-
CONTENTS
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . 1
Chapter 1. Functions. . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Chapter 2. Limits and Continuity. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . 43
Chapter 3. The Derivative. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Chapter 4. Logarithmic and Exponential Functions. . . . . . . .
. . . . . . . . . . . . . . . . . 99
Chapter 5. Analysis of Functions and Their Graphs. . . . . . . .
. . . . . . . . . . . . . . . . 139
Chapter 6. Applications of the Derivative. . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . 177
Chapter 7. Integration. . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
Chapter 8. Applications of the Definite Integralin Geometry,
Science, and Engineering. . . . . . . . . . . . . . . . . . . . . .
. . . 256
Chapter 9. Principles of Integral Evaluation. . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . 292
Chapter 10. Mathematical Modeling with Differential Equations. .
. . . . . . . . . . . .343
Chapter 11. Infinite Series. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
Chapter 12. Analytic Geometry in Calculus. . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . .408
Chapter 13. Three-Dimensional Space; Vectors. . . . . . . . . .
. . . . . . . . . . . . . . . . . . . 448
Chapter 14. Vector-Valued Functions. . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . 490
Chapter 15. Partial Derivatives. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 524
Chapter 16. Multiple Integrals. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . .573
Chapter 17. Topics in Vector Calculus. . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . 608
Appendix A. Real Numbers, Intervals, and Inequalities. . . . . .
. . . . . . . . . . . . . . . . . 640
Appendix B. Absolute Value. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 647
Appendix C. Coordinate Planes and Lines. . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . 650
Appendix D. Distance, Circles, and Quadratic Equations. . . . .
. . . . . . . . . . . . . . . . . 658
Appendix E. Trigonometry Review. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . 668
Appendix F. Solving Polynomial Equations. . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . 674
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CALCULUS:
A New Horizon from Ancient Roots
EXERCISE SET FOR INTRODUCTION
1. (a) x = 0.123123123 . . .; 1000x = 123.123123123 . . . = 123
+ x; 999x = 123; x =123
999=
41
333
(b) x = 12.7777 . . .; 10x = 127.7777 . . ., so 9x = 10x− x =
115; x = 1159
(c) x = 38.07818181 . . .; 100x = 3807.818181 . . .; 99x =
3769.74;
x =3769.74
99=
376974
9900=
41886
1100=
20943
550
(d) 0.4296000 . . . = 0.4296 =4296
10000=
537
1250
2. (a) π is irrational, and thus has a nonrepeating decimal
expansion, whereas22
7= 3.
repeats︷ ︸︸ ︷142857 . . .
(b)22
7> π
3. (a)223
71<
333
106<
63
25
(17 + 15
√5
7 + 15√
5
)<
355
113<
22
7(b)
63
25
(17 + 15
√5
7 + 15√
5
)
(c)333
106(d)
63
25
(17 + 15
√5
7 + 15√
5
)
4. (a) If r is the radius, then D = 2r so
(8
9D
)2=
(16
9r
)2=
256
81r2. The area of a circle of radius r is
πr2 so 256/81 was the approximation used for π.
(b) 256/81 ≈ 3.16049, 22/7 ≈ 3.14268, and π ≈ 3.14159 so 256/81
is worse than 22/7.
5. The first series, taken to ten terms, adds to 3.0418; the
second, as printed, adds to 3.1416.
6. (a)1
9= 0.111111 . . . =
1
10+
1
100+
1
1000+
1
10000+
1
100000+
1
1000000+ . . .
(b)2
27= 0.185185 . . . =
1
10+
8
100+
5
1000+
1
10000+
8
100000+
5
1000000+ . . .
(c)14
45= 0.311111 . . . =
3
10+
1
100+
1
1000+
1
10000+
1
100000+
1
1000000+ . . .
7. (a)7
11= 0.636363 . . . =
6
10+
3
100+
6
1000+
3
10000+
6
100000+
3
1000000+ . . .
(b)8
33= 0.242424 . . . =
2
10+
4
100+
2
1000+
4
10000+
2
100000+
4
1000000+ . . .
(c)5
12= 0.416666 . . . =
4
10+
1
100+
6
1000+
6
10000+
6
100000+
6
1000000+ . . .
8. (a) 1, 2, 1.75, 1.7321 (b) 1, 3, 2.33, 2.238, 2.2361
9. (a) 1, 4, 2.875, 2.6549, 2.6458 (b) 1, 25.5, 13.7, 8.69,
7.22, 7.0726, 7.0711
10. (a) Let x1 =12 (a+ b), x2 =
12 (a+ x1), x3 =
12 (a+ x2), etc. Then b > x1 > x2 > · · · > xn−1
> xn > a
so all the xi’s are distinct, there are infinitely many of them
and they all lie between a and b.
(b) x = 0.99999 . . ., 10x = 9.99999 . . ., 9x = 9, x = 1
(c) (1.999999 . . .)/2 = 0.999999 . . . = 1; yes it is
consistent, as all three are equal.
(d) 10x = 9 + x, so x = 9/9 = 1. They are equal.
1
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CHAPTER 1
Functions
EXERCISE SET 1.1
1. (a) around 1943 (b) 1960; 4200
(c) no; you need the year’s population (d) war; marketing
techniques
(e) news of health risk; social pressure, antismoking campaigns,
increased taxation
2. (a) 1989; $35,600 (b) 1983; $32,000 (c) the first two years;
the curve is steeper (downhill)
3. (a) −2.9,−2.0, 2.35, 2.9 (b) none (c) y = 0(d) −1.75 ≤ x ≤
2.15 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2
4. (a) x = −1, 4 (b) none (c) y = −1(d) x = 0, 3, 5 (e) ymax = 9
at x = 6; ymin = −2 at x = 0
5. (a) x = 2, 4 (b) none (c) x ≤ 2; 4 ≤ x (d) ymin = −1; no
maximum value
6. (a) x = 9 (b) none (c) x ≥ 25 (d) ymin = 1; no maximum
value
7. (a) Breaks could be caused by war, pestilence, flood,
earthquakes, for example.
(b) C decreases for eight hours, takes a jump upwards, and then
repeats.
8. (a) Yes, if the thermometer is not near a window or door or
other source of sudden temperaturechange.
(b) The number is always an integer, so the changes are in
movements (jumps) of at least one unit.
9. (a) If the side adjacent to the building has length x then L
= x + 2y. Since A = xy = 1000,L = x+ 2000/x.
(b) x > 0 and x must be smaller than the width of the
building, which was not given.
(c) 120
8020 80
(d) Lmin ≈ 89.44
10. (a) V = lwh = (6− 2x)(6− 2x)x (b) From the figure it is
clear that 0 < x < 3.
(c) 20
00 3
(d) Vmax ≈ 16
2
-
3 Chapter 1
11. (a) V = 500 = πr2h so h =500
πr2. Then
C = (0.02)(2)πr2 + (0.01)2πrh = 0.04πr2 + 0.02πr500
πr2
= 0.04πr2 +10
r; Cmin ≈ 4.39 at r ≈ 3.4, h ≈ 13.8.
7
41.5 6
(b) C = (0.02)(2)(2r)2 + (0.01)2πrh = 0.16r2 +10
r. Since
0.04π < 0.16, the top and bottom now get more weight.Since
they cost more, we diminish their sizes in the solution,and the
cans become taller.
7
41.5 5.5
(c) r ≈ 3.1, h ≈ 16.0, C ≈ 4.76
12. (a) The length of a track with straightaways of length L and
semicircles of radius r isP = (2)L+ (2)(πr) ft. Let L = 360 and r =
80 to get P = 720 + 160π = 1222.65 ft.Since this is less than 1320
ft (a quarter-mile), a solution is possible.
(b) P = 2L+ 2πr = 1320 and 2r = 2x+ 160, soL= 12 (1320− 2πr) =
12 (1320− 2π(80 + x)) = 660− 80π − πx.
450
00 100
(c) The shortest straightaway is L = 360, so x = 15.49 ft.
(d) The longest straightaway occurs when x = 0, so L = 660− 80π
= 408.67 ft.
EXERCISE SET 1.2
1. (a) f(0) = 3(0)2 − 2 = −2; f(2) = 3(2)2 − 2 = 10; f(−2) =
3(−2)2 − 2 = 10; f(3) = 3(3)2 − 2 = 25;f(√
2) = 3(√
2)2 − 2 = 4; f(3t) = 3(3t)2 − 2 = 27t2 − 2
(b) f(0) = 2(0) = 0; f(2) = 2(2) = 4; f(−2) = 2(−2) = −4; f(3) =
2(3) = 6; f(√
2) = 2√
2;
f(3t) = 1/3t for t > 1 and f(3t) = 6t for t ≤ 1.
2. (a) g(3) =3 + 1
3− 1 = 2; g(−1) =−1 + 1−1− 1 = 0; g(π) =
π + 1
π − 1 ; g(−1.1) =−1.1 + 1−1.1− 1 =
−0.1−2.1 =
1
21;
g(t2 − 1) = t2 − 1 + 1t2 − 1− 1 =
t2
t2 − 2
(b) g(3) =√
3 + 1 = 2; g(−1) = 3; g(π) =√π + 1; g(−1.1) = 3; g(t2 − 1) = 3
if t2 < 2 and
g(t2 − 1) =√t2 − 1 + 1 = |t| if t2 ≥ 2.
3. (a) x 6= 3 (b) x ≤ −√
3 or x ≥√
3
-
Exercise Set 1.2 4
(c) x2 − 2x + 5 = 0 has no real solutions so x2 − 2x + 5 is
always positive or always negative. Ifx = 0, then x2 − 2x+ 5 = 5
> 0; domain: (−∞,+∞).
(d) x 6= 0 (e) sinx 6= 1, so x 6= (2n+ 12 )π,n = 0,±1,±2, . .
.
4. (a) x 6= −75
(b) x− 3x2 must be nonnegative; y = x− 3x2 is a parabola that
crosses the x-axis at x = 0, 13
and
opens downward, thus 0 ≤ x ≤ 13
(c)x2 − 4x− 4 > 0, so x
2 − 4 > 0 and x − 4 > 0, thus x > 4; or x2 − 4 < 0
and x − 4 < 0, thus
−2 < x < 2(d) x 6= −1(e) cosx ≤ 1 < 2, 2− cosx > 0,
all x
5. (a) x ≤ 3 (b) −2 ≤ x ≤ 2 (c) x ≥ 0 (d) all x (e) all x
6. (a) x ≥ 23
(b) −32≤ x ≤ 3
2(c) x ≥ 0 (d) x 6= 0 (e) x ≥ 0
7. (a) yes (b) yes
(c) no (vertical line test fails) (d) no (vertical line test
fails)
8. The sine of θ/2 is (L/2)/10 (side opposite over hypotenuse),
so that L = 20 sin(θ/2).
9. The cosine of θ is (L− h)/L (side adjacent over hypotenuse),
so h = L(1− cos θ).
10. T
t
11.
t
h 12.
5 10 15t
w
13. (a) If x < 0, then |x| = −x so f(x) = −x + 3x + 1 = 2x +
1. If x ≥ 0, then |x| = x sof(x) = x+ 3x+ 1 = 4x+ 1;
f(x) =
{2x+ 1, x < 04x+ 1, x ≥ 0
(b) If x < 0, then |x| = −x and |x− 1| = 1− x so g(x) = −x+
1− x = 1− 2x. If 0 ≤ x < 1, then|x| = x and |x− 1| = 1− x so
g(x) = x+ 1− x = 1. If x ≥ 1, then |x| = x and |x− 1| = x− 1so g(x)
= x+ x− 1 = 2x− 1;
g(x) =
1− 2x, x < 01, 0 ≤ x < 12x− 1, x ≥ 1
14. (a) If x < 5/2, then |2x−5| = 5−2x so f(x) = 3+(5−2x) =
8−2x. If x ≥ 5/2, then |2x−5| = 2x−5so f(x) = 3 + (2x− 5) = 2x−
2;
f(x) =
{8− 2x, x < 5/22x− 2, x ≥ 5/2
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5 Chapter 1
(b) If x < −1, then |x− 2| = 2− x and |x+ 1| = −x− 1 so g(x)
= 3(2− x)− (−x− 1) = 7− 2x. If−1 ≤ x < 2, then |x− 2| = 2− x and
|x+ 1| = x+ 1 so g(x) = 3(2− x)− (x+ 1) = 5− 4x. Ifx ≥ 2, then |x−
2| = x− 2 and |x+ 1| = x+ 1 so g(x) = 3(x− 2)− (x+ 1) = 2x− 7;
g(x) =
7− 2x, x < −15− 4x, −1 ≤ x < 22x− 7, x ≥ 2
15. (a) V = (8−2x)(15−2x)x (b) −∞ < x < +∞,−∞ < V <
+∞ (c) 0 < x < 4(d) minimum value at x = 0 or at x = 4;
maximum value somewhere in between (can be approxi-
mated by zooming with graphing calculator)
16. (a) x = 3000 tan θ (b) θ 6= nπ+π/2 for n an integer, −∞ <
n
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Exercise Set 1.3 6
3. (b) and (c) are good; (a) is very bad.
