y=2sin(-1/2x-π/6)
=-2sin(1/2x+π/6)
函數(shù)sin(1/2x+π/6)的遞增區(qū)間是
y=-2sin(1/2x+π/6)的遞減區(qū)間
函數(shù)sin(1/2x+π/6)的遞減區(qū)間是
y=-2sin(1/2x+π/6)的遞增區(qū)間
由2kπ-π/2≤1/2x+π/6≤2kπ+π/2
得4kπ-4π/3≤x≤4kπ+2π/3,k∈Z
取k=0,得到在【0,2π】上的單調(diào)遞減區(qū)間
[0,2π/3],
由2kπ+π/2≤1/2x+π/6≤2kπ+3π/2
得4kπ+2π/3≤x≤4kπ+8π/3,k∈Z
取k=0,得到在【0,2π】上的單調(diào)遞增區(qū)間
[2π/3,2π]
根據(jù)上面的結(jié)論知
x=2π/3時,y取得最小值-2
x=0時,y=-1,x=2π時,y=1
∴ymax=1
∴函數(shù)的值域為[-2,1]很感謝親，還有這個，能幫我嗎，我會特別特別感激你的??！ y=1/2cos(-2x+π/3）的單調(diào)區(qū)間及值域y=1/2cos(-2x+π/3）=1/2cos(2x-π/3) ∵x∈R ∴值域為[-1/2,1/2] 由2kπ≤2x-π/3≤2kπ+π得kπ+π/6≤x≤kπ+2π/3,k∈Z∴函數(shù)遞減區(qū)間為[kπ+π/6,kπ+2π/3],k∈Z 由2kπ-π≤2x-π/3≤2kπ得kπ-π/3≤x≤kπ+π/6,k∈Z∴函數(shù)遞增區(qū)間為[kπ-π/3,kπ+π/6],k∈Z