Re: [考古] 反函數
※ 引述《verysong (verysong)》之銘言：
: 1. f(x) = (3-x)^2 限制f之定義域 定義一新函數F
: 使得f, F一對一 ,求F的反函數
: f(2) -1
: 2. f(x) = x^3 + x 在區間 [1,2]時為一對一 求∫ f (x) dx 之值
: f(1)
: 解答是51/4 請問對還是錯？
: 虛心請益
f(1) = 1 + 1 = 2 , f(2) = 8 + 2 = 10
f(2) -1
∫ f (x) dx
f(1)
-1 |f(2) f(2) -1
= (x)(f (x)) | - ∫ x d(f (x))
|f(1) f(1)
f(2) -1 -1
= (f(2))(2) - (f(1))(1) - ∫ f(f (x)) d(f (x))
f(1)
2
= (10)*(2) - (2)*(1) - ∫ f(y) dy
1
-1 -1
(令 y = f (x) , 則 x = f(2) => y = f (f(2)) = 2
-1
x = f(1) => y = f (f(1)) = 1 )
2
= 20 - 2 - ∫ y^3 + y dy
1
y^4 y^2 |2
= 18 - (----- + -----) |
4 2 |1
1 1
= 18 - ((4 + 2) - (--- + ---))
4 2
1 1
= 18 - 4 - 2 + --- + ---
4 2
3 51
= 12 + --- = ----
4 4
因此答案沒錯
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◆ From: 61.66.173.21
※ 編輯: LuisSantos 來自: 61.66.173.21 (07/03 00:11)
推
07/03 00:13, , 1F
07/03 00:13, 1F
→
07/03 00:14, , 2F
07/03 00:14, 2F
→
07/03 00:15, , 3F
07/03 00:15, 3F
推
07/03 00:16, , 4F
07/03 00:16, 4F
推
07/03 00:50, , 5F
07/03 00:50, 5F
推
07/03 15:06, , 6F
07/03 15:06, 6F
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