When data points “curve up,” the vertical tance gets larger for the same horizontal dis-tance.. A BRIEF PREVIEW OF CALCULUS 5It would be difficult to measure lengths with aratio of 12√ 2
Trang 1Chapter 0
Preliminaries
the Cartesian Plane
1 3x + 2 < 11
3x < 9
x < 3(−∞, 3)
2 4x + 1 < −5
4x < −6
x < −32
−∞, −32
3 2x − 3 < −7
2x < −4
x < −2(−∞, −2)
4 3x − 1 < 9
3x < 10
x < 103
−∞,103
5 4 − 3x < 6
− 3x < 2
x > −23
7 4 ≤ x + 1 < 7
3 ≤ x < 6[3, 6)
8 −1 < 2 − x < 3
− 3 < −x < 1
3 > x > −1
− 1 < x < 3(−1, 3)
11 x2+ 3x − 4 > 0(x + 4)(x − 1) > 0
- (x + 4)
|0
−4
- (x − 1)
|01
- (x + 4)(x − 1)
|0
−4
+
|01
x < −4 or x > 1,(−∞, −4) ∪ (1, ∞)
12 x2+ 4x + 3 < 0(x + 3)(x + 1) < 0
- (x + 3)
|0
−3
- (x + 1)
|0
−1
- (x + 3)(x + 1)
|0
−3
+
|0
−1
Trang 2−3 < x < −1(−3, −1)
13 x2− x − 6 < 0
(x − 3)(x + 2) < 0
- (x − 3)
|03
- (x + 2)
|0
−2
- (x − 3)(x + 2)
|03
+
|0
15 3x2+ 4 > 0
3x2> −4
x2> −43This is always true (−∞, ∞)
is always positive (−∞, ∞)
17 |x − 3| < 4
−4 < x − 3 < 4
−1 < x < 7(−1, 7)
18 |2x + 1| < 1
−1 < 2x + 1 < 1
−2 < 2x < 0
−1 < x < 0(−1, 0)
19 |3 − x| < 1 −1 < 3 − x < 1
− 4 < −x < −2
4 > x > 2
2 < x < 4(2, 4)
20 |3 + x| > 1
3 + x < −1 or 3 + x > 1
x < −4 or x > −2(−∞, −4) ∪ (−2, ∞)
21 |2x + 1| > 22x + 1 < 2 or 2x + 1 > 22x < −3 2x > 1
x < −3
2 or x >
12
−∞, −32
−2
- (x − 2)
|02
−2
|x2+
x < −2 or x > 2(−∞, −2) ∪ (2, ∞)
x + 1 > 0
x > −1(−1, ∞)
Trang 30.1 THE REAL NUMBERS AND THE CARTESIAN PLANE 3
25 x
2− x − 2(x + 4)2 > 0(x − 2)(x + 1)(x + 4)2 > 0
- (x − 2)
|02
- (x + 1)
|0
−1
- (x + 4)2
|0
−4
(x − 2)(x + 1)(x + 4)2
-|02
|0
−1
−
|x
−4+
x < −4 or −4 < x < −1 or x > 2(−∞, −4) ∪ (−4, −1) ∪ (2, ∞)
26 3 − 2x
(x + 1)2 < 0
- (3 − 2x)
|0
3 2
- (x + 1)2
|0
−1
-3 − 2x(x + 1)2
|0
3 2
−
|x
−1
x > 32
- (x + 1)3
|0
−1
-−8x(x + 1)3
|00
−
|x
−1
x < −1 or x > 0(−∞, −1) ∪ (0, ∞)
28 x − 2(x + 2)3 > 0
- (x − 2)
|02
- (x + 2)3
|0
−2
-x − 2(x + 2)3
|02
|x
−2+
=√
1 + 25 =√
26d{(3, 4), (0, 6)} =p(0 − 3)2+ (6 − 4)2
=√
9 + 4 =√
13(√
13)2+ (√
13)2= (√
26)2This statement is true, so yes, this is a righttriangle
36 d{(0, 2), (4, 8)} =p(4 − 0)2+ (8 − 2)2
=√
16 + 36 =√
52d{(0, 2), (−2, 12)} =p(−2 − 0)2+ (12 − 2)2
=√
4 + 100 =√
104d{(4, 8), (−2, 12)} =p(−2 − 4)2+ (12 − 8)2
=√
36 + 16 =√
52
Trang 4(√52)2+ (√
=p[−4 − (−2)]2+ (13 − 3)2
=√
4 + 100 =√
104d{(2, 9), (−4, 13)} =p(−4 − 2)2+ (13 − 9)2
=√
36 + 16 =√
52(√
=√
1 + 25 =√
26d{(0, 6), (−3, 8)} =p(−3 − 0)2+ (8 − 6)2
=√
9 + 4 =√
13(√
13)2+ (√
13)2= (√
26)2This statement is true, so yes, this is a righttriangle
2
1,600
800
0 0
5 4 2,800
3
2,400 2,000
1,200
1 400
The y-values are increasing by 550, 650, 750,
The next population is 3200 + 850 = 4050
2
33
31
5 4
The y-values are increasing by 90, 110, 130, The next population is 3490 + 150 = 3640
41
y
x (3,3910)
(2,3960)
(1,3990) (0,4000) 4,000
3,950
0 3,900
5 4 3 2 3,975
1 3,925
The y-values are decreasing by 10, 30, 50, The next population is 3910 − 70 = 3840
42
(2,2100) (1,2200) (0,2100) y
x (3,1700)
5 4 3 2 1 0
Possible answer:
The increments in y are +100, −100, −400
A reasonable next increment would be −900 for
