luoo 發問時間: 科學數學 · 6 年前

函數y=f(x)=√3sin(x+π/6)-2sinx¸請問

函數y=f(x)=√3sin(x+π/6)-2sinx¸請問下列哪些敘述是正確的(請列出計算過程)?

(1)函數f(x)的週期為π

(2)函數f(x)的振幅為1

(3)函數f(x)的最大值為√7

(4)在0≤x≤π/2時,y=f(x)的圖形為遞減

(5)把y=sinx的圖形向右平移2π/3後,可得y=f(x)的圖形

2 個解答

評分
  • 麻辣
    Lv 7
    6 年前
    最佳解答

    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)

  • 6 年前

    第5個選項不是正確的喔!

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