求y=3sin(3x-π 4)的周期-搜答案-智慧作业帮

π/2+2kπ再问：换元法有没有？再答：令3x+π/4=t，y=2sint的递减区间是：π/2+2kπ

y=-1/2sin(2/3x-π/4)所以y和sin(2/3x-π/4)单调性相反sinx的增区间是(2kπ-π/2,2kπ+π/2)减区间是(2kπ+π/2,2kπ+3π/2)所以sin(2/3x-

y=3sin(3x+π/4)单调增区间是:2kPai-Pai/2

y=2sin(3x+π/4)依题意-π/2+2kπ

,有公式的.2拍除以X前面的系数,答案是T=6拍

y=-3sin(2x-Pai/4)定义域是R最小正周期T=2Pai/2=Pai单调增区间是2kPai+Pai/2

设t=2x+π/4,则y=3sin(2x+π/4)为y=3sint由正弦函数的单调性质,可知正弦函数的单调递增区间为[2kπ-π/2,2kπ+π/2]所以y=3sint的单调递增区间为[2kπ-π/2

同增异减因为log1/3X在（0,+无穷）上单调递减；故若要单调增区间,则求sin(x+π/4)的单调递减区间,为（π/2+2πk,3/2π+2πk)（k=0,+-1,+-2,+-3,……）两区域相交

只有老师的答案,和第三步你后面自己想的是正确的.你是没有分清楚复合函数的增减区间的确定是跟什么有关.复合函数的增函数区间y=f(g(x))的递增区间==>>g(x)递增区间,f(x)递增区间同时满足=

复合函数应该用链式法则求导：若y=g(u),u=f(x),则dy/dx=dy/du*du/dxy=sin^5xdy/dx=dsin^5x/dsinx*dsinx/dx=5sin⁴xcosx

sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3（1/2-cos^2y）+sin^2y=1.5-3cos^2y）+sin^2y又有sin^2y+c

设对称轴为x=m则3m+π/4=kπ+π/2则3m=kπ+π/4∴m=kπ/3+π/12∴对称轴为x=kπ/3+π/12,k∈Z设对称中心为（m,0）则3m+π/4=kπ则3m=kπ-π/4∴m=kπ

y=sin(2x+3/4π)单调增区间:2kπ-π/2

y=3sin(2x)-1就是正弦曲线.如图：再问：函数应该是y=3sin(2x+π/4)-1..打错了.麻烦再画一下可以么...？再答：那就是在X轴方向上平移一下而已。如图：

你可以令0再问：麻烦写一下具体过程啊啊拜托。急再答：增：2kπ

(-π/8,-1)、(π/8,2)、(3π/8,-1)、(5π/8,-4)、(7π/8,-1)

y是2次复合y=u^2u=sinvv=3x+π/4y'=2u*u'*v'∴y'=2sin(3x+π/4)*cos(3x+π/4)*(3x+π/4)'=2sin(3x+π/4)*cos(3x+π/4)*

y＝2sin(2x+3π/4)可从y=sinx的单调性求得.(注意此一点)而y=sinx在[π/2+2kπ,3π/2+2kπ],k为整数上为递减令:π/2+2kπ≤2x+3π/4≤3π/2+2kπ,k

max:2x+π/4=2kπ+π/22x=2kπ+π/4x=kπ+π/8min:2x+π/4=2kπ-π/22x=2kπ-3π/4x=kπ-3π/8

y=sinx增区间[2kπ-π/2,2kπ+π/2]所以本题,2kπ-π/2≤π/4+2x≤2kπ+π/2kπ-3π/8