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函数y=sin(-3x+π/4),x∈R,在什么区间上是增函数
为什么我算后是[π/4+2/3kπ,7/12π+2/3kπ],可答案是[-π/12+2/3kπ,π/4+2/3kπ],
人气:108 ℃ 时间:2020-03-29 01:34:45
解答
因为函数y=sinx在[-π/2+2kπ,π/2+2kπ],所以-π/2+2kπ《-3x+π/4《π/2+2kπ,得-3π/4+2kπ《-3x《π/4+2kπ,解得-π/12-2/3kπ《x《π/4-2/3kπ,因为k属于整数,所以π/12+2/3kπ《x《π/4+2/3kπ是-π/12啊。。。。因为函数y=sinx在[-π/2+2kπ,π/2+2kπ],所以-π/2+2kπ《-3x+π/4《π/2+2kπ,得-3π/4+2kπ《-3x《π/4+2kπ,解得-π/12-2/3kπ《x《π/4-2/3kπ,因为k属于整数,所以-π/12+2/3kπ《x《π/4+2/3kπ打漏了,对不起
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