Question

这题当x属于0到兀/2时值域

已知函数f(x)=Asin(wx+φ)(A>0,w>0,|φ|<π/2)的图像过点P(π/12,0)
2013-06-12 | 分享
且图像上与点P最近的一个最低点是Q(-π/6,-2)
（1）求函数f（x）的解析式

Answer 1

f(x)=Asin(wx+φ) 图像过点P(π/12,0)：
Asin(wπ/12+φ)=0
sin(wπ/12+φ)=0
与P点最近的最低点为Q(-π/6,-2)
A＞0，A=2
w>0，xP-xQ=(2π/w)/4，π/12+π/6 = π/(2w)，w=2
f(x) = 2sin(2x+φ)
f(π/12) = 2sin(π/6+φ)=0；f(-π/6) = 2sin(-π/3+φ)=-2；|φ|<π/2
φ = -π/6
f(x) = 2sin(2x-π/6)

Answer 2

f(x)=Asin(wx+φ) 图像过点P(π/12,0)：
Asin(wπ/12+φ)=0
sin(wπ/12+φ)=0
与P点最近的最低点为Q(-π/6,-2)
A＞0，A=2
w>0，xP-xQ=(2π/w)/4，π/12+π/6 = π/(2w)，w=2
f(x) = 2sin(2x+φ)
f(π/12) = 2sin(π/6+φ)=0；f(-π/6) = 2sin(-π/3+φ)=-2；|φ|