Explanation Given: The equation is written below. x 2 e − x + x 2 − 2 = 0 The interval is [ − 3 , 0 ] . 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: The given equations can be written in the form f ( x ) = 0 , where f ( x ) is shown below. f ( x ) = x 2 e − x + x 2 − 2 ...... ( 1 ) Differentiate the function f ( x ) with respect to x as shown below. f ′ ( x ) = ( 2 x e − x − x 2 e − x ) + ( 2 x ) f ′ ( x ) = 2 x e − x − x 2 e − x + 2 x ...... ( 2 ) For the interval [ − 3 , 0 ] , obtain the value of f ( x ) at − 3 and 0 as follows: f ( − 3 ) = ( − 3 ) 2 e − ( − 3 ) + ( − 3 ) 2 − 2 ≈ 187.7698 ( Positive ) f ( 0 ) = ( 0 ) 2 e − ( 0 ) + ( 0 ) 2 − 2 = − 2 ( Negative ) The value of f ( − 3 ) and f ( 0 ) are in the opposite sign. So, there exist a solution for the equation f ( x ) = 0 in the interval ( − 3 , 0 ) . Consider an initial value − 3 for c 1 and obtain the value of c 2 as follows: c 2 = ( − 3 ) − f ( − 3 ) f ′ ( − 3 ) ( c 2 = c 1 − f ( c 1 ) f ′ ( c 1 ) , c 1 = − 3 ) = − 3 − 187

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 14E

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The solution for x2e−x+x2−2=0 in the interval [−3,0].

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Chapter 12.6, Problem 13E

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Calculus with Applications (11th Edition)

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The solution for x 2 e − x + x 2 − 2 = 0 in the interval [ − 3 , 0 ] .

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The solution for x2e−x+x2−2=0 in the interval [−3,0].

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Chapter 12 Solutions