函數(shù)f(x)=1+2sin(2x-π/3)單調(diào)區(qū)間

數(shù)學(xué)人氣：322 ℃時間：2020-01-26 14:11:49

優(yōu)質(zhì)解答

解當(dāng)2kπ-π/2≤2x-π/3≤2kπ+π/2,k屬于Z時,f(x)=1+2sin(2x-π/3)增函數(shù),
即當(dāng)kπ-1π/12≤x≤kπ+5π/12,k屬于Z時,f(x)=1+2sin(2x-π/3)增函數(shù)
即f(x)=1+2sin(2x-π/3)的增區(qū)間是[kπ-1π/12,2kπ+5π/12],k屬于Z時.
當(dāng)2kπ+π/2≤2x-π/3≤2kπ+3π/2,k屬于Z時,f(x)=1+2sin(2x-π/3)減函數(shù),
即當(dāng)kπ+5π/12≤x≤kπ+11π/12,k屬于Z時,f(x)=1+2sin(2x-π/3)減函數(shù)
即f(x)=1+2sin(2x-π/3)的減區(qū)間是[kπ+5π/12,2kπ+11π/12],k屬于Z時.