Calculus: Early Transcendentals 9/e
Metric Version [Stewart], Chapter 1-16
SOLUTIONS
,TABLE OF CONTENTS
1. FUNCTIONS AND MODELS.
2. LIMITS AND DERIVATIVES.
3. DIFFERENTIATION RULES.
4. APPLICATIONS OF DIFFERENTIATION.
5. INTEGRALS.
6. APPLICATIONS OF INTEGRATION.
7. TECHNIQUES OF INTEGRATION.
8. FURTHER APPLICATIONS OF INTEGRATION.
9. DIFFERENTIAL EQUATIONS.
10. PARAMETRIC EQUATIONS AND POLAR COORDINATES.
11. SEQUENCES, SERIES, AND POWER SERIES.
12. VECTORS AND THE GEOMETRY OF SPACE.
13. VECTOR FUNCTIONS.
14. PARTIAL DERIVATIVES.
15. MULTIPLE INTEGRALS.
16. VECTOR CALCULUS.
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1 FUNCTIONS AND MODELS
1.1 Four Ways to Represent a Function
√ √
1. The functions ( ) = + 2 − and ( ) = + 2 − give exactly the same output values for every input value, so
and are equal.
2 — ( −1)
2. ( ) = = = for − 1 6= 0, so and [where ( ) = ] are not equal because (1) is undefined and
−1 −1
(1) = 1.
3. (a) The point (−2 2) lies on the graph of , so (−2) = 2. Similarly, (0) = −2, (2) = 1, and (3) 2 5.
(b) Only the point (−4 3) on the graph has a -value of 3, so the only value of for which ( ) = 3 is −4.
(c) The function outputs ( ) are never greater than 3, so ( ) ≤ 3 for the entire domain of the function. Thus, ( ) ≤ 3 for
−4 ≤ ≤ 4 (or, equivalently, on the interval [−4 4]).
(d) The domain consists of all -values on the graph of : { | −4 ≤ ≤ 4} = [−4 4]. The range of consists of all the
-values on the graph of : { | −2 ≤ ≤ 3} = [−2 3].
(e) For any 1 2 in the interval [0 2], we have ( 1) ( 2). [The graph rises from (0 −2) to (2 1).] Thus, ( ) is increasing
on [0 2].
4. (a) From the graph, we have (−4) = −2 and (3) = 4.
(b) Since (−3) = −1 and (−3) = 2, or by observing that the graph of is above the graph of at = −3, (−3) is larger
than (−3).
(c) The graphs of and intersect at = −2 and = 2, so ( ) = ( ) at these two values of .
(d) The graph of lies below or on the graph of for −4 ≤ ≤ −2 and for 2 ≤ ≤ 3. Thus, the intervals on which
( ) ≤ ( ) are [−4 −2] and [2 3].
(e) ( ) = −1 is equivalent to = −1, and the points on the graph of with -values of −1 are (−3 −1) and (4 −1), so the
solution of the equation ( ) = −1 is = −3 or = 4.
(f) For any 1 2 in the interval [−4 0], we have ( 1) ( 2). Thus, ( ) is decreasing on [−4 0]. (g) The
domain of is { | −4 ≤ ≤ 4} = [−4 4]. The range of is { | −2 ≤ ≤ 3} = [−2 3].
(h) The domain of is { | −4 ≤ ≤ 3} = [−4 3]. Estimating the lowest point of the graph of as having coordinates
(0 0 5), the range of is approximately { | 0 5 ≤ ≤ 4} = [0 5 4].
5. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 −85). The highest point occurs at about (17
115). Thus, the range of the vertical ground acceleration is −85 ≤ ≤ 115. Written in interval notation, the range is
[−85 115].
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10 ¤ CHAPTER 1 FUNCTIONS AND MODELS
6. Example 1: A car is driven at 60 mi h for 2 hours. The distance
traveled by the car is a function of the time . The domain of the
function is { | 0 ≤ ≤ 2}, where is measured in hours. The
range
of the function is { | 0 ≤ ≤ 120}, where is measured in miles.
Example 2: At a certain university, the number of students on
campus at any time on a particular day is a function of the time
after midnight. The domain of the function is { | 0 ≤ ≤ 24}, where
is
measured in hours. The range of the function is { | 0 ≤ ≤ },
where is an integer and is the largest number of students on
campus at once.
