微分應用,,

(a)
f(x)=x^3+hx^2+kx+2
f'(x)=3x^2+2hx+k
因圖形有兩相異轉各點
f'(x)有2實根
判別式>0
4h^2-12k>0
h^2>3k
(b)
(i)
f(x)的圖形經過(-2,0)
f(-2)=0
(-2)^3+h(-2)^2+k(-2)+2=0
-8+4h-2k+2=0
4h-6=2k
k=2h-3
(ii)
因
h^2>3k. k=2h-3
h^2>3(2h-3)
h^2-6h+9>0
(h-3)^2>0
h<-3 或 h>3
若h是整數﹐則h=4or5
(iii)
當h=4,k=2h-3=5
f(x)=x^3+4x^2+5x+2
f'(x)=3x^2+8x+5
令f'(x)=0
(3x+5)(x+1)=0
x=-5/3 或 -1
f''(x)=6x+8
f''(-5/3)=-2
f''(-1)=2
所以
(-5/3,0.148148)是極大點
(-1,0)是極小點


2007-04-22 18:15:01 補充：
補充由因f(x)與直線y=2只交於(0,2)1點令f(x)=x^3+hx^2+kx+2=2得x(x^2+hx+k)=0所以h^2-4k&lt;0h^2&lt;4k合併得3k&lt;4k

2007-04-22 19:07:48 補充：
我(b) (ii) 做漏了一個部份h^2&lt;4k. k=2h-3h^2&lt;4(2h-3)h^2-8h+12&lt;0(h-2)(h-6)&lt;02&lt;6合併h=4,5

2007-04-22 19:11:55 補充：
我最初做到h^2&gt;3k﹐跟住以為你4k是打錯的所以b(ii)做了h&gt;3就硬寫h=4,5 (我以為題目有規定h值範圍而你冇打)不過後來發現自己冇用到只交於(0,2)1點這個條件肯定有些地方做漏最後才想到可以用來證明h^2&lt;4k

函數f(x)=x3次+hx2次+kx+2(h,k為常數)的圖形有兩相異轉各點,且與直線y=2只交於(0,2)1點

a.證明3k<4k
f(x) = x^3 + hx^2 + kx + 2
f'(x) = 3x^2 + 2hx + k
when f'(x) = 0
(2h)^2 - 4(3)(k) > 0
h^2 - 3k > 0
h^2 > 3k

also,
when f(x) = 2
x^3 + hx^2 + kx = 0
x(x^2 + hx + k) = 0
h^2 - 4k < 0
h^2 < 4k

so
3k < h^2 < 4k


b.已知f(x)的圖形經過(-2,0)

i.以h表k
f(-2) = 0
-8 + 4h - 2k + 2 = 0
k = 2h - 3

ii.若h為一整數,利用a.bi的結果證明h=4or5
3k < h^2 < 4k
h^2 > 3(2h - 3)
h^2 - 6h + 9 > 0
(h - 3)^2 > 0
h < -3 or h > 3
and
h^2 < 4(2h - 3)
h^2 - 8h + 12 < 0
(h - 2)(h - 6) < 0
2 < h < 6
since h > 3
3 < h < 6
as h is an integer,h = 4 or h = 5


iii.當h=4,求f(x)圖形的極大點及極小點
when h = 4,k = 5
f(x) = x^3 + 4x^2 + 5x + 2
f'(x) = 3x^2 + 8x + 5
f''(x) = 6x + 8
when f'(x) = 0
3x^2 + 8x + 5 = 0
(3x + 5)(x + 1) = 0
x = -1 or x = -5/3
y = 0 or y = 0.148
f''(-1) = 2
>0
f''(-5/3) = -2
<0
(-1,0) is minimum and (-5/3,0.148) is maximum.