Question

已知函数fx=3sin(2x-π/4)求(1)fx最小正周期(2)求fx在区间[π/6.3π/4]的值域

已知函数fx=3sin(2x-π/4)求
(1)fx最小正周期
(2)求fx在区间[π/6.3π/4]的值域

Answer 1

解：因为函数f(x) = 3sin(2x-π/4)

(1)T=2π/2=π
(2)因为x∈[π/6，3π/4]

所以：2x-π/4∈[π/12，5π/4]
所以sin(2x-π/4)∈[-√2/2,1]
所以3sin(2x-π/4)∈[-3√2/2,3]
即函数值域为：[-3√2/2,3]