12
13
14
15
y
-2 -1 0 1 2x
4. (b) and (c) are good; (a) is very bad.
-14
-13
-12
-11
y
-2 -1 0 1 2x
5. [−3, 3]× [0, 5]
2
4
y
-3 -2 -1 1 2 3x
6. [−4, 2]× [0, 3]
1
2
y
-3 -2 -1 1x
7. (a) window too narrow, too short (b) window wide enough, but
too short
(c) good window, good spacing
-500
-400
-300
-200
-100
y
-5 5 10 15 20x
(d) window too narrow, too short
(e) window too narrow, too short
8. (a) window too narrow (b) window too short
(c) good window, good tick spacing
-250
-200
-150
-100
-50
50
y
-16 -12 -8 -4 4x
(d) window too narrow, too short
(e) shows one local minimum only, window too narrow, too
short
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7 Chapter 1
9. [−5, 14]× [−60, 40]
-60
-40
-20
20
40
y
-5 5 10x
10. [6, 12]× [−100, 100]
-100
-50
50
100
y
8 10 12x
11. [−0.1, 0.1]× [−3, 3]
-3
-2
-1
1
2
3
y
-0.1x
0.1
12. [−1000, 1000]× [−13, 13]
-10
-5
5
10
y
-1000x
1000
13. [−250, 1050]× [−1500000, 600000]
-500000
y
-1000 1000x
14. [−3, 20]× [−3500, 3000]
-2000
-1000
1000
5 10 15x
y
15. [−2, 2]× [−20, 20]
-20
-10
10
20
y
x-2 -1 1 2
16. [1.6, 2]× [0, 2]
0.5
1
1.5
2
y
1.6 1.7 1.8 1.9 2x
17. depends on graphing utility
-6
-4
-2
2
4
6
y
-4 -2 2 4x
18. depends on graphing utility
-6
-4
-2
2
4
6
y
-4 -2 2 4x
-
Exercise Set 1.3 8
19. (a) f(x) =√
16− x2
1234
-4 -2 2 4x
y
(b) f(x) = −√
16− x2
-4-3-2-1
x
y
-4 -2 2 4
(c)
-4
-2
2
4
x
y
-4 -2 2 4
(d)
1
2
3
4
1 2 3 4
y
(e) No; the vertical line test fails.
20. (a) y = ±3√
1− x2/4 (b) y = ±√x2 + 1
-4
-2
2
4
-4 -2 2 4x
y
21. (a)
-1 1x
y
1
(b)
1x
1
2
y (c)
-1 1x
-1
y
(d)
2ππ
1
y
x
(e)
π
1
y
x−π
—1
(f)
1x
1
-1
y
-1
22.1
x
y
-1 1
-1
23. The portions of the graph of y = f(x) which lie below the
x-axis are reflected over the x-axis to give thegraph of y =
|f(x)|.
-
9 Chapter 1
24. Erase the portion of the graph of y = f(x) which lies in the
left-half plane and replace it with the reflectionover the y-axis
of the portion in the right-half plane (symmetry over the y-axis)
and you obtain the graphof y = f(|x|).
25. (a) for example, let a = 1.1
µ
y
(b)
1
2
3
y
1 2 3x
26. They are identical.
x
y
27.
5
10
15
y
-1 1 2 3x
28. This graph is very complex. We show three views, small (near
the origin), medium and large:
(a) y
x
-1
-1 2
(b)
-50
y
x10
(c)
1000
y
40x
29. (a)
0.5
1
1.5
y
-3 -2 -1 1 2 3x
(b)
-1
-0.5
0.5
1
-3 -2 -1 1 2 3x
y
(c)
0.5
1
1.5
y
-1 1 2 3x
(d)
0.5
1
1.5
-2 -1 1x
y
-
Exercise Set 1.3 10
30. y
1
2x
31. (a) stretches or shrinks the graph in they-direction; flips
it if c changes sign
-2
2
4c = 2
c = 1
c = -1.5
-1 1x
y
(b) As c increases, the parabola moves downand to the left. If c
increases, up and right.
2
4
6
8
-2 -1 1 2x
c = 2
c = 1
c = -1.5
y
(c) The graph rises or falls in the y-direction with changes in
c.
2
4
6
8
-2 -1 1 2x
c = +2
c = .5
c = -1
y
32. (a)y
—2
2
1 2x
(b) x-intercepts at x = 0, a, b. Assume a < b and let a
approach b. The two branches of the curvecome together. If a moves
past b then a and b switch roles.
-3
-2
-1
1
2
3
1 2 3x
a = 0.5b = 1.5
y
-3
-2
-1
1
2
3
1 2 3x
a = 1b = 1.5
y
-3
-2
-1
1
2
3
1 2 3x
a = 1.5b = 1.6
y
-
11 Chapter 1
33. The curve oscillates between the linesy = x and y = −x with
increasing ra-pidity as |x| increases.
-30
-20
-10
10
20
30
y
-30 -20 -10 10 20 30x
34. The curve oscillates between the linesy = +1 and y = −1,
infinitely many timesin any neighborhood of x = 0.
y
—1
4—2
—1
x
35. Plot f(x) on [−10, 10]; then on [−1, 0], [−0.7,−0.6],
[−0.65,−0.64], [−0.646,−0.645]; for the otherroot use [4, 5], [4.6,
4.7], [4.64, 4.65], [4.645, 4.646]; roots −0.6455, 4.6455.
36. Plot f(x) on [−10, 10]; then on [−4,−3], [−3.7,−3.6],
[−3.61,−3.60], [−3.606,−3.605]; for the otherroot use [3, 4], [3.6,
3.7], [3.60, 3.61], [3.605, 3.606]; roots 3.6055,−3.6055.
EXERCISE SET 1.4
1. (a)
-1
0
1
y
-1 1 2x
(b)2
1
y
1 2 3x
(c)
1
y
-1 1 2x
(d)
2
y
-4 -2 2x
2. (a)
x
y
1-2
-2
(b)
x
y
3
2
2
(c)
x2
y
-1
3
1
(d)
x
y
1
1-1
-
Exercise Set 1.4 12
3. (a)
-1
1
y
-2 -1 1 2x
(b)
-1
1
y
-0.5 0.5 1 1.5x
(c)
-1
1
y
-1 1 2 3x
(d)
-1
1
y
-1 1 2 3x
4.
1
y
-2 2x
5. Translate right 2 units, and up one unit.
10
-2 2 4 6x
y
6. Translate left 1 unit, reflect over x-axis, andtranslate up 2
units.
y
x
–2
1
7. Translate left 1 unit, stretch vertically bya factor of 2,
reflect over x-axis, translatedown 3 units.
-100
-80
-60
-40
-20
-8 -6 -4 -2 2 4 6x
y
8. Translate right 3 units, compress verticallyby a factor of 12
, and translate up 2 units.
y
x
2
4
9. y = (x + 3)2 − 9; translate left 3 units anddown 9 units.
-5
5
10
15
-8 -6 -4 -2 2x
y
-
13 Chapter 1
10. y = (x+ 3)2 − 19; translate left 3 unitsand down 19
units.
y
x
–5
–4
11. y = −(x − 1)2 + 2; translate right 1 unit,reflect over
x-axis, translate up 2 units.
-6
-4
-2
2
-2 -1 1 2 3 4x
y
12. y = 12 [(x− 1)2 + 2]; translate left 1 unitand up 2 units,
compress vertically bya factor of 12 .
y
x
1
2
1
13. Translate left 1 unit, reflect over x-axis,translate up 3
units.
1
2
2 4 6 8 10 12x
y
14. Translate right 4 units and up 1 unit.
y
x
1
4
4 10
15. Compress vertically by a factor of 12 ,translate up 1
unit.
2
y
1 2 3x
16. Stretch vertically by a factor of√
3 andreflect over x-axis.
y
x
-1
2
17. Translate right 3 units.
-10
10
y
2 4 6x
-
Exercise Set 1.4 14
18. Translate right 1 unit and reflectover x-axis.
y
x
-4
2
-2 2
19. Translate left 1 unit, reflect over x-axis,translate up 2
units.
-8-6-4-2
2468
1012
y
-4 -3 -2 -1 1 2x
20. y = 1− 1/x; reflect over x-axis, translateup 1 unit.
y
x
-5
5
2
21. Translate left 2 units and down 2 units.
-2
-4 -2x
y
22. Translate right 3 units, reflect over x-axis,translate up 1
unit.
y
x
-1
1
5
23. Stretch vertically by a factor of 2, translateright 1 unit
and up 1 unit.
y
x
2
4
2
24. y = |x− 2|; translate right 2 units.
y
x
1
2
2 4
25. Stretch vertically by a factor of 2, reflectover x-axis,
translate up 2 units.
-1
1
2
3
4
-2 2x
y
-
15 Chapter 1
26. Translate right 2 units and down 3 units.
y
x
–2
2
27. Translate left 1 unit and up 2 units.
1
2
3
y
-3 -2 -1 1x
28. Translate right 2 units, reflect over x-axis.
y
x
–1
1
4
29. (a)2
y
-1 1x
(b) y =
{0 if x ≤ 0
2x if 0 < x
30. y
x
–5
2
31. x2 + 2x+ 1, all x; 2x− x2 − 1, all x; 2x3 + 2x, all x;
2x/(x2 + 1), all x
32. 3x− 2 + |x|, all x; 3x− 2− |x|, all x; 3x|x| − 2|x|, all x;
(3x− 2)/|x|, all x 6= 0
33. 3√x− 1, x ≥ 1;
√x− 1, x ≥ 1; 2x− 2, x ≥ 1; 2, x > 1
34. (2x2 + 1)/x(x2 + 1), all x 6= 0; −1/x(x2 + 1), all x 6= 0;
1/(x2 + 1), all x 6= 0; x2/(x2 + 1), all x 6= 0
35. (a) 3 (b) 9 (c) 2 (d) 2
36. (a) π − 1 (b) 0 (c) −π2 + 3π − 1 (d) 1
-
Exercise Set 1.4 16
37. (a) t4 + 1 (b) t2 + 4t+ 5 (c) x2 + 4x+ 5 (d)1
x2+ 1
(e) x2 + 2xh+ h2 + 1 (f) x2 + 1 (g) x+ 1 (h) 9x2 + 1
38. (a)√
5s+ 2 (b)√√
x+ 2 (c) 3√
5x (d) 1/√x
(e) 4√x (f) 0 (g) 1/ 4
√x (h) |x− 1|
39. 2x2−2x+1, all x; 4x2 +2x, all x 40. 2− x6, all x; −x6 + 6x4
− 12x2 + 8, all x
41. 1− x, x ≤ 1;√
1− x2, |x| ≤ 1 42.√√
x2 + 3− 3, |x| ≥√
6;√x, x ≥ 3
43.1
1− 2x , x 6=1
2, 1; − 1
2x− 1
2, x 6= 0, 1 44. x
x2 + 1, x 6= 0; 1
x+ x, x 6= 0
45. x−6 + 1 46.x
x+ 1
47. (a) g(x) =√x, h(x) = x+ 2 (b) g(x) = |x|, h(x) = x2 − 3x+
5
48. (a) g(x) = x+ 1, h(x) = x2 (b) g(x) = 1/x, h(x) = x− 3
49. (a) g(x) = x2, h(x) = sinx (b) g(x) = 3/x, h(x) = 5 +
cosx
50. (a) g(x) = 3 sinx, h(x) = x2 (b) g(x) = 3x2 + 4x, h(x) =
sinx
51. (a) f(x) = x3, g(x) = 1 + sinx, h(x) = x2 (b) f(x) =√x, g(x)
= 1− x, h(x) = 3√x
52. (a) f(x) = 1/x, g(x) = 1− x, h(x) = x2 (b) f(x) = |x|, g(x)
= 5 + x, h(x) = 2x
53. y
x
-4
-3
-2
-1
1
2
-3 -2 -1 1 2 3
54. {−2,−1, 0, 1, 2, 3}
55. Note that f(g(−x)) = f(−g(x)) = f(g(x)), so f(g(x)) is
even.
f (g(x))
x
y
1–3 –1–1
–3
1
-
17 Chapter 1
56. Note that g(f(−x)) = g(f(x)), so g(f(x)) is even.
x
y
–1
–2
1
3–1
1
3
–3
g( f (x))
57. f(g(x)) = 0 when g(x) = ±2, so x = ±1.4; g(f(x)) = 0 when
f(x) = 0, so x = ±2.
