Question

已知函数f(x)=sin(1/2x+π/4),求函数在区间[-2π,2π]上的单调区间,周期

Answer 1

f(x)=sin(1/2x+π/4)的周期为T=2π/(1/2)=4π,

递增区间为-π/2+2kπ≤1/2x+π/4≤π/2+2kπ,

即为-3π/2+4kπ≤x≤π/2+4kπ,

当k=0时,得求函数在区间[-2π,2π]上的单调递增区间[-3π/2,π/2],

而剩下的[-2π,-3π/2]U[π/2,2π]则为递减区间.

O(∩_∩)O~