Question

设函数f(x)=丨2x+1丨－丨x－3丨. 解不等式f（x）≥4; 求函数y=f(x)的...

设函数f(x)=丨2x+1丨－丨x－3丨.
解不等式f（x）≥4;
求函数y=f(x)的最小值.

Answer 1

1:(1)当x≥3时
f(x)=2x+1-(x-3)=x+4
f(x)≥4
x≥0
(2)当x≤-1/2时
f(x)=-2x-1+x-3=-x-4
f(x)≥4
x≤-8
(3)当-1/2<x<3时
f(x)=2x+1+x-3=3x-2
f(x)≥4
x≥2
所以f（x）≥4的解集是：x≥2或x≤-8
2::(1)当x≥3时
f(x)=2x+1-(x-3)=x+4
f(x)min=3+4=7
(2)当x≤-1/2时
f(x)=-2x-1+x-3=-x-4
f(x)min=-7/2
(3)当-1/2<x<3时
f(x)=2x+1+x-3=3x-2
f(x)min>-7/2
所以函数y=f(x)的最小值是-7/2，x=-1/2时取得最小值

Answer 2

f(x)=|2x+1|-|x-3|>=4
(1)x<=-1/2:
-(2x+1)+(x-3)>=4
-x>=8, x<=-8,即解是x<=-8
(2)-1/2=<x<=3:
2x+1+(x-3)>=4
3x>=6, x>=2,即解是2=<x<=3
(3)x>=3:
2x+1-(x-3)>=4
x>=0,即解是x>=3
即解是x<=-8或x>=2

( 2)
f(x)=-x-4,(x<=-1/2), f(x)>=-3.5
=3x-2,(-1/2<=x<=3), -3.5<=f(x)<=7
=x+4,(x>=3) ,f(x)>=7
故最小值是:-3.5