题目

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. 答案：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（-∞,-]∪［,+∞.

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