已知函数fx等于asin括号wx加
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f(x)=(m-1)x^2+3x+(2-n)为奇函数则常数项和偶次方项为零2-n=0,m-1=0m=1,n=2
f(5π/12)=Asin(5π/12+π/4)=Asin(2π/3)=A*√3/2,(√为根号)=3/2A=√3f(θ)+f(-θ)=3/2√3sin(θ+π/4)+√3sin(-θ+π/4)=3/
解由题知A=3T=4(π/2-(-π/2))=4π又由T=2π/w故2π/w=4π故w=1/2故f(x)=3sin(1/2x+φ)其图像过点(-π/2,3)知3sin(1/2x(-π/2)+φ)=3即
A=1/23/4T=(π/12+π/6)所以T=π/3W=2π/T=2π/(π/3)=6所以y=1/2sin(6x+∮)6(-π/6)+∮=0所以∮=πy=1/2sin(6x+π)ymax=1/2ym
T=2π/3=2π/ω,∴ω=3.∵最小值为﹣2,∴A=2.将﹙5π/9,0﹚代入函数,可得:2sin(5π/9×3+φ)=0,解得:φ=kπ-5π/3.∵φ的绝对值<π,∴﹣π<φ<π,即:∵﹣π<
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12π/2w=π→w=1.若a>0,则f(x)最大值是a+a/2+b=7/4即(3/2)a+b=7/4.f(x)最小值是-a+a/2+b=3/4即(-1/2)a+b=3/4.解方程组得:a=1/2;b
fx=2sin(wx+6/π)得到sin(wx+6/π)=√2/2令wx1+6/π=π/4wx2+6/π=3π/4则x2-x1=π两式相减得到w=1/2再问:为什么设π/4和3π/4呢?再答:这个是随
A=2T=4*[π/6-(-π/6)]=4π/3w=2π/(4π/3)=1.5f(x)=2sin(1.5x+φ)2sin(1.5*π/6+φ)=2π/6+φ=π/2φ=π/3f(x)=2sin(1.5
已知函数fx=Asin(wx+α)+1(w>0.A>00
解析:因为f(x)=Asin(wx+φ)(A>0,w>0,0w=2所以,f(x)=2sin(2x+φ)==>f(π/12)=2sin(π/6+φ)=2==>φ=π/3所以,f(x)=2sin(2x+π
(1)fx=Asin(wx+π/4)(其中x∈R,A>0,w>0)的最大值为2,最小正周期为8.那么A=2,2π/w=8∴w=π/4∴f(x)=2sin(π/4x+π/4)(2)两点P、Q的横坐标依次
我已经算出函数y=f(x)+f(x+2)的简式y=2根号2cosπ/4x求当x∈[-6,-2/3]函数y的最大值与最小值以及相应的x值解析:∵y=2√2cos(π/4x)∴函数y周期为T=8,所以,当
已知函数fx=Asin(wx+φ),其中w>0.1)当A=w=2,φ=π/6时,函数g(x)=f(x)-m在[0,π/2]上有两个零点,求m的范围.2)当A=1,φ=π/6时,若函数fx图像的相邻两条
根据周期为π,可得w为2.由f(π/4)=Asin(2*π/4+ψ)+n=Asin(π/2+ψ)+n=Acosψ+n=√3+1,由fx的最大值为3可得A+n=3可得n=1,A=2,ψ=π/6所以,f(
已知函数fx=Asin(wx+)+B的一系列对应值如下表X-π/6π/35π/64π/311π/67π/317π/6Y-1131-113(1)根据表格提供的数据求函数y=f(x)的解析式(2)若对任意
f(x)=2sin(x-π/3)f(kx)=2sin(kx-π/3)T=2π/|k|=2π/3==>|k|=3∵ksin(-3x-π/3)=m/2∵x∈[0,4π/9)∴-3x∈(-4π/3,0]∴-
f'(x)=2x+a>0x>-a/2-a/2=-2a=4
∵相邻两个交点之间的距离为π/2∴函数半个周期为π/2∴函数周期为π∴w=2π/π=2又∵最低点为M(2π/3,-2)∴A=2∴fx=2sin(2x+φ)再将为M(2π/3,-2)代入函数中则φ=π/