Chapter 1 Solutions.pdf
Chapter 1 Solutions.pdf
Chapter 1 Solutions.pdf
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CHAPTER 1<br />
Introduction to Calculus<br />
Review of Prerequisite Skills, pp. 2–3<br />
1. a. m 5 27 2 5<br />
6 2 2<br />
523<br />
b. m 5 4 2 (24)<br />
21 2 3<br />
522<br />
c. m 5 4 2 0<br />
1 2 0<br />
5 4<br />
d. m 5 4 2 0<br />
21 2 0<br />
524<br />
e. m 5 4 2 4.41<br />
22 2 (22.1)<br />
524.1<br />
f.<br />
m 5 21 4 2 1 4<br />
7<br />
4 2 3 4<br />
2 2 4<br />
5<br />
1<br />
52 1 2<br />
2. a. Substitute the given slope and y-intercept into<br />
y 5 mx 1 b.<br />
y 5 4x 2 2<br />
b. Substitute the given slope and y-intercept into<br />
y 5 mx 1 b.<br />
y 522x 1 5<br />
c. The slope of the line is<br />
m 5 12 2 6<br />
4 2 (21)<br />
5 6 5<br />
The equation of the line is in the form<br />
y 2 y 1 5 m(x 2 x 1 ). The point is (21, 6) and<br />
m 5 6 5.<br />
The equation of the line is y 2 6 5 6 5(x 1 1) or<br />
y 5 6 5(x 1 1) 1 6.<br />
8 2 4<br />
d. m 5<br />
26 2 (22)<br />
521<br />
y 2 4 521(x 2 (22))<br />
y 2 4 52x 2 2<br />
x 1 y 2 2 5 0<br />
e.<br />
f.<br />
3. a.<br />
b.<br />
c.<br />
d.<br />
4. a.<br />
b.<br />
x 523<br />
y 5 5<br />
f(2) 526 1 5<br />
521<br />
f(2) 5 (8 2 2)(6 2 6)<br />
5 0<br />
f(2) 523(4) 1 2(2) 2 1<br />
529<br />
f(2) 5 (10 1 2) 2<br />
5 144<br />
f(210) 5 210<br />
100 1 4<br />
f(23) 5 23<br />
9 1 4<br />
52 3 13<br />
c. f(0) 5 0<br />
0 1 4<br />
5 0<br />
d. f(10) 5 10<br />
100 1 4<br />
5 5<br />
52<br />
"3 2 x, if x , 0<br />
5. f(x) 5 •<br />
"3 1 x, if x $ 0<br />
a. f(233) 5 6<br />
b. f(0) 5 "3<br />
c. f(78) 5 9<br />
d. f(3) 5 "6<br />
1<br />
, if 23 , t , 0<br />
t<br />
6. s(t) 5 μ<br />
5, if t 5 0<br />
t 3 , if t . 0<br />
a. s(22) 52 1 2<br />
b. s(21) 521<br />
52 5<br />
52<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-1
c. s(0) 5 5<br />
d. s(1) 5 1<br />
e. s(100) 5 100 3 or 10 6<br />
7. a. (x 2 6)(x 1 2) 5 x 2 2 4x 2 12<br />
b. (5 2 x)(3 1 4x) 5 15 1 17x 2 4x 2<br />
c. x(5x 2 3) 2 2x(3x 1 2) 5 5x 2 2 3x 2 6x 2 2 4x<br />
52x 2 2 7x<br />
d. (x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)<br />
5 x 2 1 2x 2 3 2 (2x 2 1 x 2 10)<br />
5 2x 2 1 x 1 7<br />
e. (a 1 2) 3 5 (a 1 2)(a 1 2)(a 1 2)<br />
5 (a 2 1 4a 1 4)(a 1 2)<br />
5 a 3 1 6a 2 1 12a 1 8<br />
f. (9a 2 5) 3 5 (9a 2 5)(9a 2 5)(9a 2 5)<br />
5 (81a 2 2 90a 1 25)(9a 2 5)<br />
5 729a 3 2 1215a 2 1 675a 2 125<br />
8. a. x 3 2 x 5 x(x 2 2 1)<br />
