Question

若函数f（x）=x3+ax2+x-7在R上单调递增，则实数a的取值范围是？

请详细，谢谢各位大虾

Answer 1

设f(x)=x^3+ax^2+x-7,函数的导函数f’(x)=3x^2+2ax+1.若函数在R上单调递增，则导函数的函数值在R上不为负，即f’(x)≥0在R上恒成立。∵f’(x)为二次函数，∴判别式Δ=4a^2-12≤0，解得：-√3≤a≤√3.

Answer 2

f'(x)=3x^2+2ax+1 令f'(x)在R上恒不小于0 4a^2-12<0
-3^1/2<a<3^1/2

Answer 3

导数没学过？a属于R