Example 3: A certain employee is paid $8 00 per hour and pay
works a maximum of 30 hours per week. The number of hours 240
238
worked is rounded down to the nearest quarter of an hour. 236
This employee’s gross weekly pay is a function of the number
4
of hours worked . The domain of the function is [0 30] and the
2
range of the function is 0 0.25 0.50 0.75 29.50 29.75 30 hours
{0 2 00 4 00 238 00 240 00}.
7. We solve 3 −5 = 7 for : 3 − 5 = 7 ⇔ −5 = −3 +7 ⇔ = 3 − 7 . Since the equation determines exactly
5 5
one value of for each value of , the equation defines as a function of .
2 2
8. We solve 3 − 2 = 5 for : 3 − 2 = 5 ⇔ −2 = −3 2 + 5 ⇔ = 3 2 − 5 . Since the equation determines
2 2
exactly one value of for each value of , the equation defines as a function of .
√
2 2 2 2 2 2
9. We solve + ( − 3) = 5 for : + ( − 3) = 5 ⇔ ( − 3) = 5 − ⇔ −3 = ± 5− 2 ⇔
√
= 3 ± 5 − 2. Some input values correspond to more than one output . (For instance, = 1 corresponds to = 1 and
to = 5.) Thus, the equation does not define as a function of .
10. We solve 2 + 5 2 = 4 for : 2 + 5 2 = 4 ⇔ 5 2 + (2 ) − 4 = 0 ⇔
2 √ √
(2 ) − 4(5)(−4) 2 + 20
−2 ± = −2 ± 4 2 + 80 − ±
=
(using the quadratic formula). Some input
= 2(5) 10 5
values correspond to more than one output . (For instance, = 4 corresponds to = −2 and to = 2 5.) Thus, the
equation does not define as a function of .
3 3 √
11. We solve ( + 3) + 1 = 2 for : ( + 3) + 1 = 2 ⇔ ( + 3)3 = 2 − 1 ⇔ + 3 = 3 2 −1 ⇔
√
= −3 + 3 2 − 1. Since the equation determines exactly one value of for each value of , the equation defines as a
function of .
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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤
11
12. We solve 2 − | | = 0 for : 2 − | | = 0 ⇔ | | = 2 ⇔ = ±2 . Some input values correspond to more than one
output . (For instance, = 1 corresponds to = −2 and to = 2.) Thus, the equation does not define as a function
of .
13. The height 60 in ( = 60) corresponds to shoe sizes 7 and 8 ( = 7 and = 8). Since an input value corresponds to
more than output value , the table does not define as a function of .
14. Each year corresponds to exactly one tuition cost . Thus, the table defines as a function of .
15. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence,
the curve fails the Vertical Line Test.
16. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−2 2] and
the range is [−1 2].
17. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−3 2] and
the range is [−3 −2) ∪ [−1 3].
18. No, the curve is not the graph of a function since for = 0, ±1, and ±2, there are infinitely many points on the curve.
19. (a) When = 1950, ≈ 13 8◦C, so the global average temperature in 1950 was about
13 8◦C. (b) When = 14 2◦C, ≈ 1990.
(c) The global average temperature was smallest in 1910 (the year corresponding to the lowest point on the
graph) and largest in 2000 (the year corresponding to the highest point on the graph).
(d) When = 1910, ≈ 13 5◦C, and when = 2000, ≈ 14 4◦C. Thus, the range of is about [13 5, 14 4].
20. (a) The ring width varies from near 0 mm to about 1 6 mm, so the range of the ring width function is approximately [0 1 6].
(b) According to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled
again into the late 1800s, and has been steadily warming since then. In the mid-19th century, there was
variation that could have been associated with volcanic eruptions.
21. The water will cool down almost to freezing as the ice melts. Then,
when the ice has melted, the water will slowly warm up to room
temperature.
22. The temperature of the pie would increase rapidly, level off to
oven temperature, decrease rapidly, and then level off to room
temperature.
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12 ¤ CHAPTER 1 FUNCTIONS AND MODELS
23. (a) The power consumption at 6 AM is 500 MW which is obtained by reading the value of power when = 6 from
the graph. At 6 PM we read the value of when = 18 obtaining approximately 730 MW
(b) The minimum power consumption is determined by finding the time for the lowest point on the graph, = 4 or 4
AM. The maximum power consumption corresponds to the highest point on the graph, which occurs just before
= 12 or right before noon. These times are reasonable, considering the power consumption schedules of most
individuals and businesses.
