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
2. f(x) = xe^(-x^2)這樣
1 個解答
- WilliamLv 78 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)