Question

已知a∈R,函数f(x)=x²|x-a|.

(1)当a=2时,求使f(x)＞x成立的x的集合;

(2)求f(x)在区间［1,2］上的最小值.

Answer 1

解:(1)当a=2时,f(x)=x²|x-2|,?

当x＜2时,f(x)=x²(2-x)＞x,解得x＜0.                                                                ?

当x≥2时,f(x)=x²(x-2)＞x,解得x＞1+2,?

∴所求x的集合为{x|x＜0或x＞1+2}.                                                                      ?

(2)设此最小值为M,?

①当a≤1时,在区间［1,2］上,f(x)=x³-ax²,?

因为f′(x)=3x²-2ax=3x(x-a)＞0(x∈［1,2］),??

所以f(x)在［1,2］上递增.?

∴M=f(1)=1-a.                                                                                                        ?

②当1＜a≤2时,在区间［1,2］上,?

f(x)=x²|x-a|≥0,且f(a)=0.?

∴M=f(a)=0.                                                                                                           ?

③当a＞2时,在区间［1,2］上,f(x)=ax²-x³,?

f′(x)=2ax-3x²=3x(a-x),?

若a≥3,在区间(1,2)内有f′(x)＞0,从而f(x)在［1,2］上递增,

∴M=f(1)=a-1.?

若2＜a＜3,则1＜a＜2,当x∈(1,a)时,f′(x)＞0;??

当x∈(a,2)时,f′(x)＜0,故f(x)在［1,a］上递增,在区间［a,2］上递减,因此M=min{f(1),f(2)}.?

当f(1)＜f(2),即a-1＜4(a-2),即＜a＜3时,M=f(1)=a-1.?

当f(1)≥f(2),即a-1≥4(a-2),即2＜a≤时,M=f(2)=4(a-2).                             ?

综合上述,所求函数f(x)的最小值?

m=