Question

求函数y=√2sin(2x+π/4),x∈[0,π/2]的值域

Answer 1

函数最小正周期是π
单调区间为
当2kπ-π/2<=2x+π/4<=2kπ+π/2
2kπ-π<=2x<=2kπ+π/4
kπ-π/2<=x<=kπ+π/8时单增
当2kπ+π/2<=2x+π/4<=2kπ+3π/2
2kπ+π/8<=2x<=2kπ+5π/4
kπ+π/8<=x<=kπ+5π/8时单减
x∈[0,π/2]=[0,π/8]和[π/8,π/2]的并集
当x∈[0,π/8]
ymin=√2sinπ/4=1
ymax=√2sinπ/2=根号2
当x∈[π/8,π/2]时
ymin=√2sin5π/4=1
ymax=√2sinπ/2=根号2
所以值域是[1,根号2]

Answer 2

y=√2sin(2x+π/4),x∈[0,π/2]
当x=π/8时，2x+π/4=π/2；y取最大值=√2
当x=π/2时，2x+π/4=5π/4;y取最小值=-1
所以。函数的值域为[-1,√2]

Answer 3

定义域为2x+π/4在[π/4,5π/4] ，sin(2x+π/4)在-[√2/2,1],所以值域维[-1,√2]