Section.7 Partial Fractions Section.7 Partial Fractions N You should know how to decompose a rational function into partial fractions. D (a) If the fraction is improper, divide to obtain N D p N D (a) where p is a polnomial. (b) Factor the denominator completel into linear and irreducible quadratic factors. (c) For each factor of the form p q m, the partial fraction decomposition includes the terms p q p q... m p q. m (d) For each factor of the form a b c n, the partial fraction decomposition includes the terms You should know how to determine the values of the constants in the numerators. (a) Set a b c a b c... n n a b c. n N D partial fraction decomposition. (b) Multipl both sides b D to obtain the basic equation. (c) For distinct linear factors, substitute the zeros of the distinct linear factors into the basic equation. (d) For repeated linear factors, use the coefficients found in part (c) to rewrite the basic equation. Then use other values of to solve for the remaining coefficients. (e) For quadratic factors, epand the basic equation, collect like terms, and then equate the coefficients of like terms. Solutions to Odd-Numbered Eercises. Matches (b).. Matches (d) 5. 7 7 7. 0 0 0 9. 5 5 5 5. 0 0 0. D E
Part I Solutions to Odd-Numbered Eercises and Practice Tests 5. Let : Let : 7. Let 0: Let : 9. Let : Let 0:. Let : Let :. Let 0: Let : Let : 0 5 5 5. Let 0: Let : Let : 5 5 7. Let : 9 Let 0: 0 0 9 9
Section.7 Partial Fractions 9., 0, and. Therefore,,, and 0.. 0,, and 0. Therefore,,, and.. Using the first and third equation, we have 0 and 0; b subtraction, 0. Using the second and fourth equation, we have D and D 0; b subtraction, D, so D. Substituting 0 for and for D in the first and second equations, we have 0 and so and,. D D D D D D 0, D, 0, and 0 D D
Part I Solutions to Odd-Numbered Eercises and Practice Tests 5. D D D D D D 0, 0 D,, and 0 D. Using the first and third equations, we have 0 and ; b subtraction,, so. Using the second and fourth equations, we have D 0 and D 0; b subtraction D 0, so D 0. Substituting for and 0 for D in the first and second equations, we have and 0, so and. D 7. 5 5, 0, and 5. Subtracting both sides of the second equation from the first gives ; combining this with the third equation gives and. Since, we also have 0. 5 9.. 5 9 9 Let : Let : 7 7 5 7 9 7 7
Section.7 Partial Fractions 5. Let : Let 0: Let : 0 So, and. 5. 5 5 Let : 9 Let : 5 7. 9. D D Let : 0 D Let 0: Let : 0 Equating coefficients of like powers: 0 0 0 D D 0 5. 5 5 5 0 5 5 0 Let : Let : 9 5 5
Part I Solutions to Odd-Numbered Eercises and Practice Tests 5. ( Let 0: Let : 0 0 0 Vertical asmptotes: 0 Vertical asmptote: 0 Vertical asmptote: and The combination of the vertical asmptotes of the terms of the decomposition are the same as the vertical asmptotes of the rational function. 55. 9 ( Let : Let : 0 5 9 5 9 5 Vertical asmptotes: ± Vertical asmptote: Vertical asmptote: The combination of the vertical asmptotes of the terms of the decomposition are the same as the vertical asmptotes of the rational function.
Section.7 Partial Fractions 7 57. (a) 000 77 7 7, 0 < Let Let 000 7 7 0,000 : 5 7 7 7 000 7 : 500 5 000 000 000 000 000 000 77 7 7 7 7, 0 < (b) ma 000 (c) 000 7 000 Yma min 7 Ymin 0 00 (d) ma 0.5 00F min 0.5.7F 59. False. The partial fraction decomposition is 0 0 0.. a a a a Let a: a a Let a: a a a a a a, a is a constant.. a a a a Let 0: a a Let a: a a a a a 5. f 9 7. Intercepts: 0,,, 0,, 0 Graph rises to the left and rises to the right. f Intercepts: 0, 0,, 0 Graph rises to the left and falls to the right. 5 0 5 9. f 5 -intercepts: -intercept:, 0,, 0 0, 5 0 5 0 5 0 5 0 5 Vertical asmptote: 5 Slant asmptote: No horizontal asmptote.