Section 8: Applications of Rational Functions

Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

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Section 7.8 Applications of Rational Functions 731 Version: Fall 2007 7.8 Applications of Rational Functions In this section, we will investigate the use of rational functions in several applications. Number Problems We start by recalling the definition of the reciprocal of a number. Definition 1. For any nonzero real number a , the reciprocal of a is the number 1 /a . Note that the product of a number and its reciprocal is always equal to the number 1. That is, a · 1 a = 1 . For example, the reciprocal of the number 3 is 1 / 3 . Note that we simply “invert” the number 3 to obtain its reciprocal 1 / 3 . Further, note that the product of 3 and its reciprocal 1 / 3 is 3 · 1 3 = 1 . As a second example, to find the reciprocal of 3 / 5 , we could make the calculation 1 3 5 = 1 ÷ 3 5 = 1 · 5 3 = 5 3 , but it’s probably faster to simply “invert” 3 / 5 to obtain its reciprocal 5 / 3 . Again, note that the product of 3 / 5 and its reciprocal 5 / 3 is 3 5 · 5 3 = 1 . Let’s look at some applications that involve the reciprocals of numbers. l⚏ Example 2. The sum of a number and its reciprocal is 29 / 10 . Find the number(s). Let x represent a nonzero number. The reciprocal of x is 1 /x . Hence, the sum of x and its reciprocal is represented by the rational expression x + 1 /x . Set this equal to 29 / 10 . x + 1 x = 29 10 To clear fractions from this equation, multiply both sides by the common denominator 10 x . Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/ 1
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732 Chapter 7 Rational Functions Version: Fall 2007 10 x x + 1 x = 29 10 10 x 10 x 2 + 10 = 29 x This equation is nonlinear (it has a power of x larger than 1), so make one side equal to zero by subtracting 29 x from both sides of the equation. 10 x 2 29 x + 10 = 0 Let’s try to use the ac -test to factor. Note that ac = (10)(10) = 100 . The integer pair {− 4 , 25 } has product 100 and sum 29 . Break up the middle term of the quadratic trinomial using this pair, then factor by grouping. 10 x 2 4 x 25 x + 10 = 0 2 x (5 x 2) 5(5 x 2) = 0 (2 x 5)(5 x 2) = 0 Using the zero product property, either 2 x 5 = 0 or 5 x 2 = 0 . Each of these linear equations is easily solved. x = 5 2 or x = 2 5 Hence, we have two solutions for x . However, they both lead to the same number- reciprocal pair. That is, if x = 5 / 2 , then its reciprocal is 2 / 5 . On the other hand, if x = 2 / 5 , then its reciprocal is 5 / 2 . Let’s check our solution by taking the sum of the solution and its reciprocal. Note that 5 2 + 2 5 = 25 10 + 4 10 = 29 10 , as required by the problem statement. Let’s look at another application of the reciprocal concept. l⚏ Example 3. There are two numbers. The second number is 1 larger than twice the first number. The sum of the reciprocals of the two numbers is 7 / 10 . Find the two numbers. Let x represent the first number. If the second number is 1 larger than twice the first number, then the second number can be represented by the expression 2 x + 1 .
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