Question

求函数f(x)=|x-2|-|x+1|的最值

Answer 1

x<-1时
f(x)=2-x-(-x-1)=3
-1=<x<=2时
f(x)=2-x-(x+1)=1-2x
单调(递减
f(-1)=3 f(2)=-3所以-3<f(x)<3
x>2时
f(x)=x-2-(x+1)=-3
综上所述 f(x)max=3 f(x)min=-3

Answer 2

1

Answer 3

x<-1时，f(x)=3
-1≤x<2时，f(x)=-2x+1
x≥2时，f(x)=-3
所以f(x)的最小值为-3，最大值为3

Answer 4

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

1、当 x < -1时

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

= -(x - 2)-【-(x+1)】

= -x +2 + x +1
= 3

2、当 -1 ≤ x < 2 时

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

= - (x - 2)-(x + 1)

= -x + 2 - x -1

= 1-2x

3、当 x ≥ 2 时

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

= (x - 2)-(x + 1)

= x -2 - x - 1

=-3
综上所述，-3≤f（x）≤3