24. Runner A won the race, reaching the finish line at 100 meters in about 15 seconds, followed by runner B with a time of about
19 seconds, and then by runner C who finished in around 23 seconds. B initially led the race, followed by C,
and then A. C then passed B to lead for a while. Then A passed first B, and then passed C to take the lead
and finish first. Finally,
B passed C to finish in second place. All three runners completed the race.
25. Of course, this graph depends strongly 26. The summer solstice (the longest day of the year) is
on the geographical location! around June 21, and the winter solstice (the
shortest day) is around December 22. (Exchange
the dates for the southern hemisphere.)
27. As the price increases, the amount sold decreases. 28. The value of the car decreases fairly rapidly initially, then
somewhat less rapidly.
29.
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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤
13
30. (a) (b)
(c) (d)
31. (a) (b) 9:00 AM corresponds to = 9. When = 9,
the temperature is about 74◦F.
32. (a) (b) The blood alcohol concentration rises rapidly, then
slowly decreases to near zero.
33. ( ) = 3 2 − + 2.
(2) = 3(2)2 − 2 + 2 = 12 − 2 + 2 = 12.
(−2) = 3(−2)2 − (−2) + 2 = 12 + 2 + 2 = 16.
( ) = 3 2 − + 2.
(− ) = 3(− )2 − (− ) + 2 = 3 2 + + 2.
( + 1) = 3( + 1)2 − ( + 1) + 2 = 3( 2 + 2 + 1) − − 1 + 2 = 3 2 + 6 + 3 − + 1 = 3 2 + 5 + 4. 2 ( )= 2· ( ) =
2 2
2(3 − + 2) = 6 −2 + 4.
(2 ) = 3(2 )2 −(2 ) + 2 = 3(4 2) −2 + 2 = 12 2 −2 + 2.
[continued]
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14 ¤ CHAPTER 1 FUNCTIONS AND MODELS
( 2) = 3( 2)2 −( 2) + 2 = 3( 4) − 2 + 2 = 3 4 − 2 + 2.
2 2
[ ( )] = 3 2 − + 2 = 3 2 − + 2 3 2 − + 2
= 9 4 − 3 3 + 6 2 − 3 3 + 2 − 2 + 6 2 − 2 + 4 = 9 4 − 6 3 + 13 2 − 4 + 4
( + ) = 3( + )2 −( + ) + 2 = 3( 2 + 2 + 2) − − + 2 = 3 2 + 6 + 3 2 − − + 2.
34. ( ) = √ .
+1
0
= 0.
(0) = √
0 +1
3 3
= .
(3) = √ 2
3+1
5
.
5 ( )= 5· √ =√
1 1 + 1 1 + 14 2
(4 ) = · (4 ) = · √ =√ .
2 2 2 4 +1 4 +1
2 2 2
( 2) = √ ; [ ( )]2 = √ = .
2+1 +1 +1
( + )
( + )= + .
= √
( + )+1 + +1
( − ) −
( − )= = √ .
(
− )+1 − +1
35. ( ) = 4 + 3 − 2, so (3 + ) = 4 + 3(3 + ) − (3 + )2 = 4 + 9 + 3 − (9 + 6 + 2
) = 4 − 3 − 2,
2
(3 + ) − (3) (4 − 3 − ) − 4 (−3 − )
and = = = 3 − − .
3
36. ( ) = , so ( + ) = ( + )3 = 3
+3 2
+3 2
+ 3,
3 2 2 3 3 2 2
( + )− ( ) ( +3 +3 + )− (3 +3 + ) 2 2
and = = =3 +3 + .
1 1 −
−
37. ( ) = , so
1
( )− ( ) = − = − 1( − ) = − 1 .
− = = − − ( − ) ( − )
√ √
( ) − (1) √
38. ( ) = + 2, so
= + 2 − 3. . Depending upon the context, this may be considered simplified.