5 x(x 1 1)(x 2 1)<br />
b.<br />
c.<br />
d.<br />
x 2 1 x 2 6 5 (x 1 3)(x 2 2)<br />
2x 2 2 7x 1 6 5 (2x 2 3)(x 2 2)<br />
x 3 1 2x 2 1 x 5 x(x 2 1 2x 1 1)<br />
5 x(x 1 1)(x 1 1)<br />
e. 27x 3 2 64 5 (3x 2 4)(9x 2 1 12x 1 16)<br />
f. 2x 3 2 x 2 2 7x 1 6<br />
x 5 1 is a zero, so x 2 1 is a factor. Synthetic or<br />
long division yields<br />
2x 3 2 x 2 2 7x 1 6 5 (x 2 1)(2x 2 1 x 2 6)<br />
5 (x 2 1)(2x 2 3)(x 1 2)<br />
5xPR 0 x $256<br />
9. a.<br />
b. 5xPR6<br />
c. 5xPR 0 x 2 16<br />
d.<br />
e.<br />
5xPR 0 x 2 06<br />
2x 2 2 5x 2 3 5 (2x 1 1)(x 2 3)<br />
e xPR ` x 2 2 1 2 , 3 f<br />
f. 5xPR 0 x 2 25, 22, 16<br />
10. a. h(0) 5 2, h(1) 5 22.1<br />
average rate of change 5 22.1 2 2<br />
1 2 0<br />
5 20.1 ms ><br />
b. h(1) 5 22.1, h(2) 5 32.4<br />
32.4 2 22.1<br />
average rate of change 5<br />
2 2 1<br />
5 10.3 ms ><br />
11. a. The average rate of change during the second<br />
hour is the difference in the volume at t 5 120 and<br />
t 5 60 (since t is measured in minutes), divided by<br />
the difference in time.<br />
V(120) 2 V(60)<br />
120 2 60<br />
b. To estimate the instantaneous rate of change in<br />
volume after exactly 60 minutes, calculate the average<br />
rate of change in volume from minute 59 to minute 61.<br />
V(61) 2 V(59) 1186.56 2 1213.22<br />
8<br />
61 2 59<br />
2<br />
5213.33 L>min<br />
c. The instantaneous rate of change in volume is<br />
negative for 0 # t # 120 because the volume of<br />
water in the hot tub is always decreasing during that<br />
time period, a negative change.<br />
12. a., b.<br />
y<br />
8<br />
The slope of the tangent line is 28.<br />
c. The instantaneous rate of change in f(x) when<br />
x 5 5 is 28.<br />
1.1 Radical Expressions:<br />
Rationalizating Denominators, p. 9<br />
1. a. 2"3 1 4<br />
b. "3 2 "2<br />
c. 2"3 1 "2<br />
d. 3"3 2 "2<br />
e. "2 1 "5<br />
f. 2"5 2 2"2<br />
"3 1 "5<br />
2. a.<br />
? "2<br />
"2 "2<br />
5<br />
"6 1 "10<br />
2<br />
2"3 2 3"2<br />
b.<br />
"2<br />
5 2"6 2 6<br />
2<br />
5 "6 2 3<br />
–2<br />
4<br />
–4<br />
–8<br />
5 0 2 1200<br />
60<br />
5220 L>min<br />
0<br />
? "2<br />
"2<br />
2<br />
4 6<br />
x<br />
1-2 <strong>Chapter</strong> 1: Introduction to Calculus
4"3 1 3"2<br />
c.<br />
2"3<br />
5<br />
5 4 1 "6<br />
2<br />
3"5 2 "2<br />
d.<br />
2"2<br />
2"5<br />
b.<br />
2"5 1 3"2<br />
5<br />
"3 2 "2<br />
c.<br />
"3 1 "2<br />
d.<br />
5<br />
5<br />
2"3 2 "2<br />
e.<br />
5"2 1 "3<br />
