f(x)=(1/2)ax^2+2x-lnx(x>0).(1)a=3,则f'(x)=3x+2-1/x=(3x^2+2x-1)/x=(x+1)(3x-1)/x.函数f(x)的递减区间是(0,1/3),递增区间是(1/3,+无穷).(2)f'(x)=ax+2-1/x=(ax^2+2x-1)/x.a>=0时符合题意.a0,a>-1.此时ax^2+2x-1=0...