Explanation Given: The function is y = x − 2 3 , the value of x is 1.2 and the value of a is 1. Concept used: The n th Taylor polynomial centered at x = a for the function is T n ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) 1 + f ″ ( a ) 2 ! ( x − a ) 2 + ⋯ + f n ( a ) n ! ( x − a ) n ....... (1) If a = 0 , then T n is also called the n th Maclaurin polynomial. Calculation: Consider, the given function y = x − 2 3 as f ( x ) = x − 2 3 . Choose n = 2 and differentiate the function f ( x ) = x − 2 3 up to degree 2 as shown below. The first derivative for the function f ( x ) = x − 2 3 is calculated as, f ′ ( x ) = d d x ( x − 2 3 ) = − 2 3 x − 2 3 − 1 = − 2 3 x − 5 3 The secondderivative for the function f ( x ) = x − 2 3 is calculated as, f ″ ( x ) = d d x ( − 2 3 x − 5 3 ) = ( − 2 3 ) ( − 5 3 ) x − 5 3 − 1 = 10 9 x − 8 3 In order to obtain the Taylor polynomial of degree 2 , use the first two derivative of f ( x ) evaluated at x = 1 as shown in Table 1. Derivative of function f ( x ) Value of derivative at x = 1 f ( x ) = x − 2 3 f ( 1 ) = 1 f ′ ( x ) = − 2 3 x − 5 3 f ′ ( 1 ) = − 2 3 f ″ ( x ) = 10 9 x − 8 3 f ″ ( 1 ) = 10 9 Table 1 The Taylor polynomial of degree 2 is, T 2 ( x ) = f ( a ) + f ′ ( a ) 1 ! ( x − a ) 1 + f ″ ( a ) 2 ! ( x − a ) 2 Substitute 1 for a in above equation

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 9.4, Problem 27E

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To find: The error |f(x)−T2(x)| at x=1.2 for the function y=x−23 .

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Chapter 9.4, Problem 26E

Chapter 9.4, Problem 28E

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Calculus - Standalone book

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The error | f ( x ) − T 2 ( x ) | at x = 1.2 for the function y = x − 2 3 .

Solution Summary: The author calculates the Taylor polynomial of degree 2 using the first two derivatives of f(x) evaluated at x=1.

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To find: The error |f(x)−T2(x)| at x=1.2 for the function y=x−23 .

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Chapter 9 Solutions