Question

设x是实数,f(x)=|x+1|+|x-1|+|x-2|,求f(x)的最小值

Answer 1

,f(x)=|x+1|+|x-1|+|x-2|
(1)当x>=2
f(x)=x+1+x-1+x-2
=3x-2
f(x)的取值范围是x>=4
(2)当1<=x<2
f(x)=x+1+x-1+2-x
=x+2
f(x)的取值范围是3<=x<4
(3)当-1<=x<1
f(x)=x+1+1-x+2-x
=4-x
f(x)的取值范围是3<x<=5
(4)当x<=-1
f(x)=-x-1+1-x+2-x
=2-x
f(x)的取值范围是x>=3
f(x)的最小值是3

Answer 2

f(x)=|x+1|+|x-1|+|x-2|>=|x+1|+|1-x|+|x-2|=|x+1+1-x+x-2|=|x|>=0