摘要:15.已知函数f(x)=ax-x4.x∈[.1].A.B是其图象上不同的两点.若直线AB的斜率k总满足≤k≤4.则实数a的值是 . 答案: 解析:f ′(x)=a-4x3.x∈[.1].由题意得≤a-4x3≤4.即4x3+≤a≤4x3+4在x∈[.1]上恒成立.求得≤a≤.则实数a的值是.
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