Question

f（x）=xlnx/（x+1） （1）若任取x∈[1,+∞﹚,f（x）≤m（x-1）恒成立,求m范围

Answer 1

1、即,使xlnx/(x^2-1)≤m在x∈[1,+∞﹚上恒成立

令g(x)=xlnx/(x^2-1)

g'(x)=[(lnx+1)(x^2-1)-2x^2lnx]/(x^2-1)^2=[-x^2lnx+x^2-lnx-1]/(x^2-1)^2

令h(x)=-x^2lnx+x^2-lnx-1

h'(x)=-2xlnx+x-1/x

h''(x)=-2lnx+1/(x^2)-1

h'''(x)=-2/x-2/x^3=-(2/x)*(1+1/x^2)