−1 −1
Note: We may also rationalize the numerator:
√ √ √ √ √ √
+2 − 3 +2 − 3 +2 + 3 ( +2) − 3
= ·√ √ = √ √
−1 −1 +2+ 3 ( − 1) −2 + 3
−1 1
= √ √ =√ √
( − 1) −2+ 3 + 2 + 3
2 2
39. ( ) = ( + 4) ( − 9) is defined for all except when 0 = − 9 ⇔ 0 = ( + 3)( − 3) ⇔ = −3 or 3, so the domain is { ∈ |
6= −3 3} = (−∞ −3) ∪ (−3 3) ∪ (3 ∞).
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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤
15
2+ 1 2
40. The function ( ) = is defined for all values of except those for which
+ 4 − 21 = 0 ⇔
2 +4 − 21
( + 7)( − 3) = 0 ⇔ = −7 or = 3. Thus, the domain is { ∈ | 6= −7 3} = (−∞ −7) ∪ (−7 3) ∪ (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 [−2 3].
√
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 ∞).
+1 1 1
44. ( ) = =0 ⇔
1 is defined when +1 6= 0 [ 6= −1] and 1 + 6= 0. Since 1 +
1+ +1
+1
+1
1 = −1 ⇔ 1 = − − 1 ⇔
= −2, the domain is { | 6= −2, 6= −1} = (−∞ −2) ∪(−2 −1) ∪(−1 ∞).
+1
√ √ √ √ √
45. ( ) = 2 − is defined when ≥ 0 and 2 − ≥ 0. Since 2 − ≥0 ⇔ 2 ≥ ⇔ ≤2 ⇔
0 ≤ ≤ 4, the domain is [0 4].
√ 2
46. The function ( ) = 2 − 4 − 5 is defined when − 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 ∞).
√ √ 2 2 2 2
47. ( ) = 4 − 2 . Now = 4− 2 ⇒ =4− ⇔ + = 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 ≥ | |} = [−2 2]. From the graph, the range is 0
≤ ≤ 2, or [0 2].
2 —4
48. The function ( ) = is defined when − 2 6= 0 ⇔ 6= 2, so the
−2
domain is { | 6= 2} = (−∞ 2) ∪ (2 ∞). On its domain,
2
— 4 ( − 2)( + 2)
( )= = = + 2. Thus, the graph of is the
−2 −2
line = +2 with a hole at (2 4).
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Metric Version [Stewart], Chapter 1-16
SOLUTIONS
,TABLE OF CONTENTS
1. FUNCTIONS AND MODELS.
2. LIMITS AND DERIVATIVES.
3. DIFFERENTIATION RULES.
4. APPLICATIONS OF DIFFERENTIATION.
5. INTEGRALS.
6. APPLICATIONS OF INTEGRATION.
7. TECHNIQUES OF INTEGRATION.
8. FURTHER APPLICATIONS OF INTEGRATION.
9. DIFFERENTIAL EQUATIONS.
10. PARAMETRIC EQUATIONS AND POLAR COORDINATES.
11. SEQUENCES, SERIES, AND POWER SERIES.
12. VECTORS AND THE GEOMETRY OF SPACE.
13. VECTOR FUNCTIONS.
14. PARTIAL DERIVATIVES.
15. MULTIPLE INTEGRALS.
16. VECTOR CALCULUS.
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1 FUNCTIONS AND MODELS
1.1 Four Ways to Represent a Function
√ √
1. The functions ( ) = + 2 − and ( ) = + 2 − give exactly the same output values for every input value, so
and are equal.
2 — ( −1)
2. ( ) = = = for − 1 6= 0, so and [where ( ) = ] are not equal because (1) is undefined and
−1 −1
(1) = 1.
3. (a) The point (−2 2) lies on the graph of , so (−2) = 2. Similarly, (0) = −2, (2) = 1, and (3) 2 5.
(b) Only the point (−4 3) on the graph has a -value of 3, so the only value of for which ( ) = 3 is −4.
(c) The function outputs ( ) are never greater than 3, so ( ) ≤ 3 for the entire domain of the function. Thus, ( ) ≤ 3 for
−4 ≤ ≤ 4 (or, equivalently, on the interval [−4 4]).
(d) The domain consists of all -values on the graph of : { | −4 ≤ ≤ 4} = [−4 4]. The range of consists of all the
-values on the graph of : { | −2 ≤ ≤ 3} = [−2 3].
(e) For any 1 2 in the interval [0 2], we have ( 1) ( 2). [The graph rises from (0 −2) to (2 1).] Thus, ( ) is increasing
on [0 2].