a quadratic pattern, giving a next population
of 1700 − 900 = 800
43 When a calculation is done using a finite ber of digits, the result can only contain a fi-nite number of digits Therefore, the result can
num-be written as a quotient of two integers (wherethe denominator is a power of 10), so the resultmust be rational
44 When data points “curve up,” the vertical tance gets larger for the same horizontal dis-tance This corresponds to larger increases inconsecutive y-values
dis-45
1 2
7
− 2 3
12
1 2
7 ≈ 0.013 or 1.3%
46 If x is the ratio of consecutive string lengths,then x12is the ratio of string lengths spaced 12strings (or 1 octave) apart Therefore, x12= 2
Trang 50.2 A BRIEF PREVIEW OF CALCULUS 5
It would be difficult to measure lengths with aratio of 12√
2 because irrational numbers can bedifficult to locate on a number line
47
0.551 0.5680.587 0.5930.404 0.4140.538 0.5560.605 0.615
Calculus
1 Yes The slope of the line joining the points
(2, 1) and (0, 2) is −1
2, which is also the slope
of the line joining the points (0, 2) and (4, 0)
2 No The slope of the line joining the points
(3, 1) and (4, 4) is 3, while the slope of the linejoining the points (4, 4) and (5, 8) is 4
3 No The slope of the line joining the points
(4, 1) and (3, 2) is −1, while the slope of theline joining the points (3, 2) and (1, 3) is −1
2.
4 No The slope of the line joining the points
(1,2) and (2,5) is 3, but the slope of the linejoining the points (2, 5) and (4, 8) is 3
In exercises 11-16, the equation of theline is given along with the graph Anypoint on the given line will suffice for asecond point on the line
-4
x
3 2 1 -1
-2 -3
4 3 2 1 0 -1 -2
13 y = 1(0, 1) is a second point on the line
y
Trang 617 Parallel Both have slope 3
18 Neither Slopes are 2 and 4
19 Perpendicular Slopes are −2 and 1
2;theirproduct is -1
20 Neither Slopes are 2 and −2
21 Perpendicular Slopes are 3 and −1
3;theirproduct is -1
22 Parallel Both have slope −1
29 Yes, passes vertical line test
30 Yes, passes vertical line test
31 No The vertical line x = 0 meets the curvetwice; nearby vertical lines meet it three times
32 No, does not pass vertical line test
33 Both: This is clearly a cubic polynomial, andalso a rational function because it can be writ-ten as f (x) = x
37 Neither: Contains square root
38 Neither: Contains exponent 2
x ≤ −2 or x ≥ 3 and x 6= 5(−∞, −2] ∪ [3, 5) ∪ (5, ∞)
41 The function under square root should be negative and the denominator should be non-zero
non-3 − 2x > 0 when x < 3
2.The domain of f is
−∞,32
42 The function is defined for all vales of 3 +√
xbut, 3 +√
x is well defined for x ≥ 0
The domain of f is [0, ∞)
Trang 70.2 A BRIEF PREVIEW OF CALCULUS 7
43 Negatives are permitted inside the cube root
There are no restrictions, so the domain is(−∞, ∞) or all real numbers
44 We need the numerator function under square
root be non-negative x2 − 4 ≥ 2, when
|x| ≥ 2 Also the denominator cannot be zero
7) ∪(−1 −√
= 12
f 12
=
1
2+ 11
=
r1
2+ 1 =
r3
2 =
√62
= 313
= 9
