(I)若a∈R且a≠0,求函数f(x)=ax2+x﹣a的“局部对称点”;
(II)若函数f(x)=4x﹣m•2x+1+m2﹣3在R上有局部对称点,求实数m的取值范围.
代入f(﹣x)=﹣f(x),得ax2+x﹣a+ax2﹣x﹣a=0,即ax2﹣a=0(a≠0),
∴x=±1,
∴函数f(x)=ax2+x﹣a的局部对称点是±1;
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