Explanation Given: The function is written below: f ( x ) = x 4 − 3 x 3 + 6 x − 1 ...... ( 1 ) Concept: For a closed interval [ a , b ] , if f ( a ) and f ( b ) are of opposite sign then there must exist one value of c where f ( c ) = 0 . This number c is a zero of function f . Newton’s Method: If f ' ( c n ) ≠ 0 , then c n + 1 = c n − f ( c n ) f ′ ( c n ) , where c n is the current approximate zero of the function f and c n + 1 is the upcoming approximate zero of the function f in the interval [ a , b ] . Calculation: Differentiate the function f ( x ) = x 4 − 3 x 3 + 6 x − 1 with respect to x as shown below. f ′ ( x ) = ( 4 x 3 ) − 3 ( 3 x 2 ) + 6 ( 1 ) f ′ ( x ) = 4 x 3 − 9 x 2 + 6 ...... ( 2 ) Differentiate the function f ′ ( x ) = 4 x 3 − 9 x 2 + 6 with respect to x as shown below. f ″ ( x ) = 4 ( 3 x 2 ) − 9 ( 2 x ) f ″ ( x ) = 12 x 2 − 18 x ...... ( 3 ) Consider f ′ ( x ) as g ( x ) and f ″ ( x ) as g ′ ( x ) for simplicity. g ( x ) = 4 x 3 − 9 x 2 + 6 Consider − 1 for x and evaluate the value of g ( − 1 ) as follows: g ( − 1 ) = 4 ( − 1 ) 3 − 9 ( − 1 ) 2 + 6 = − 7 ( Negative ) Consider 0 for x and evaluate the value of g ( 0 ) as follows: g ( 0 ) = 4 ( 0 ) 3 − 9 ( 0 ) 2 + 6 = 6 ( Positive ) So, there exist a solution in the interval ( − 1 , 0 ) for g ( x ) = 0 . Consider an initial value − 1 for c 1 and obtain the value of c 2 as follows: c 2 = ( − 1 ) − g ( − 1 ) g ′ ( − 1 ) ( c 2 = c 1 − g ( c 1 ) g ′ ( c 1 ) , c 1 = − 1 ) = − 1 − − 7 30 ≈ − 0.77 Similarly calculate the next approximations of c , till two subsequent approximations will agree to the nearest hundredth as shown in Table 1. g ( c n ) g ′ ( c n ) c n + 1 = c n − g ( c n ) g ′ ( c n ) g ( c 1 ) = − 7 g ′ ( c 1 ) = 30 c 2 = c 1 − g ( c 1 ) g ′ ( c 1 ) = − 1 − − 7 30 ≈ − 0.77 g ( c 2 ) = − 1.162 g ′ ( c 2 ) = 20.975 c 3 = c 2 − g ( c 2 ) g ′ ( c 2 ) = − 0.77 − − 1.162 20.975 ≈ − 0.71 g ( c 3 ) = 0.03146 g ′ ( c 3 ) = 18.829 c 4 = c 3 − g ( c 3 ) g ′ ( c 3 ) = − 0.71 − 0.03146 18.829 ≈ − 0.71 Table 1 From Table 1, it is observed that c 3 and c 4 are approximately equal nearest to the hundredth that is − 0.71 . The value of the function f ″ ( x ) at − 0.71 is calculated as follows: f ″ ( − 0.71 ) = 12 ( − 0.71 ) 2 − 18 ( − 0.71 ) = 18.829 ( Positive ) The value of f ″ ( − 0.71 ) is positive. Therefore, the graph of f ( x ) has a relative minimum at x = − 0.71 . Consider 1 for x and evaluate the value of g ( 1 ) as follows: g ( 1 ) = 4 ( 1 ) 3 − 9 ( 1 ) 2 + 6 = 1 ( Positive ) Consider 1.5 for x and evaluate the value of g ( 1.5 ) as follows: g ( 1.5 ) = 4 ( 1.5 ) 3 − 9 ( 1.5 ) 2 + 6 = − 0.75 ( Negative ) So, there exist a solution in the interval ( 1 , 1.5 ) for g ( x ) = 0 . Consider an initial value 1 for c 1 and obtain the value of c 2 as follows: c 2 = ( 1 ) − g ( 1 ) g ′ ( 1 ) ( c 2 = c 1 − g ( c 1 ) g ′ ( c 1 ) , c 1 = 1 ) = 1 − 1 6 ≈ 1

Calculus with Applications (11th Edition)

11th Edition

ISBN: 9780321979421

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 12.6, Problem 29E

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The critical points for the function f(x)=x4−3x3+6x−1 with approximation to nearest hundredth and state whether the critical point leads to a relative maximum or a relative minimum.

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Chapter 12.6, Problem 28E

Chapter 12.6, Problem 30E

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Chapter 12 Solutions

Calculus with Applications (11th Edition)

Ch. 12.1 - Find the first four terms of the sequence having...Ch. 12.1 - Prob. 2YT Ch. 12.1 - Prob. 3YT Ch. 12.1 - Prob. 4YT Ch. 12.1 - Prob. 5YT Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E Ch. 12.1 - List the first n terms of the geometric sequence...Ch. 12.1 - Prob. 5E

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The critical points for the function f ( x ) = x 4 − 3 x 3 + 6 x − 1 with approximation to nearest hundredth and state whether the critical point leads to a relative maximum or a relative minimum.

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The critical points for the function f(x)=x4−3x3+6x−1 with approximation to nearest hundredth and state whether the critical point leads to a relative maximum or a relative minimum.

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Chapter 12 Solutions