Question

设f(x)=丨x丨+2丨x-a丨(a>0) 1..当a=1时,解不等式f(x)≤4

Answer 1

f(x)=丨x丨+2丨x-1丨，在a小于0，f(x)=2-3x；在小于1大于0，f(x)=2-x；在大于1，f(x)=3x-2

在直角坐标系中画出以上图像，并画出y=4，在其以下的部分则为解。可以得出解为x小于等于2，且大于等于-三分之二。

Answer 2

f(x)=|x|+2|x-1|<=4

(1)x>=1:
x+2(x-1)<=4
3x<=6
x<=2
所以有1<=x<=2
(2)0<=x<1:
x+2(1-x)<=4
-x<=2
x>=-2
故有0<=x<1
(3)x<0:
-x+2(1-x)<=4
-3x<=2
x>=-2/3
故有-2/3<=x<0
综上所述,解是-2/3<=x<=2

Answer 3

解：当x<0时，-x-2x+2<=4∴x>=-2/3∴-2/3<=x<0
当0<=x<1时，x-2x+2<=4∴x>=-2∴0<=x<1
当x>=1时，x+2x-2<=4∴x<=2∴1<=x<=2
综上，-2/3<=x<=2