5<br />
5<br />
12 1 3"6<br />
6<br />
5 3"10 2 2<br />
4<br />
5 "5 1 "2<br />
20 2 6"10<br />
20 2 18<br />
5 10 2 3"10<br />
5 3 1 2"6 1 2<br />
3 2 2<br />
5 5 1 2"6<br />
44 2 22"5<br />
11<br />
5 4 2 2"5<br />
? "3<br />
"3<br />
? "2<br />
"2<br />
3 "5 1 "2<br />
3. a.<br />
?<br />
"5 2 "2 "5 1 "2<br />
3("5 1 "2)<br />
5<br />
3<br />
2"5 2 8<br />
2"5 1 3 ? 2"5 2 3<br />
2"5 2 3<br />
20 2 22"5 1 24<br />
20 2 9<br />
?<br />
2"5 2 3"2<br />
2"5 2 3"2<br />
?<br />
"3 2 "2<br />
"3 2 "2<br />
?<br />
5"2 2 "3<br />
5"2 2 "3<br />
10"6 2 6 2 10 1 "6<br />
50 2 3<br />
11"6 2 16<br />
47<br />
3"3 2 2"2<br />
f.<br />
3"3 1 2"2<br />
5<br />
5<br />
"5 2 1<br />
4. a. ? "5 1 1<br />
4 "5 1 1<br />
5 2 1<br />
5<br />
4("5 1 1)<br />
1<br />
5<br />
!5 1 1<br />
2 2 3"2<br />
b. ? 2 1 3"2<br />
2 2 1 3"2<br />
4 2 18<br />
5<br />
2(2 1 3"2)<br />
27<br />
5<br />
2 1 3"2<br />
"5 1 2<br />
c.<br />
2"5 2 1 ? "5 2 2<br />
"5 2 2<br />
5 2 4<br />
5<br />
10 2 5"5 1 2<br />
1<br />
5<br />
12 2 5!5<br />
8"2<br />
5. a.<br />
"20 2 "18<br />
5<br />
5<br />
5 8"10 1 24<br />
8"2<br />
b.<br />
2"5 2 3"2<br />
5<br />
5<br />
27 2 12"6 1 8<br />
27 2 8<br />
35 2 12"6<br />
19<br />
8"40 1 8"36<br />
20 2 18<br />
16"10 1 48<br />
2<br />
16"10 1 48<br />
20 2 18<br />
16"10 1 48<br />
2<br />
?<br />
3"3 2 2"2<br />
3"3 2 2"2<br />
?<br />
"20 1 "18<br />
"20 1 "18<br />
?<br />
2"5 1 3"2<br />
2"5 1 3"2<br />
5 8"10 1 24<br />
c. The expressions in the two parts are equivalent.<br />
The radicals in the denominator of part a. have been<br />
simplified in part b.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-3
2"2<br />
6. a.<br />
2"3 2 "8<br />
5 4"6 1 8<br />
6 2 8<br />
5 22"3 2 4<br />
2"6<br />
b.<br />
2"27 2 "8<br />
5<br />
5<br />
5<br />
2"2<br />
c.<br />
"16 2 "12<br />
5<br />
5<br />
3"2 1 2"3<br />
d.<br />
"12 2 "8<br />
5<br />
5<br />
e.<br />
5<br />
12"15 1 15"10<br />
5 2<br />
2<br />
"18 1 "12<br />
f.<br />
"18 2 "12<br />
5<br />
5<br />
4"162 1 2"48<br />
54 2 8<br />
36"2 1 8"3<br />
46<br />
18"2 1 4"3<br />
23<br />
2"2<br />
4 2 2"3 ? 4 1 2"3<br />
4 1 2"3<br />
8"2 1 4"6<br />
16 2 12<br />
5 2"2 1 "6<br />
3"24 1 12 1 12 1 2"24<br />
12 2 8<br />
24 1 15"3<br />
4<br />
3!5 4!3 1 5!2<br />
?<br />
4!3 2 5!2 4!3 1 5!2<br />
12"15 1 15"10<br />
48 2 50<br />
18 1 2"216 1 12<br />
18 2 12<br />
30 1 12"6<br />
6<br />
5 5 1 2"6<br />
?<br />
2"3 1 "8<br />
2"3 1 "8<br />
?<br />