58. f(g(x)) = 0 at x = −1 and g(f(x)) = 0 at x = −1
59.3(x+ h)2 − 5− (3x2 − 5)
h=
6xh+ 3h2
h= 6x+ 3h
60.(x+ h)2 + 6(x+ h)− (x2 + 6x)
h=
2xh+ h2 + 6h
h= 2x+ h+ 6
61.1/(x+ h)− 1/x
h=x− (x+ h)xh(x+ h)
=−1
x(x+ h)
62.1/(x+ h)2 − 1/x2
h=x2 − (x+ h)2x2h(x+ h)2
= − 2x+ hx2(x+ h)2
63. (a) the origin (b) the x-axis (c) the y-axis (d) none
64. (a)
x
y (b)
x
y (c)
x
y
65. (a)x −3 −2 −1 0 1 2 3
f(x) 1 −5 −1 0 −1 −5 1 (b)x −3 −2 −1 0 1 2 3
f(x) 1 5 −1 0 1 −5 -1
66. (a)
x
y (b)
x
y
67. (a) even (b) odd (c) odd (d) neither
68. neither; odd; even
69. (a) f(−x) = (−x)2 = x2 = f(x), even (b) f(−x) = (−x)3 = −x3
= −f(x), odd(c) f(−x) = | − x| = |x| = f(x), even (d) f(−x) = −x+
1, neither
(e) f(−x) = (−x)3 − (−x)
1 + (−x)2 = −x3 + x
1 + x2= −f(x), odd
(f) f(−x) = 2 = f(x), even
-
Exercise Set 1.4 18
70. (a) x-axis, because x = 5(−y)2 + 9 gives x = 5y2 + 9(b)
x-axis, y-axis, and origin, because x2 − 2(−y)2 = 3, (−x)2 − 2y2 =
3, and
(−x)2 − 2(−y)2 = 3 all give x2 − 2y2 = 3(c) origin, because
(−x)(−y) = 5 gives xy = 5
71. (a) y-axis, because (−x)4 = 2y3 + y gives x4 = 2y3 + y
(b) origin, because (−y) = (−x)3 + (−x)2 gives y =
x
3 + x2
(c) x-axis, y-axis, and origin because (−y)2 = |x| − 5, y2 = | −
x| − 5, and (−y)2 = | − x| − 5 all givey2 = |x| − 5
72. 3
–3
-4 4
73. 2
-2
-3 3
74. (a) Whether we replace x with −x, y with −y, or both, we
obtain the same equation, so by Theorem1.4.3 the graph is symmetric
about the x-axis, the y-axis and the origin.
(b) y = (1− x2/3)3/2
(c) For quadrant II, the same; for III and IV use y =
−(1−x2/3)3/2. (For graphing it may be helpfulto use the tricks that
precede Exercise 29 in Section 1.3.)
75.
1
2
3
4
5
y
-1 1 2 3 4x
76. y
x
2
2
77. (a)
1
y
Cx
O c o
(b)
2
y
Ox
C c o
-
19 Chapter 1
78. (a)
x1
y
1
-1
-1
(b)
x1
y
1
-1
2 3
(c)
x1
y
1
-1
3
(d)
x
y
1
-1
π/2−π
79. Yes, e.g. f(x) = xk and g(x) = xn where k and n are
integers.
80. If x ≥ 0 then |x| = x and f(x) = g(x). If x < 0 then f(x)
= |x|p/q if p is even and f(x) = −|x|p/q if pis odd; in both cases
f(x) agrees with g(x).
EXERCISE SET 1.5
1. (a)3− 00− 2 = −
3
2,3− (8/3)
0− 6 = −1
18,
0− (8/3)2− 6 =
2
3
(b) Yes; the first and third slopes above are negative
reciprocals of each other.
2. (a)−1− (−1)−3− 5 = 0,
−1− 35− 7 = 2,
3− 37− (−1) = 0,
−1− 3−3− (−1) = 2
(b) Yes; there are two pairs of equal slopes, so two pairs of
parallel lines.
3. III < II < IV < I
4. III < IV < I < II
5. (a)1− (−5)1− (−2) = 2,
−5− (−1)−2− 0 = 2,
1− (−1)1− 0 = 2. Since the slopes connecting all pairs of points
are
equal, they lie on a line.
(b)4− 2−2− 0 = −1,
2− 50− 1 = 3,
4− 5−2− 1 =
1
3. Since the slopes connecting the pairs of points are not
equal, the points do not lie on a line.
6. The slope, m = −2, is obtained from y − 5x− 7 , and thus y −
5 = −2(x− 7).
(a) If x = 9 then y = 1. (b) If y = 12 then x = 7/2.
7. The slope, m = 3, is equal toy − 2x− 1 , and thus y − 2 =
3(x− 1).
(a) If x = 5 then y = 14. (b) If y = −2 then x = −1/3.
-
Exercise Set 1.5 20
8. (a) Compute the slopes:y − 0x− 0 =
1
2or y = x/2. Also
y − 5x− 7 = 2 or y = 2x−9. Solve simultaneously
to obtain x = 6, y = 3.
9. (a) The first slope is2− 01− x and the second is
5− 04− x . Since they are negatives of each other we get
2(4− x) = −5(1− x) or 7x = 13, x = 13/7.
10. (a) 27◦ (b) 135◦ (c) 63◦ (d) 91◦
11. (a) 153◦ (b) 45◦ (c) 117◦ (d) 89◦
12. (a) m = tanφ = −√
3/3, so φ = 150◦ (b) m = tanφ = 4, so φ = 76◦
13. (a) m = tanφ =√
3, so φ = 60◦ (b) m = tanφ = −2, so φ = 117◦
14. y = 0 and x = 0 respectively 15. y = −2x+ 46
0-1 1
16. y = 5x− 32
–8
–1 2
17. Parallel means the lines have equal slopes, soy = 4x+ 7.
12
0-1 1
18. The slope of both lines is −3/2, so y − 2 = (−3/2)(x−
(−1)),or 3x+ 2y = 1
4
–4
-2 1
19. The negative reciprocal of 5 is −1/5, so y = −15x+ 6. 12
0-9 9
-
21 Chapter 1
20. The slope of x− 4y = 7 is 1/4 whose negative reciprocalis
−4, so y − (−4) = −4(x− 3) or y + 4x = 8.
9
–3
0 18
21.y − (−7)x− 1 =
4− (−7)2− 1 , or y = 11x− 18
7
-9
0 4
22.y − 1
x− (−2) =6− 1
−3− (−2) , or y = −5x− 915
–5
–4 0
23. (a) m1 = m2 = 4, parallel (b) m1 = 2 = −1/m2,
perpendicular(c) m1 = m2 = 5/3, parallel
(d) If A 6= 0 and B 6= 0 then m1 = −A/B = −1/m2, perpendicular;
if A = 0 or B = 0 (not both)then one line is horizontal, the other
vertical, so perpendicular.
(e) neither
24. (a) m1 = m2 = −5, parallel (b) m1 = 2 = −1/m2,
perpendicular(c) m1 = −4/5 = −1/m2, perpendicular (d) If B 6= 0, m1
= m2 = −A/B; if B = 0 both
are vertical, so parallel
(e) neither
25. (a) m = (0− (−3))/(2− 0)) = 3/2 so y = 3x/2− 3(b) m = (−3−
0)/(4− 0) = −3/4 so y = −3x/4
26. (a) m = (0− 2)/(2− 0)) = −1 so y = −x+ 2(b) m = (2− 0)/(3−
0) = 2/3 so y = 2x/3
27. (a) The velocity is the slope, which is5− (−4)10− 0 = 9/10
ft/s.
(b) x = −4(c) The line has slope 9/10 and passes through (0,−4),
so has equation x = 9t/10 − 4; at t = 2,
x = −2.2.(d) t = 80/9
-
Exercise Set 1.5 22
28. (a) v =5− 14− 2 = 2 m/s (b) x− 1 = 2(t− 2) or x = 2t− 3 (c)
x = −3
29. (a) The acceleration is the slope of the velocity, so a =3−
(−1)
1− 4 = −4
3ft/s2.
(b) v − 3 = −43
(t− 1), or v = −43t+
13
3(c) v =
13
3ft/s
30. (a) The acceleration is the slope of the velocity, so a =0−
510− 0 = −
5
10=−12
ft/s2.
(b) v = 5 ft/s (c) v = 4 ft/s (d) t = 4 s
31. (a) It moves (to the left) 6 units with velocity v = −3
cm/s, then remains motionless for 5 s, thenmoves 3 units to the
left with velocity v = −1 cm/s.
(b) vave =0− 910− 0 = −
9
10cm/s
(c) Since the motion is in one direction only, the speed is the
negative of the velocity, so
save =9
10cm/s.
32. It moves right with constant velocity v = 5 km/h; then
accelerates; then moves with constant, thoughincreased, velocity
again; then slows down.
33. (a) If x1 denotes the final position and x0 the initial
position, then v = (x1−x0)/(t1− t0) = 0 mi/h,since x1 = x0.
(b) If the distance traveled in one direction is d, then the
outward journey took t = d/40 h. Thus
save =total dist
total time=
2d
t+ (2/3)t=
80t
t+ (2/3)t= 48 mi/h.
(c) t+ (2/3)t = 5, so t = 3 and 2d = 80t = 240 mi round trip
34. (a) down, since v < 0 (b) v = 0 at t = 2 (c) It’s
constant at 32 ft/s2.
35. (a) v
t
20
40
60
80
100
20 40 60 80 100 120
(b) v =
10t if 0≤ t ≤ 10100 if 10≤ t ≤ 100600− 5t if 100≤ t ≤ 120
36. x
t
-
23 Chapter 1
37. (a) y = 20 − 15 = 5 when x = 45, so 5 = 45k,k = 1/9, y =
x/9
(b) y
x
0.2
0.4
0.6
2 4 6
(c) l = 15 + y = 15 + 100(1/9) = 26.11 in. (d) If ymax = 15 then
solve 15 = kx = x/9 forx = 135 lb.
38. (a) Since y = 0.2 = (1)k, k = 1/5 and y = x/5 (b) y
x
1
2 4 6
(c) y = 3k = 3/5 so 0.6 ft. (d) ymax = (1/2)3 = 1.5 so solve 1.5
= x/5 forx = 7.5 tons
39. Each increment of 1 in the value of x yields the increment
of 1.2 for y, so the relationship is linear. Ify = mx+ b then m =
1.2; from x = 0, y = 2, follows b = 2, so y = 1.2x+ 2
40. Each increment of 1 in the value of x yields the increment
of −2.1 for y, so the relationship is linear.If y = mx+ b then m =
−2.1; from x = 0, y = 10.5 follows b = 10.5, so y = −2.1x+ 10.5
41. (a) With TF as independent variable, we haveTC − 100TF −
212
=0− 10032− 212 , so TC =
5
9(TF − 32).
(b) 5/9 (c) Set TF = TC =5
9(TF − 32) and solve for TF : TF = TC = −40◦ (F or C).
(d) 37◦ C
42. (a) One degree Celsius is one degree Kelvin, so the slope is
the ratio 1/1 = 1. Thus TC = TK−273.15.(b) TC = 0− 273.15 =
−273.15◦ C
43. (a)p− 1h− 0 =
5.9− 150− 0 , or p = 0.098h+ 1 (b) when p = 2, or h = 1/0.098 ≈
10.20 m
44. (a)R− 123.4T − 20 =
133.9− 123.445− 20 , so R = 0.42T + 115.
(b) T = 32.38◦C
45. (a)r − 0.80t− 0 =
0.75− 0.804− 0 , so r = −0.0125t+ 0.8
(b) 64 days
46. (a) Let the position at rest be y0. Then y0 +y = y0 +kx;
with x = 11 we get y0 +kx = y0 +11k = 40,and with x = 24 we get y0
+ kx = y0 + 24k = 60. Solve to get k = 20/13 and y0 = 300/13.
(b) 300/13 + (20/13)W = 30, so W = (390− 300)/20 = 9/2 g.
-
Exercise Set 1.6 24
47. (a) For x trips we have C1 = 2x andC2 = 25 + x/4
C
x
20
40
60
5 10 15 20 25 30
(b) 2x = 25 + x/4, or x = 100/7, so the com-muter pass becomes
worthwhile at x = 15.
48. If the student drives x miles, then the total costs would be
CA = 4000 + (1.25/20)x andCB = 5500 + (1.25/30)x. Set 4000 + 5x/80
= 5500 + 5x/120 and solve for x = 72, 000 mi.
49. (a) H ≈ 20000/110 ≈ 181
(b) One light year is 9.408× 1012 km and t = dv
=1
H=
1
20km/s/Mly=
9.408× 1018km20km/s
= 4.704× 1017 s = 1.492× 1010 years.(c) The Universe would be
even older.
