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標題:
求函數f(x)的解析式
發問:
最佳解答:
2x * f(x) + (x - 3) * f(1/(1-x)) = 4x2 - 10x - 1/2 ... (1) 令 y = 1/(1-x) , 則 1/(1-y) = (x-1)/x = z , 那麼 1/(1-z) = x。 分別以y , z 代入(1) 得 :2x * f(x) + (x - 3) * f(y) = 4x2 - 10x - 1/2 ... (1)2y * f(y) + (y - 3) * f(z) = 4y2 - 10y - 1/2 ... (2)2z * f(z) + (z - 3) * f(x) = 4z2 - 10z - 1/2 ... (3) (1) * 2y - (2) * (x - 3) 得 : 4xy * f(x) - (x - 3)(y - 3) * f(z) = 2y(4x2 - 10x - 1/2) - (x - 3)(4y2 - 10y - 1/2) ...(4) (4) * 2z + (3) * (x - 3)(y - 3) 得 :8xyz * f(x) + (x - 3)(y - 3)(z - 3) * f(x) = yz(16x2 - 40x - 2) - z(x - 3)(8y2 - 20y - 1) + (x - 3)(y - 3)(4z2 - 10z - 1/2) 注意 y = 1/(1-x) , z = (x-1)/x , xyz = - 1 , xy = x/(1-x) , yz = - 1/x , zx = x - 1 。 - 8 * f(x) + (x - 3) (2-3x)/(x-1) (-2x-1)/x * f(x) = (- 16x + 40 + 2/x) - (x - 1 - 3(x-1)/x)(8/(1-x)2 - 20/(1-x) - 1) + (x - 3) (2-3x)/(x-1) (4(x-1)2/x2 - 10(x-1)/x - 1/2) (6x3 - 27x2 + 9x + 6) / (x(x-1)) * f(x) = (9x?- 93x3 + 180x2 - 18x - 48) / (2x2(x - 1)) f(x) = (3x?- 31x3 + 60x2 - 6x - 16) / (4x?- 18x3 + 6x2 + 4x) 2012-11-15 20:59:59 補充： 不肯定~我覺得只有這個解~ 那個網我看不懂,不過你小心打上連結會被系統扣分~
其他解答:
How come your command is not deleted?? I got few hundred points deleted because of posting links. Stupid YAHOO server.|||||?雨後陽光?，你咁肯定這條函數方程只有這個解？ 不怕好似f(1/x) = xf(x)（http://math.stackexchange.com/questions/28737）咁樣有無限多個解，且任意部分為某種類的函數？
求函數f(x)的解析式
發問:
此文章來自奇摩知識+如有不便請留言告知
若2x * f(x) + (x - 3) * f(1/(1-x)) = 4x^2 - 10x - 1/2，求f(x)。最佳解答:
2x * f(x) + (x - 3) * f(1/(1-x)) = 4x2 - 10x - 1/2 ... (1) 令 y = 1/(1-x) , 則 1/(1-y) = (x-1)/x = z , 那麼 1/(1-z) = x。 分別以y , z 代入(1) 得 :2x * f(x) + (x - 3) * f(y) = 4x2 - 10x - 1/2 ... (1)2y * f(y) + (y - 3) * f(z) = 4y2 - 10y - 1/2 ... (2)2z * f(z) + (z - 3) * f(x) = 4z2 - 10z - 1/2 ... (3) (1) * 2y - (2) * (x - 3) 得 : 4xy * f(x) - (x - 3)(y - 3) * f(z) = 2y(4x2 - 10x - 1/2) - (x - 3)(4y2 - 10y - 1/2) ...(4) (4) * 2z + (3) * (x - 3)(y - 3) 得 :8xyz * f(x) + (x - 3)(y - 3)(z - 3) * f(x) = yz(16x2 - 40x - 2) - z(x - 3)(8y2 - 20y - 1) + (x - 3)(y - 3)(4z2 - 10z - 1/2) 注意 y = 1/(1-x) , z = (x-1)/x , xyz = - 1 , xy = x/(1-x) , yz = - 1/x , zx = x - 1 。 - 8 * f(x) + (x - 3) (2-3x)/(x-1) (-2x-1)/x * f(x) = (- 16x + 40 + 2/x) - (x - 1 - 3(x-1)/x)(8/(1-x)2 - 20/(1-x) - 1) + (x - 3) (2-3x)/(x-1) (4(x-1)2/x2 - 10(x-1)/x - 1/2) (6x3 - 27x2 + 9x + 6) / (x(x-1)) * f(x) = (9x?- 93x3 + 180x2 - 18x - 48) / (2x2(x - 1)) f(x) = (3x?- 31x3 + 60x2 - 6x - 16) / (4x?- 18x3 + 6x2 + 4x) 2012-11-15 20:59:59 補充： 不肯定~我覺得只有這個解~ 那個網我看不懂,不過你小心打上連結會被系統扣分~
其他解答:
How come your command is not deleted?? I got few hundred points deleted because of posting links. Stupid YAHOO server.|||||?雨後陽光?，你咁肯定這條函數方程只有這個解？ 不怕好似f(1/x) = xf(x)（http://math.stackexchange.com/questions/28737）咁樣有無限多個解，且任意部分為某種類的函數？
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