Stitz-Zeager_College_Algebra_e-book

In figure 5 of the x 19 paper he graphs a scatter

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Unformatted text preview: ) 2x2 − 1 3x − 2 ÷2 x2 − 1 x −1 Solution. 1. To find the domain of f , we proceed as we did in Section 1.5: we find the zeros of the denominator and exclude them from the domain. Setting x + 1 = 0 results in x = −1. Hence, 232 Rational Functions our domain is (−∞, −1) ∪ (−1, ∞). The expression f (x) is already in the form requested and when we check for common factors among the numerator and denominator we find none, so we are done. 2. Proceeding as before, we determine the domain of g by solving x + 1 = 0. As before, we find the domain of g is (−∞, −1) ∪ (−1, ∞). To write g (x) in the form requested, we need to get a common denominator 3 x+1 2 3 − 1 x+1 (2)(x + 1) 3 − (1)(x + 1) x + 1 (2x + 2) − 3 x+1 2x − 1 x+1 g (x) = 2 − = = = = This formula is also completely simplified. 3. The denominators in the formula for h(x) are both x2 − 1 whose zeros are x = ±1. As a result, the domain of h is (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). We now proceed to simplify h(x). Since we have the same denominator in both terms, we subtract the numerators. We then factor the resulting numerator and denominator, and cancel out the common factor. h(x) = = = = = = = 2x2 − 1 3x − 2 −2 x2 − 1 x −1 2x2 − 1 − (3x...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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