2"27 1 "8<br />
2"27 1 "8<br />
?<br />
"12 1 "8<br />
"12 1 "8<br />
?<br />
"18 1 "12<br />
"18 1 "12<br />
"a 2 2<br />
7. a.<br />
a 2 4 ? "a 1 2<br />
"a 1 2<br />
a 2 4<br />
5<br />
(a 2 4)("a 2 2)<br />
1<br />
5<br />
"a 2 2<br />
"x 1 4 2 2<br />
b.<br />
? "x 1 4 1 2<br />
x "x 1 4 1 2<br />
x 1 4 2 4<br />
5<br />
x("x 1 4 1 2)<br />
x<br />
5<br />
x("x 1 4 1 2)<br />
1<br />
5<br />
"x 1 4 2 2<br />
!x 1 h 2 !x<br />
c.<br />
?<br />
h<br />
x 1 h 2 x<br />
5<br />
5<br />
5<br />
1.2 The Slope of a Tangent, pp. 18–21<br />
1. a. m 5 28 2 7<br />
23 2 2<br />
5 3<br />
b.<br />
hA!x 1 h 1 !xB<br />
h<br />
hA!x 1 h 1 !xB<br />
1<br />
!x 1 h 1 !x<br />
m 5 27 2 2 3 2<br />
7<br />
2 2 1 2<br />
5 210 2<br />
6<br />
2<br />
52 5 3<br />
21 2 (22.6)<br />
c. m 5<br />
1.5 2 6.3<br />
!x 1 h 1 !x<br />
!x 1 h 1 !x<br />
52 1 3<br />
2. a. The slope of the given line is 3, so the slope<br />
of a line perpendicular to the given line is 2 1 3.<br />
b. 13x 2 7y 2 11 5 0<br />
27y 5213x 2 11<br />
y 5 13<br />
7 x 1 11<br />
7<br />
13<br />
The slope of the given line is 7 , so the slope of a line<br />
perpendicular to the given line is 213.<br />
7<br />
1-4 <strong>Chapter</strong> 1: Introduction to Calculus
3. a.<br />
y 2 (24) 5 7 (x 2 (24))<br />
17<br />
17y 1 68 5 7x 1 28<br />
7x 2 17y 2 40 5 0<br />
y<br />
4<br />
–2<br />
b. The slope and y-intercept are given.<br />
y 5 8x 1 6<br />
y<br />
8<br />
–4<br />
c. (0, 23), (5, 0)<br />
m 5 0 2 (23)<br />
5 2 0<br />
5 3 5<br />
m 5 25 3 2 (24)<br />
5<br />
3 2 (24)<br />
2<br />
–2<br />
–4<br />
5<br />
0<br />
–2<br />
7<br />
3<br />
17<br />
3<br />
5 7<br />
17<br />
y 2 0 5 3 (x 2 5)<br />
5<br />
3x 2 5y 2 15 5 0<br />
4<br />
–4<br />
–8<br />
0<br />
2<br />
4 6<br />
2<br />
4<br />
x<br />
x<br />
d. The line is a vertical line because both points<br />
have the same x-coordinate.<br />
x 5 5<br />
y<br />
4<br />
4. a.<br />
b.<br />
c.<br />
d.<br />
–2<br />
–2<br />
4<br />
2<br />
–2<br />
–4<br />
0<br />
2<br />
–2<br />
–4<br />
0<br />
y<br />
2<br />
2<br />
(5 1 h) 3 2 125<br />
h<br />
5 (5 1 h 2 5)((5 1 h)2 1 5(5 1 h) 1 25)<br />
h<br />
5 h(75 1 15h 1 h2 )<br />
h<br />
5 75 1 15h 1 h 2<br />
(3 1 h) 4 2 81<br />
h<br />
5 ((3 1 h)2 2 9)((3 1 h) 2 1 9)<br />
h<br />
5 (9 1 6h 1 h2 2 9)(9 1 6h 1 h 2 1 9)<br />
h<br />
5 (6 1 h)(18 1 6h 1 h 2 )<br />
5 108 1 54h 1 12h 2 1 h 3<br />
1<br />
1 1 h 2 1<br />
5 1 2 1 2 h<br />
h h(1 1 h) 52 1<br />