51 The only constraint we know is that the width
should not be negative, so a reasonable domainwould be {x|x > 0}
52 Width can be anywhere from 0 to 200 feet Areasonable domain is {x|0 ≤ x ≤ 200}
53 We know that x should not be negative, that
x must be an integer and that x cannot ceed the number of candy bars made, i.e.,
ex-a reex-asonex-able domex-ain would be {x|0 ≤ x ≤number made, xan integer}
54 Cost could be any positive value A reasonabledomain is {x|x > 0}
55 Answers vary There may well be a positivecorrelation (more study hours = better grade),but not necessarily a functional relation.Youmay study the same amount of time for twodifferent exams and get different grades
56 Answers vary Evidence supports a ship
relation-57 Answers vary While not denying a negativecorrelation (more exercise = less weight), thereare too many other factors (metabolic rate,diet) to be able to quantify a person’s weight as
a function just of the amount of exercise.Twopeople may exercise the same amount of timebut have different weights
58 Answers vary Objects of all weights fall at thesame speed unless friction affects them differ-ently
59 A flat interval corresponds to an interval ofconstant speed; going up means that the speed
is increasing while the graph going down meansthat the speed is decreasing It is likely thatthe bicyclist is going uphill when the graph isgoing down and going downhill when the graph
Trang 870 Quadratic formula gives
x = −4 ±p42− 4(2)(−1)
−2 ±√62
71 x3− 3x2+ 2x = x(x2− 3x + 2)
= x(x − 2)(x − 1),
so the zeros are x = 0, 1, and 2
72 x3− 2x2− x + 2 = (x − 2)(x − 1)(x + 1),
so the zeros are x = −1, 1, and 2
73 With t = x3, x6+ x3− 2 becomes t2+ t − 2 and
factors as (t + 2)(t − 1) The expression is zeroonly if one of the factors is zero, i.e., if t = 1
or t = −2 With x = t1/3, the first occurs only
if x = (1)1/3 = 1 The latter occurs only if
x = (−2)1/3, about −1.2599
74 x3+ x2− 4x − 4 = (x − 2)(x + 1)(x + 2),
so the zeros are x = −2, −1, and 2
75 If B(h) = −1.8h + 212, then we can solve
B(h) = 98.6 for h as follows:
98.6 = −1.8h + 2121.8h = 113.4
h = 113.41.8 = 63This altitude (63,000 feet above sea-level, morethan double the height of Mt Everest) would
be the elevation at which we humans boil alive
in our skins Of course the cold of space andthe near-total lack of external pressure createadditional complications which we shall not try
to analyze
76 Let x represent compression and L(x) sent spin rate Given the points (120, 9100)and (60, 10,000), the linear function is
repre-y = −15(x − 60) + 10, 000The spin rate of a 90-compression ball is 9550,and the spin rate of a 100-compression ball is9400
77 This is a two-point line-fitting problem If apoint is interpreted as
(R, T ) =(chirp rate, temperature),then the two given points are (160, 79) and(100, 64)
The slope being 79 − 64
78 From problem 77 we know the temperature is
a function of chirping rate, T (r) = 1
4r + 39,where r is measured in chirps per minute Thenumber of chirps in 15 seconds will then be1
