Question

求函数y | x+3|+ |x-5 |的值域

Answer 1

解y =| x+3|+ |x-5 |
≥| x+3-（x-5） |
=/8/
=8
故y =| x+3|+ |x-5 |的值域为[8，正无穷大）。

Answer 2

由绝对值的几何意义 即某点X在数轴上与-3和5的距离之和。
显然y≥8

Answer 3

Answer 4

y= | x+3|+ |x-5 |
case 1: x<-3
y= | x+3|+ |x-5 |
=-(x+3)-(x-5)
=-2x+2
y的值域 (8,+∞)

case 2: x=-3
y= | x+3|+ |x-5 |
=8

case 3: -3<x<5
y= | x+3|+ |x-5 |
=(x+3)-(x-5)
=8

case 4: x=5
y= | x+3|+ |x-5 |
=8

case 5: x>5
y= |x+3|+ |x-5 |
=(x+3)+(x-5)
=2x-2
y的值域 (8,+∞)

y= | x+3|+ |x-5 |的值域
case 1 or case 2 or case 3 or case 4 or case 5
y的值域 [8,+∞)

Answer 5

y =| x+3|+ |x-5 |的值域
[8，+∞）