4. (a) From the graph, we have (−4) = −2 and (3) = 4.
(b) Since (−3) = −1 and (−3) = 2, or by observing that the graph of is above the graph of at = −3, (−3) is larger
than (−3).
(c) The graphs of and intersect at = −2 and = 2, so ( ) = ( ) at these two values of .
(d) The graph of lies below or on the graph of for −4 ≤ ≤ −2 and for 2 ≤ ≤ 3. Thus, the intervals on which
( ) ≤ ( ) are [−4 −2] and [2 3].
(e) ( ) = −1 is equivalent to = −1, and the points on the graph of with -values of −1 are (−3 −1) and (4 −1), so the
solution of the equation ( ) = −1 is = −3 or = 4.
(f) For any 1 2 in the interval [−4 0], we have ( 1) ( 2). Thus, ( ) is decreasing on [−4 0]. (g) The
domain of is { | −4 ≤ ≤ 4} = [−4 4]. The range of is { | −2 ≤ ≤ 3} = [−2 3].
(h) The domain of is { | −4 ≤ ≤ 3} = [−4 3]. Estimating the lowest point of the graph of as having coordinates
(0 0 5), the range of is approximately { | 0 5 ≤ ≤ 4} = [0 5 4].
5. From Figure 1 in the text, the lowest point occurs at about ( ) = (12 −85). The highest point occurs at about (17
115). Thus, the range of the vertical ground acceleration is −85 ≤ ≤ 115. Written in interval notation, the range is
[−85 115].
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10 ¤ CHAPTER 1 FUNCTIONS AND MODELS
6. Example 1: A car is driven at 60 mi h for 2 hours. The distance
traveled by the car is a function of the time . The domain of the
function is { | 0 ≤ ≤ 2}, where is measured in hours. The
range
of the function is { | 0 ≤ ≤ 120}, where is measured in miles.
Example 2: At a certain university, the number of students on
campus at any time on a particular day is a function of the time
after midnight. The domain of the function is { | 0 ≤ ≤ 24}, where
is
measured in hours. The range of the function is { | 0 ≤ ≤ },
where is an integer and is the largest number of students on
campus at once.
Example 3: A certain employee is paid $8 00 per hour and pay
works a maximum of 30 hours per week. The number of hours 240
238
worked is rounded down to the nearest quarter of an hour. 236
This employee’s gross weekly pay is a function of the number
4
of hours worked . The domain of the function is [0 30] and the
2
range of the function is 0 0.25 0.50 0.75 29.50 29.75 30 hours
{0 2 00 4 00 238 00 240 00}.
7. We solve 3 −5 = 7 for : 3 − 5 = 7 ⇔ −5 = −3 +7 ⇔ = 3 − 7 . Since the equation determines exactly
5 5
one value of for each value of , the equation defines as a function of .
2 2
8. We solve 3 − 2 = 5 for : 3 − 2 = 5 ⇔ −2 = −3 2 + 5 ⇔ = 3 2 − 5 . Since the equation determines
2 2
exactly one value of for each value of , the equation defines as a function of .
√
2 2 2 2 2 2
9. We solve + ( − 3) = 5 for : + ( − 3) = 5 ⇔ ( − 3) = 5 − ⇔ −3 = ± 5− 2 ⇔
√
= 3 ± 5 − 2. Some input values correspond to more than one output . (For instance, = 1 corresponds to = 1 and
to = 5.) Thus, the equation does not define as a function of .
10. We solve 2 + 5 2 = 4 for : 2 + 5 2 = 4 ⇔ 5 2 + (2 ) − 4 = 0 ⇔
2 √ √
(2 ) − 4(5)(−4) 2 + 20
−2 ± = −2 ± 4 2 + 80 − ±
=
(using the quadratic formula). Some input
= 2(5) 10 5
values correspond to more than one output . (For instance, = 4 corresponds to = −2 and to = 2 5.) Thus, the
equation does not define as a function of .
3 3 √
11. We solve ( + 3) + 1 = 2 for : ( + 3) + 1 = 2 ⇔ ( + 3)3 = 2 − 1 ⇔ + 3 = 3 2 −1 ⇔
√
= −3 + 3 2 − 1. Since the equation determines exactly one value of for each value of , the equation defines as a
function of .
