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**Unformatted text preview: **− 2)
x2 − 1
2x2 − 1 − 3x + 2
x2 − 1
2x2 − 3x + 1
x2 − 1
(2x − 1)(x − 1)
(x + 1)(x − 1)
$
(2x − 1)$$$
(x − 1)
$
(x + 1)$$$
(x − 1)
2x − 1
x+1 4.1 Introduction to Rational Functions 233 4. To ﬁnd the domain of r, it may help to temporarily rewrite r(x) as
2x2 − 1
2
r (x) = x − 1
3x − 2
x2 − 1
We need to set all of the denominators equal to zero which means we need to solve not only
2
−2
x2 − 1 = 0, but also 3x−1 = 0. We ﬁnd x = ±1 for the former and x = 3 for the latter. Our
x2
2
domain is (−∞, −1) ∪ −1, 3 ∪ 2 , 1 ∪ (1, ∞). We simplify r(x) by rewriting the division as
3
multiplication by the reciprocal and then simplifying
r(x) =
= 2x2 − 1 3x − 2
÷2
x2 − 1
x −1
2x2 − 1 x2 − 1
·
x2 − 1 3x − 2 = 2x2 − 1 x2 − 1
(x2 − 1) (3x − 2) = 2x2 − 1 $$− $
x2 $ 1
$
x2 $$
$$− 1 (3x − 2) = 2x2 − 1
3x − 2 $ A few remarks about Example 4.1.1 are in order. Note that the expressions for f (x), g (x) and
h(x) work out to be the same. However, only two of these functions are actually equal. Recall that
functions are ultimately sets of ordered pairs,1 and so for two functions to be equal, they need,
among other things, to have the same domain. Since f (x) = g (x) and f and g have the same
domain, they are equal functions. Even though the formula h(x) is the same as f (x), the domain
of h is diﬀerent than the domain of f , and thus they are diﬀerent functions.
x−
We now turn our attention to th...

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