設f(x)=ax+b (a不等於0).. 且f(1)=-1,f(2)=2,求f(4)?
請給我解題過程..感謝!
2005-08-09 08:10:11 · 7 個解答 · 發問者 Mraz 1 in 教育與參考 ➔ 考試
f(1) = -1
a(1) + b = -1
a+b = -1 --- (1)
f(2) = 2
a(2) + b = 2
2a+b = 2
b = 2-2a --- (2)
Put (2) into (1),
a + (2-2a) = -1
a = 3
b = 2 - 2(3) => b = -4
f(4) = a(4) + b = 3(4) + (-4) = 8
2005-08-09 08:13:40 · answer #1 · answered by ? 6 · 0⤊ 0⤋
寫的好亂看不懂
2005-08-10 12:19:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋
f(x)=ax+b (a不等於0)..
f(1)=-1 => a+b= -1......A
f(2)=2 => 2a+b=2......B
B式-A式 => a=3......C
C式代入A式 => b= - 4
所以f(x)=3x- 4
得 f(4)=12 - 4= 8
答: f(4)=8
2005-08-09 08:22:29 · answer #3 · answered by ilovelinichenhuge 3 · 0⤊ 0⤋
f(x) = ax + b (a不等於0)
f(1) = - 1
f(2) = 2
解：
f(1) = - 1代入f(x) = ax + b 可得 a + b = - 1
f(2) = 2 代入f(x) = ax + b 可得 2a + b = 2
由以上兩式解聯立方程式
下面式子 減 上面式子 可得 a = 3 代回上面式子 可得 b = - 4
則 f(x) = 3x - 4
所以f(4) = 3*4 -4 = 8
2005-08-09 08:17:46 · answer #4 · answered by 立威 1 · 0⤊ 0⤋
because that f(x)=ax+b(and a is not 0).if f(1)=-1, so, a+b=-1if f(2)=2, so, 2a+b=2(2a+b)-(a+b)=3=a, and b=-4that f(4)=4(3)+(-4)=12-4=8.so f(4) = 8.
2005-08-09 08:16:03 · answer #5 · answered by lian 7 · 0⤊ 0⤋
這是一個聯立方程式
f(1)=-1
f(2)=2
-1=a+b
2=2a+b
得a=3 b= - 4
所以
f(4)=3x-4=8
2005-08-09 12:15:50 補充：
為什麼有人答案寫錯來登啊
2005-08-09 08:14:58 · answer #6 · answered by Anonymous · 0⤊ 0⤋
x=1 代入f(x)=ax+b 得f(1)=a+b =-1---[1]
x=2 代入f(x)=ax+b 得f(2)=2a+b =2------[2]
[1][2]聯立得a=3 b=- 4
f(x)=3x- 4
f(4)=3*4- 4=12-4=8
2005-08-09 08:13:53 · answer #7 · answered by 宛儀 3 · 0⤊ 0⤋