EXERCISE SET 1.6
1. (a) y = 3x+ b (b) y = 3x+ 6
(c)
-10
-5
5
10
y
-2 -1 1 2x
y = 3x + 6
y = 3x + 2
y = 3x - 4
2. Since the slopes are negative reciprocals, y = −13x+ b.
3. (a) y = mx+ 2 (b) m = tanφ = tan 135◦ = −1, so y = −x+
2(c)
-1
1
2
3
4
5
-2 -1 1 2x
y
m = –1
m = 1
m = 1.5
4. (a) y = mx (b) y = m(x− 1)(c) y = −2 +m(x− 1) (d) 2x+ 4y =
C
-
25 Chapter 1
5. (a) The slope is −1.
-3-2-1
12345
-2 -1 1 2x
y
(b) The y-intercept is y = −1.
-6
-4
-2
2
4
-2 -1 1 2x
y
(c) They pass through the point (−4, 2).y
x
-2
2
4
6
-6 -4 -2
(d) The x-intercept is x = 1.
-3
-2
-1
1
2
3
0.5 1 1.5 2x
y
6. (a) horizontal lines
x
y
(b) The y-intercept is y = −1/2.
x
y
- 1
2
2
(c) The x-intercept is x = −1/2.
x
y
1
1
(d) They pass through (−1, 1).
x
y
- 2
1
1
( -1 ,1 )
7. Let the line be tangent to the circle at the point (x0, y0)
where x20 + y
20 = 9. The slope of the tangent
line is the negative reciprocal of y0/x0 (why?), so m = −x0/y0
and y = −(x0/y0)x + b. Substitutingthe point (x0, y0) as well as y0
= ±
√9− x20 we get y = ±
9− x0x√9− x20
.
8. Solve the simultaneous equations to get the point (−2, 1/3)
of intersection. Then y = 13
+m(x+ 2).
-
Exercise Set 1.6 26
9. The x-intercept is x = 10 so that with depreciation at 10%
per year the final value is always zero, andhence y = m(x− 10). The
y-intercept is the original value.
y
2 4 6 8 10x
10. A line through (6,−1) has the form y+1 = m(x−6). The
intercepts are x = 6+1/m and y = −6m−1.Set −(6+1/m)(6m+1) = 3, or
36m2 +15m+1 = (12m+1)(3m+1) = 0 with roots m = −1/12,−1/3;thus y +
1 = −(1/3)(x− 6) and y + 1 = −(1/12)(x− 6).
11. (a) VI (b) IV (c) III (d) V (e) I (f) II
12. In all cases k must be positive, or negative values would
appear in the chart. Only kx−3 decreases, sothat must be f(x).
Next, kx2 grows faster than kx3/2, so that would be g(x), which
grows faster thanh(x) (to see this, consider ratios of successive
values of the functions). Finally, experimentation (aspreadsheet is
handy) for values of k yields (approximately) f(x) = 10x−3, g(x) =
x2/2, h(x) = 2x1.5.
13. (a)
-30
-20
-10
10
20
30
-2 -1 1 2x
y
-60
-50
-40
-30
-20
-10
-2 -1 1 2x
y
(b)
-4
-2
2
4
y
-4 -2 2 4x
2
4
6
8
10
y
-4 -2 2 4x
(c)
-2
-1
1
2
-1 1 2 3x
y
0.5
1
1.5
y
1 2 3x
-
27 Chapter 1
14. (a)
x
y
2
40
- 2
- 4 0
x
y
2
40
- 2
80
(b)
x
y
2
4
- 2
- 4
x
y
2
- 2
- 2
- 4
(c)
x
y
- 1 - 2
4
- 3
x
y
- 1
- 2
15. (a)
-10
-5
5
10
y
-2 -1 1 2x
(b)
-2
2
4
6
y
-2 -1 1 2x
(c)
-10
-5
5
10
y
-2 -1 1 2x
16. (a)
x
y
2
3
(b)
x
y
2
2
(c)
x
y
2
2
17. (a)
2
4
6
8
y
-3 -2 -1 1x
(b)
-80-60-40-20
20406080
y
-1 1 2 3 4 5x
-
Exercise Set 1.6 28
(c)
-50
-40
-30
-20
-10
y-5 -4 -3 -2 -1 1 2 3
x
(d)
-40
-20
20
40
y
1 2 3 4 5x
18. (a)
x
y
1
1
(b)
x
y
1
1
- 2
(c)
x
y
2
4
- 2
- 4
(d)
x
y
- 2
4
- 6
19. (a)
-1.5
-1
-0.5
0.5
1
1.5
y
-3 -2 -1 1x
(b)
-1
-0.5
0.5
1
y
1 2 3 4 5 6x
(c)
-30
-20
-10
10
20
30
y
-1 1 2 3x
(d)
-30
-20
-10
10
20
30
y
-2 -1 1 2x
20. (a)
-10
10
y
1 3 4x
(b)
-30
-20
-10
10
20
30
y
-1 1 2 3x
-
29 Chapter 1
(c)
-106
106
y
-1 1x
(d)
5
10
15
y
-4 -2 2x
21.
5
10
15
y
-4 -2 2x
22. (a)
0.5
1
1.5
y
-2 2x
(b)
0.5
1
1.5
y
-2 2x
23. t = 0.445√d
2.3
00 25
24. (a) t = 0.373r1.5 (b) 238,000 km (c) 1.89 days
25. (a) N·m (b) 20 N·m
(c) V (L) 0.25 0.5 1.0 1.5 2.0
P (N/m2) 80× 103 40× 103 20× 103 13.3× 103 10× 103
26. If the side of the square base is x and the height of the
container is y then V = x2y = 100; minimizeA = 2x2 + 4xy = 2x2 +
400/x. A graphing utility with a zoom feature suggests that the
solution is a
cube of side 10013 cm.
27. (a) F = k/x2 so 0.0005 = k/(0.3)2 andk = 0.000045 N·m2.
(b) 0.000005 N
-
Exercise Set 1.6 30
(c)
d5 10
F
10-5
(d) When they approach one another, the forcebecomes infinite;
when they get far apart ittends to zero.
28. (a) 2000 = C/(4000)2, so C = 3.2× 1010 lb·mi2 (b) W =
C/50002 = (3.2 × 1010)/(25 × 106) =1280 lb.
(c)
5000
10000
15000
20000
W
2000 4000 6000 8000 10000x
(d) No, but W is very small when x is large.
29. (a) II; y = 1, x = −1, 2 (b) I; y = 0, x = −2, 3(c) IV; y =
2 (d) III; y = 0, x = −2
30. The denominator has roots x = ±1, so x2−1 is the
denominator. To determine k use the point (0,−1)to get k = 1, y =
1/(x2 − 1).
31. Order the six trigonometric functions as sin, cos, tan, cot,
sec, csc:
(a) pos, pos, pos, pos, pos, pos (b) neg, zero, undef, zero,
undef, neg
(c) pos, neg, neg, neg, neg, pos (d) neg, pos, neg, neg, pos,
neg
(e) neg, neg, pos, pos, neg, neg (f) neg, pos, neg, neg, pos,
neg
32. (a) neg, zero, undef, zero, undef, neg (b) pos, neg, neg,
neg, neg, pos
(c) zero, neg, zero, undef, neg, undef (d) pos, zero, undef,
zero, undef, pos
(e) neg, neg, pos, pos, neg, neg (f) neg, neg, pos, pos, neg,
neg
33. (a) sin(π − x) = sinx; 0.588 (b) cos(−x) = cosx; 0.924(c)
sin(2π + x) = sinx; 0.588 (d) cos(π − x) = − cosx; −0.924(e) sin 2x
= ±2 sinx
√1− sin2 x; use the +
sign for x small and positive; 0.951(f) cos2 x = 1− sin2 x;
0.654
34. (a) sin(3π + x) = − sinx; −0.588 (b) cos(−x− 2π) = cosx;
0.924(c) sin(8π + x) = sinx ; 0.588 (d) sin(x/2) = ±
√(1− cosx)/2; use the nega-
tive sign for x small and negative; −0.195
(e) cos(3π + 3x) = −4 cos3 x+ 3 cosx; −0.384 (f) tan2 x = 1−
cos2 x
cos2 x; 0.172
35. (a) −a (b) b(c) −c (d) ±
√1− a2
(e) −b (f) −a(g) ±2b
√1− b2 (h) 2b2 − 1
(i) 1/b (j) −1/a(k) 1/c (l) (1− b)/2
-
31 Chapter 1
36. (a) The distance is 36/360 = 1/10th of a greatcircle, so it
is (1/10)2πr = 2, 513.27 mi.
(b) 36/360 = 1/10
37. If the arc length is 1, then solve the ratiox
1=
2πr
29.5to get x ≈ 80, 936 km.
38. The distance travelled is equal to the length of that
portion of the circumference of the wheel whichtouches the road,
and that is the fraction 225/360 of a circumference, so a distance
of (225/360)(2π)3 =11.78 ft
39. The second quarter revolves twice (720◦) about its own
center.
40. Add r to itself until you exceed 2πr; since 6r < 2πr <
7r, you can cut off 6 pieces of pie, but there’snot enough for a
full seventh piece. We conclude that there is no exact solution of
the equation‘One pie = 2πr’.
41. (a) y = 3 sin(x/2) (b) y = 4 cos 2x (c) y = −5 sin 4x
42. (a) y = 1 + cosπx (b) y = 1 + 2 sinx (c) y = −5 cos 4x
43. (a) y = sin(x+ π/2)
(b) y = 3 + 3 sin(2x/9)
(c) y = 1 + 2 sin(2(x− π/4))
44. V = 120√
2 sin(120πt)
45. (a) 3, π/2, 0
y
x
-2
1
3
π/2
(b) 2, 2, 0
-2
2
2 4x
(c) 1, 4π, 0
y
x
1
2
3
2π 4π 6π
46. (a) 4, π, 0
y
x
-4
-2
2
π/4 3π/4 5π/4 7π/4
(b) 1/2, 2π/3, π/3
y
x
-0.2
0.4
π/3 2π/3 π
(c) 4, 6π,−6πy
x
-4
-2
2
4
3π/2 9π/2 15π/2 21π/2
47. (a) A sin(ωt+ θ) = A sin(ωt) cos θ +A cos(ωt) sin θ = A1
sin(ωt) +A2 cos(ωt)
(b) A1 = A cos θ,A2 = A sin θ, so A =√A21 +A
22 and θ = tan
−1(A2/A1).
(c) A = 5√
13/2, θ = tan−11
2√
3;
x =5√
13
2sin
(2πt+ tan−1
1
2√
3
) 10
-10
^ 6
-
Exercise Set 1.7 32
48. three; x = 0, x = ±1.89553
–3
-3 3
EXERCISE SET 1.7
1. (a) x+ 1 = t = y − 1, y = x+ 2
2
4
6
y
2 4
t=0
t=1
t=2
t=3
t=4
t=5
x
(c) t 0 1 2 3 4 5x −1 0 1 2 3 4y 1 2 3 4 5 6
2. (a) x2 + y2 = 1
1
y
-1 1xt=1
t=.5
t=.75 t=.25
t=0
(c) t 0 0.2500 0.50 0.7500 1x 1 0.7071 0.00 −0.7071 −1y 0 0.7071
1.00 0.7071 0
3. t = (x+ 4)/3; y = 2x+ 10
-8 6
12
x
y
4. t = x+ 3; y = 3x+ 2,−3 ≤ x ≤ 0
x
y
(0, 2)
(-3, -7)
5. cos t = x/2, sin t = y/5;x2/4 + y2/25 = 1
-5 5
-5
5
x
y
6. t = x2; y = 2x2 + 4, x ≥ 0
1
4
8
x
y
-
33 Chapter 1
7. cos t = (x − 3)/2, sin t = (y − 2)/4;(x− 3)2/4 + (y − 2)2/16
= 1
7-2
6
x
y
8. sec2 t − tan2 t = 1;x2 − y2 = 1,x ≤ −1 and y ≥ 0
-1x
y
9. cos 2t = 1− 2 sin2 t;x = 1− 2y2,−1 ≤ y ≤ 1
-1 1
-1
1x
y
10. t = (x− 3)/4; y = (x− 3)2 − 9
x
y
(3, -9)
11. x/2 + y/3 = 1, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3
x
y
2
3
12. y = x− 1, x ≥ 1, y ≥ 0
1
1
x
y
13. x = 5 cos t, y = −5 sin t, 0 ≤ t ≤ 2π5
-5
-7.5 7.5
14. x = cos t, y = sin t, π ≤ t ≤ 3π/21
–1
-1 1
15. x = 2, y = t
2
-1
0 3
16. x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ 2π3
–3
-2 2
-
Exercise Set 1.7 34
17. x = t2, y = t, −1 ≤ t ≤ 11
-1
0 1
18. x = 1 + 4 cos t, y = −3 + 4 sin t, 0 ≤ t ≤ 2π2
-8
-7 8
19. (a) IV, because x always increases whereas y oscillates.
(b) II, because (x/2)2 + (y/3)2 = 1, an ellipse.
(c) V, because x2 + y2 = t2 increases in magnitude while x and y
keep changing sign.
(d) VI; examine the cases t < −1 and t > −1 and you see
the curve lies in the first, second andfourth quadrants only.
