f(x)=x³-px²-qx
f'(x)=3x²-2px-q
f'(1)=0
3*1²-2p*1-q=0
3-2p-q=0.....(1)
1³-p*1²-q*1=0
1-p-q=0......(2)
(2)-(1): -2+p=0
p=2
1-2-q=0
q=-1
f(x)=x³-2x²+x
f'(x)=3x²-4x+1
极值:f'(x)=0
3x²-4x+1=0
(3x-1)(x-1)=0
3x-1=0
x=1/3
x-1=0
x=1
f(1/3)=(1/3)³-2*(1/3)²+1/3=1/27-2/9+1/3=4/27
f(1)=1³-2*1²+1=0
极大值:f(1/3)=4/27
函数图象过点(1,0),带入得p+q=1
切点在(1,0),所以y的导数在该点值为0,得到2p+q=3
联立得到p=2,q=-1