#### Transcript 有理函數的圖形 - Proera

```微積分數位化教材
C. L. Lang, Department of Applied Mathematics, I-Shou University
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C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑝(𝑥)
𝑓 𝑥 =
𝑞(𝑥)


C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑝(𝑥)
𝑓 𝑥 =
𝑞(𝑥)
𝑥 ∈ 𝑅| 𝑞(𝑥) ≠ 0
C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥2 + 𝑥 − 2
𝑓 𝑥 =
𝑥2 − 1
𝑥 ∈ 𝑅| 𝑥 ≠ ±1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥2 + 𝑥 − 2
𝑓 𝑥 =
𝑥2 + 1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥2 + 𝑥 − 2
𝑓 𝑥 =
𝑥2 − 1
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C. L. Lang, Department of Applied Mathematics, I-Shou University
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1.
2.

C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥3 − 1
𝑓 𝑥 = 2
𝑥 −1

𝑥 3 − 1 (𝑥 − 1)(𝑥 2 + 𝑥 + 1)
=
2
𝑥 −1
(𝑥 − 1)(𝑥 + 1)
𝑓 𝑥 的定義域為 𝑥 ∈ 𝑅| 𝑥 ≠ ±1 ，其圖形為
C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥3 − 1
𝑓 𝑥 = 2
𝑥 −1


C. L. Lang, Department of Applied Mathematics, I-Shou University
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1.

𝑥3 − 1
𝑓 𝑥 = 2
𝑥 −1

2.

C. L. Lang, Department of Applied Mathematics, I-Shou University
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1.
2.

C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑝(𝑥)
𝑓 𝑥 =
𝑞(𝑥)

3𝑥 2 + 𝑥 − 1
𝑓 𝑥 =
𝑥2 − 1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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3𝑥 2 + 𝑥 − 1
𝑓 𝑥 =
𝑥2 − 1

C. L. Lang, Department of Applied Mathematics, I-Shou University
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C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥2 + 𝑥 − 2
𝑓 𝑥 =
𝑥2 − 1

C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥2 + 𝑥 − 2
𝑓 𝑥 =
𝑥2 − 1


C. L. Lang, Department of Applied Mathematics, I-Shou University
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3
(1, )
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C. L. Lang, Department of Applied Mathematics, I-Shou University
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deg 𝑝(𝑥) > deg 𝑞(𝑥)由除法原理可得
𝑝 𝑥 = 𝑔 𝑥 𝑞 𝑥 + 𝑟(𝑥)

𝑝(𝑥)
𝑟(𝑥)
𝑓 𝑥 =
=𝑔 𝑥 +
。
𝑞(𝑥)
𝑞(𝑥)

𝑟(𝑥)

𝑞(𝑥)

x 越來越大時有理函數 𝑓 𝑥 的圖形越來越接近

C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑥4 − 𝑥2 + 𝑥 + 2
𝑓 𝑥 =
𝑥2 − 1
𝑥4 − 𝑥2 + 𝑥 + 2
𝑥+2
2
=𝑥 + 2
2
𝑥 −1
𝑥 −1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝑝(𝑥)
𝑞(𝑥)

𝑝(𝑥)
𝑞(𝑥)

C. L. Lang, Department of Applied Mathematics, I-Shou University
