Question

证明f(x)=x+1/x在(1,+∞)内是增函数

Answer 1

令x2＞x1＞1
f(x2)-f(x1) = (x2+1/x2) - (x1+1/x1) = (x2-x1) + (1/x2-1/x1) = (x2-x1) - (x2-x1)/(x1x2)
= (x2-x1)(x1x2-1)/(x1x2)
∵x2＞x1，∴x2-x1＞弯和0
∵x2＞x1＞1，∴x1x2＞枝绝1，∴(x1x2-1)/埋搭盯(x1x2)＞0
∴f(x2)-f(x1) = (x2-x1)(x1x2-1)/(x1x2)＞0，得证

Answer 2

当f'(x)=1-1/(x*x).令f'(x)=0.当(x>0)时，x=1。x>喊御1时，f'(x)>0.从而f(x)=x+1/x在(1,+∞)内昌渗模是增函耐缓数。

Answer 3

设孙消x2>x1>1
那么f(x2)-f(x1)=x2-x1+(1/X2-1/X1)
=(x2-x1)(1-1/x1x2)
x2>x1>1
x1x2>1那么1-1/x1x2>0 x2-x1>0
所以御前f(x2)-f(x1)>0 即 f(x2)>f(x1)

对于x2>x1 有f(x2)>f(x1)
所以 f(x)=x+1/x在(1,+∞)内是增函数镇凯清

Answer 4

f(x)=x+1/x
f'(x)=1-1/x^2>0
(x属于派升(1,+∞)时稿氏)
所以
f(x)=x+1/x在(1,+∞)内是增函键羡散数.