(e) III because y > 0.
(f) I; since x and y are bounded, the answer must be I or II;
but as t runs, say, from 0 to π, x goesdirectly from 2 to −2, but y
goes from 0 to 1 to 0 to −1 and back to 0, which describes I butnot
II.
20. (a) from left to right
(b) counterclockwise
(c) counterclockwise
(d) As t travels from −∞ to −1, the curve goes from (near) the
origin in the third quadrant andtravels up and left. As t travels
from −1 to +∞ the curve comes from way down in the secondquadrant,
hits the origin at t = 0, and then makes the loop clockwise and
finally approaches theorigin again as t→ +∞.
(e) from left to right
(f) Starting, say, at (1, 0), the curve goes up into the first
quadrant, loops back through the originand into the third quadrant,
and then continues the figure-eight.
21. (a) 14
0-35 8
(b) t 0 1 2 3 4 5x 0 5.5 8 4.5 −8 −32.5y 1 1.5 3 5.5 9 13.5
(c) x = 0 when t = 0, 2√
3. (d) for 0 < t < 2√
2 (e) at t = 2
22. (a) 5
0-2 14
(b) y is always ≥ 1 since cos t ≤ 1
(c) greater than 5, since cos t ≥ −1
-
35 Chapter 1
23. (a) 3
-5
0 20
(b) o
O
-1 1
24. (a) 1.7
-1.7
-2.3 2.3
(b)
-10 10
^
6
25. (a)x− x0x1 − x0
=y − y0y1 − y0
(b) Set t = 0 to get (x0, y0); t = 1 for (x1, y1).
(c) x = 1 + t, y = −2 + 6t (d) x = 2− t, y = 4− 6t
26. (a) x = −3− 2t, y = −4 + 5t (b) x = at, y = b(1− t)
27. (a) |R−P |2 = (x−x0)2 +(y−y0)2 = t2[(x1−x0)2 +(y1−y0)2] and
|Q−P |2 = (x1−x0)2 +(y1−y0)2,so r = |R− P | = |Q− P |t = qt.
(b) t = 1/2 (c) t = 3/4
28. x = 2 + t, y = −1 + 2t(a) (5/2, 0) (b) (9/4,−1/2) (c) (11/4,
1/2)
29. The two branches corresponding to −1 ≤ t ≤ 0 and 0 ≤ t ≤ 1
coincide.
30. (a) Eliminatet− t0t1 − t0
to obtainy − y0x− x0
=y1 − y0x1 − x0
.
(b) from (x0, y0) to (x1, y1) (c) x = 3− 2(t− 1), y = −1 + 5(t−
1)5
-2
0 5
31. (a)x− ba
=y − dc
(b)
1
2
3
y
1 2 3x
-
Exercise Set 1.7 36
32. (a) If a = 0 the line segment is vertical; if c = 0 it is
horizontal.
(b) The curve degenerates to the point (b, d).
33.
0.5
1
1.5
2
y
0.5 1x
34. x= 1/2− 4t, y = 1/2 for 0≤ t≤ 1/4x=−1/2, y = 1/2− 4(t− 1/4)
for 1/4≤ t≤ 1/2x=−1/2 + 4(t− 1/2), y =−1/2 for 1/2≤ t≤ 3/4x= 1/2, y
=−1/2 + 4(t− 3/4) for 3/4≤ t≤ 1
35. (a) x = 4 cos t, y = 3 sin t (b) x = −1 + 4 cos t, y = 2 + 3
sin t(c) 3
-3
-4 4
5
-1
-5 3
36. (a) t = x/(v0 cosα), so y = x tanα− gx2/(2v20 cos2
α).(b)
2000
4000
6000
8000
10000
12000
40000 80000x
y
37. (a) From Exercise 36, x = 400√
2t, y = 400√
2t− 4.9t2. (b) 16,326.53 m (c) 65,306.12 m
38. (a) 15
–15
-25 25
(b) 15
–15
-25 25
(c) 15
–15
-25 25
a = 3, b = 2
15
–15
-25 25
a = 2, b = 3
15
–15
-25 25
a = 2, b = 7
-
37 Chapter 1
39. Assume that a 6= 0 and b 6= 0; eliminate the parameter to
get (x−h)2/a2 + (y−k)2/b2 = 1. If |a| = |b|the curve is a circle
with center (h, k) and radius |a|; if |a| 6= |b| the curve is an
ellipse with center(h, k) and major axis parallel to the x-axis
when |a| > |b|, or major axis parallel to the y-axis when|a|
< |b|.
(a) ellipses with a fixed center and varying axes of
symmetry
(b) (assume a 6= 0 and b 6= 0) ellipses with varying center and
fixed axes of symmetry(c) circles of radius 1 with centers on the
line y = x− 1
40. Refer to the diagram to get bθ = aφ, θ = aφ/b but θ − α = φ+
π/2so α = θ − φ− π/2 = (a/b− 1)φ− π/2x= (a− b) cosφ− b sinα
= (a− b) cosφ+ b cos(a− bb
)φ,
y= (a− b) sinφ− b cosα= (a− b) sinφ− b sin
(a− bb
)φ.
x
y
θ
αφφaθba − b
41. (a)
-a
a
-a a
y
x
(b) Use b = a/4 in the equations of Exercise 40 to get
x =3
4a cosφ+
1
4a cos 3φ, y =
3
4a sinφ− 1
4a sin 3φ;
but trigonometric identities yield cos 3φ = 4 cos3 φ− 3 cosφ,
sin 3φ = 3 sinφ− 4 sin3 φ,so x = a cos3 φ, y = a sin3 φ.
(c) x2/3 + y2/3 = a2/3(cos2 φ+ sin2 φ) = a2/3
42.
-50
0
50
y
-3 -2 -1x
a = −2
-50
50
y
-2 -1x
a = −1-1
1
y
-1 1x
a = 0
-50
50
y
1 2x
a = 1
-50
0
50
y
1 2 3x
a = 2
-
Supplementary Exercises 1 38
CHAPTER 1 SUPPLEMENTARY EXERCISES
1. 1940-45; the greatest five-year slope
2. (a) f(−1) = 3.3, g(3) = 2 (b) x = −3, 3(c) x < −2, x >
3 (d) the domain is −5 ≤ x ≤ 5 and the range is
−5 ≤ y ≤ 4(e) the domain is −4 ≤ x ≤ 4.1, the range is
−3 ≤ y ≤ 5(f) f(x) = 0 at x = −3, 5; g(x) = 0 at
x = −3, 2
3.
40
50
60
70
0 2 4 6t
T 4. x
5 8 13t
5. If the side has length x and height h, then V = 8 = x2h, so h
= 8/x2. Then the cost C = 5x2 +2(4)(xh) = 5x2 + 64/x.
6. Assume that the paint is applied in a thin veneer of uniform
thickness, so that the quantity of paintto be used is proportional
to the area covered. If P is the amount of paint to be used, P =
kπr2. Theconstant k depends on physical factors, such as the
thickness of the paint, absorption of the wood,etc.
7. y
x-1
5
-5 5
8. Suppose the radius of the uncoated ball is r and that of the
coated ball is r+ h. Then the plastic has
volume equal to the difference of the volumes, i.e. V =4
3π(r+ h)3 − 4
3πr3 =
4
3πh[3r2 + 3rh+ h2] in3.
9. (a) The base has sides (10− 2x)/2 and 6− 2x, and the height
is x, so V = (6− 2x)(5− x)x ft3.(b) From the picture we see that x
< 5 and 2x < 6, so 0 < x < 3.
(c) 3.57 ft ×3.79 ft ×1.21 ft
10. {x 6= 0} and ∅ (the empty set)
11. impossible; we would have to solve 2(3x− 2)− 5 = 3(2x− 5)−
2, or −9 = −17
12. (a) (3− x)/x (b) no; f(g(x)) can be defined at x = 1,whereas
g, and therefore f ◦ g, requiresx 6= 1
13. 1/(2− x2) 14. g(x) = x2 + 2x
15. x −4 −3 −2 −1 0 1 2 3 4f(x) 0 −1 2 1 3 −2 −3 4 −4g(x) 3 2 1
−3 −1 −4 4 −2 0
(f ◦ g)(x) 4 −3 −2 −1 1 0 −4 2 3(g ◦ f)(x) −1 −3 4 −4 −2 1 2 0
3
-
39 Chapter 1
16. (a) y = |x− 1|, y = |(−x)− 1| = |x+ 1|,y = 2|x+ 1|, y = 2|x+
1| − 3,y = −2|x+ 1|+ 3
(b) y
x
-1
1
3
-3 -1 2
17. (a) even × odd = odd (b) a square is even(c) even + odd is
neither (d) odd × odd = even
18. (a) y = cosx− 2 sinx cosx = (1− 2 sinx) cosx, so x =
±π2,±3π
2,π
6,
5π
6,−7π
6,−11π
6
(b) (±π2, 0), (±3π
2, 0), (
π
6,√
3/2), (5π
6,−√
3/2), (−7π6,−√
3/2), (−11π6,√
3/2)
19. (a) If x denotes the distance from A to the base of the
tower, and y the distance from B to thebase, then x2 + d2 = y2.
Moreover h = x tanα = y tanβ, so d2 = y2 − x2 = h2(cot2 β − cot2
α),
h2 =d2
cot2 β − cot2 α =d2 sin2 α sin2 β
sin2 α cos2 β − cos2 α sin2 β . The trigonometric identity
sin(α+ β) sin(α− β) = sin2 α cos2 β − cos2 α sin2 β yields h = d
sinα sinβ√sin(α+ β) sin(α− β)
.
(b) 295.72 ft.
20. (a)
-20
20
40
60
y
100 200 300t
(b) when2π
365(t − 101) = 3π
2, or t = 374.75,
which is the same date as t = 9.75, so dur-ing the night of
January 10th-11th
(c) from t = 0 to t = 70.58 and from t = 313.92 to t = 365 (the
same date as t = 0) , for a total ofabout 122 days
21. C is the highest nearby point on the graph; zoom to find
that the coordinates of C are (2.0944, 1.9132).Similarly, D is the
lowest nearby point, and its coordinates are (4.1888, 1.2284).
Since f(x) = 12x−sinxis an odd function, the coordinates of B are
(−2.0944,−1.9132) and those of A are (−4.1888,−1.2284).
22. Let y = A + B sin(at + b). Since the maximum and minimum
values of y are 35 and 5, A + B = 35and A − B = 5, so A = 20, B =
15. The period is 12 hours, so 12a = 2π and a = π/6. Themaximum
occurs at t = 2, so 1 = sin(2a+ b) = sin(π/3 + b), π/3 + b = π/2, b
= π/2− π/3 = π/6 andy = 20 + 15 sin(πt/6 + π/6).
23. (a) The circle of radius 1 centered at (a, a2); therefore,
the family of all circles of radius 1 withcenters on the parabola y
= x2.
(b) All parabolas which open up, have latus rectum equal to 1
and vertex on the line y = x/2.
24. (a) x = f(1− t), y = g(1− t) 25.
-2
-1
1
2
y
-2 -1 1 2x
-
Supplementary Exercises 1 40
26. Let y = ax2 + bx + c. Then 4a + 2b + c = 0, 64a + 8b + c =
18, 64a − 8b + c = 18, from which b = 0and 60a = 18, or finally y
=
3
10x2 − 6
5.
27.
-2
-1
1
2
y
-1 1 2x
28. (a) R = R0 is the R-intercept, R0k is the slope,and T = −1/k
is the T -intercept
(b) −1/k = −273, or k = 1/273
(c) 1.1 = R0(1 + 20/273), or R0 = 1.025 (d) T = 126.55◦C
29. d =√
(x− 1)2 + (√x− 2)2;d = 9.1 at x = 1.358094
1
2
y
1 2x
30. d =√
(x− 1)2 + 1/x2;d = 0.82 at x = 1.380278
0.8
1
1.2
1.4
1.6
1.8
2
0.5 1 1.5 2 2.5 3
y
31. w = 63.9V , w = 63.9πh2(5/2− h/3); h = 0.48 ft when w = 108
lb
32. (a)
1000
2000
3000
4000
W
1 2 3 4 5h
(b) w = 63.9πh2(5/2 − h/3); at h = 5/2,w = 2091.12 lb
33. (a)
50
100
150
200
N
10 20 30 40 50t
(b) N = 80 when t = 9.35 yrs
(c) 220 sheep
-
41 Chapter 1
34. (a) T
v
10
20
(b) T = 17◦F, 27◦F, 32◦F
35. (a)
-20
20
WCI
10 20 30 40 50v
(b) T = 3◦F, −11◦F, −18◦F, −22◦F(c) v = 35, 19, 12, 7 mi/h
36. The domain is the set of all x, the range is −0.1746 ≤ y ≤
0.1227.
37. The domain is the set −0.7245 ≤ x ≤ 1.2207, the range is
−1.0551 ≤ y ≤ 1.4902.
