已知函数f(x)=㏑x-(x+a)/﹙x-1﹚,(a为常数)若f(x)在[2,+∞)上单调递增,则实数a的取值范围为?
问题描述:
已知函数f(x)=㏑x-(x+a)/﹙x-1﹚,(a为常数)若f(x)在[2,+∞)上单调递增,则实数a的取值范围为?
答
f(x)=lnx - (x+a)/(x-1)
f'(x)=1/x- [(x-1)-(x+a)]/(x-1)^2
=1/x + [1+a]/(x-1)^2
=(x-1)^2/x(x-1)^2 + [x+ax]/x(x-1)^2
=[x^2+1-2x+x+ax]/x(x-1)^2
=[x^2+(a-1)x+1]/x(x-1)^2
令f'(x)