Question

已知函数f(x)=Asin(ωx+φ),x∈R，(其中A＞0,ω＞0,0＜φ＜π/2)的周期为π且图象上的一个最低点为M（2∏

已知函数f(x)=Asin(ωx+φ),x∈R，(其中A＞0,ω＞0,0＜φ＜π/2)的周期为π且图象上的一个最低点为M（2∏／3，－2）。求F(X)的解析式，当X∈[0，∏／12]时，F（X）的最值？

Answer 1

因为周期为π,则T=2π/ω=π
ω=2
所以 f(x)=Asin(2x+φ)
因为最低点为M（2∏／3，－2）
则最底点是sin(2*2π/3+φ)=sin(4π/3+φ)=-1
则4π/3+φ=2kπ-π/2
φ=2kπ-π/2-4π/3=2kπ-11π/6=2kπ-2π+π/6=2(k-1)π+π/6
因为0＜φ＜π/2
所以φ=π/6
因为sin(2x+π/6)=-1
则-A=-2
A=2
所以f(x)=2sin(2x+π/6)

当-1<=sin(2x+π/6)<=1
时
2kπ-π/2<=2x+π/6<=2kπ+π/2
2kπ-2π/3<=2x<=2kπ+π/3
kπ-π/3<=x<=kπ+π/6
所以当x=π/6时有最大值f(π/6)=2
因为|0-π/6|=π/6
|π/6-π/12|=π/12
π/6>π/12
x=0离x=π/6比x=π/12离x=π/6要远些
所以当x=0时有最小值f(0)=1