Question

求函数y=2sin(π/4-2x)的单调递减区间

Answer 1

y=-2sin(2x-π/4)
y递减则sin(2x-π/4)递增

sinx增区间是(2kπ-π/2,2kπ+π/2)
则2kπ-π/2<2x-π/4<2kπ+π/2
2kπ-π/4<2x<2kπ+3π/4
kπ-π/8<x<kπ+3π/8
所以减区间是(kπ-π/8，kπ+3π/8)

追问

需要等于吗

追答

不用

Answer 2

函数y=sin(π/4-2x)的单调递减区间
y=sin(π/4-2x)
=sin[-(2x-π/4)]
=-sin(2x-π/4)
要求y的单调递减区间，只需要求sin(2x-π/4)的单调递增区间即可
kπ-π哗鸡糕课蕹酒革旬宫莫/2≤2x-π/4≤kπ+π/2
kπ/2-π/8≤x≤kπ/2+3π/8
k∈z

Answer 3

函数y=2sin(π/4-2x)的单调递减区间
就是函数y=2sin(2x-π/4)的单调递增区间
函数y=2sin(2x-π/4)的单调递增区间
2kπ-π/2<=2x-π/4<=2kπ+π/2
[kπ-π/8，kπ+3π/8]

函数y=2sin(π/4-2x)的单调递减区间 是
[kπ-π/8，kπ+3π/8]

Answer 4

y‘ = 2sin(π/4-2x)' <= 0
2cos(π/4-2x)*2<=0
cos(π/4-2x)<=0
2nπ+ π/2 <=π/4-2x <= 2nπ +3π/2
-π/4 - 2nπ >= 2x >= -5π/4 - 2nπ
nπ -π/8 > x >nπ - 5π/4
及单调递减区间为： [nπ-5π/8, nπ-π/8]