求函数的极值?

y = x³/(x - 1)²
dy/dx = [3x²(x-1)² - 2x³(x-1)]/(x-1)⁴
= [3x²(x-1) - 2x³]/(x-1)³
= (x³-3x²)/(x-1)³
= x²(x-3)/(x-1)³
令dy/dx = 0
得x₁= x₂= 0, x₃= 3

∵dy/dx = (x³-3x²)/(x-1)³
∴d²y/dx²= [(3x²-6x)(x-1)³ - 3(x³-3x²)(x-1)²]/(x-1)^6
= [(3x²-6x)(x-1) - 3(x³-3x²)]/(x-1)⁴
= x[(3x-6)(x-1) - 3(x²-3x)]/(x-1)⁴
= x(3x²-9x+6-3x²+9x)/(x-1)⁴
= 6x/(x-1)⁴
当x=0时，d²y/dx²=0，拐点
当x=3时，d²y/dx²=18/16=9/8>0，极小值点
y(min)=3³/(3 - 1)²=27/4
当x=1,y=∞,x=1是垂直渐近线。

解：求导得，y'=[(x-3)*x^2]/(x-1)^3.===>y'(0)=y'(3)=0.===>ymin=y(3)=27/4.ymax=+∞.