Question

设函数f(x)=ax-a/x-2lnx

⑴若f'(2)=0,求f(x)的单调区间

⑵若f（x）在定义域上是增函数，求a的取值范围

Answer 1

f(x)=ax-a/x-2lnx,x>0,
f'(x)=a+a/x^2-2/x,
f'(2)=5a/4-1=0,a=4/5.
∴f'(x)=[(4/5)x^2-2x+(4/5)]/x^2=(4/5)(x-2)(x-1/2)/x^2,
1/2<x<2时f'(x)<0,f(x)是减函数；0<x<1/2或x>2时f'(x)>0,f(x)是增函数。
(2)f（x）在定义域上是增函数,
∴f'(x)≥0,a+a/x^2≥2/x,
a≥2x/(x^2+1)=2/(x+1/x),
x+1/x≥2,当x=1时取等号，
∴2x/(x^2+1)<=1,
∴a≥1,为所求.