f(x)=x
x^2-(p+1)x+q=0有唯一解是x=2,即有(x-2)^2=0
x^2-4x+4=0
对比得到p=3,q=4
故有f(x)=x^2-3x+4
f(x-1)=(x-1)^2-3(x-1)+4=x^2-5x+8<=x
x^2-6x+8<=0
(x-2)(x-4)<=0
2<=x<=4
f(x)=(x-3/2)^2-9/4+4=(x-3/2)^2+7/4
在[2,4]上是增函数,则有最大值是f(4)=16-12+4=8,最小值是f(2)=4-6+4=2