4r, and the temperature may conveniently befound by adding 39
79 Her winning percentage is calculated by theformula P = 100w
t , where P is the winningpercentage, w is the number of games wonand t is the total number of games Plugging
in w = 415 and t = 415 + 120 = 535, wefind her winning percentage is approximately
P ≈ 77.57, so we see that the percentage played is rounded up from the actual percent-age Let x be the number of games won in arow If she doesn’t lose any games, her newwinning percentage will be given by the for-mula P = 100(415 + x)
dis-535 + x . In order to haveher winning percentage displayed as 80%, sheonly needs a winning percentage of 79.5 orgreater Thus, we must solve the inequality79.5 ≤ 100(415 + x)
535 + x :
Trang 90.3 GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS 9
79.5 ≤ 100(415 + x)
535 + x79.5(535 + x) ≤ 41500 + 100x42532.5 + 79.5x ≤ 41500 + 100x1032.5 ≤ 20.5x
50.4 ≤ x(In the above, we are allowed to multiply bothsides of the inequality by 535 + x because weassume x (the number of wins in a row) is pos-itive.) Thus she must win at least 50.4 times
in a row to get her winning percentage to play as 80% Since she can’t win a fraction of
dis-a gdis-ame, she must win dis-at ledis-ast 51 gdis-ames in dis-arow
Computer Algebra Systems
1 (a) Intercepts: x = ±1, y = −1 Minimum
2.0
1.2 0.8 0.4
8 6
−20
3 16
−16 y
5 0
Trang 102.5 0
5.0
50
25
−25 0.0
−5
2
15 10
0 0
5.0
50 25
−25 0.0
(b) Intercepts: x ≈ 0.475, −1.395,
y = −1 Minimum at (approximately)(−1/√3
−1
−2 y
−3
−2
2 1
Trang 110.3 GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS 11
3 x
s
24 +√176
10 , while the twolocal minima occur at x =
s
24 +√17610and x = −
s
24 −√176
16 20
−4 4
2 1
−0.3
1.01 1.04
9 (a) Intercepts: y = −3 (no x-intercepts) No
extrema Horizontal asymptote y = 0
Vertical asymptote x = 1
Trang 120
−2 8
8
4
4 0
−8
−6 1
10
6
5 2
10 20
−5
−20
x
5 4 3 30
−10
0 20
10 12
−8 0 4 2
−30
−10
y x
0, −12
.Horizontal asymptote: y = 0
Vertical asymptotes: x = ±2
Trang 130.3 GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS 13
−3
−5
x
−4 y
−4
3 5
−10
8
y
−2 8
−6 0
Trang 140.2 0.3
−10 x
5 0.9
0
0.8 0.7
0.5
0.0
10
−5 0.1
−15
0.4
15
0.6 y
(b) No x-intercept y-intercept at y = 2
3.Maximum at (0, 2
3).
Horizontal asymptote y = 0
0.3
10 0.4
15 x
0.6
−10 −5
0.1
5 0.2 y
0.8
0.0
0.5 0.7
14 (a) Intercepts: x = −2, y = −1
3. No trema Horizontal asymptotes at y = 0
ex-Vertical asymptotes at x = −3 and x = 2
−3
−5
x
−4 y
−4
3 5
maxi-Horizontal asymptote y = 0
Vertical asymptotes x = −3 and x = −1
5.0
2.5 7.5
5.0 y
−2.5 x
15 (a) Intercepts: x = 0, y = 0 No extrema
Horizontal asymptotes: x = ±3
No vertical asymptotes
y
8 2
2
−8
−8
6 10
16 (a) Intercepts: x = 0, y = 0 No extrema
Horizontal asymptotes: y = ±2
No vertical asymptotes
Trang 150.3 GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS 15
−5
3 0
5 2
−1
−1
−4
4 2
0 5
−2 x
1
−5 y
22 Vertical asymptotes where
x2− 9 = 0 ⇒ x = ±3
23 A window with −0.1 ≤ x ≤ 0.1 and
−0.0001 ≤ y ≤ 0.0001 shows all details
0.0
