f(x)=3x^2/(3x-2)
=(1/3)[(3x+2)(3x-2)+4]/(3x-2)
=(1/3)(3x+2)+(4/3)/(3x-2)
=(1/3)(3x-2)+(4/3)/(3x-2)+4/3
>=2√(1/3)(4/3)+4/3=8/3
(1/3)(3x-2)=(4/3)/(3x-2)时,取得最小值
(3x-2)^2=4,x>2/3
x=4/3时有最小值
(2/3,4/3)减区间,(4/3,+∞)增区间