✔ 最佳答案
y(x)=√3sin(x+π/6)-2sin(x),哪些敘述是正確的?列出計算過程!
(1)函數y(x)的週期為πy(-π) = √3sin(-π+π/6) - 2sin(-π)= -√3sin(π-π/6) + 2sin(π)= -√3sin(π/6) + 0= -√3/2
y(0) = √3sin(π/6) + 2sin(0)= √3/2 + 0= √3/2=\= y(-π)= (X)
(2)函數y(x)的振幅為1y(x) = sin(x+2π/3) ;;; by (5)y'(x) = cos(x+2π/3)= 0= cos(π/2) or cos(3π/2)x = (π/2, 3π/2) - 2π/3= -π/6 or 5π/6
y"(x) = -sin(x+2π/3)
y"(-π/6) = -sin(-π/6+2π/3)= -sin(π/2)= -1 < 0
y"(5π/6) = -sin(5π/6+2π/3)= -sin(3π/2)= 1 > 0
max = y(-π/6)= sin(-π/6 + 2π/3)= sin(π/2)= 1
min = sin(5π/6)= sin(5π/6+2π/3)= sin(3π/2)= -1
Amplitude = 1 = (O)
(3)函數y(x)的最大值為√7 = (X) ;;; by (2)
(4)在0≤x≤π/2時,y=f(x)的圖形為遞減
y(x) = sin(x+2π/3) ;;; by (5)
y(0) = sin(2π/3) = √3/2
y(π/6) = sin(π/6+2π/3)= sin(5π/6)= 1/2
y(π/3) = sin(π/3+2π/3)= sin(π)= 0= (O)
(5)把y=sinx的圖形向右平移2π/3後,可得y=f(x)的圖形 y(x) = √3sin(x+π/6) - 2sin(x)= √3[sin(x)*cos(π/6) + cos(x)*sin(π/6)] - 2sin(x)= √3[(√3/2)sin(x) + (1/2)cos(x)] - 2sin(x)= (3/2)sin(x) + (√3/2)cos(x) - 2sin(x)= (√3/2)cos(x) - (1/2)sin(x)= sin(π/3 - x)= sin(π-x-2π/3)= sin(x+2π/3)= (O)
Answers = (2),(4),(5)