標題:
發問:
(1) If f(x) = x(x+1), find the values of(a) f(2) + [f(-3)]^2 (2) It is given that f(x) = (x+3)(x+2) - k^2 and f(k) = 2k.(a) Find the value of k.(b) If f(x+2) - f(x-2) = kx - 32 , find the value of k.(c) Hence, solve for x if f(x+2) = f(x-2)+40 3. It is given that f(x) = (x+2)(x-2) + ax + b .... 顯示更多 (1) If f(x) = x(x+1), find the values of (a) f(2) + [f(-3)]^2 (2) It is given that f(x) = (x+3)(x+2) - k^2 and f(k) = 2k. (a) Find the value of k. (b) If f(x+2) - f(x-2) = kx - 32 , find the value of k. (c) Hence, solve for x if f(x+2) = f(x-2)+40 3. It is given that f(x) = (x+2)(x-2) + ax + b . If f(2) = 2 and f(-2) =4, find the values of a and b.
最佳解答:
(1) If f ( x ) = x ( x + 1 ), find the values of (a) f ( 2 ) + [ f ( - 3 ) ] ^ 2 = 2 ( 2 + 1 ) + [ ( - 3 ) ( - 3 + 1 ) ] ^ 2 = 6 + [ ( - 3 ) ( - 2 ) ] ^ 2 = 6 + 6 ^ 2 = 6 + 3 6 = 4 2 2a) f ( x ) = ( x + 3 ) ( x + 2 ) - k ^ 2 f ( k ) = ( k + 3 ) ( k + 2 ) - k ^ 2 2 k = k ^ 2 + 5 k + 6 - k ^2 - 6 = 3 k k = - 2 b) f ( x + 2 ) - f ( x - 2 ) = k x - 3 2 ( x + 2 + 3 ) ( x + 2 + 2 ) - k ^ 2 - [ ( x - 2 + 3 ) ( x - 2 + 2 ) - k ^ 2 ] = k x - 3 2 ( x + 5 ) ( x + 4 ) - ( x + 1 ) ( x ) = k x - 3 2 x ^ 2 + 9 x + 2 0 - x ^ 2 - x = k x - 3 2 8 x + 2 0 = k x - 3 2 8 x - k x = - 5 2 k x - 8 x = 5 2 k x = 5 2 + 8 x k = ( 5 2 / x ) + 8 c) f ( x + 2 ) = f ( x - 2 ) + 4 0 f ( x + 2 ) - f ( x - 2 ) = 4 0 compare with b) 4 0 = k x - 3 2 7 2 = [ ( 5 2 / x ) + 8 ] x 7 2 = 5 2 + 8 x 2 0 = 8 x x = 5 / 2 3. f ( x ) = ( x + 2 ) ( x - 2 ) + ax + b f ( 2 ) = ( 2 + 2 ) ( 2 - 2 ) + 2 a + b 2 = 2 a + b---------(1) f ( - 2 ) = ( - 2 + 2 ) ( - 2 - 2 ) - 2 a + b 4 = - 2 a + b---------(2) ( 1 ) + ( 2 ) 2 + 4 = 2 a - 2 a + b + b 6 = 2 b b = 3 for b=3 2 = 2 a + 3 - 1 = 2 a a = - 1 / 2 so a=-1/2 , b = 3
其他解答:
此文章來自奇摩知識+如有不便請留言告知
(40) mathematics (functions) f.4發問:
(1) If f(x) = x(x+1), find the values of(a) f(2) + [f(-3)]^2 (2) It is given that f(x) = (x+3)(x+2) - k^2 and f(k) = 2k.(a) Find the value of k.(b) If f(x+2) - f(x-2) = kx - 32 , find the value of k.(c) Hence, solve for x if f(x+2) = f(x-2)+40 3. It is given that f(x) = (x+2)(x-2) + ax + b .... 顯示更多 (1) If f(x) = x(x+1), find the values of (a) f(2) + [f(-3)]^2 (2) It is given that f(x) = (x+3)(x+2) - k^2 and f(k) = 2k. (a) Find the value of k. (b) If f(x+2) - f(x-2) = kx - 32 , find the value of k. (c) Hence, solve for x if f(x+2) = f(x-2)+40 3. It is given that f(x) = (x+2)(x-2) + ax + b . If f(2) = 2 and f(-2) =4, find the values of a and b.
最佳解答:
(1) If f ( x ) = x ( x + 1 ), find the values of (a) f ( 2 ) + [ f ( - 3 ) ] ^ 2 = 2 ( 2 + 1 ) + [ ( - 3 ) ( - 3 + 1 ) ] ^ 2 = 6 + [ ( - 3 ) ( - 2 ) ] ^ 2 = 6 + 6 ^ 2 = 6 + 3 6 = 4 2 2a) f ( x ) = ( x + 3 ) ( x + 2 ) - k ^ 2 f ( k ) = ( k + 3 ) ( k + 2 ) - k ^ 2 2 k = k ^ 2 + 5 k + 6 - k ^2 - 6 = 3 k k = - 2 b) f ( x + 2 ) - f ( x - 2 ) = k x - 3 2 ( x + 2 + 3 ) ( x + 2 + 2 ) - k ^ 2 - [ ( x - 2 + 3 ) ( x - 2 + 2 ) - k ^ 2 ] = k x - 3 2 ( x + 5 ) ( x + 4 ) - ( x + 1 ) ( x ) = k x - 3 2 x ^ 2 + 9 x + 2 0 - x ^ 2 - x = k x - 3 2 8 x + 2 0 = k x - 3 2 8 x - k x = - 5 2 k x - 8 x = 5 2 k x = 5 2 + 8 x k = ( 5 2 / x ) + 8 c) f ( x + 2 ) = f ( x - 2 ) + 4 0 f ( x + 2 ) - f ( x - 2 ) = 4 0 compare with b) 4 0 = k x - 3 2 7 2 = [ ( 5 2 / x ) + 8 ] x 7 2 = 5 2 + 8 x 2 0 = 8 x x = 5 / 2 3. f ( x ) = ( x + 2 ) ( x - 2 ) + ax + b f ( 2 ) = ( 2 + 2 ) ( 2 - 2 ) + 2 a + b 2 = 2 a + b---------(1) f ( - 2 ) = ( - 2 + 2 ) ( - 2 - 2 ) - 2 a + b 4 = - 2 a + b---------(2) ( 1 ) + ( 2 ) 2 + 4 = 2 a - 2 a + b + b 6 = 2 b b = 3 for b=3 2 = 2 a + 3 - 1 = 2 a a = - 1 / 2 so a=-1/2 , b = 3
其他解答:
文章標籤
全站熱搜
留言列表
