Question

函数y=sin（π/4-2x)的单调增区间是？ 要过程

Answer 1

y=sin(π/4-2x)
y=-sin(2x-π/4) 这时注意前面已经有个负号了，图像就颠倒了。

根据颠倒过来的图像得: 2Kπ+π<= 2x-π/4 <=2kπ+2π
2Kπ+π+π/4<= 2x <= 2kπ+2π+π/4
2kπ+5π/4<= 2x <= 2kπ+9π/4
kπ+5π/8<= x <= kπ+9π/8
所以函数的增区间为｛x丨kπ+5π/8<= x <= kπ+9π/8 ，k∈Z }

Answer 2

sinx的单增区间是2kπ-π/2~~2kπ+π/2 所以 2kπ-π/2《π/4-2x《2kπ+π/2
解出为-kπ-π/8~~-kπ+3/8 π （k为整数）

Answer 3

2kπ-π/2≤π/4-2x≤2kπ+π/2
2kπ-π/2-π/4≤-2x≤2kπ+π/2-π/4
2kπ-3π/4≤-2x≤2kπ+π/4
kπ-3π/8≤-x≤kπ+π/8
-kπ-π/8≤x≤-kπ+3π/8
kπ-π/8≤x≤kπ+3π/8(k∈R)
区间是＜kπ-π/8，kπ+3π/8＞(k∈R)

Answer 4

解：2kπ-π/2<π/4-2x<2kπ+π/2
即 kπ+π/8<x<kπ+3π/8
本题也可以求导解，y'=-2cos(π/4-2x) 令y'=0 可解的两个极值x1=π/8 x2=3π/8
so: kπ+π/8<x<kπ+3π/8