Question

三角函数 已知f（x）=根号2/2sin（2x+π/4）+1/2,当x∈[-π/8,3π/8]时,求f（x）的单调区间

Answer 1

∵f（x）=√2/2sin（2x+π/4）+1/2,

当2x+π/4∈[-π/2+2kπ,π/2+2kπ]时单调递增

当2x+π/4∈[π/2+2kπ,3π/2+2kπ]时单调递减

∴x∈[-3π/8+kπ,π/8+kπ]时单调递增

x∈[π/8+kπ,5π/8+kπ]时单调递减

∴f(x)在[-π/8,π/8]上单调递增

在[π/8,3π/8]上单调递减