若f(x)=Asin(wx ) 1(w>0
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/05 07:16:27
设函数f(x)=Asin(wx+q),(A=/0,w>0,-pai/2
A影响值域,w影响周期,f是函数左右移动
已知函数f(x)=Asin(ωx+φ)图像如图求f(x).答:f(x)=2sin(150°x+90°)由图可见,A=2f(x)=2sin(ωx+φ)f(2)=2sin(2ω+φ)=√3-->2ω+φ=
sinx函数图像在0-π之两个区间在x轴上方观测f(x)0点之后两区间在x轴位置,上方则A为正,下方则A为负.
A=22sinφ=√3φ=π/3w*(5π/6)+π/3=π或w(5π/6)+π/3=2πw=4/5w=2f(x)=2sin【(4/5)x+π/3】f(x)=2sin(2x+π/3)-π/2
最大值是3,则A=3.函数周期是π,则2π/w=π,w=2.f(x)=3sin(2x+α)当x=π/6时f(x)取得最大值3,则3=3sin(π/3+α),π/3+α=π/2,α=π/6.∴f(x)=
1、sin3π/4=sinπ/4cosπ/4cosa-sinπ/4sina=cos(π/4+a)=0a=π/42、两对称轴相距为半个周期,所以周期为2π/3w=3所以解析式为f(x)=Asin(3x+
f(x)=Asin(wx+∮)=Asinw(x+∮/w)w>0,0
1.由题知f(x)为正弦函数图像变化而来,故由y轴右侧的第一个最大值点和最小值点分别为(a,2)和(a+3π,-2)可知A=2又因f(0)=Asinp=2sinp=1或-1得sinp=1/2或-1/2
(1)1/4T=π/6T=2π/3w=2π/T=3A=2所以现在方程为f(x)=2sin(3x+φ)将(π/12,2)代入π/4+φ=π/2φ=π/4方程为f(x)=2sin(3x+π/4)第二问图就
(1)函数最小正周期为:T=2(5π/8-π/8)=πω=T/(2π)=2函数最大值和最小值差为4,所以2A=4,A=22sin(2×π/8+φ)=2sinπ/2π/4+φ=π/2φ=π/4(2)f(
已知函数f(x)=Asin(wx+a)(A>0,w>0,-π/20,w>0,-π/2π/3+a=π/2==>a=π/6∴f(x)=3sin(2x+π/6)单调增区间:2kπ-π/2x0=0==>2x0
由图可知,最大值为√2,则A=√2周期为T=(6+2)*2=16,则T=2π/w=>w=2π/T=2π/16=π/8x=2时取得最大值,则2w+a=π/4+a=π/2=>a=π/4∴函数解析式为f(x
f(x)=(1+cos2x)/2+1/2*sin2x=1/2*(sin2x+cos2x)+1/2=1/2*√2(√2/2*sin2x+√2/2cos2x)+1/2=√2/2*(sin2xcosπ/4+
由图可得A=2当x=0时,y=√3即√3=2sinbsinb=√3/2b=π/3或2π/3当y=0时,x=2π/9即2sin(2πw/9+b)=0sin(2πw/9+b)=02πw/9+b=π或2πw
只想问fai啊?简单说,就是求出w之后把M带进去啊,因为0再问:我就是想问一下,带进去以后sin(4π/3+fai)=-14π/3+fai=2kπ-π/2为什么是等于2kπ-π/2而不是3π/2,而有
∵f(π/3+x)=f(π/3-x)∴Asin(πω/3+xω+y)=Asin(πω/3-xω+y)∴πω/3+y=nπ/2(n=1、3、5、7、9、……)g(π/3)=Acos(πω/3+y)=Ac
周期T=2π/w=π所以w=2;把点M代入函数f(x)=Asin(4π/3+y)=-2;又因为M为最低点,所以A=2,sin(4π/3+y)=-1得出y=π/6;得出f(x)=2sin(2x+π/6)
T=π,w=2A=2,B=1Φ=-π/6f(x)=2sin(2x-π/6)+1f(kx)=2sin(2kx-π/6)+1周期为2π/32k=3k=3/2f(kx)=2sin(3x-π/6)+1x∈[0
(1)由函数的最小值为-2及A>0得:A=2;又函数经过(0,跟号3),所以,sin(φ)=根号3得:φ=π/3+2kπ或φ=2π/3+2kπ(k为整数)由|φ|0得:A=2;又函数经过(0,跟号3)