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 𝑞 𝑥
= (𝑎1 𝑥 + 𝑏1 )𝑛1 ⋯ (𝑎𝑘 𝑥 + 𝑏𝑘 )𝑛𝑘
(𝑐1 𝑥 2 + 𝑑1 𝑥 + 𝑒1 )𝑚1 ⋯ (𝑐𝑙 𝑥 2 + 𝑑𝑙 𝑥 + 𝑒𝑙 )𝑚𝑙


𝑝(𝑥)
𝑞(𝑥)
𝑘
𝑛𝑖
𝐴𝑖𝑗
=
𝑗
𝑖=1
𝑗=1 (𝑎𝑖 𝑥 + 𝑏𝑖 )
𝑙
𝑚𝑖
𝐶𝑖𝑗 𝑥 + 𝐷𝑖𝑗
+
2
𝑗
𝑖=1
𝑗=1 (𝑐𝑖 𝑥 + 𝑑𝑖 𝑥 + 𝑒𝑖 )
C. L. Lang, Department of Applied Mathematics, I-Shou University
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2𝑥
𝑥2 − 1
𝑥 2 − 1 = (𝑥 − 1)(𝑥 + 1)
2𝑥
𝐴
𝐵
=
+
2
𝑥 −1 𝑥−1 𝑥+1

2𝑥 = 𝐴(𝑥 + 1) + 𝐵(𝑥 − 1)
C. L. Lang, Department of Applied Mathematics, I-Shou University
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𝐴=𝐵=1
2𝑥
1
1
=
+
2
𝑥 −1 𝑥−1 𝑥+1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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3𝑥 2 − 2𝑥 + 5
𝑥3 − 1
𝑥 3 − 1 = (𝑥 − 1)(𝑥 2 + 𝑥 + 1)
3𝑥 2 − 2𝑥 + 5
𝐴
𝐵𝑥 + 𝐶
=
+ 2
3
𝑥 −1
𝑥−1 𝑥 +𝑥+1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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3𝑥 2 − 2𝑥 + 5
= 𝐴(𝑥 2 + 𝑥 + 1) + (𝐵𝑥 + 𝐶)(𝑥 − 1)
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3𝑥 2 − 2𝑥 + 5
2
𝑥−3
=
+ 2
3
𝑥 −1
𝑥−1 𝑥 +𝑥+1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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4𝑥
𝑥3 − 𝑥2 − 𝑥 + 1
𝑥 3 − 𝑥 2 − 𝑥 + 1 = (𝑥 − 1)2 (𝑥 + 1)
4𝑥
𝐴
𝐵
𝐶
=
+
+
3
2
2
𝑥 − 𝑥 − 𝑥 + 1 𝑥 − 1 (𝑥 − 1)
𝑥+1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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4𝑥 = 𝐴(𝑥 2 − 1) + 𝐵(𝑥 + 1) + 𝐶(𝑥 − 1)2

𝐶 = −1 ，代入 𝑥 = 0 得 −𝐴 + 2 − 1 = 0得 𝐴 =
1
4𝑥
1
2
1
=
+
−
3
2
2
𝑥 − 𝑥 − 𝑥 + 1 𝑥 − 1 (𝑥 − 1)
𝑥+1
C. L. Lang, Department of Applied Mathematics, I-Shou University
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2𝑥 4 − 𝑥 3 + 2𝑥 2 + 1
𝑥 5 − 𝑥 4 + 2𝑥 3 − 2𝑥 2 + 𝑥 − 1

𝑥 5 − 𝑥 4 + 2𝑥 3 − 2𝑥 2 + 𝑥 − 1 = (𝑥 − 1)(𝑥 2 + 1)2
 所以
2𝑥 4 − 𝑥 3 + 2𝑥 2 + 1
(𝑥 − 1)(𝑥 2 + 1)2
𝐴
𝐵𝑥 + 𝐶
𝐷𝑥 + 𝐸
=
+ 2
− 2
𝑥 − 1 𝑥 + 1 (𝑥 + 1)2

C. L. Lang, Department of Applied Mathematics, I-Shou University
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2𝑥 4 − 𝑥 3 + 2𝑥 2 + 1
= 𝐴 𝑥 2 + 1 2 + (𝐵𝑥 + 𝐶)(𝑥 − 1) + (𝐷𝑥 + 𝐸)(𝑥
− 1)(𝑥 2 + 1)
 比較係數解方程組得
𝐴 = 𝐵 = 1, 𝐷 = −1, 𝐶 = 𝐸 = 0
2𝑥 4 − 𝑥 3 + 2𝑥 2 + 1
𝑥 5 − 𝑥 4 + 2𝑥 3 − 2𝑥 2 + 𝑥 − 1
1
𝑥
𝑥
=
+ 2
− 2
𝑥 − 1 𝑥 + 1 (𝑥 + 1)2

C. L. Lang, Department of Applied Mathematics, I-Shou University
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I-Shou University Department of Applied
Mathematics, C. L. Lang
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