求函数y=2sin(2x+π⼀4)+1单调递减区间以及最值 要过程

2025-06-20 07:57:54
推荐回答(3个)
回答1:

y=2sin(2x+π/4)+1
y=sinx单调递减区间是 【2kπ-π/2,2kπ+π/2】
所以2kπ-π/2<=2x+π/4<=2kπ+π/2
kπ-3π/8<=x<=kπ+π/8
单调递减区间【kπ-3π/8,kπ+π/8】k∈Z
sin(2x+π/4)=1 有最大值 ymax=3
sin(2x+π/4)=-1 有最小值ymin=-1

回答2:

y=2sin(2x+π/4)+1
当2x+π/4∈(2kπ+π/2,2kπ+3π/2)时,y单调减,
此时2x∈(2kπ+π/4,2kπ+5π/4)
x∈(kπ+π/8,kπ+5π/8)
即单调减区间(kπ+π/8,kπ+5π/8),其中k∈Z

-1≤sin(2x+π/4)≤1
-2≤2sin(2x+π/4)≤2
-1≤2sin(2x+π/4)+1≤3
最小值-1,最大值3

回答3:

y=2sin(2x+π/4)+1单调递减区间
2kπ-π/2<=2x+π/4<=2kπ+π/2
即单调递减区间:
kπ-3π/8<=x<=kπ+π/8,k∈z
y最大值=3
y最小值=-1