Explanation Formula used: Taylor series expansion of a function: f ( x ) = f ( a ) + f ′ ( a ) ( x − a ) + f ″ ( a ) ( x − a ) 2 2 ! + … . Proof: Consider the function f ( x ) = 2 x 3 + x 2 − 3 x + 5 with x = 1 . The derivatives and its values as shown below table. f ( x ) = 2 x 3 + x 2 − 3 x + 5 f ( 1 ) = 5 f ′ ( x ) = 6 x 2 + 2 x − 3 f ′ ( 1 ) = 5 f ″ ( x ) = 12 x + 2 f ″ ( 1 ) = 14 f ‴ ( x ) = 12 f ‴ ( 1 ) = 12 The Taylor series of a function as follows

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 30.5, Problem 33E

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To show: The polynomial 2x3+x2−3x+5 can be written as

2(x−1)3+7(x−1)2+5(x−1)+5.

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Chapter 30.5, Problem 31E

Chapter 30.5, Problem 34E

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Chapter 30 Solutions

Basic Technical Mathematics

Ch. 30.1 - Prob. 1PE Ch. 30.1 - Prob. 2PE Ch. 30.1 - Prob. 1E Ch. 30.1 - Prob. 2E Ch. 30.1 - Prob. 3E Ch. 30.1 - Prob. 4E Ch. 30.1 - Prob. 5E Ch. 30.1 - Prob. 6E Ch. 30.1 - Prob. 7E Ch. 30.1 - Prob. 8E

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The polynomial 2 x 3 + x 2 − 3 x + 5 can be written as 2 ( x − 1 ) 3 + 7 ( x − 1 ) 2 + 5 ( x − 1 ) + 5 .

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To show: The polynomial 2x3+x2−3x+5 can be written as 2(x−1)3+7(x−1)2+5(x−1)+5.

Want to see the full answer?

Chapter 30 Solutions

To show: The polynomial 2x3+x2−3x+5 can be written as

2(x−1)3+7(x−1)2+5(x−1)+5.