Chapter 7 Rational Functions - College of the Redwoods

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698 Chapter 7 Rational Functions 1 x − 1 1 − 1 = − x − 1 · x x 2 x 2 (x + 1)(x − 1) = −x − 1 · x xx (x + 1)(x − 1) Hence, 1 x − 1 1 − 1 = − x x + 1 . x 2 Now, let’s try the problem a second time, multiplying numerator and denominator by x 2 to clear fractions from both the numerator and denominator. ( ) ( 1 1 x − 1 x x) 2 x 2 − (1)x 2 1 x − 1 1 − 1 x 2 = (1 − 1 x 2 ) x 2 = ( ) 1 (1)x 2 − x 2 x 2 = x − x2 x 2 − 1 The order in the numerator of the last fraction intimates that a sign change would be helpful. Negate the numerator and fraction bar, factor, then cancel common factors. 1 x − 1 1 − 1 = − x2 − x x 2 − 1 x 2 x(x − 1) x(x − 1) = − = − (x + 1)(x − 1) (x + 1)(x − 1) = − x x + 1 This is precisely the same answer found with the first technique. To list the restrictions, we must make sure that no values of x make any denominator equal to zero, at the beginning of the problem, in the body of our work, or in the final answer. In the original problem, if x = 0, then both 1/x and 1/x 2 are undefined, so x = 0 is a restriction. In the body of our work, the factors x + 1 and x − 1 found in various denominators make x = −1 and x = 1 restrictions. No other denominators supply restrictions that have not already been listed. Hence, for all x other than −1, 0, and 1, the left-hand side of 1 x − 1 1 − 1 = − x x + 1 x 2 is identical to the right-hand side. Again, the calculator’s table utility provides ample evidence of this fact in the screenshots shown in Figure 1. Note the ERR (error) messages at each of the restricted values of x, but also note the perfect agreement of Y1 and Y2 at all other values of x. (8) Let’s look at another example, an important example involving function notation. Version: Fall 2007
Section 7.6 Complex Fractions 699 (a) (b) (c) Figure 1. Using the table feature of the graphing calculator to check the identity in (8). ◮ Example 9. Given that f(x) = 1 x , simplify the expression List all restrictions. f(x) − f(2) . x − 2 Remember, f(2) means substitute 2 for x. f(2) = 1/2, so f(x) − f(2) x − 2 = Because f(x) = 1/x, we know that 1 x − 1 2 x − 2 . To clear the fractions from the numerator, we’d use a common denominator of 2x. There are no fractions in the denominator that need clearing, so the common denominator for numerator and denominator is 2x. Multiply numerator and denominator by 2x. f(x) − f(2) x − 2 = ( 1 x − 1 2) 2x (x − 2)2x = ( 1 x) 2x − (x − 2)2x ( 1 2) 2x Negate the numerator and fraction bar, then cancel common factors. f(x) − f(2) x − 2 = − x − 2 2x(x − 2) = − x − 2 2x(x − 2) = − 1 2x = 2 − x 2x(x − 2) In the original problem, we have a denominator of x − 2, so x = 2 is a restriction. If the body of our work, there is a fraction 1/x, which is undefined when x = 0, so x = 0 is also a restriction. The remaining denominators provide no other restrictions. Hence, for all values of x except 0 and 2, the left-hand side of is identical to the right-hand side. f(x) − f(2) x − 2 = − 1 2x Version: Fall 2007
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7 Rational Functions In this chapte
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Section 7.1 Introducing Rational Fu
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Section 7.2 Reducing Rational Funct
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Section 7.3 Graphing Rational Funct
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Chapter 7 Rational Functions - College of the Redwoods

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