函数fx=2mx^2-2(4-m)x+1,gx=mx,对任一实数,fx与gx的值至少有一个为正数,
则m的取值范围是多少
人气:121 ℃ 时间:2019-12-21 22:25:51
解答
我先算下大于0小于8求采纳要过程额假设m大于0则gx在x大于0的时候大于0,则此时只要小于0时fx大于0时即可,因为x=0时fx=1>0 所以只需对称轴>0 或对称轴<0同时判别式<0 当m小于0时 画图可得不可能恒成立求采纳
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