
f(x1)>f(x2)
f(x)=x+1/x在区间(0,1]上是减函数

Answer 2

取任意实数x1,x2, 且满丛埋足0<x1<x2≤1，
f(x1)-f(x2)=x1+1/x1-x2-1/x2
=(x1x2+x2-x1x2-x1)/x1x2
=(x2-x1)/x1x2>0
∵f(x1)>辩知f(x2) 又∵x1<x2
∴函数f(x)=x+1/x在渗灶蚂区间(0,1]上是减函