Question

f(x)=（a+1）lnx+ax^2+1,a=4丨x1-x2丨,求a的取值范围

Answer 1

导数f'(x)=（a+1）/x+2ax,

因为a0,

所以f'(x)x2,则f(x1)=4(x1-x2),亦即f(x2)+4x2>=f(x1)+4x1,

设g(x)=f(x)+4x,则g(x)在（0,+oo）上递减,

所以g'(x)