復(fù)合函數(shù)f(y)=y²-y-2, y=cosx
cosx在[0+2kπ,π+2kπ]減,（π+2kπ,2π+2kπ]增
f(y)是典型的二次函數(shù),y定義域[-1,+1],f(y)在[-1,-1/2]減,（-1/2,+1]增
cos(2π/3+2kπ)=cos(4π/3+2kπ)=-1/2,cos(π+2kπ)=-1,cos(0+2kπ)=1
∴x∈(2π/3+2kπ,4π/3+2kπ)時(shí),y∈(-1,-1/2),f(x)減
x∈(－2π/3+2kπ,2π/3+2kπ)時(shí),y∈(-1/2,+1),f(x)增
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畫下圖就明白,我也不知道數(shù)有沒錯(cuò),我沒算,方法這樣應(yīng)該沒錯(cuò)咯