若f(x)=ax^3+lnx则f'(x)=3ax^2+1/x (x>0)若曲线f(x)=ax^3+lnx存在垂直于y轴的切线则f'(x)=3ax^2+1/x=0(x>0)有解f'(x)=3ax^2+1/x=03ax^3+1=0x^3=-1/(3a)因为x>0 所以x^3=-1/(3a)>0实数a的取值范围是(-∞,0)...