f(x)=sinwx cos(wx pai 6)的图像上相邻两条对称轴间
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![f(x)=sinwx cos(wx pai 6)的图像上相邻两条对称轴间](/uploads/image/f/575351-71-1.jpg?t=f%28x%29%3Dsinwx+cos%28wx+pai+6%29%E7%9A%84%E5%9B%BE%E5%83%8F%E4%B8%8A%E7%9B%B8%E9%82%BB%E4%B8%A4%E6%9D%A1%E5%AF%B9%E7%A7%B0%E8%BD%B4%E9%97%B4)
∂w/∂x=f1(x+y+z,xyz)+f2(x+y+z,xyz)*yz∂2w/∂x∂z=f11+f12*xy+y*f2+yz*(f21+f
&=π/2,w=2.f(x)=sin(2x+π/2)=cos2x,偶函数,关于点M(3π/4,0)对称,且在[0,π/2]上是单调递减函数.
f(x)的函数错了吧应该是x1/x吧再问:检查过了,没有错啊。。。我在百度上找得到1、(1)和1、(2)但是1、(2)我看不懂再答:这就是考一个勾形函数
函数f(x)=sin(ωx+φ)(w>0,0≤φ≤π)是R上的偶函数∴f(-x)=f(x)→sin(-wx+φ)=sin(wx+φ)→-sinωxcosφ=sinωxcosφsinωx不恒等于0,∴c
f(x)=√2/2sin(2ωx+π/4)-1/2+1/2=√2/2sin(2ωx+π/4)∴2π/2ω=2πω=1/2f(A)=√2/2sin(A+π/4)=√2/2∴sin(A+π/4)=1∴A+
向量a⊥向量bcosφ=sinφφ=pi/4周期为piw=2f(x)=sin(2x+pi/4)g(x)=sin(2(x-6)+pi/4)=sin(2x-12+pi/4)-pi/26-3pi/8
若f(u,v,w)=(u-v)^w+w^(u+v)f(x+y,x-y,xy)=[(x+y)-(x-y)]^xy+(xy)^[(x+y)+(x-y)]=(2y)^xy+(xy)^2x
再问:为何会是这样算的呢?麻烦您解释一下再答:1,xyz都是自变量2,乘法的求导法则
f(x)=sin(2x+w)是偶函数,则sin(2x+w)=sin(-2x+w)2x+w+(-2x+w)=2kπ+π(因x是变量,故不可能有2x+w=2kπ+(-2x+w))那么w=(k+1/2)π,
(1)f(0)=sinw=0w=kpai(2)f(x)=f(-x)sin(2x+w)=sin(-2x+w)sin2xcosw+cos2xsinw=sin(-2x)cosw+cos(-2x)sinwsi
解题思路:本试题考查了三角函数的图像与性质的综合运用。解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.co
(1)解析:∵函数f(x)=cos(wx+f)(w>0,-π/2<f<0)的最小正周期为π∴w=2π/π=2,f(x)=cos(2x+f)∵f(π/4)=√3/2f(π/4)=cos
f是一个二元函数,f1表示f对其第一个自变量的偏导数,f2是f对其第二个自变量的偏导数.如果引入中间变量u=xy,v=yz,则f1与f2还可以写作fu,fv.上面的三个偏导数是用复合函数的求导法则得到
【参考答案】w=3/2由函数图象知,当x=2π/3时,取得最大值;当x=0时,取得最小值.∴函数半个周期是2π/3-0=2π/3函数最小正周期是2×(2π/3)=4π/3即T=2π/w=4π/3∴w=
f(x)=f(-x)sin(x+w)+sqrt(3)*cos(x-w)=sin(w-x)+sqrt(3)*cos(x+w)2(cos30度cos(x+w)-sin30度sin(x+w))=2(cos3
函数f(x)=sin(wx+φ)(w>0,|φ|0,|φ|φ=2π/3f(x)=sin(2x-2π/3+φ)=-sin2x==>φ=-π/3∵|φ|x=kπ+5π/122x-π/3=2kπ-π/2==
f(x)=sin(wx+φ){w>0,|φ|0,|φ|φ=π/6∴f(x)=sin(2x+π/6)2x+π/6=2kπ==>x=kπ-π/12;2x+π/6=2kπ+π==>x=kπ+5π/12;点对
(1)a=2,w=2f(x)是偶函数故f(0)=2或-2所以sinf=1或-1所以f=π/2+kπ(k是整数)0
f=x+1f+u=2x+3f+u+c=3x+8f+u+c+k=4x+15f(f,u,c,k)=(x+1)(2x+3)(3x+8)(4x+15)
周期为2π,w>0,则w=1f(x)=sin(wx+a)(w>0,0