Explanation Given: The function is f ( x ) = x 3 − 10 , x 0 = 2 . Formula used: Newton method: x n + 1 = x n − f ( x n ) f ′ ( x n ) Calculation: The function is f ( x ) = x 3 − 10 , x 0 = 2 . We have to find the value of x 1 , x 2 and x 3 . Find the derivative of function. f ′ ( x ) = 3 x 2 Use the formula: x n + 1 = x n − f ( x n ) f ′ ( x n ) Compute the successive approximations. First, find the value of x 1 . x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) = x 0 − x 0 3 − 10 3 x 0 2 Substitute 2 for x 0 in the above function. x 1 = 2 − ( 2 ) 3 − 10 3 ( 2 ) 2 = 2 − 8 − 10 12 = 2 + 1 6 = 2.16667 Now, find the value of x 2 . x 2 = x 1 − f ( x 1 ) f ′ ( x 1 ) = x 1 − x 1 3 − 10 3 x 1 2 Substitute 2

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 4.7, Problem 3E

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Use Newton method to f and initial guess x0 to find x1,x2 and x3.

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Chapter 4.7, Problem 2E

Chapter 4.7, Problem 4E

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٣:٥٣ النموذج الاول . . . O O O بشما ند الحمر الحمر الجمهورية الجنية وزارة التربية والتعليم اليوم التاريخ اللجنة العليا للاختبارات الزمن اختبار مادة الجبر والهندسة لجنة المطابع السرية المركزية للشهادة الثانوية العامة (القسم العلمي) الفترة %97 (1) ظلل في ورقة الإجابة الدائرة التي تحتوي على الحرف ( ص ) للإجابة الصحيحة والحرف ( خ ) للإجابة الخطأ بحسب رقم الفقرة لكل مما يأتي ( درجة لكل فقرة ) )1 ) 2 ) 3 ) 4 ) بؤرة القطع س" = ١٢ ص هي ( ۲ ) طول المحور الأصغر للقطع ٩ س + ص = ٩ يساوي 6 وحدات طول . ) إذا كان & عدد مركب ، 181 + 11 = ٦ ، فإن ١١ = ٣ . ) إذا كان م + ۳ ت = ۲ + ت ب م ، ب دع ، فإن م + ب = 5 ( ) إذا كان & = ۱ + ٣ ت ، فإن ٠ = ١٠ . 6 ( - ) إذا كان ٥٠ - ٣ - ١٢٠ ٤ - ٣ ، فإن قيمة ٧ = ٥ . 1 ) = N ) إذا كان ح هو الحد الخالي من س في المفكوك ( س + v. N 8 ( ( قيمة المقدار , = + ۱ ، * . . + ، فإن قيمة ٧ = ١٦ . ۱ + 9 ( ) المستقيمان المقاربان للقطع الذي معادلته س" = ١ هما ص = : ۹ 10 ( ) إذا كان ٥ + س = ٢٤ ، فإن قيمة س = - 1 س 11 ( ) إذا كانت النسبة بين الحدين الأوسطين تساوي 9 في المفكوك ( س + - ) ،…

Chapter 4 Solutions

CALCULUS (CLOTH)

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E Ch. 4.1 - Prob. 11E Ch. 4.1 - Prob. 12E Ch. 4.1 - Prob. 13E Ch. 4.1 - Prob. 14E Ch. 4.1 - Prob. 15E Ch. 4.1 - Prob. 16E Ch. 4.1 - Prob. 17E Ch. 4.1 - Prob. 18E Ch. 4.1 - Prob. 19E Ch. 4.1 - Prob. 20E Ch. 4.1 - Prob. 21E Ch. 4.1 - Prob. 22E Ch. 4.1 - Prob. 23E Ch. 4.1 - Prob. 24E Ch. 4.1 - Prob. 25E Ch. 4.1 - Prob. 26E Ch. 4.1 - Prob. 27E Ch. 4.1 - Prob. 28E Ch. 4.1 - Prob. 29E Ch. 4.1 - Prob. 30E Ch. 4.1 - Prob. 31E Ch. 4.1 - Prob. 32E Ch. 4.1 - Prob. 33E Ch. 4.1 - Prob. 34E Ch. 4.1 - Prob. 35E Ch. 4.1 - Prob. 36E Ch. 4.1 - Prob. 37E Ch. 4.1 - Prob. 38E Ch. 4.1 - Prob. 39E Ch. 4.1 - Prob. 40E Ch. 4.1 - Prob. 41E Ch. 4.1 - Prob. 42E Ch. 4.1 - Prob. 43E Ch. 4.1 - Prob. 44E Ch. 4.1 - Prob. 45E Ch. 4.1 - Prob. 46E Ch. 4.1 - Prob. 47E Ch. 4.1 - Prob. 48E Ch. 4.1 - Prob. 49E Ch. 4.1 - Prob. 50E Ch. 4.1 - Prob. 51E Ch. 4.1 - Prob. 52E Ch. 4.1 - Prob. 53E Ch. 4.1 - Prob. 54E Ch. 4.1 - Prob. 55E Ch. 4.1 - Prob. 56E Ch. 4.1 - 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Use Newton method to f and initial guess x 0 to find x 1 , x 2 and x 3 .

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Chapter 4 Solutions