求函数y=根号2sin(2x-π/4)的单调递减区间,最小正周期和最值
人气:470 ℃ 时间:2019-11-24 20:25:36
解答
∵y=√2sin(x-π/4)
∴最小正周期为:[0,2π];
最大值√2,最小值-√2;
∴在x∈[2kπ+3π/4,2kπ+5π/4]上,y=√2sin(x-π/4)是单调递减;
在x∈[2kπ+5π/24,2k(π+2π+π/4]上y=√2sin(x-π/4)是单调递增.是y=√2sin(2x-π/4)
推荐
猜你喜欢
- A、B都是正整数,如果A除以B等于10,那么A、B的最小公倍数是几谢谢了,
- The words of his old teacher left a ______ impression on his mind.He is still influenced by them.
- 有关交通安全的作文600字
- 举例说出“经过两点且只有一条直线”和“两点之间,线段最短”这两个结论在实际生活中的运用
- 在某张月历上,一个月内是五个星期日正好在同列,且它们的日期数之和是80,则本月第一个星期一是__号?
- each other for a change是什么意思
- "Bill,is this_____bike?" "Yes,it is."
- 翻译 Happiness is not wishing for what we dont have,but enjoying what we do possess