Tiffany 發問時間: 科學數學 · 8 年前

微積分高手請進來~~

Find the local extreme values of the following functions

1. f(x) = x^3-3x^2-9x+2

2. f(x) = xe^-x^2

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3.Find values of a,b,c and d so that the function f(x) = ax^3+bx^2+cx+d has a relative minimum at (0,1) and a relative maximum at (1,2).

幫我解這三題的過程和答案,感謝

已更新項目:

2. f(x) = xe^(-x^2)這樣

1 個解答

評分
  • 8 年前
    最佳解答

    1.f(x)=x^3-3x^2-9x+2

    f'(x)=3x^2-6x-9=3(x-3)(x+1)

    f'(x)=0 -> x=3 or x=-1

    f"(x)=6x-6

    f"(3)=12>0 f"(-1)=-12<0

    So we have relative min at x=3, f(3)=-25

    and relative max at x=-1, f(-1)=7

    2.題目好像有打錯

    3.f(x)=ax^3+bx^2+cx+d

    f(0)=d=1..................(1)

    f(1)=a+b+c+d=a+b+c+1=2

    so a+b+c=1 ...............(2)

    f'(x)=3ax^2+2bx+c

    f'(0)=c=0 ................(3)

    f'(1)=3a+2b+c=3a+2b=0 ....(4)

    from (1)(2)(3)(4) we get

    a=-2,b=3,c=0,d=1

    2013-02-16 16:38:10 補充:

    2.f(x)=xe^(-x^2)

    f'(x)=xe^(-x^2)(-2x)+e^(-x^2)=(-2x^2)e^(-x^2)+e^(-x^2)=(1-2x^2)e^(-x^2)

    f'(x)=0 -> x= +/- 1/√2

    f"(x)=[e^(-x^2)](-4x)+(1-2x^2)[e^(-x^2)](-2x)=(4x^3-6x)e^(-x^2)

    f"(1/√2)=(-4/√2)e^(-x^2) < 0

    f"(-1/√2)=(4√2)e^(-x^2) > 0

    2013-02-16 16:38:26 補充:

    so we have relative max at x=1/√2 that f(1/√2)=(1/√2)e^(-1/2)

    and relative min at x= -1/√2 that f(-1/√2)=(-1/√2)e^(-1/2)

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