Question

已知函数f(x)=Asin(wx+φ)(A>0,w>0,│φ│<1/2)的周期为π,最低点M(2π/3，-3） 求函数y=f（x）+f(x+π/4

Answer 1

已知函数f(x)=Asin(wx+φ)(A>0,w>0,│φ│<1/2)的周期为π,最低点M(2π/3，-3）求函数y=f（x）+f(x+π/4)

解析：∵函数f(x)=Asin(wx+φ)(A>0,w>0,│φ│<1/2)的周期为π
∴w=2π/π=2==>f(x)=Asin(2x+φ)
∵最低点M(2π/3，-3）
∴f(2π/3)=3sin(4π/3+φ)=-3==>4π/3+φ=3π/2==>φ=π/6
∴f(x)=3sin(2x+π/6)
∵函数y=f（x）+f(x+π/4)=3sin(2x+π/6)+3sin(2x+π/2+π/6)
=3√2sin(2x+π/6+π/4) =3√2sin(2x+5π/12)

追问

y=f（x）+f(x+π/4)的最大值及对应x的值

追答

2x+5π/12=2kπ+π/2==>x=kπ+π/24，最大的值为3√2