38. (a) The potato is done in the interval27.65 < t <
32.71.
(b) 91.54 min.
39. (a)
5
10
15
20
25
v
1 2 3 4 5t
(b) As t→∞, (0.273)t → 0, and thusv → 24.61 ft/s.
(c) For large t the velocity approaches c. (d) No; but it comes
very close (arbitrarilyclose).
(e) 3.013 s
CHAPTER 1 HORIZON MODULE
1. (a) 0.25, 6.25× 10−2, 3.91× 10−3, 1.53× 10−5, 2.32× 10−10,
5.42× 10−20, 2.94× 10−39, 8.64× 10−78,7.46× 10−155, 5.56× 10−309;1,
1, 1, 1, 1, 1, 1, 1, 1, 1;
4, 16, 256, 65536, 4.29× 109, 1.84× 1019, 3.40× 1038, 1.16×
1077, 1.34× 10154, 1.80× 10308
2. 2, 2.25, 2.2361111, 2.23606798, 2.23606798, . . .
3. (a)1
2,
1
4,
1
8,
1
16,
1
32,
1
64(b) yn =
1
2n
4. (a) yn+1 = 1.05yn
(b) y0 =$1000, y1 =$1050, y2 =$1102.50, y3 =$1157.62, y4
=$1215.51, y5 =$1276.28
(c) yn+1 = 1.05yn for n ≥ 1 (d) yn = (1.05)n1000; y15
=$2078.93
-
Horizon Module 1 42
5. (a) x1/2, x1/4, x1/8, x1/16, x1/32 (b) They tend to the
horizontal line y = 1, witha hole at x = 0.
1.8
00 3
6. (a)1
2
2
3
3
5
5
8
8
13
13
21
21
34
34
55
55
89
89
144
(b) 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ;
each new numerator is the sum of the previous two
numerators.
(c)144
233,
233
377,
377
610,
610
987,
987
1597,
1597
2584,
2584
4181,
4181
6765,
6765
10946,
10946
17711
(d) F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2 for n ≥ 2.(e) the positive
solution
7. (a) y1 = cr, y2 = cy1 = cr2, y3 = cr
3, y4 = cr4
(b) yn = crn
(c) If r = 1 then yn = c for all n; if r < 1 then yn tends to
zero; if r > 1, then yn gets ever larger(tends to +∞).
8. The first point on the curve is (c, kc(1 − c)), so y1 = kc(1
− c) and hence y1 is the first iterate. Thepoint on the line to the
right of this point has equal coordinates (y1, y1), and so the
point above it onthe curve has coordinates (y1, ky1(1− y1)); thus
y2 = ky1(1− y1), and y2 is the second iterate, etc.
9. (a) 0.261, 0.559, 0.715, 0.591, 0.701
(b) It appears to approach a point somewhere near 0.65.
-
CHAPTER 2
Limits and Continuity
EXERCISE SET 2.1
1. (a) −1 (b) 3 (c) does not exist(d) 1 (e) −1 (f) 3
2. (a) 2 (b) 0 (c) does not exist
(d) 2 (e) 0 (f) 2
3. (a) 1 (b) 1 (c) 1 (d) 1 (e) −∞ (f) +∞
4. (a) 3 (b) 3 (c) 3 (d) 3 (e) +∞ (f) +∞
5. (a) 0 (b) 0 (c) 0 (d) 3 (e) +∞ (f) +∞
6. (a) 2 (b) 2 (c) 2 (d) 3 (e) −∞ (f) +∞
7. (a) −∞ (b) +∞ (c) does not exist(d) undef (e) 2 (f) 0
8. (a) +∞ (b) +∞ (c) +∞ (d) undef (e) 0 (f) −1
9. (a) −∞ (b) −∞ (c) −∞ (d) 1 (e) 1 (f) 2
10. (a) 1 (b) −∞ (c) does not exist(d) −2 (e) +∞ (f) +∞
11. (a) 0 (b) 0 (c) 0
(d) 0 (e) does not exist (f) does not exist
12. (a) 3 (b) 3 (c) 3
(d) 3 (e) does not exist (f) 0
13. for all x0 6= −4 14. for all x0 6= −6, 3
15. (a) At x = 3 the one-sided limits fail to exist.
(b) At x = −2 the two-sided limit exists but is not equal to F
(−2).(c) At x = 3 the limit fails to exist.
16. (a) At x = 2 the two-sided limit fails to exist.
(b) At x = 3 the two-sided limit exists but is not equal to F
(3).
(c) At x = 0 the two-sided limit fails to exist.
43
-
Exercise Set 2.1 44
17. (a) 2 1.5 1.1 1.01 1.001 0 0.5 0.9 0.99 0.9990.1429 0.2105
0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.3337
1
00 2
The limit is 1/3.
(b) 2 1.5 1.1 1.01 1.001 1.00010.4286 1.0526 6.344 66.33 666.3
6666.3
50
01 2
The limit is +∞.
(c) 0 0.5 0.9 0.99 0.999 0.9999−1 −1.7143 −7.0111 −67.001 −667.0
−6667.0
0
-50
0 1The limit is −∞.
18. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.250.5359
0.5132 0.5001 0.5000 0.5000 0.4999 0.4881 0.4721
0.6
0-0.25 0.25
The limit is 1/2.
-
45 Chapter 2
(b) 0.25 0.1 0.001 0.00018.4721 20.488 2000.5 20001
100
00 0.25
The limit is +∞.
(c) −0.25 −0.1 −0.001 −0.0001−7.4641 −19.487 −1999.5 −20000
0
-100
-0.25 0The limit is −∞.
19. (a) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.252.7266
2.9552 3.0000 3.0000 3.0000 3.0000 2.9552 2.7266
3
2-0.25 0.25
The limit is 3.
(b) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.0011 1.7552
6.2161 54.87 541.1 −0.1415 −4.536 −53.19 −539.5
60
-60
-1.5 0
The limit does not exist.
-
Exercise Set 2.1 46
20. (a) 0 −0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.0011.5574
1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.0000
1.5
1-1.5 0
The limit is 1.
(b) −0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.1 0.251.9794 2.4132
2.5000 2.5000 2.5000 2.5000 2.4132 1.9794
2.5
2-0.25 0.25
The limit is 5/2.
21. (a) −100,000,000 −100,000 −1000 −100 −10 10 100 10002.0000
2.0001 2.0050 2.0521 2.8333 1.6429 1.9519 1.9950
100,000 100,000,0002.0000 2.0000
40
-40-14 6
asymptote y = 2 as x→ ±∞
(b) −100,000,000 −100,000 −1000 −100 −10 10 100 100020.0855
20.0864 20.1763 21.0294 35.4013 13.7858 19.2186 19.9955
100,000 100,000,00020.0846 20.0855
70
0-160 160
asymptote y = 20.086.
-
47 Chapter 2
(c) −100,000,000 −100,000 −1000 −100 −10 10 100 1000 100,000
100,000,000−100,000,001 −100,000 −1001 −101.0 −11.2 9.2 99.0 999.0
99,999 99,999,999
50
–50
-20 20
no horizontal asymptote
22. (a) −100,000,000 −100,000 −1000 −100 −10 10 100 1000 100,000
100,000,0000.2000 0.2000 0.2000 0.2000 0.1976 0.1976 0.2000 0.2000
0.2000 0.2000
0.2
-1.2
-10 10asymptote y = 1/5 as x→ ±∞
(b) −100,000,000 −100,000 −1000 −100 −10 10 1000.0000 0.0000
0.0000 0.0000 0.0016 1668.0 2.09× 1018
1000 100,000 100,000,0001.77× 10301 ? ?
10
0-6 6
asymptote y = 0 as x → −∞, none asx→ +∞
(c) −100,000,000 −100,000 −1000 −100 −10 10 1000.0000 0.0000
0.0008 −0.0051 −0.0544 −0.0544 −0.0051
1000 100,000 100,000,0000.0008 0.0000 0.0000
1.1
-0.3
-30 30
asymptote y = 0 as x→ ±∞
23. (a) limx→0+
sinx
x(b) lim
x→0+x− 1x+ 1
(c) limx→0−
(1 + 2x)1/x
-
Exercise Set 2.2 48
24. (a) limx→0+
cosx
x(b) lim
x→0+1
x+ 1(c) lim
x→0−
(1 +
2
x
)x25. (a) y
x
(b) yes; for example f(x) = (sinx)/x
26. (a) no
(b) yes; tanx and secx at x = nπ + π/2, and cotx and cscx at x =
nπ, n = 0,±1,±2, . . .
29. (a) The plot over the interval [−a, a] becomes subject to
catastrophic subtraction if a is small enough(the size depending on
the machine).
(c) It does not.
EXERCISE SET 2.2
1. (a) −6 (b) 13 (c) −8 (d) 16 (e) 2 (f) −1/2(g) The limit
doesn’t exist because the denominator tends to zero but the
numerator doesn’t.
(h) The limit doesn’t exist because the denominator tends to
zero but the numerator doesn’t.
2. (a) 0
(b) The limit doesn’t exist because lim f doesn’t exist and lim
g does.
(c) 0 (d) 3 (e) 0
(f) The limit doesn’t exist because the denominator tends to
zero but the numerator doesn’t.
(g) The limit doesn’t exist because√f(x) is not defined for 0 ≤
x < 2.
(h) 1
3. (a) 7 (b) −3 (c) π (d) −6 (e) 36 (f) −∞
4. (a) 1 (b) −1 (c) 1 (d) −1 (e) 1 (f) −1
5. 0 6. 3/4 7. 8 8. −3 9. 4
10. 12 11. −4/5 12. 0 13. 3/2 14. 4/3
15. 0 16. 0 17. 0 18. 5/3 19. −√
5
20. 3√
3/2 21. 1/√
6 22.√
5 23.√
3 24. −1/√
6
25. +∞ 26.√
3 27. does not exist 28. −∞ 29. −∞
30. +∞ 31. +∞ 32. does not exist 33. does not exist 34. −∞
35. +∞ 36. −∞ 37. −∞ 38. does not exist 39. −1/7
40. +∞ 41. 6 42. +∞ 43. +∞ 44. 4
-
49 Chapter 2
45. +∞ 46. +∞ 47. −∞ 48. +∞
49. (a) 2 (b) 2 (c) 2
50. (a) −2 (b) 0 (c) does not exist
51. (a) 3 (b) y
x
4
1
52. (a) −6 (b) F (x) = x− 3
53. (a) Theorem 2.2.2(a) doesn’t apply; moreover one cannot
add/subtract infinities.
(b) limx→0+
(1
x− 1x2
)= lim
x→0+
(x− 1x2
)= −∞
54. limx→0−
(1
x+
1
x2
)= lim
x→0−x+ 1
x2= +∞ 55. lim
x→0x
x(√x+ 4 + 2
) = 14
56. limx→0
x2
x(√x+ 4 + 2
) = 057. lim
x→+∞(√x2 + 3− x)
√x2 + 3 + x√x2 + 3 + x
= limx→+∞
3√x2 + 3 + x
= 0
58. limx→+∞
(√x2 − 3x− x)
√x2 − 3x+ x√x2 − 3x+ x
= limx→+∞
−3x√x2 − 3x+ x
= −3/2
59. limx→+∞
(√x2 + ax− x
) √x2 + ax+ x√x2 + ax+ x
= limx→+∞
ax√x2 + ax+ x
= a/2
60. limx→+∞
(√x2 + ax−
√x2 + bx
) √x2 + ax+√x2 + bx√x2 + ax+
√x2 + bx
= limx→+∞
(a− b)x√x2 + ax+
√x2 + bx
=a− b
2
61. limx→+∞
p(x) = (−1)n∞ and limx→−∞
p(x) = +∞
62. If m > n the limits are both zero. If m = n the limits
are both 1. If n > m the limits are (−1)n+m∞and +∞,
respectively.
63. If m > n the limits are both zero. If m = n the limits
are both equal to am, the leading coefficient ofp. If n > m the
limits are ±∞ where the sign depends on the sign of am and whether
n is even orodd.
64. (a) p(x) = q(x) = x (b) p(x) = x, q(x) = x2
(c) p(x) = x2, q(x) = x (d) p(x) = x+ 3, q(x) = x
65. The left and/or right limits could be plus or minus
infinity; or the limit could exist, or equal anypreassigned real
number. For example, let q(x) = x− x0 and let p(x) = a(x− x0)n
where n takes onthe values 0, 1, 2.
-
Exercise Set 2.3 50
66. If m > n the limit is zero. If m = n the limit is cm/dm.
If n > m the limit is ±∞, where the signdepends on the signs of
cn and dm.
