求函数y=根号2sin(2x-π/4)的单调递减区间,最小正周期和最值

∵y=√2sin(x-π/4)
∴最小正周期为：[0,2π]；
最大值√2,最小值-√2；
∴在x∈[2kπ+3π/4,2kπ+5π/4]上,y=√2sin(x-π/4)是单调递减；
在x∈[2kπ+5π/24,2k(π+2π+π/4]上y=√2sin(x-π/4)是单调递增.是y=√2sin(2x-π/4)