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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤
11
12. We solve 2 − | | = 0 for : 2 − | | = 0 ⇔ | | = 2 ⇔ = ±2 . Some input values correspond to more than one
output . (For instance, = 1 corresponds to = −2 and to = 2.) Thus, the equation does not define as a function
of .
13. The height 60 in ( = 60) corresponds to shoe sizes 7 and 8 ( = 7 and = 8). Since an input value corresponds to
more than output value , the table does not define as a function of .
14. Each year corresponds to exactly one tuition cost . Thus, the table defines as a function of .
15. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence,
the curve fails the Vertical Line Test.
16. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−2 2] and
the range is [−1 2].
17. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−3 2] and
the range is [−3 −2) ∪ [−1 3].
18. No, the curve is not the graph of a function since for = 0, ±1, and ±2, there are infinitely many points on the curve.
19. (a) When = 1950, ≈ 13 8◦C, so the global average temperature in 1950 was about
13 8◦C. (b) When = 14 2◦C, ≈ 1990.
(c) The global average temperature was smallest in 1910 (the year corresponding to the lowest point on the
graph) and largest in 2000 (the year corresponding to the highest point on the graph).
(d) When = 1910, ≈ 13 5◦C, and when = 2000, ≈ 14 4◦C. Thus, the range of is about [13 5, 14 4].
20. (a) The ring width varies from near 0 mm to about 1 6 mm, so the range of the ring width function is approximately [0 1 6].
(b) According to the graph, the earth gradually cooled from 1550 to 1700, warmed into the late 1700s, cooled
again into the late 1800s, and has been steadily warming since then. In the mid-19th century, there was
variation that could have been associated with volcanic eruptions.
21. The water will cool down almost to freezing as the ice melts. Then,
when the ice has melted, the water will slowly warm up to room
temperature.
22. The temperature of the pie would increase rapidly, level off to
oven temperature, decrease rapidly, and then level off to room
temperature.
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12 ¤ CHAPTER 1 FUNCTIONS AND MODELS
23. (a) The power consumption at 6 AM is 500 MW which is obtained by reading the value of power when = 6 from
the graph. At 6 PM we read the value of when = 18 obtaining approximately 730 MW
(b) The minimum power consumption is determined by finding the time for the lowest point on the graph, = 4 or 4
AM. The maximum power consumption corresponds to the highest point on the graph, which occurs just before
= 12 or right before noon. These times are reasonable, considering the power consumption schedules of most
individuals and businesses.
24. Runner A won the race, reaching the finish line at 100 meters in about 15 seconds, followed by runner B with a time of about
19 seconds, and then by runner C who finished in around 23 seconds. B initially led the race, followed by C,
and then A. C then passed B to lead for a while. Then A passed first B, and then passed C to take the lead
and finish first. Finally,
B passed C to finish in second place. All three runners completed the race.
25. Of course, this graph depends strongly 26. The summer solstice (the longest day of the year) is
on the geographical location! around June 21, and the winter solstice (the
shortest day) is around December 22. (Exchange
the dates for the southern hemisphere.)
27. As the price increases, the amount sold decreases. 28. The value of the car decreases fairly rapidly initially, then
somewhat less rapidly.
29.
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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤
13
30. (a) (b)
(c) (d)
31. (a) (b) 9:00 AM corresponds to = 9. When = 9,
the temperature is about 74◦F.
32. (a) (b) The blood alcohol concentration rises rapidly, then
slowly decreases to near zero.
33. ( ) = 3 2 − + 2.
(2) = 3(2)2 − 2 + 2 = 12 − 2 + 2 = 12.
(−2) = 3(−2)2 − (−2) + 2 = 12 + 2 + 2 = 16.
( ) = 3 2 − + 2.
(− ) = 3(− )2 − (− ) + 2 = 3 2 + + 2.
( + 1) = 3( + 1)2 − ( + 1) + 2 = 3( 2 + 2 + 1) − − 1 + 2 = 3 2 + 6 + 3 − + 1 = 3 2 + 5 + 4. 2 ( )= 2· ( ) =
2 2
2(3 − + 2) = 6 −2 + 4.
(2 ) = 3(2 )2 −(2 ) + 2 = 3(4 2) −2 + 2 = 12 2 −2 + 2.
[continued]
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14 ¤ CHAPTER 1 FUNCTIONS AND MODELS
( 2) = 3( 2)2 −( 2) + 2 = 3( 4) − 2 + 2 = 3 4 − 2 + 2.
