令f(x)=y
f'(x)=2(x-3)(x-2)+(x-3)²
=(x-3)(2(x-2)+(x-3))
=(x-3)(3x-7)
若f'(x)=0,则x=3或x=7/3
x0,则x=3时,f(x)取极小值
f(3)=0
则y极小值为x=3时,y=0
极大值为x=7/3时,y=4/27