f(x)=ln(x)/(x-1)²
f'(x)=[(x-1)/x-2lnx]/(x-1)²
驻点:(x₀-1)/x₀-2lnx₀=0 x₀∈(0,½)
极值f(x₀)=ln(x₀)/(x₀-1)²=[½(x₀-1)/x₀]/(x₀-1)²=1/[2x₀(x₀-1)]
令g(x)=1/[2x(x-1)]
g'(x)=-½[x(x-1)]'/[x(x-1)]² =-½(2x-1)/[x(x-1)]²
驻点:x₀=½ 左+右- 为极大值点 极大值g(½)=-2
x₀∈(0,½)→极值f(x₀)=g(x)<g(½)=-2
追问请问为什么x0在0和1/2之间啊