Question

已知函数y=f(x)的导数f'(x)=x²(x–1),则该函数有几个极值点？

Answer 1

令f'(x)=0，得x=0或x=1
x<0时，f'(x)<0；
0<x<1时，f'(x)<0；
f(x)在x<1时递减。
x>1时，f'(x)>0，f(x)递增
故f(x)只有一个极值点，即极小值点x=1

Answer 2

令 f(x) = (x–1)(x-2)²(x-3)³ = y
lny = ln(x-1) + 2ln(x-2) + 3ln(x-3)
1/y * y ′ = 1/(x-1) + 2/(x-2) + 3/(x-3)
y ′ = y * {1/(x-1) + 2/(x-2) + 3/(x-3)}
= (x–1)(x-2)²(x-3)³ * {1/(x-1) + 2/(x-2) + 3/(x-3)}
= (x-2)²(x-3)³ + 2(x–1)(x-2)(x-3)³ + 3 (x–1)(x-2)²(x-3)²
f ′(1) = (1-2)²(1-3)³ + 2(1–1)(1-2)(1-3)³ + 3 (1–1)(1-2)²(1-3)²
= (-1)² * (-2)³ + 0 + 0
= - 8