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Answer: Setting up partial fraction decomposition Give the

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 7E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 7E

Setting up partial fraction decomposition Give the appropriate farm of the partial fraction decomposition for the following functions.

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Step 1 of 3

Problem 7ESetting up partial fraction decomposition Give the appropriate farm of the partial fraction decomposition for the following functions. Answer;Step-1Proper fraction definition ; In a rational fraction , if the degree of f(x) < the degree of g(x) , then the rational fraction is called a proper fraction. The sum of two proper fractions is a proper fraction. Example; Partial fractions Depending upon the nature of factors of Denominator ;1. When the denominator has non-repeated linear factors; A non - repeated linear factor (x-a) of denominator corresponds a partial fraction of the form , where A is a constant to be determined’ If g(x) = (x-a)(x-b)(x-c)............(x-n), then we assume that = ++ +...............+Where A, B, C,............N are constants which can be determined by equating the numerator of L.H.Sto the numerator of R.H.S , and substituting x = a,b ,c ….n.Step-2 The given fraction is ; The given fraction is form , and f(x) < g(x). Therefore , the given fraction is a proper fraction , and the denominator has non- repeated linear factors. Thus , from the above step the given fraction can be written as; = = = , since (a+b)(a-b) = = Therefore , = = +…………….(1) = Thus , = = Equating the numerator of L.H.S to the numerator of R.H.S Then , ( x) = A(x+4) +B(x-4) X = (A+B)x+ (A-B) A+B = 1 , and (A-B) = 0.So , A = B Solving A+B=1, A- B = 0 , then B = , and A+B= 1 A = 1 -B = 1 -= A = ………..(2) Therefore, from(1), (2) = = = + , since from(1) = [+] Therefore , = [+]

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Chapter 7.4, Problem 7E is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Answer: Setting up partial fraction decomposition Give the