1 1 h<br />
3(1 1 h) 2 2 3<br />
h<br />
4 6<br />
4 6<br />
5 3((1 1 h)2 2 1)<br />
h<br />
5 3(1 1 2h 1 h2 2 1)<br />
h<br />
x<br />
x<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-5
e.<br />
f.<br />
3<br />
4 1 h 2 3 4<br />
5<br />
h<br />
21<br />
2 1 h 1 1 2<br />
5<br />
h<br />
5<br />
5 1<br />
4 1 2h<br />
"h 2 1 5h 1 4 2 2<br />
b.<br />
5 h2 1 5h 1 4 2 4<br />
h<br />
"5 1 h 2 "5 5 1 h 2 5<br />
c.<br />
5<br />
h h("5 1 h 1 "5)<br />
1<br />
5<br />
"5 1 h 1 "5<br />
6. a. P(1, 3), Q(1 1 h, f(1 1 h)), f(x) 5 3x 2<br />
m 5 3(1 1 h)2 2 3<br />
h<br />
5 6 1 3h<br />
b. P(1, 3), Q(1 1 h, (1 1 h) 3 1 2)<br />
m 5 (1 1 h)3 1 2 2 3<br />
h<br />
5 1 1 3h 1 3h2 1 h 3 2 1<br />
h<br />
5 3 1 3h 1 h 2<br />
c. P(9, 3), Q(9 1 h, "9 1 h)<br />
m 5 "9 1 h 2 3<br />
h<br />
1<br />
5<br />
"9 1 h 1 3<br />
5 3(2h 1 h2 )<br />
h<br />
5 6 1 3h<br />
12 2 12 2 3h<br />
4(4 1 h)<br />
h<br />
5 23<br />
4(4 1 h)<br />
22 1 2 1 h<br />
2(2 1 h)<br />
h<br />
h<br />
2h(2 1 h)<br />
"16 1 h 2 4 16 1 h 2 16<br />
5. a.<br />
5<br />
h h("16 1 h 1 4)<br />
1<br />
5<br />
"16 1 h 1 4<br />
h("h 2 1 5h 1 4 1 2)<br />
h 1 5<br />
5<br />
"h 2 1 5h 1 4 1 2<br />
? "9 1 h 1 3<br />
"9 1 h 1 3<br />
7. a.<br />
b. 12<br />
c. (2, 8), ((2 1 h), (2 1 h) 3 )<br />
m 5 (2 1 h)3 2 8<br />
2 1 h 2 2<br />
5 8 1 12h 1 6h2 1 h 3 2 8<br />
h<br />
5 12 1 6h 1 h 2<br />
d. m 5 lim(12 1 6h 1 h 2 )<br />
hS0<br />
5 12<br />
e. They are the same.<br />
f.<br />
y<br />
12<br />
–4<br />
P Q Slope of Line PQ<br />
(2, 8) (3, 27) 19<br />
(2, 8) (2.5, 15.625) 15.25<br />
(2, 8) (2.1, 9.261) 12.61<br />
(2, 8) (2.01, 8.120 601) 12.060 1<br />
(2, 8) (1, 1) 7<br />
(2, 8) (1.5, 3.375) 9.25<br />
(2, 8) (1.9, 6.859) 11.41<br />
(2, 8) (1.99, 7.880 599) 11.940 1<br />
–2<br />
8<br />
4<br />
–4<br />
0<br />
8. a. y 5 3x 2 , (22, 12)<br />
3(22 1 h) 2 2 12<br />
m 5 lim<br />
hS0 h<br />
12 2 12h 1 3h 2 2 12<br />
5 lim<br />
hS0 h<br />
5 lim(212 1 3h)<br />
hS0<br />
5212<br />
b. y 5 x 2 2 x at x 5 3, y 5 6.<br />
(3 1 h) 2 2 (3 1 h) 2 6<br />
m 5 lim<br />
hS0<br />
h<br />
9 1 6h 1 h 2 2 3 2 h 2 6<br />
5 lim<br />
hS0<br />
h<br />
5 lim(5 1 h)<br />
hS0<br />
5 5<br />
2<br />
4<br />
x<br />
1-6 <strong>Chapter</strong> 1: Introduction to Calculus
c. at x 522, y 528.<br />
(22 1 h) 3 1 8<br />
m 5 lim<br />
hS0 h<br />
28 1 12h 2 6h 2 1 h 3 1 8<br />
5 lim<br />
hS0<br />
h<br />
5 lim(12 2 6h 1 h 2 )<br />
5 12<br />
9. a. y 5 "x 2 2; (3, 1)<br />
