已知函数f(x)=(1)求f(x)的值域.(2)设函数g(x)=ax-2,x∈［-2,2］.若对于任意x

(1)求f(x)的值域.

(2)设函数g(x)=ax-2,x∈［-2,2］.若对于任意x₁∈［-2,2］,总存在x₀∈［-2,2］,使得g(x₀)=f(x₁)成立,求实数a的取值范围.

解:(1)当x∈［-2,-1］时,f(x)=x+在［-2,-1］上是增函数,

此时f(x)∈［-,-2］;                                                     

当x∈［-1,］时,f(x)=-2;                                                  

当x∈［,2］时,f(x)=x-在［,2］上是增函数,此时f(x)∈［-,］.         

所以f(x)的值域为［-,-2］∪［-,］.                                    

(2)①若a=0,g(x)=-2,对于任意x₁∈［-2,2］,f(x₁)∈［-,-2］∪［-,］,不存在x₀∈［-2,2］,使得g(x₀)=f(x₁)成立.                                                       

②若a＞0,g(x)=ax-2在［-2,2］上是增函数,g(x)∈［-2a-2,2a-2］.                    

任给x₁∈［-2,2］,f(x₁)∈［-,-2］∪［-,］,若存在x₀∈［-2,2］,使得g(x₀)=f(x₁)成立,

则［-,-2］∪［-,］［-2a-2,2a-2］.

∴                                                       

∴a≥.                                                                

③若a＜0,g(x)=ax-2在［-2,2］上是减函数,g(x)∈［2a-2,-2a-2］.                   

同理,可得                                                   

∴a≤-.                                                                

综上,实数a的取值范围是（-∞,-]∪［,+∞.