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Lecture 11

# MTH 162 Lecture 11: 6.3 Partial Fractions Notes Premium

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School
University of Miami
Department
Mathematics
Course
MTH 162
Professor
Pachero
Semester
Fall

Description
MTH 162 Calculus II 6.3 Partial Fractions Notes L. Sterling Abstract Provide a generalization to each of the key terms listed in this section. Rational Functions p(x) ▯ Rational functions are functions that are in the common form of R(x) = q(x) which shows where both p(x) and q(x) are polynomial functions while it’s only q(x) can’t be a zero polynomial. – This would generally mean the following: q (x) 6= 0 ▯ When you are trying to ﬁgure out the domain, besides that fact that q (x) 6= 0, then you ﬁnd out that it [the domain] will be all real numbers. ▯ If you have any rational function that is in the general form of R(x) = q(x) in its lowest form, then you’ll have R will have a vertical asymptote for each and every value of x for which q(x) = 0. Steps to Partial Fractions ▯ Partial fractions deals with writing ration functions as the sum of the simpler fractions and these are the steps to partial fractions: – These are the steps when the rational function is actually while it’s numerator’s degree is less than the denominator’s degree: ▯ Factor q(x), which is the denominator. ▯ Create a sum of fractions with the all of the numerators are all diﬀerent variables with the denominators are actually each of those factors. ▯ Create fractions with factor’s increasing powers in there are any repeated factors. ▯ Multiply by q(x), which is the denominator. ▯ Set the numerator, which is p(x), to 0. ▯ Solve for all variables. – These are the steps when the rational function is actually while it’s numerator’s degree is greater than the denominator’s degree: ▯ Perform any long division that can come in the form of the following: p(x) = f (x) + r (x) q (x) q (x) r(x) ▯ Find the remainder’s, which is q(x) partial function. 1 Partial Fraction Example: Decomposing Decompose the following into partial functions: x R(x) = (x + 1)(x ▯ 4) x (x+1)(x▯4) x a0 a1 (x▯4)(x+1)= x▯4+ x+1 x(x▯4)(x+1) a0(x▯4)(x+1) a1(x▯4)(x+1) (x▯4)(x+1) = x▯4 + x+1 x = a0(x + 1) + a1(x ▯ 4) (▯1) = a0((▯1) + 1) + a1((▯1) ▯ 4) ▯1 = a (0) + a (▯5) 0 1 ▯1 = ▯5a 1 ▯1 ▯5a ▯5 = ▯51 1 a1= 5 x = a0(x + 1) + a1(x ▯ 4) 4 = a0(1 + 4) + a1(4 ▯ 4) 4 = a0(5) + a1(0) 4 = 5a 0 4= 5a0 5 4 a0= 5
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