2 0 6
x 10
24 A window with −4 ≤ x ≤ 12 and
−1600 ≤ y ≤ 2000 shows all details
y 500
−500
−1,500
2,000 1,500 1,000
0
−1,000
x
10.0 7.5 5.0 2.5 0.0
−2.5
25 A window with −15 ≤ x ≤ 15 and
−80 ≤ y ≤ 80 shows all details
0
15 10 0
50 25
Trang 1627 A window with −10 ≤ x ≤ 10 and −5 ≤ y ≤ 2
shows all details
−1
10
−2 2
28 A window with −10 ≤ x ≤ 10 and −11 ≤ y ≤ 2
shows all details
−4
−2.5
−7.5
−8 x
10 8 6 4 2 0
−2
y 0.0
1.0 1.05 1.15 0.05
The blow-up makes it appear that there aretwo intersection points Solving algebraically,
1 = (x − 1) (x + 1)2= x2− 1 (x + 1)
= x3+ x2− x − 1Hence x3+ x2− x − 2 = 0 By solver or spread-sheet, this equation has only the one solution
−1.5
−0.5 0.0
The graph shows the only intersection near
Trang 170.3 GRAPHING CALCULATORS AND COMPUTER ALGEBRA SYSTEMS 17
0.0 0
−5
−2 y
1 0 0
−1 2
Graph shows no intersections
37 Calculator shows zeros at approximately
y = 2(x − 1)2+ 3 if the latter is drawn on agraphing window centered at the point (1, 3)with
1 − 5√
2 ≤ x ≤ 1 + 5√
2, −7 ≤ y ≤ 13
44 The graph of y = x4 is below the graph of
y = x2 when −1 ≤ x ≤ 1, and above it when
x > 1 Both graphs have roughly the same ward parabola shape, but y = x4 is flatter atthe bottom
up-45 py2 is the distance from (x, y) to the x-axisq
x2+ (y − 2)2 is the distance from (x, y) tothe point (0, 2) If we require that these be thesame, and we square both quantities, we have
y2= x2+ (y − 2)2
y2= x2+ y2− 4y + 44y = x2+ 4
y = 1
4x
2+ 1
Trang 18In this relation, we see that y is a quadraticfunction of x The graph is commonly known
= 45◦
(b) π3
180◦
=2π3(d) (30◦) π
180◦
= π6
4 (a) 40◦ π
180◦ =
2π9(b) 80◦ π
180◦ =
4π9(c) 450◦ π
180◦ =
5π2(d) 390◦ π
180◦ =
13π6
5 2 cos (x) − 1 = 0 when cos (x) = 1/2 This curs whenever x = π
oc-3 + 2kπ or x = −
π
3 + 2kπfor any integer k
6 2 sin x + 1 = 0 when sin x = −1
2 This occurswhenever x = −π
6+ 2kπ or x = −
5π
6 + 2kπ forany integer k
7 √
2 cos (x) − 1 = 0 when cos (x) = 1/√
2 Thisoccurs whenever x = π
4+2kπ or x = −
π
4+2kπfor any integer k
8 2 sin x −√
3 = 0 when sin x =
√3
2 This occurswhenever x = π
3 + 2kπ or x =
2π
3 + 2kπ forany integer k
9 sin2x−4 sin x+3 = (sin x − 1) (sin x − 3) whensin x = 1 (sin x 6= 3 for any x) This occurswhenever x = π
2 + 2kπ for any integer k.
10 sin2x−2 sin x−3 = (sin x − 3) (sin x + 1) whensin x = −1 (sin x 6= 3 for any x) sin x = −1whenever x = 3π
2 + 2kπ for any integer k.
11 sin2x + cos x − 1 = 1 − cos2x + cos x − 1
= (cos x) (cos x − 1) = 0when cos x = 0 or cos x = 1 This occurswhenever x = π
2 + kπ or x = 2kπ for anyinteger k
12 Use the sine double angle formula to get
2 sin x cos x − cos x = (2 sin x − 1) cos x = 0then (2 sin x − 1) = 0 whenever x = π
6 + 2kπ
or x = 5π
6 + 2kπ and cos x = 0 whenever
x = π
2 + kπ for any integer k.
13 cos2x + cos x = (cos x) (cos x + 1) = 0 whencos x = 0 or cos x = −1 this occurs whenever
x = π
2 + kπ or x = π + 2kπ for any integer k.
14 sin2x − sin x = sin x (sin x − 1) = 0 whenever
x = kπ or x = π
2 + 2kπ for any integer k.
15 The graph of f (x) = sin 3x