{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Stitz-Zeager_College_Algebra_e-book

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}