Question

已知函数f(x)Asin(ωx+φ)(A>0，ω>0，|φ|<π/2)的部分图象如图所示（1）求f（

已知函数f(x)Asin(ωx+φ)(A>0，ω>0，|φ|<π/2)的部分图象如图所示（1）求f（x）的解析式（2）写出f（x）的递增区间

Answer 1

解析：由图可知，A=√2，T/2=6-(-2)=8==>T=16
w=2π/T=π/8
所以，f(x)=√2sin(π/8x+φ)
f(-2)=√2sin(π/8*(-2)+φ)=0==>φ=π/4
f(x)=√2sin(π/8x+π/4)
2kπ-π/2<=π/8x+π/4<=2kπ+π/2==>16k-6<=x<=16k+2，单调增区间

Answer 2

(1)从-2到6的长是半个周期，所以T=2(6-(-2))=16, 16=2π/ω, ω=π/8, A=V2,(表示根号2)
函数式为y=V2sin(πx/8+φ), 把点(2,V2)代入， V2=V2sin(π/4+φ), sin(π/4+φ)=1，
π/4+φ=π/2, φ=π/4, 所以 y=V2sin(πx/8+π/4)
(2) 2kπ-π/2<=πx/8+π/4<=2kπ+π/2, 16k<=x<=16k+2 (k为整数) 为单调增区间