已知函数f(x)=|2x-1|+2,g(x)=-|x+2|+3解不等式:g(x)大于等于-2
人气:141 ℃ 时间:2019-10-10 05:50:39
解答
你的题中有f(x)什么事啊?
g(x)>=-2
-|x+2|+3>=-2
|x+2|呃 这个也不清楚 是一同学在考试,发过来的题目还有第二问呢题(2):当x属于R时,f(x)-g(x)大于等于m+2恒成立,求实数m的取值范围 帮帮忙给解一下呗f(x)-g(x)=)2x-1|+2+|x+2|-3>=m+2|2x-1|+|x-2|>=m+3即求|2x-1|+|x-2|的最小值.x>=2时,|2x-1|+|x-2|=2x-1+x-2=3x-3>=3*2-3=31/23/2x<=1/2时,|2x-1|+|x-2|=1-2x+2-x=3-3x>=3/2可见|2x-1|+|x-2|的最小值为:3即m+3<=3,m<=0
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