f '(x)=(x-m)²+2x(x-m)因为在x=1处取得极小值所以f '(1)=0即(1-m)²+2(1-m)=0(1-m)(1-m+2)=0(1-m)(3-m)=0解得m=1或m=3①当m=1时,f '(x)=(x-1)²+2x(x-1)=(x-1)(3x-1),若1/3＜x＜1时,f '(x)＜0若x＞1时,...