Question

f(x)=2x+1/x-1在(1,正无穷)上为单调增函数

Answer 1

命题错误,应为单调减
（一）
f(x) = (2x+1)/(x-1) = {2(x-1)+3}/(x-1) = 2 + 3/(x-1),x≠1
在(1,正无穷,)x-1单调增,3/(x-1)单调减,2 + 3/(x-1)单调减
（二）
令1＜x1＜x2
f(x2) - f(x1)
= { 2 + 3/(x2-1) } - { 2 + 3/(x1-1) }
= 3/(x2-1)} - 3/(x1-1)
= 3{(x1-1)-(x2-1)}/{(x1-1)(x2-1)}
= 3(x1-x2) /{(x1-1)(x2-1)}
∵1＜x1＜x2
∴x1-x2＜0,x1-1＞0,x2-2＞0
∴f(x2) - f(x1) = 3(x1-x2) /{(x1-1)(x2-1)}＜0
∴f(x2) ＜ f(x1),得证
（三）
f(x) = (2x+1)/(x-1) = {2(x-1)+3}/(x-1) = 2 + 3/(x-1)
f'(x) = -3/(x-1)^2＜0
∴f(x)在(1,正无穷)上为单调减