"3 1 h 2 2 2 1<br />
m 5 lim<br />
hS0 h<br />
5 1 2<br />
b. y 5 "x 2 5 at x 5 9, y 5 2<br />
"9 1 h 2 5 2 2<br />
m 5 lim<br />
hS0 h<br />
5 lim £ "4 1 h 2 2 3 "4 1 h 1 2<br />
hS0 h "4 1 h 1 2 §<br />
1<br />
5 lim<br />
hS0 "4 1 h 1 2<br />
5 1 4<br />
c. y 5 "5x 2 1 at x 5 2, y 5 3<br />
"10 1 5h 2 1 2 3<br />
m 5 lim<br />
hS0 h<br />
"9 1 5h 2 3 "9 1 5h 1 3<br />
5 lim £ 3<br />
hS0 h "9 1 5h 1 3 §<br />
5<br />
5 lim<br />
hS0 "9 1 5h 1 3<br />
5 5 6<br />
hS0<br />
5 lim £ "1 1 h 2 1 3 "1 1 h 1 1<br />
hS0 h "1 1 h 1 1 §<br />
1<br />
5 lim<br />
hS0 "1 1 h 1 1<br />
10. a. y 5 8 at (2, 4)<br />
x<br />
8<br />
2 1 h 2 4<br />
m 5 lim<br />
hS0 h<br />
24<br />
5 lim<br />
hS0 2 1 h<br />
522<br />
b. y 5 8 at x 5 1; y 5 2<br />
3 1 x<br />
8<br />
4 1 h 2 2<br />
m 5 lim<br />
hS0 h<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
5 lim<br />
52 1 2<br />
c. y 5 1 at x 5 3; y 5 1 x 1 2<br />
5<br />
m 5 lim<br />
hS0 h<br />
21<br />
5 lim<br />
hS0 5(5 1 h)<br />
52 1<br />
10<br />
11. a. Let y 5 f(x).<br />
f(2) 5 (2) 2 2 3(2) 5 4 2 6 522<br />
f(2 1 h) 5 (2 1 h) 2 2 3(2 1 h)<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 2 is<br />
f(2 1 h) 2 f(2)<br />
m 5 lim<br />
hS0 h<br />
(2 1 h) 2 2 3(2 1 h) 2 (22)<br />
5 lim<br />
hS0<br />
h<br />
4 1 4h 1 h 2 2 6 2 3h 1 2<br />
5 lim<br />
hS0<br />
h<br />
h 2 1 h<br />
5 lim<br />
hS0 h<br />
5 lim (h 1 1)<br />
hS0<br />
hS0<br />
22<br />
4 1 h<br />
1<br />
5 1 h 2 1 5<br />
5 0 1 1<br />
5 1<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 x 2 2 3x at x 5 2 is 1.<br />
b. f(22) 5 4<br />
22 522 4<br />
f(22 1 h) 5<br />
22 1 h<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 522 is<br />
f(22 1 h) 2 f(22)<br />
m 5 lim<br />
hS0 h<br />
4<br />
22 1 h<br />
5 lim<br />
2 (22)<br />
hS0 h<br />
4<br />
22 1 h<br />
5 lim<br />
1 2<br />
hS0 h<br />
5 lim c 4 2 4 1 2h ? 1<br />
hS0 22 1 h h d<br />
2h<br />
5 lim c<br />
hS0 22 1 h ? 1 h d<br />
y 5 x 3 1-7
2<br />
5 lim<br />
hS0 22 1 h<br />
2<br />
5<br />
22 1 0<br />
521<br />
Therefore, the slope of the tangent to f(x) 5 4 at<br />
x<br />
x 522 is 21.<br />
c. Let y 5 f(x).<br />
f(1) 5 3(1) 3 5 3<br />
f(1 1 h) 5 3(1 1 h) 3<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 1 is<br />
f(1 1 h) 2 f(1)<br />
m 5 lim<br />
hS0 h<br />
3(1 1 h) 3 2 3<br />
5 lim<br />
hS0 h<br />
Using the binomial formula to expand (1 1 h) 3 (or<br />
one could simply expand using algebra), the slope m is<br />
