设函数f(x)=丨x-1丨-丨x+2丨解不等式f(x)≤2(2)若不等式f(x)≤丨a-1丨对x∈R恒成立,求a的取值范围
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![设函数f(x)=丨x-1丨-丨x+2丨解不等式f(x)≤2(2)若不等式f(x)≤丨a-1丨对x∈R恒成立,求a的取值范围](/uploads/image/z/11950844-68-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%E4%B8%A8x-1%E4%B8%A8-%E4%B8%A8x%2B2%E4%B8%A8%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28x%29%E2%89%A42%EF%BC%882%EF%BC%89%E8%8B%A5%E4%B8%8D%E7%AD%89%E5%BC%8Ff%EF%BC%88x%EF%BC%89%E2%89%A4%E4%B8%A8a-1%E4%B8%A8%E5%AF%B9x%E2%88%88R%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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设函数f(x)=丨x-1丨-丨x+2丨解不等式f(x)≤2(2)若不等式f(x)≤丨a-1丨对x∈R恒成立,求a的取值范围
设函数f(x)=丨x-1丨-丨x+2丨解不等式f(x)≤2
(2)若不等式f(x)≤丨a-1丨对x∈R恒成立,求a的取值范围
设函数f(x)=丨x-1丨-丨x+2丨解不等式f(x)≤2(2)若不等式f(x)≤丨a-1丨对x∈R恒成立,求a的取值范围
(1)当x≤-2时,f(x)=1-x+x+2=3>2,不合题意;
当-2
(1) |x-1|+|x+2|≤2
由几何意义可得[-3/2,+∞)
(2) f(x)max=3≤|a-1|
a-1≤-3或a-1≥3
a≤-2或a≥4