（1）∵f（x）=ax（x-2）2=ax3-4ax2+4ax,∴f′（x）=3ax2-8ax+4a．由f′（x）=0,得3ax2-8ax+4a=0．∵a≠0,∴3x2-8x+4=0．解得x=2或x=2/3∵a＞0,∴x＜2/3或x＞2时,f′（x）＞0；2/3＜x＜2时,f′（x）＜0．∴当x=2/3时...