已知f(x) =sin(2x+π/6)+3/2,x∈R.求函数fx的单调减区间
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![已知f(x) =sin(2x+π/6)+3/2,x∈R.求函数fx的单调减区间](/uploads/image/z/6912842-50-2.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%EF%BC%89+%3Dsin%EF%BC%882x%2B%CF%80%2F6%EF%BC%89%2B3%2F2%2Cx%E2%88%88R.%E6%B1%82%E5%87%BD%E6%95%B0fx%E7%9A%84%E5%8D%95%E8%B0%83%E5%87%8F%E5%8C%BA%E9%97%B4)
已知f(x) =sin(2x+π/6)+3/2,x∈R.求函数fx的单调减区间
已知f(x) =sin(2x+π/6)+3/2,x∈R.求函数fx的单调减区间
已知f(x) =sin(2x+π/6)+3/2,x∈R.求函数fx的单调减区间
f(x) =sin(2x+π/6)+3/2
2kπ+π/2≤2x+π/6≤2kπ+3π/2 k∈Z
2kπ+π/6≤x≤2kπ+2π/3
因此,f(x)的单减区间为:[2kπ+π/6,2kπ+2π/3] (k∈Z)
π/6+kπ <= x <= 2π/3+kπ
基础题!!!。。。。解答给好评,谢谢基础题!自己想!努力!