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=0
3ax^3+1=0
x^3=-1/(3a)
因为x>0 所以x^3=-1/(3a)>0
实数a的取值范围是(-∞,0)
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=0
3ax^3+1=0
x^3=-1/(3a)
因为x>0 所以x^3=-1/(3a)>0
实数a的取值范围是(-∞,0)