已知函数
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.
(1)若f
-1(mx
2+mx+1)的定义域为R,求实数m的取值范围;
(2)当x∈[-1,1]时,求函数y=f
2(x)-2af(x)+3的最小值g(a).
(3)是否存在实数m>n>3,使得g(x)的定义域为[n,m],值域为[n
2,m
2],若存在,求出m、n的值;若不存在,则说明理由.
考点分析:
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