(Ⅰ)∵f(x)=mx3-(2+)x2+4x+1,∴f′(x)=mx2-(4+m)x+4=(mx-4)(x-1) 1)若m>4,则0<<1,此时x∈(-∞,)∪(1,+∞)都有f/(x)>0,x∈(,1), 有f′(x)<0,∴f(x)的单调递增区间为(-∞,]和[[1,+∞); 2)若m=4,则f′(x)=4(x-1)2≥0,∴f(x)的单调递增区间为(-∞,+∞). (Ⅱ)当m<0时,f/(x)=mx2-(4+m)x+4=m(x-)(x-1)且<1 ∴当2≤x≤3时,都有f′(x)<0 ∴此时f(x)在[2,3]上单调递减,∴f(x)max=f(2)=m+1 又g(x)=mx+5在[2,3]上单调递减,∴g(x)min=g(3)=3m+5 ∴m+1-3m-5≤1,解得m≥-,又m<0, 所以-≤m<0 |