Question

一道数学题，急需在线等

如果定义域是[-π/12，π/12]，求y=-2sin(2x-π/3)+1的值域

Answer 1

定义域x∈[-π/12，π/12]
则 2x∈[-π/6,π/6]
则 2x-π/3∈[-π/2,-π/6]
所以 sin（2x-π/3）∈[-1.-1/2]
所以 -2sin（2x-π/3）∈[1.2]
所以-2sin(2x-π/3)+1∈[2,3]
所以这个函数的值域就是[2,3]

Answer 2

y=-2sin(2x-π/3)+1
∵x ∈[-π/12，π/12]
∴2x-π/3 ∈[-π/2，-π/6]
∴-1≤sin(2x-π/3) ≤-1/2
1 ≤ -2 sin(2x-π/3) ≤ 2
2 ≤ -2 sin(2x-π/3) +1 ≤3
值域[2,3]

Answer 3

定义域x∈[-π/12，π/12]
则 2x∈[-π/6,π/6]
则 2x-π/3∈[-π/2,-π/6]
所以 sin（2x-π/3）∈[-1.-1/2]
所以 -2sin（2x-π/3）∈[-2.-1]
所以-2sin(2x-π/3)+1∈[-1,0]
所以这个函数的值域就是[-1,0]
y=-2sin(2x-π/3)+1
∵x ∈[-π/12，π/12]
∴2x-π/3 ∈[-π/2，-π/6]
∴-1≤sin(2x-π/3) ≤-1/2
1 ≤ -2 sin(2x-π/3) ≤ 2
2 ≤ -2 sin(2x-π/3) +1 ≤3
值域[2,3]

Answer 4

[2,3]