当x∈【0,2π/3】时,y=2sin(2x+π/3)的单调区间和值域.最好有解释,

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 16:34:57
当x∈【0,2π/3】时,y=2sin(2x+π/3)的单调区间和值域.最好有解释,
xN0_+q8b ^pЛBF2TpRf4͈;HZWZG7p"Y߷;%mOqcWC?X7޵J g6[ԛsԾ7r]##`2Nu>z\Xyto=+AEy} Afk &>t54yhZPJA0SJBeI(C8SBiZ2 ڐ6}03@#:Б-y^**d|H|5bMhӅ CN! dH#ə+HQRDZcxBo;#)ϜH3-dJ.jp{,/hA79l CB2r>k'8B?W͑rArqCS|*""

当x∈【0,2π/3】时,y=2sin(2x+π/3)的单调区间和值域.最好有解释,
当x∈【0,2π/3】时,y=2sin(2x+π/3)的单调区间和值域.最好有解释,

当x∈【0,2π/3】时,y=2sin(2x+π/3)的单调区间和值域.最好有解释,
x∈【0,2π/3】时,
2x+π/3∈【π/3,5π/3】
y=2sin(2x+π/3)
2x+π/3∈【π/3,π/2)和【3π/2,5π/3)时单调增,此时x∈【0,π/12),【7π/12,2π/3)
2x+π/3∈【π/2,3π/2)时单调减,此时x∈【π/12,7π/12)
单调增区间【0,π/12),【7π/12,2π/3)
单调减区间【π/12,7π/12)
2x+π/3=π/2时有最大值2
2x+π/3=3π/2时有最小值-2
值域【-2,2】

x∈【0,2π/3】,2x+π/3∈【π/3,5π/3】
因此
当x∈【0,π/12】,2x+π/3∈【π/3,π/2】,函数单增
当x∈【π/12,7π/12】,2x+π/3∈【π/3,3π/2】,函数单减
当x∈【7π/12,2π/3】,2x+π/3∈【3π/2,5π/3】,函数单增

嗯嗯 就是这样的

x∈[0,2π/3],2x+π/3∈[π/3,5π/3],根据y=sinx的增减性分析可知,
当2x+π/3∈[π/3,π/2],即x∈[0,π/12],函数单增;
当2x+π/3∈[π/3,3π/2],即x∈[π/12,7π/12],函数单减;
当2x+π/3∈[3π/2,5π/3],即x∈[7π/12,2π/3],函数单增。