Question

已知函數(shù)f(x)=|x-1|+|x+3|.

(1)求x的取值范圍,使f(x)為常數(shù)函數(shù).

(2)若關(guān)于x的不等式f(x)-a≤0有解,求實(shí)數(shù)a的取值范圍.

 

Answer 1

(1) x∈[-3,1] (2) a≥4

【解析】(1)f(x)=|x-1|+|x+3|=

則當(dāng)x∈[-3,1]時,f(x)為常數(shù)函數(shù).

(2)方法一:如圖,結(jié)合(1)知函數(shù)f(x)的最小值為4,

∴實(shí)數(shù)a的取值范圍為a≥4.

方法二:|x-1|+|x+3|≥|x-1-(x+3)|;

∴|x-1|+|x+3|≥4,

等號當(dāng)且僅當(dāng)x∈[-3,1]時成立.

得函數(shù)f(x)的最小值為4,則實(shí)數(shù)a的取值范圍為a≥4.