2 2
[ ( )] = 3 2 − + 2 = 3 2 − + 2 3 2 − + 2
= 9 4 − 3 3 + 6 2 − 3 3 + 2 − 2 + 6 2 − 2 + 4 = 9 4 − 6 3 + 13 2 − 4 + 4
( + ) = 3( + )2 −( + ) + 2 = 3( 2 + 2 + 2) − − + 2 = 3 2 + 6 + 3 2 − − + 2.
34. ( ) = √ .
+1
0
= 0.
(0) = √
0 +1
3 3
= .
(3) = √ 2
3+1
5
.
5 ( )= 5· √ =√
1 1 + 1 1 + 14 2
(4 ) = · (4 ) = · √ =√ .
2 2 2 4 +1 4 +1
2 2 2
( 2) = √ ; [ ( )]2 = √ = .
2+1 +1 +1
( + )
( + )= + .
= √
( + )+1 + +1
( − ) −
( − )= = √ .
(
− )+1 − +1
35. ( ) = 4 + 3 − 2, so (3 + ) = 4 + 3(3 + ) − (3 + )2 = 4 + 9 + 3 − (9 + 6 + 2
) = 4 − 3 − 2,
2
(3 + ) − (3) (4 − 3 − ) − 4 (−3 − )
and = = = 3 − − .
3
36. ( ) = , so ( + ) = ( + )3 = 3
+3 2
+3 2
+ 3,
3 2 2 3 3 2 2
( + )− ( ) ( +3 +3 + )− (3 +3 + ) 2 2
and = = =3 +3 + .
1 1 −
−
37. ( ) = , so
1
( )− ( ) = − = − 1( − ) = − 1 .
− = = − − ( − ) ( − )
√ √
( ) − (1) √
38. ( ) = + 2, so
= + 2 − 3. . Depending upon the context, this may be considered simplified.
−1 −1
Note: We may also rationalize the numerator:
√ √ √ √ √ √
+2 − 3 +2 − 3 +2 + 3 ( +2) − 3
= ·√ √ = √ √
−1 −1 +2+ 3 ( − 1) −2 + 3
−1 1
= √ √ =√ √
( − 1) −2+ 3 + 2 + 3
2 2
39. ( ) = ( + 4) ( − 9) is defined for all except when 0 = − 9 ⇔ 0 = ( + 3)( − 3) ⇔ = −3 or 3, so the domain is { ∈ |
6= −3 3} = (−∞ −3) ∪ (−3 3) ∪ (3 ∞).
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SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤
15
2+ 1 2
40. The function ( ) = is defined for all values of except those for which
+ 4 − 21 = 0 ⇔
2 +4 − 21
( + 7)( − 3) = 0 ⇔ = −7 or = 3. Thus, the domain is { ∈ | 6= −7 3} = (−∞ −7) ∪ (−7 3) ∪ (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 [−2 3].
√
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 ∞).
+1 1 1
44. ( ) = =0 ⇔
1 is defined when +1 6= 0 [ 6= −1] and 1 + 6= 0. Since 1 +
1+ +1
+1
+1
1 = −1 ⇔ 1 = − − 1 ⇔
= −2, the domain is { | 6= −2, 6= −1} = (−∞ −2) ∪(−2 −1) ∪(−1 ∞).
+1
√ √ √ √ √
45. ( ) = 2 − is defined when ≥ 0 and 2 − ≥ 0. Since 2 − ≥0 ⇔ 2 ≥ ⇔ ≤2 ⇔
0 ≤ ≤ 4, the domain is [0 4].
√ 2
46. The function ( ) = 2 − 4 − 5 is defined when − 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 ∞).
√ √ 2 2 2 2
47. ( ) = 4 − 2 . Now = 4− 2 ⇒ =4− ⇔ + = 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 ≥ | |} = [−2 2]. From the graph, the range is 0
≤ ≤ 2, or [0 2].
2 —4
48. The function ( ) = is defined when − 2 6= 0 ⇔ 6= 2, so the
−2
domain is { | 6= 2} = (−∞ 2) ∪ (2 ∞). On its domain,
2
— 4 ( − 2)( + 2)
( )= = = + 2. Thus, the graph of is the
−2 −2
line = +2 with a hole at (2 4).
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