=
√ + 2 − √ 3
− 1
. Depending upon the context, this may be considered simplified.
Note: We may also rationalize the numerator:
√ + 2 − √ 3
− 1
=
√ + 2 − √ 3
− 1
·
√ + 2 + √ 3
√ + 2 + √ 3 =
( + 2) − 3
( − 1) √ − 2 +
√ 3
=
− 1
( − 1) √ − 2 +
√ 3 =
1
√ + 2 + √ 3
39. () = ( + 4)( 2 − 9) is defined for all except when 0 = 2 − 9 ⇔ 0 = ( + 3)( − 3) ⇔ = −3 or 3, so the
domain is { ∈ | 6= −33} = (−∞−3) ∪ (−33) ∪ (3∞).
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.1 FOURWAYS TO REPRESENTA FUNCTION
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15
40. The function () =
2 + 1
2 + 4 − 21
is defined for all values of except those for which 2 + 4 − 21 = 0 ⇔
( + 7)( − 3) = 0 ⇔ = −7 or = 3. Thus, the domain is { ∈ | 6= −73} = (−∞−7) ∪ (−73) ∪ (3∞).
41. () =
3
√ 2 − 1 is defined for all real numbers. In fact
3
(), where () is a polynomial, is defined for all real numbers.
Thus, the domain is or (−∞∞).
42. () =
√ 3 − − √ 2 + is defined when 3 − ≥ 0
⇔ ≤ 3 and 2 + ≥ 0 ⇔ ≥ −2. Thus, the domain is
−2 ≤ ≤ 3, or [−23].
43. () = 1
4
√ 2
− 5 is defined when 2 − 5 0 ⇔ ( − 5) 0. Note that 2 − 5 6= 0 since that would result in
division by zero. The expression ( − 5) is positive if 0 or 5. (See Appendix A for methods for solving
inequalities.) Thus, the domain is (−∞0) ∪ (5∞).
44. () =
+ 1
1 +
1
+ 1
is defined when + 1 6= 0 [ 6= −1] and 1 +
1
+ 1
6= 0. Since 1 +
1
+ 1
= 0 ⇔
1
+ 1
= −1 ⇔ 1 = −−1 ⇔ = −2, the domain is { | 6= −2, 6= −1} = (−∞−2)∪(−2−1)∪(−1∞).
45. () =
2 − √ is defined when ≥ 0 and 2 − √ ≥ 0. Since 2 − √ ≥ 0
⇔ 2 ≥
√
⇔
√ ≤ 2
⇔
0 ≤ ≤ 4, the domain is [04].
46. The function () =
√ 2
− 4 − 5 is defined when 2 − 4 − 5 ≥ 0 ⇔ ( + 1)( − 5) ≥ 0. The polynomial
() = 2 − 4 − 5 may change signs only at its zeros, so we test values of on the intervals separated by = −1 and
= 5: (−2) = 7 0, (0) = −5 0, and (6) = 7 0. Thus, the domain of , equivalent to the solution intervals
of () ≥ 0, is { | ≤ −1 or ≥ 5} = (−∞−1] ∪ [5∞).
47. () =
√ 4 − 2 . Now = √ 4 − 2
⇒ 2 = 4 − 2 ⇔ 2 + 2 = 4, so
the graph is the top half of a circle of radius 2 with center at the origin. The domain
is
| 4 − 2
≥ 0 =
| 4 ≥ 2
= { | 2 ≥ ||} = [−22]. From the graph,
the range is 0 ≤ ≤ 2, or [02].
48. The function () =
2 − 4
− 2
is defined when − 2 6= 0 ⇔ 6= 2, so the
domain is { | 6= 2} = (−∞2) ∪ (2∞). On its domain,
() =
2 − 4
− 2
=
( − 2)( + 2)
− 2
= + 2. Thus, the graph of is the
line = + 2 with a hole at (24).
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 FUNCTIONSAND MODELS
49. () =
2
+ 2 if 0
if ≥ 0
(−3) = (−3) 2 + 2 = 11, (0) = 0, and (2) = 2.