Question

已知a∈R,函數(shù)f(x)=x²|x-a|.

(1)當(dāng)a=2時,求使f(x)＞x成立的x的集合;

(2)求f(x)在區(qū)間［1,2］上的最小值.

Answer 1

解:(1)當(dāng)a=2時,f(x)=x²|x-2|,?

當(dāng)x＜2時,f(x)=x²(2-x)＞x,解得x＜0.                                                                ?

當(dāng)x≥2時,f(x)=x²(x-2)＞x,解得x＞1+2,?

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

(2)設(shè)此最小值為M,?

①當(dāng)a≤1時,在區(qū)間［1,2］上,f(x)=x³-ax²,?

因為f′(x)=3x²-2ax=3x(x-a)＞0(x∈［1,2］),??

所以f(x)在［1,2］上遞增.?

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

②當(dāng)1＜a≤2時,在區(qū)間［1,2］上,?

f(x)=x²|x-a|≥0,且f(a)=0.?

∴M=f(a)=0.                                                                                                           ?

③當(dāng)a＞2時,在區(qū)間［1,2］上,f(x)=ax²-x³,?

f′(x)=2ax-3x²=3x(a-x),?

若a≥3,在區(qū)間(1,2)內(nèi)有f′(x)＞0,從而f(x)在［1,2］上遞增,

∴M=f(1)=a-1.?

若2＜a＜3,則1＜a＜2,當(dāng)x∈(1,a)時,f′(x)＞0;??

當(dāng)x∈(a,2)時,f′(x)＜0,故f(x)在［1,a］上遞增,在區(qū)間［a,2］上遞減,因此M=min{f(1),f(2)}.?

當(dāng)f(1)＜f(2),即a-1＜4(a-2),即＜a＜3時,M=f(1)=a-1.?

當(dāng)f(1)≥f(2),即a-1≥4(a-2),即2＜a≤時,M=f(2)=4(a-2).                             ?

綜合上述,所求函數(shù)f(x)的最小值?

m=