高考问答欧嘉璐尼简约沙发:高一数学高手唰唰
来源：百度文库 编辑：高考问答 时间：2024/05/03 02:58:36
函数y＝2sin(-3x+π/4）的单调递增区间是＿＿
＋＋详细解析过程..答得好的加分..
＋＋详细解析过程..答得好的加分..
2kπ-π/2<=-3x+π/4<=2kπ+π/2
2kπ-3π/4<=-3x<=2kπ+π/4
2kπ/3-π/4<=-x<=2kπ/3+π/12
2kπ/3-π/12<=x<=2kπ/3+π/4
所以[2kπ/3-π/12,2kπ/3+π/4] k属于Z
2kπ+π/2<=3x-π/4<=2kπ+3π/2
2kπ/3+π/4<=x<=2kπ/3+7π/12
所以[2kπ/3+π/4,2kπ/3+7π/12] k属于Z
用画图的方法也能看出来
因为sin（-x）= -sinx
原式y＝2sin(-3x+π/4）
变为y＝-2sin(3x-π/4）
画出图像y＝sinx，y轴不变，x变为原来的1/3倍，得到y＝sin3x，
将其向右平移π/12，得到y＝sin(3x-π/4）求增减区间，不用再管y轴，这个图像就可以看出来了
[2kπ/3+5π/12,2kπ/3+3π/4]
方法应该没错~答案是蒙滴~~
第一种;2kπ-π/2<=-3x+π/4<=2kπ+π/2
2kπ-3π/4<=-3x<=2kπ+π/4
2kπ/3-π/4<=-x<=2kπ/3+π/12
2kπ/3-π/12<=x<=2kπ/3+π/4
所以[2kπ/3-π/12,2kπ/3+π/4] k属于Z
第二种;2kπ+π/2<=3x-π/4<=2kπ+3π/2
2kπ/3+π/4<=x<=2kπ/3+7π/12
所以[2kπ/3+π/4,2kπ/3+7π/12] k属于Z