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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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