求函数f(x)=（根号2）sin(2x-π/4)在区间[π/8,3π/4]上的最小值和最大值

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f(x)=√2sin(2x-π/4)
x∈[π/8,3π/4]
∴2x∈[π/4,3π/2]
2x-π/4∈[0,5π/4]
∴sin(2x-π/4)∈[-√2/2,1]
∴√2sin(2x-π/4)∈[-1,√2]
最大值=√2
最小值=-1