求函数y=2sin(2x+π/4)+1单调递减区间以及最值

问题描述:

求函数y=2sin(2x+π/4)+1单调递减区间以及最值

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

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

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