EXERCISE SET 2.3
1. (a) |f(x)− f(0)| = |x+ 2− 2| = |x| < 0.1 if and only if
|x| < 0.1(b) |f(x)− f(3)| = |(4x− 5)− 7| = 4|x− 3| < 0.1 if
and only if |x− 3| < (0.1)/4 = 0.0025(c) |f(x)− f(4)| = |x2 −
16| < ² if |x− 4| < δ. We get f(x) = 16 + ² = 16.001 at x =
4.000124998,
which corresponds to δ = 0.000124998; and f(x) = 16 − ² = 15.999
at x = 3.999874998, forwhich δ = 0.000125002. Use the smaller δ:
thus |f(x)− 16| < ² provided |x− 4| < 0.000125 (tosix
decimals).
2. (a) |f(x)− f(0)| = |2x+ 3− 3| = 2|x| < 0.1 if and only if
|x| < 0.05(b) |f(x)− f(0)| = |2x+ 3− 3| = 2|x| < 0.01 if and
only if |x| < 0.005(c) |f(x)− f(0)| = |2x+ 3− 3| = 2|x| <
0.0012 if and only if |x| < 0.0006
3. (a) x1 = (1.95)2 = 3.8025, x2 = (2.05)
2 = 4.2025
(b) δ = min ( |4− 3.8025|, |4− 4.2025| ) = 0.1975
4. (a) x1 = 1/(1.1) = 0.909090 . . . , x2 = 1/(0.9) = 1.111111 .
. .
(b) δ = min( |1− 0.909090|, |1− 1.111111| ) = 0.0909090 . .
.
5. |2x− 8| = 2|x− 4| < 0.1 if |x− 4| < 0.05, δ = 0.05
6. |x/2 + 1| = (1/2)|x− (−2)| < 0.1 if |x+ 2| < 0.2, δ =
0.2
7. |7x+ 5− (−2)| = 7|x− (−1)| < 0.01 if |x+ 1| < 1700
, δ =1
700
8. |5x− 2− 13| = 5|x− 3| < 0.01 if |x− 3| < 1500
, δ =1
500
9.
∣∣∣∣x2 − 4x− 2 − 4∣∣∣∣ = ∣∣∣∣x2 − 4− 4x+ 8x− 2
∣∣∣∣ = |x− 2| < 0.05 if |x− 2| < 0.05, δ = 0.0510.
∣∣∣∣x2 − 1x+ 1 − (−2)∣∣∣∣ = ∣∣∣∣x2 − 1 + 2x+ 2x+ 1
∣∣∣∣ = |x+ 1| < 0.05 if |x+ 1| < 0.05, δ = 0.0511. if δ
< 1 then
∣∣x2 − 16∣∣ = |x− 4||x+ 4| < 9|x− 4| < 0.001 if |x− 4|
< 19000
, δ =1
9000
12. if δ < 1 then |√x− 3|∣∣∣∣√x+ 3√x+ 3
∣∣∣∣ = |x− 9||√x+ 3| < |x− 9|√8 + 3 < 14 |x− 9| < 0.001
if |x− 9| < 0.004, δ = 0.00413. if δ ≤ 1 then
∣∣∣∣ 1x − 15∣∣∣∣ = |x− 5|5|x| ≤ |x− 5|20 < 0.05 if |x− 5|
< 1, δ = 1
14. |x− 0| = |x| < 0.05 if |x| < 0.05, δ = 0.05 15. |3x−
15| = 3|x− 5| < ² if |x− 5| < 13², δ = 13²
16. |(4x− 5)− 7| = |4x− 12| = 4|x− 3| < ² if |x− 3| < 14²,
δ = 14²
17. |2x− 7− (−3)| = 2|x− 2| < ² if |x− 2| < 12², δ =
12²
-
51 Chapter 2
18. |2− 3x− 5| = 3|x+ 1| < ² if |x+ 1| < 13², δ = 13²
19.
∣∣∣∣x2 + xx − 1∣∣∣∣ = |x| < ² if |x| < ², δ = ² 20. ∣∣∣∣x2
− 9x+ 3 − (−6)
∣∣∣∣ = |x+ 3| < ² if |x+ 3| < ², δ = ²21. if δ < 1 then
|2x2 − 2| = 2|x− 1||x+ 1| < 6|x− 1| < ² if |x− 1| < 16², δ
= min(1, 16²)
22. if δ < 1 then |x2 − 5− 4| = |x− 3||x+ 3| < 7|x− 3|
< ² if |x− 3| < 17², δ = min(1, 17²)
23. if δ <1
6then
∣∣∣∣ 1x − 3∣∣∣∣ = 3|x−
1
3|
|x| < 18|x−1
3| < ² if |x− 1
3| < 1
18², δ = min( 16 ,
118²)
24. If δ <1
2and |x− (−2)| < δ then −5
2< x < −3
2, x+ 1 < −1
2, |x+ 1| > 1
2; then∣∣∣∣ 1x+ 1 − (−1)
∣∣∣∣ = |x+ 2||x+ 1| < 2|x+ 2| < ² if |x+ 2| < 12², δ =
min( 12 , 12²)25. |√x− 2| =
∣∣∣∣(√x− 2)√x+ 2√x+ 2∣∣∣∣ = ∣∣∣∣ x− 4√x+ 2
∣∣∣∣ < 12 |x− 4| < ² if |x− 4| < 2², δ = 2²26. |
√x+ 3− 3|
∣∣∣∣√x+ 3 + 3√x+ 3 + 3∣∣∣∣ = |x− 6|√x+ 3 + 3 ≤ 13 |x− 6| < ²
if |x− 6| < 3², δ = 3²
27. |f(x)− 3| = |x+ 2− 3| = |x− 1| < ² if 0 < |x− 1| <
², δ = ²
28. If δ < 1 then |(x2 + 3x− 1)− 9| = |(x− 2)(x+ 5)| <
8|x− 2| < ² if |x− 2| < 18², δ = min (1, 18²)
29. (a) |f(x)− L| = 1x2
< 0.1 if x >√
10, N =√
10
(b) |f(x)− L| = | xx+ 1
− 1| = | 1x+ 1
| < 0.01 if x+ 1 > 100, N = 99
(c) |f(x)− L| =∣∣∣∣ 1x3∣∣∣∣ < 11000 if |x| > 10, x <
−10, N = −10
(d) |f(x)− L| =∣∣∣∣ xx+ 1 − 1
∣∣∣∣ = ∣∣∣∣ 1x+ 1∣∣∣∣ < 0.01 if |x+ 1| > 100, −x− 1 >
100, x < −101, N = −101
30. (a)
∣∣∣∣ 1x3∣∣∣∣ < 0.1, x > 101/3, N = 101/3 (b) ∣∣∣∣ 1x3
∣∣∣∣ < 0.01, x > 1001/3, N = 1001/3(c)
∣∣∣∣ 1x3∣∣∣∣ < 0.001, x > 10, N = 10
31. (a)x21
1 + x21= 1− ², x1 = −
√1− ²²
;x22
1 + x22= 1− ², x2 =
√1− ²²
(b) N =
√1− ²²
(c) N = −√
1− ²²
32. (a) x1 = −1/²3; x2 = 1/²3 (b) N = 1/²3 (c) N = −1/²3
33.1
x2< 0.01 if |x| > 10, N = 10
-
Exercise Set 2.3 52
34.1
x+ 2< 0.005 if |x+ 2| > 200, x > 198, N = 198
35.
∣∣∣∣ xx+ 1 − 1∣∣∣∣ = ∣∣∣∣ 1x+ 1
∣∣∣∣ < 0.001 if |x+ 1| > 1000, x > 999, N = 99936.
∣∣∣∣4x− 12x+ 5 − 2∣∣∣∣ = ∣∣∣∣ 112x+ 5
∣∣∣∣ < 0.1 if |2x+ 5| > 110, 2x > 105, N = 52.537.
∣∣∣∣ 1x+ 2 − 0∣∣∣∣ < 0.005 if |x+ 2| > 200, −x− 2 >
200, x < −202, N = −202
38.
∣∣∣∣ 1x2∣∣∣∣ < 0.01 if |x| > 10, −x > 10, x < −10, N
= −10
39.
∣∣∣∣4x− 12x+ 5 − 2∣∣∣∣ = ∣∣∣∣ 112x+ 5
∣∣∣∣ < 0.1 if |2x+ 5| > 110, −2x− 5 > 110, 2x <
−115, x < −57.5, N = −57.540.
∣∣∣∣ xx+ 1 − 1∣∣∣∣ = ∣∣∣∣ 1x+ 1
∣∣∣∣ < 0.001 if |x+ 1| > 1000, −x− 1 > 1000, x <
−1001, N = −100141.
∣∣∣∣ 1x2∣∣∣∣ < ² if |x| > 1√² , N = 1√² 42.
∣∣∣∣ 1x∣∣∣∣ < ² if |x| > 1² , −x > 1² , x < −1² , N
= −1²
43.
∣∣∣∣ 1x+ 2∣∣∣∣ < ² if |x+ 2| > 1² , −x− 2 < 1² , x >
−2− 1² , N = −2− 1²
44.
∣∣∣∣ 1x+ 2∣∣∣∣ < ² if |x+ 2| > 1² , x+ 2 > 1² , x >
1² − 2, N = 1² − 2
45.
∣∣∣∣ xx+ 1 − 1∣∣∣∣ = ∣∣∣∣ 1x+ 1
∣∣∣∣ < ² if |x+ 1| > 1² , x > 1² − 1, N = 1² − 146.
∣∣∣∣ xx+ 1 − 1∣∣∣∣ = ∣∣∣∣ 1x+ 1
∣∣∣∣ < ² if |x+ 1| > 1² , −x− 1 > 1² , x < −1− 1² ,
N = −1− 1²47.
∣∣∣∣4x− 12x+ 5 − 2∣∣∣∣ = ∣∣∣∣ 112x+ 5
∣∣∣∣ < ² if |2x+5| > 11² , −2x−5 > 11² , 2x < −11²
−5, x < −112²− 52 , N = −52− 112²48.
∣∣∣∣4x− 12x+ 5 − 2∣∣∣∣ = ∣∣∣∣ 112x+ 5
∣∣∣∣ < ² if |2x+ 5| > 11² , 2x > 11² − 5, x > 112² −
52 , N = 112² − 5249. (a)
1
x2> 100 if |x| < 1
10(b)
1
|x− 1| > 1000 if |x− 1| <1
1000
(c)−1
(x− 3)2 < −1000 if |x− 3| <1
10√
10(d) − 1
x4< −10000 if x4 < 1
10000, |x| < 1
10
50. (a)1
(x− 1)2 > 10 if and only if |x− 1| <1√10
(b)1
(x− 1)2 > 1000 if and only if
|x− 1| < 110√
10
(c)1
(x− 1)2 > 100000 if and only if |x− 1| <1
100√
10
51. if M > 0 then1
(x− 3)2 > M , 0 < (x− 3)2 <
1
M, 0 < |x− 3| < 1√
M, δ =
1√M
-
53 Chapter 2
52. if M < 0 then−1
(x− 3)2 < M , 0 < (x− 3)2 < − 1
M, 0 < |x− 3| < 1√
−M, δ =
1√−M
53. if M > 0 then1
|x| > M , 0 < |x| <1
M, δ =
1
M
54. if M > 0 then1
|x− 1| > M , 0 < |x− 1| <1
M, δ =
1
M
55. if M < 0 then − 1x4
< M , 0 < x4 < − 1M
, |x| < 1(−M)1/4 , δ =
1
(−M)1/4
56. if M > 0 then1
x4> M , 0 < x4 <
1
M, x <
1
M 1/4, δ =
1
M 1/4
57. if x > 2 then |x+ 1− 3| = |x− 2| = x− 2 < ² if 2 <
x < 2 + ², δ = ²
58. if x < 1 then |3x+ 2− 5| = |3x− 3| = 3|x− 1| = 3(1− x)
< ² if 1− x < 13², 1− 13² < x < 1, δ = 13²
59. if x > 4 then√x− 4 < ² if x− 4 < ²2, 4 < x <
4 + ²2, δ = ²2
60. if x < 0 then√−x < ² if −x < ²2, −²2 < x < 0,
δ = ²2
61. if x > 2 then |f(x)− 2| = |x− 2| = x− 2 < ² if 2 <
x < 2 + ², δ = ²
62. if x < 2 then |f(x)− 6| = |3x− 6| = 3|x− 2| = 3(2− x)
< ² if 2− x < 13², 2− 13² < x < 2, δ = 13²
63. (a) if M < 0 and x > 1 then1
1− x < M , x− 1 < −1
M, 1 < x < 1− 1
M, δ = − 1
M
(b) if M > 0 and x < 1 then1
1− x > M , 1− x <1
M, 1− 1
M< x < 1, δ =
1
M
64. (a) if M > 0 and x > 0 then1
x> M , x <
1
M, 0 < x <
1
M, δ =
1
M
(b) if M < 0 and x < 0 then1
x< M , −x < − 1
M,
1
M< x < 0, δ = − 1
M
65. (a) Given any M > 0 there corresponds N > 0 such that
if x > N then f(x) > M , x + 1 > M ,x > M − 1, N = M −
1.
