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**Unformatted text preview: **h. Hence,
the domain is (−∞, ∞).
4. To ﬁnd the domain of r, we notice that we have two potentially hazardous issues: not only
do we have a denominator, we have a square root in that denominator. To satisfy the square
root, we set the radicand x + 3 ≥ 0 so x ≥ −3. Setting the denominator equal to zero gives
6− √ x+3
6
62
36
33 =
=
=
=
= 0
√
x+3
√
x+3
x+3
x 2 Since we squared both sides in the course of solving this equation, we need to check our
√
√
answer. Sure enough, when x = 33, 6 − x + 3 = 6 − 36 = 0, and so x = 33 will cause
problems in the denominator. At last we can ﬁnd the domain of r: we need x ≥ −3, but
x = 33. Our ﬁnal answer is [−3, 33) ∪ (33, ∞).
2 x
5. It’s tempting to simplify I (x) = 3x = 3x, and, since there are no longer any denominators,
claim that there are no longer any restrictions. However, in simplifying I (x), we are assuming
0
x = 0, since 0 is undeﬁned.4 Proceeding as before, we ﬁnd the domain of I to be all real
numbers except 0: (−∞, 0) ∪ (0, ∞). It is worth reiter...

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