已知定义在区间上的函数f(x)=
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为奇函数且f(
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)=
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(1)求实数m,n的值;
(2)求证:函数f(x)在区间[-1,1]上是增函数.
(3)若∀x
1,x
2∈[-1,1],|f(x
1)-f(x
2)|≤t恒成立,求t的最小值.
考点分析:
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,则f(f(-
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))=
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;③y=
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