Explanation Given: The function is f ( x ) = 3 x 3 + 2 x 2 − x − 4 . 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 function f ( x ) = 3 x 3 + 2 x 2 − x − 4 . Choose j = 5 and differentiate the function f ( x ) = 3 x 3 + 2 x 2 − x − 4 up to degree 5 as shown below. The first derivative for the function f ( x ) = 3 x 3 + 2 x 2 − x − 4 is calculated as, f ′ ( x ) = d d x ( 3 x 3 + 2 x 2 − x − 4 ) = 9 x 2 + 4 x − 1 The second derivative for the function f ( x ) = 3 x 3 + 2 x 2 − x − 4 is calculated as, f ″ ( x ) = d d x ( f ′ ( x ) ) = d d x ( 9 x 2 + 4 x − 1 ) = 18 x + 4 The third derivative for the function f ( x ) = 3 x 3 + 2 x 2 − x − 4 is calculated as, f ‴ ( x ) = d d x ( f ″ ( x ) ) = d d x ( 18 x + 4 ) = 18 The fourth derivative for the function f ( x ) = 3 x 3 + 2 x 2 − x − 4 is calculated as, f ( 4 ) ( x ) = d d x ( f ‴ ( x ) ) = d d x ( 18 ) = 0 The fifth derivative for the function f ( x ) = 3 x 3 + 2 x 2 − x − 4 is calculated as, f ( 5 ) ( x ) = d d x ( f ( 4 ) ( x ) ) = d d x ( 0 ) = 0 In order to obtain the Maclaurin polynomial of degree5, use the first five derivative of f ( x ) evaluated at a = 0 as shown in Table 1. Derivative of function f ( x ) Value of derivative at a = 0 f ( x ) = 3 x 3 + 2 x 2 − x − 4 f ( 0 ) = − 4 f ′ ( x ) = 9 x 2 + 4 x − 1 f ′ ( 0 ) = − 1 f ″ ( x ) = 18 x + 4 f ″ ( 0 ) = 4 f ‴ ( x ) = 18 f ‴ ( 0 ) = 18 f ( 4 ) ( x ) = 0 f ( 4 ) ( 0 ) = 0 f ( 5 ) ( x ) = 0 f ( 5 ) ( 0 ) = 0 Table 1 The Maclaurin polynomial T 5 ( x ) by use of equation (1) can be written as, T 5 ( x ) = [ f ( 0 ) + f ' ( 0 ) 1 ! ( x − 0 ) 1 + f ' ' ( 0 ) 2 ! ( x − 0 ) 2 + f ‴ ( 0 ) 3 ! ( x − 0 ) 3 + f ( 4 ) ( 0 ) 4 ! ( x − 0 ) 4 + f ( 5 ) ( 0 ) 5 ! ( x − 0 ) 5 ] ....... (2) Substitute − 4 for f ( 0 ) , − 1 for f ′ ( 0 ) , 4 for f ″ ( 0 ) , 18 for f ‴ ( 0 ) ,0 for f ( 4 ) ( 0 ) and 0 for f ( 5 ) ( 0 ) in equation (2) to obtain the fifth Maclaurin Polynomial T 5 ( x ) as, T 5 ( x ) = − 4 + − 1 1 ! x + 4 2 ! x 2 + 18 3 ! x 3 + 0 4 ! ( x − 0 ) 4 + 0 5 ! ( x − 0 ) 5 = − 4 − 1 ⋅ x + 4 2 ⋅ 1 x 2 + 18 3 ⋅ 2 ⋅ 1 x 3 + 0 4 ! x 4 + 0 5 ! x 5 = − 4 − x + 2 x 2 + 3 x 3 The fifth Maclaurin polynomial for f ( x ) = 3 x 3 + 2 x 2 − x − 4 is − 4 − x + 2 x 2 + 3 x 3 for a = 0

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 9.4, Problem 53E

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To find: The Tj for j=1,2,3,4,5 at both a=0 and a=1 for f(x)=3x3+2x2−x−4 . Also, show that T3(x)=f(x) .

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Chapter 9.4, Problem 52E

Chapter 9.4, Problem 54E

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Chapter 9 Solutions

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The T j for j = 1 , 2 , 3 , 4 , 5 at both a = 0 and a = 1 for f ( x ) = 3 x 3 + 2 x 2 − x − 4 . Also, show that T 3 ( x ) = f ( x ) .

Solution Summary: The author analyzes the n th Taylor polynomial centered at x=a for the function f(x)=3x3+2x2x4.

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To find: The Tj for j=1,2,3,4,5 at both a=0 and a=1 for f(x)=3x3+2x2−x−4 . Also, show that T3(x)=f(x) .

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