Question

微分应用,,

Answer 1

更新1:

证明3k

更新2:

证明3k<4k 不过你都做到

劲

=D

(a) f(x)=x^3+hx^2+kx+2 f'(x)=3x^2+2hx+k 因图形有两相异转各点 f'(x)有2实根 判别式>0 4h^2-12k>0 h^2>3k (b) (i) f(x)的图形经过(-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是整数﹐则h=4or5 (iii) 当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)是极大点 (-1
0)是极小点 2007-04-22 18:15:01 补充： 补充由因f(x)与直线y=2只交于(0
2)1点令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合并得3k&lt;4k 2007-04-22 19:07:48 补充： 我(b) (ii) 做漏了一个部份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合并h=4
5 2007-04-22 19:11:55 补充： 我最初做到h^2&gt;3k﹐跟住以为你4k是打错的所以b(ii)做了h&gt;3就硬写h=4
5 (我以为题目有规定h值范围而你冇打)不过后来发现自己冇用到只交于(0
2)1点这个条件肯定有些地方做漏最后才想到可以用来证明h^2&lt;4k
函数f(x)=x3次+hx2次+kx+2(h
k为常数)的图形有两相异转各点
且与直线y=2只交于(0
2)1点 a.证明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)的图形经过(-2
0) i.以h表k f(-2) = 0 -8 + 4h - 2k + 2 = 0 k = 2h - 3 ii.若h为一整数
利用a.bi的结果证明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.当h=4
求f(x)图形的极大点及极小点 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.
参考: EASON MENSA