Question

已知函数f(x)=lnx-2(x-1)/(x+1)

已知函数f(x)=lnx-2(x-1)/(x+1)探究f(x)的单调性，并求f(x)的单调区间；(2)证明：当x>0时，不等式（x^2-1)lnx>=2(x-1)^2恒成立

Answer 1

解:由题意f'(x)=1/x-(2(x+1)-2(x-1))/(x+1)^2=1/x-4/(x+1)^2=((x+1)^2-4x)/x(x+1)^2=(x-1)^/x(x+1)^2，则在定义域x>0上f'(x)》0所以，f(x)在x>0单增