根据题意有f'(x)=0的解为x=1或x=2
f'(x)=a/x+2bx+1=(2bx^2+x+a)/x
-1/2b=1+2=3=>b=-1/6
a/2b=1*2=2=>a=-2/3
所以f'(x)=[(-1/3)x^2+x-2/3]/x=-(x-1)(x-2)/3x
(-∞,0)f'(x)>0,单增
(0,1]f'(x)<0,单减
(1,2]f'(x)>0,单增
(2,+∞)f'(x)<0,单减
所以x=1是极小值点,x=2是极大值点.