數學知識交流---值域

1)f(x) = (2x² - 3x + 87) / (4x² - 4x + 1)= (2x² - 2x + 1/2 - x + 173/2) / (4x² - 4x + 1)= 1/2 - (1/2)(2x - 173) / (4x² - 4x + 1)= 1/2 - (1/2)(2x - 1 - 172) / (2x - 1)² = 1/2 - (1/2) / (2x - 1) + 86 / (2x - 1)² , 令 k = 1 / (2x - 1) ,= 86k² - (1/2)k + 1/2 = 86(k² - 2k(1/344) + 1/344² ) + 1/2 - 86/344²= 86(k - 1/344)² + 687/1376 ≥ 687/1376即 f(x) 的值域為不小於 687/1376 的一切實數。

2) f(x) = (2x²-3x+7)/(x²-4x+4) + (x²+6x-1)/(2x²+5x-3)= 2 + (5x-1)/(x-2)² + 1/2 + (7x+1)/(2(x+3)(2x-1))= 5/(x-2) + 9/(x-2)² + (7x+1)/(2(x+3)(2x-1)) + 5/2

只考慮 y = 5/(x-2) + 9/(x-2)² + (7x+1)/(2(x+3)(2x-1)) 9/(x-2)² + 5/(x-2) + (7x+1)/(2(x+3)(2x-1)) - y = 0 因 1/(x-2) 為實數 ,
△ = 5² - 4*9 ( (7x+1)/(2(x+3)(2x-1)) - y ) ≥ 0y ≥ (7x+1)/(2(x+3)(2x-1)) - 25/36
只考慮 k = (7x+1)/((x+3)(2x-1))k(2x² + 5x - 3) = 7x + 12kx² + (5k-7)x - (3k+1) = 0 因 x 為實數 ,
△ = (5k-7)² + 4*2k(3k+1) ≥ 025k² - 70k + 49 + 24k² + 8k ≥ 049k² - 62k + 49 ≥ 0 49k² - 2(7k)(31/7) + (31/7)² + 49 ≥ 0(7k - 31/7)² + 49 ≥ 0
k 可為一切實數 ,
故 y = (7x+1)/(2(x+3)(2x-1)) - 25/36 = k/2 - 25/36 可為一切實數,
f(x) = y + 5/2 亦可為一切實數。
即 f(x) 的值域為一切實數。

2012-02-03 13:48:46 補充：
修正倒數第6 , 7行 :
49k² - 2(7k)(31/7) + (31/7)² + 49 - (31/7)² ≥ 0
(7k - 31/7)² + 1440/49 ≥ 0

k 可為一切實數

1.
Let y = f(x) ,
then (4x^2 - 4x + 1)y = 2x^2 - 3x + 87
2(2y-1)x^2 + (3-4y)x + (y-87) = 0
x is real number =>
(3-4y)^2 - 4(2)(2y-1)(y-87) >= 0
9 - 12y + 16y^2 - 8( 2y^2 - 175y + 87) >= 0
-687 + 1388y >= 0
y >= 687 / 1388

2012-02-02 18:32:22 補充：
更正 : x is real number =>
(3-4y)^2 - 4(2)(2y-1)(y-87) >= 0
9 - 24y + 16y^2 - 8( 2y^2 - 175y + 87) >= 0
9 - 24y + 1400y - 696 >= 0
y >= 687 / 1376