1、
|x-1|+|x+1|>=3
x=3
-2x>=3
x1
x-1+x+1>=3
x>=3/2
所以
x≤-3/2,x≥3/2
2、
f(x)=|x-1|+|a-x|≥|x-1+a-x|=|a-1|
所以|a-1|≥2
a-1≤-2,a-1≥2
a≤-1,a≥3
设函数f(x)=|x-1|+|x-a|.(1) 若a= -1解不等式f(x)≥3 (2)如果∀x∈R,f(x)≥2,求a的取值范围
设函数f(x)=|x-1|+|x-a|.(1) 若a= -1解不等式f(x)≥3 (2)如果∀x∈R,f(x)≥2,求a的取值范围
数学人气:654 ℃时间:2019-12-07 10:02:44
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