(b) Given any M < 0 there corresponds N < 0 such that if x
< N then f(x) < M , x + 1 < M ,x < M − 1, N = M −
1.
66. (a) Given any M > 0 there corresponds N > 0 such that
if x > N then f(x) > M , x2 − 3 > M ,x >√M + 3, N =
√M + 3.
(b) Given any M < 0 there corresponds N < 0 such that if x
< N then f(x) < M , x3 + 5 < M ,x < (M − 5)1/3, N = (M
− 5)1/3.
67. if δ ≤ 2 then |x − 3| < 2, −2 < x − 3 < 2, 1 < x
< 5, and |x2 − 9| = |x + 3||x − 3| < 8|x − 3| < ² if|x− 3|
< 18², δ = min
(2, 18²
)68. (a) We don’t care about the value of f at x = a, because
the limit is only concerned with values of
x near a. The condition that f be defined for all x (except
possibly x = a) is necessary, becauseif some points were excluded
then the limit may not exist; for example, let f(x) = x if 1/x
isnot an integer and f(1/n) = 6. Then lim
x→0f(x) does not exist but it would if the points 1/n were
excluded.
(b) when x < 0 then√x is not defined (c) yes; if δ ≤ 0.01
then x > 0, so √x is defined
-
Exercise Set 2.4 54
69. |(x3−4x+5)−2| < 0.05, −0.05 < (x3−4x+5)−2 < 0.05,
1.95 < x3−4x+5 < 2.05; x3−4x+5 = 1.95at x = 1.0616, x3 − 4x+
5 = 2.05 at x = 0.9558; δ = min (1.0616− 1, 1− 0.9558) = 0.0442
2.2
1.90.9 1.1
70.√
5x+ 1 = 3.5 at x = 2.25,√
5x+ 1 = 4.5 at x = 3.85, so δ = min(3− 2.25, 3.85− 3) =
0.755
02 4
EXERCISE SET 2.4
1. (a) no, x = 2 (b) no, x = 2 (c) no, x = 2 (d) yes
(e) yes (f) yes
2. (a) no, x = 2 (b) no, x = 2 (c) no, x = 2 (d) yes
(e) no, x = 2 (f) yes
3. (a) no, x = 1, 3 (b) yes (c) no, x = 1 (d) yes
(e) no, x = 3 (f) yes
4. (a) no, x = 3 (b) yes (c) yes (d) yes
(e) no, x = 3 (f) yes
5. (a) 3 (b) 3
6. −2/5
7. (a) y
x3
(b) y
x
1
1 3
-
55 Chapter 2
(c)y
x
-1
1
1
(d) y
x2 3
8. f(x) = 1/x, g(x) =
{0 if x = 0
sin1
xif x 6= 0
9. (a) C
t1
$4
2
(b) One second could cost you one dollar.
10. (a) no; disasters (war, flood, famine, pestilence, for
example) can cause discontinuities
(b) continuous
(c) not usually continuous; see Exercise 9
(d) continuous
11. none 12. none 13. none
14. f is not defined at x = ±1 15. f is not defined at x =
±4
16. f is not defined at x =−7±
√57
2
17. f is not defined at x = ±3
18. f is not defined at x = 0,−4 19. none
20. f is not defined at x = 0,−3
21. none; f(x) = 2x+ 3 is continuous on x < 4 and f(x) = 7
+16
xis continuous on 4 < x;
limx→4−
f(x) = limx→4+
f(x) = f(4) = 11 so f is continuous at x = 4
22. limx→1
f(x) does not exist so f is discontinuous at x = 1
23. (a) f is continuous for x < 1, and for x > 1;
limx→1−
f(x) = 5, limx→1+
f(x) = k, so if k = 5 then f is
continuous for all x
(b) f is continuous for x < 2, and for x > 2; limx→2−
f(x) = 4k, limx→2+
f(x) = 4 + k, so if 4k = 4 + k,
k = 4/3 then f is continuous for all x
24. (a) no, f is not defined at x = 2 (b) no, f is not defined
for x ≤ 2(c) yes (d) no, f is not defined for x ≤ 2
-
Exercise Set 2.4 56
25. (a) y
xc
(b) y
xc
26. (a) f(c) = limx→c
f(x)
(b) limx→1
f(x) = 2
y
x
1
-1
limx→1
g(x) = 1
y
x
1
1
(c) Define f(1) = 2 and redefine g(1) = 1.
27. (a) x = 0, limx→0−
f(x) = −1 6= +1 = limx→0+
f(x) so the discontinuity is not removable
(b) x = −3; define f(−3) = −3 = limx→−3
f(x), then the discontinuity is removable
(c) f is undefined at x = ±2; at x = 2, limx→2
f(x) = 1, so define f(2) = 1 and f becomes continuous
there; at x = −2, limx→−2
does not exist, so the discontinuity is not removable
28. (a) f is not defined at x = 2; limx→2
f(x) = limx→2
x+ 2
x2 + 2x+ 4=
1
3, so define f(2) =
1
3and f becomes
continuous there
(b) limx→2−
f(x) = 1 6= 4 = limx→2+
f(x), so f has a nonremovable discontinuity at x = 2
(c) limx→1
f(x) = 8 6= f(1), so f has a removable discontinuity at x =
1
29. (a) discontinuity at x = 1/2, not removable; atx = −3,
removable
y
x
-5
5
5
(b) 2x2 + 5x− 3 = (2x− 1)(x+ 3)
-
57 Chapter 2
30. (a) there appears to be one discontinuity nearx = −1.52
4
–4
-3 3
(b) one discontinuity at x = −1.52
31. For x > 0, f(x) = x3/5 = (x3)1/5 is the composition
(Theorem 2.4.6) of the two continuous functionsg(x) = x3 and h(x) =
x1/5 and is thus continuous. For x < 0, f(x) = f(−x) which is
the compositionof the continuous functions f(x) (for positive x)
and the continuous function y = −x. Hence f(−x) iscontinuous for
all x > 0. At x = 0, f(0) = lim
x→0f(x) = 0.
32. x4 + 7x2 + 1 ≥ 1 > 0, thus f(x) is the composition of the
polynomial x4 + 7x2 + 1, the square root√x, and the function 1/x
and is therefore continuous by Theorem 2.4.6.
33. (a) Let f(x) = k for x 6= c and f(c) = 0; g(x) = l for x 6=
c and g(c) = 0. If k = −l then f + g iscontinuous; otherwise it’s
not.
(b) f(x) = k for x 6= c, f(c) = 1; g(x) = l 6= 0 for x 6= c,
g(c) = 1. If kl = 1, then fg is continuous;otherwise it is not.
34. A rational function is the quotient f(x)/g(x) of two
polynomials f(x) and g(x). By Theorem 2.4.2 fand g are continuous
everywhere; by Theorem 2.4.3 f/g is continuous except when g(x) =
0.
35. Since f and g are continuous at x = c we know that
limx→c
f(x) = f(c) and limx→c
g(x) = g(c). In the
following we use Theorem 2.2.2.
(a) f(c) + g(c) = limx→c
f(x) + limx→c
g(x) = limx→c
(f(x) + g(x)) so f + g is continuous at x = c.
(b) same as (a) except the + sign becomes a − sign
(c)f(c)
g(c)=
limx→c
f(x)
limx→c
g(x)= lim
x→cf(x)
g(x)so
f
gis continuous at x = c
36. h(x) = f(x) − g(x) satisfies h(a) > 0, h(b) < 0. Use
the Intermediate Value Theorem or Theorem2.4.9.
37. Of course such a function must be discontinuous. Let f(x) =
1 on 0 ≤ x < 1, and f(x) = −1 on1 ≤ x ≤ 2.
38. A square whose diagonal has length r has area f(r) = r2/2.
Note thatf(r) = r2/2 < πr2/2 < 2r2 = f(2r). By the
Intermediate Value Theorem there must be a valuec between r and 2r
such that f(c) = πr2/2, i.e. a square of diagonal c whose area is
πr2/2.
39. The cone has volume πr2h/3. The function V (r) = πr2h (for
variable r and fixed h) gives the volumeof a right circular
cylinder of height h and radius r, and satisfies V (0) < πr2h/3
< V (r). By theIntermediate Value Theorem there is a value c
between 0 and r such that V (c) = πr2h/3, so thecylinder of radius
c (and height h) has volume equal to that of the cone.
40. If f(x) = x3 − 4x+ 1 then f(0) = 1, f(1) = −2. Use Theorem
2.4.9.
41. If f(x) = x3 + x2 − 2x then f(−1) = 2, f(1) = 0. Use the
Intermediate Value Theorem.
-
Exercise Set 2.4 58
42. Since limx→−∞
p(x) = −∞ and limx→+∞
p(x) = +∞ (or vice versa, if the leading coefficient of p is
negative),it follows that for M = −1 there corresponds N1 < 0,
and for M = 1 there is N2 > 0, such thatp(x) < −1 for x <
N1 and p(x) > 1 for x > N2. Choose x1 < N1 and x2 > N2
and use Theorem 2.4.9on the interval [x1, x2] to find a solution of
p(x) = 0.
43. For the negative root, use intervals on the x-axis as
follows: [−2,−1]; since f(−1.3) < 0 andf(−1.2) > 0, the
midpoint x = −1.25 of [−1.3,−1.2] is the required approximation of
the root.For the positive root use the interval [0, 1]; since
f(0.7) < 0 and f(0.8) > 0, the midpoint x = 0.75 of[0.7, 0.8]
is the required approximation.
44. x = −1.25 and x = 0.75.10
-5
-2 -1
1
-1
0.7 0.8
45. For the negative root, use intervals on the x-axis as
follows: [−2,−1]; since f(−1.7) < 0 andf(−1.6) > 0, use the
interval [−1.7,−1.6]. Since f(−1.61) < 0 and f(−1.60) > 0 the
midpointx = −1.605 of [−1.61,−1.60] is the required approximation
of the root. For the positive root usethe interval [1, 2]; since
f(1.3) > 0 and f(1.4) < 0, use the interval [1.3, 1.4]. Since
f(1.37) > 0 andf(1.38) < 0, the midpoint x = 1.375 of [1.37,
1.38] is the required approximation.
46. x = −1.605 and x = 1.375.1
-2
-1.7 -1.6
1
-0.5
1.3 1.4
47. x = 2.24.
48. Set f(x) =a
x− 1 +b
x− 3 . Since limx→1+ f(x) = +∞ and limx→3− f(x) = −∞ there exist
x1 > 1 and x2 < 3(with x2 > x1) such that f(x) > 1 for
1 < x < x1 and f(x) < −1 for x2 < x < 3. Choose x3
in (1, x1)and x4 in (x2, 3) and apply Theorem 2.4.9 on [x3,
x4].
49. The uncoated sphere has volume 4π(x−1)3/3 and the coated
sphere has volume 4πx3/3. If the volumeof the uncoated sphere and
of the coating itself are the same, then the coated sphere has
twice thevolume of the uncoated sphere. Thus 2(4π(x − 1)3/3) =
4πx3/3, or x3 − 6x2 + 6x − 2 = 0, with thesolution x = 4.847
cm.
50. Let g(t) denote the altitude of the monk at time t measured
in hours from noon of day one, and let f(t)denote the altitude of
the monk at time t measured in hours from noon of day two. Then
g(0) < f(0)and g(12) > f(12). Use Exercise 36.
51. We must show limx→c
f(x) = f(c). Let ² > 0; then there exists δ > 0 such that
if |x − c| < δ then|f(x)− f(c)| < ². But this certainly
satisfies Definition 2.3.3.
-
59 Chapter 2
EXERCISE SET 2.5
1. none 2. x = π 3. x = nπ, n = 0,±1,±2, . . .
4. x = nπ + π/2, n = 0,±1,±2, . . . 5. x = nπ, n = 0,±1,±2, . .
.
6. none 7. none 8. x = nπ + π/2, n = 0,±1,±2, . . .
9. 2nπ + π/6, 2nπ + 5π/6, n = 0,±1,±2, . . . 10. none
11. (a) sinx, x3 + 7x+ 1 (b) |x|, sinx (c) x3, cosx, x+ 1(d)
√x, 3 + x, sinx, 2x (e) sinx, sinx (f) x5 − 2x3 + 1, cosx
12. (a) Use Theorem 2.4.6. (b) g(x) = cosx, g(x) =1
x2 + 1, g(x) = x2 + 1
13. cos
(lim
x→+∞1
x
)= cos 0 = 1 14. sin
(lim
x→+∞2
x
)= sin 0 = 0
15. sin
(lim
x→+∞πx
2− 3x
)= sin
(−π
3
)= −√
3
216.
1
2limh→0
sinh
h=
1
2
17. 3 limθ→0
sin 3θ
3θ= 3 18.
(limθ→0+
1
θ
)limθ→0+
sin θ
θ= +∞
19. − limx�
