Question

已知函數(shù)=,x∈［0,1］.

       (1)求的單調(diào)區(qū)間和值域;

       (2)設(shè)a≥1,函數(shù)g(x)=x³-3a²x-2a,x∈［0,1］,若對于任意x₁∈［0,1］,總存在x₀∈［0,1］,使得g(x₀)=f(x₁)成立,求a的取值范圍.

      

Answer 1

解析：(1)對函數(shù)求導(dǎo),得?

       =?

       令=0，解得x=或x=.?

       當(dāng)x變化時,f′(x)、的變化情況如下表:

所以,當(dāng)x∈(0, )時, 是減函數(shù);當(dāng)x∈(,1)時, 是增函數(shù).?

       當(dāng)x∈［0,1］時, 的值域?yàn)椋?4,-3］.?

       (2)對函數(shù)g(x)求導(dǎo),得g′(x)=3(x²-a²).?

       因?yàn)?I >a≥1,當(dāng)x∈(0,1)時,g′(x)＜3(1-a²)≤0,?

       即當(dāng)x∈(0,1)時,g(x)為減函數(shù),從而當(dāng)x∈［0,1］時有g(shù)(x)∈［g(1),g(0)］.?

       又g(1)=1-2a-3a²,g(0)=-2a,即當(dāng)x∈［0,1］時有g(shù)(x)∈［1-2a-3a²,-2a］.?

       任給x₁∈［0,1］,f(x₁)∈［-4,-3］,存在x₀∈［0,1］使得g(x₀)=f(x₁),則［1-2a-3a²,-2a］［-4,-3］,?

       即?

       解①式得a≥1或a≤-;?

       解②式得a≤.?

       又a≥1,故a的取值范圍為1≤a≤.