求函数y=cosπ3-2x的单调递增区间求函数y=3sinπ3-π2
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/27 10:00:50
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=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-1/
y=log1/2cos(x/3+π/4)t=cos(x/3+π/4)y=log1/2(t)在定义域内是减函数要使y=log1/2cos(x/3+π/4)是减函数t=cos(x/3+π/4)必须是增函数
2kπ-π≤-2x+π/3≤2kπ2kπ-4π/3≤-2x≤2kπ-π/3kπ+π/6≤x≤kπ+2π/3
可以根据:asinα+bcosα=√(a^2+b^2)*sin(α+θ),其中θ由a,b的符号和tanθ=b/a确定具体到这题,就是:y=√[(√3+2)/2]^2+(1/2)^2*sin(2x+θ)
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=sin²x+sin2x+2cos²x=sin2x+(1+cos2x)/2+1=sin2x+1/2*cos2x+3/2=√5/2(2/√5*sin2x+1/√5*cos2x)+3
∵y=cosx+cos(x-π3)=cosx+cosxcosπ3+sinxsinπ3=32cosx+32sinx=3(cosπ6cosx+sinπ6sinx)=3cos(x-π6),∵-1≤cos(x
2kπ-π≤2x-(π/3)≤2kπ2kπ-2π/3≤2x≤2kπ+π/3kπ-π/3≤x≤kπ+π/6
解y=cos^2x-sinx+3=1-sin^2x-sinx+3=-sin^2x-sinx+4令t=sinx,由x∈[π/6,π/2],则1/2≤t≤1即y=-t^2-t+4=-(t+1/2)^2+1
1.将cos(2x-π/3)看成整体√cos(2x-π/3)的导数是1/[2√cos(2x-π/3)]将2x-π/3看成整体cos(2x-π/3)的导数是-sin(2x-π/3)2x-π/3的导数是2
解y=3cosx的对称轴方程是x=kπ.k属于Z即函数y=3cos(2x+π/4)的对称轴方程为2x+π/4=kπ,k属于Z即为x=kπ/2-π/8,k属于Z
y=3cosX-cos(2X)=3cosx-(2*(cosx^)2-1)=-2(cosx)^2+3cosx+1=-2(cosx-3/4)^2+17/8当cos=3/4时,y有最大值,为17/8当cos
y'=-sin(4-3X)*(-3)=3sin(4-3X)
y=3cos(2x+π/3).递增:2kπ-π/2
cos(2x-π/3)>0,且tanx-1≠0解cos(2x-π/3)>0得:2kπ-π/2
cos值域是【-1,1】,所以y最大1,最小-1y=1时,x/2+π/3=2kπ+π/2x=4kπ+π/3同理,y=-1时,x=4kπ-5π/3综上,x∈{x|x=4kπ+π/3,k∈Z},y最大=1
因为-2π/3<x<π,所以-2π/3<(1/2)x-π/3<π/6.所以-1/2<cos(1/2x-π/3)<=1.-√2/2<√2cos(1/2x-π/3
对称轴是cos(2x+π/3)=正负1,即2x+π/3=nπ,所以x=nπ/2-π/6,其中n是整数对称中心是cos(2x+π/3)=0,即2x+π/3=π/2+nπ,所以对称中心是(nπ/2+π/1