y=cos(π 3-2x)的单调递增区间-搜答案-智慧作业帮

这是一个复合函数,对于复合函数而言,内外相同则为增,不同则为减在考虑单调区间时,还要注意定义域外层是log1/2a=1/2,是减函数内层减区间2kπ

y=cos(π/3-x/2),x∈[-2π,2π]=cos(x/2-π/3)由2kπ-π≤x/2-π/3≤2kπ,k∈Z得2kπ-2π/3≤x/2≤2kπ+π/3,k∈Z∴4kπ-4π/3≤x≤4kπ

要求y=-cos(x/2-π/3)的单调递增区间即求y=cos(x/2-π/3)的减区间即可2kπ

2kπ-π≤-2x+π/3≤2kπ2kπ-4π/3≤-2x≤2kπ-π/3kπ+π/6≤x≤kπ+2π/3

2.如果a>0最大值为a+b=1最小值为-a+b=-7解得a=4b=-3如果a

y=cosx的递增区间是(2kπ-π,2kπ),k为整数所以y=cos2x的递增区间是(kπ-π/2,kπ).剩下是正确的

y=sin(x+π/2)cos(x+π/6)=cosx*cos(x+π/6)=cosxcosx1/2根号3+1/2cosxsinx=1/2根号3cos^2x+1/4sin2x=1/2根号3*1/2(1

y=cos(2x+π/3),x∈[-2π,2π]的单调递增区间y=cos(2x+π/3),则周期T=π设2x+π/3=X,则X∈[-11π/3,13π/3]由cosX单调递增区间为[π加减2kπ,2π

y=log1/2[cos(π/3-x/2)]的底数为1/2所以外函数为减所以当cos(π/3-x/2)为减时y递增先考虑定义域cos(π/3-x/2)>0解得x∈[-π/3-4kπ,5π/3-4kπ]

y=cos(π/3-x/2)就是y=cos(x/2-π/3)就是求cos(x/2-π/3)的递减区间x/2-π/3属于[2kπ,π+2kπ]所以x属于[2π/3+2kπ,8π/3+2kπ]

-2log《3》22、单调区间令u=-x^2-2x+3,则y=log《1/3》uy是关于u的减函数,根据二次函数的性质,在x∈(-3,-1]上,u是关于x的增

2kπ-π≤2x-(π/3)≤2kπ2kπ-2π/3≤2x≤2kπ+π/3kπ-π/3≤x≤kπ+π/6

y=cos(-2x+π/3)=cos(2x-π/3)递减区间：2kπ

cos的单调递增区间为【π/2+2kπ,3π/2+2kπ】∴π/2+2kπ＜2x-π/3＜3π/2+2kπ∴kπ+2/3π＜x＜kπ+7/6π再问：不可以根据cos的单调递增区间为【-π+2kπ，2k

y=3cos(2x+π/3).递增：2kπ-π/2

y=sin(2x+π/3)+cos(2x-π/6)=(1/2)sin2x+(√3/2)cos2x+(√3/2)cos2x+(1/2)sin2x=sin2x+√3cos2x=2sin(2x+π/3)2k

[π/6+kπ,5π/12+kπ],k属于整数Z

∵x∈（-π+2kπ,2kπ）∴2x∈（-2π+4kπ,4kπ）∴2x-π/3∈（-7/3π+4kπ,-π/3+4kπ）再问：你确定？再答：en

余弦函数在[2kπ,2kπ+π]上单调减,在[2kπ+π,2kπ+2π]上单调增所以2kπ

∵对于函数y=cos(2x-π3)的单调减区间为2kπ≤2x-π3≤2kπ+π即kπ+π6≤x≤kπ+2π3故函数f（x）的单调减区间为[kπ+π6，kπ+2π3]（k∈Z）故答案为：[kπ+π6，k