3(h 3 1 3h 2 1 3h 1 1) 2 (3)<br />
5 lim<br />
hS0<br />
h<br />
3h 3 1 9h 2 1 9h 1 3 2 3<br />
5 lim<br />
hS0<br />
h<br />
3h 3 1 9h 2 1 9h<br />
5 lim<br />
hS0 h<br />
5 lim (3h 2 1 9h 1 9)<br />
hS0<br />
5 3(0) 1 9(0) 1 9<br />
5 9<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 3x 3 at x 5 1 is 9.<br />
d. Let y 5 f(x).<br />
f(16) 5 !16 2 7 5 !9 5 3<br />
f(16 1 h) 5 !16 1 h 2 7 5 !h 1 9<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 16 is<br />
f(16 1 h) 2 f(16)<br />
m 5 lim<br />
hS0 h<br />
!h 1 9 2 3<br />
5 lim<br />
hS0 h<br />
!h 1 9 2 3<br />
5 lim<br />
? !h 1 9 1 3<br />
hS0 h !h 1 9 1 3<br />
(h 1 9) 2 9<br />
5 lim<br />
hS0 h( !h 1 9 1 3)<br />
h<br />
5 lim<br />
hS0 h( !h 1 9 1 3)<br />
1<br />
5 lim<br />
hS0 !h 1 9 1 3<br />
1<br />
5<br />
!0 1 9 1 3<br />
5 1<br />
3 1 3<br />
5 1 6<br />
Therefore, the slope of the tangent to<br />
1<br />
y 5 f(x) 5 !x 2 7 at x 5 16 is 6.<br />
e. Let y 5 f(x).<br />
f(3) 5 "25 2 (3) 2 5 !25 2 9 5 4<br />
f(3 1 h) 5 "25 2 (3 1 h) 2<br />
5 "25 2 9 2 6h 2 h 2<br />
5 "16 2 6h 2 h 2<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 3 is<br />
f(3 1 h) 2 f(3)<br />
m 5 lim<br />
hS0 h<br />
"16 2 6h 2 h 2 2 4<br />
5 lim<br />
hS0 h<br />
5 lim c "16 2 6h 2 h2 2 4<br />
hS0 h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 "25 2 (9 1 6h 1 h 2 )<br />
3 "16 2 6h 2 h2 1 4<br />
"16 2 6h 2 h 2 1 4 d<br />
16 2 6h 2 h 2 2 16<br />
h("16 2 6h 2 h 2 1 4)<br />
h(26 2 h)<br />
h("16 2 6h 2 h 2 1 4)<br />
26 2 h<br />
5 lim<br />
hS0 "16 2 6h 2 h 2 1 4<br />
26 2 0<br />
5<br />
"16 2 6(0) 2 (0) 2 1 4<br />
5 26<br />
!16 1 4<br />
5 26<br />
8<br />
52 3 4<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 "25 2 x 2 at x 5 3 is 2 3 4.<br />
f. Let y 5 f(x).<br />
f(8) 5 4 1 8<br />
8 2 2 5 12<br />
6 5 2<br />
4 1 (8 1 h)<br />
f(8 1 h) 5<br />
(8 1 h) 2 2 5 12 1 h<br />
6 1 h<br />
1-8 <strong>Chapter</strong> 1: Introduction to Calculus
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 8 is<br />
f(8 1 h) 2 f(8)<br />
m 5 lim<br />
hS0 h<br />
12 1 h<br />
6 1 h<br />
5 lim<br />
2 2<br />
hS0 h<br />
12 1 h 2 12 2 2h<br />
5 lim<br />
? 1<br />
hS0 6 1 h h<br />
2h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 21<br />
6 1 0<br />
52 1 6<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 4 1 x at x 5 8 is 2 1 x 2 2<br />
6.<br />
12.<br />
y<br />
8<br />
–4<br />
6 1 h ? 1 h<br />
21<br />
6 1 h<br />
4<br />
–4<br />
0<br />
A<br />
4<br />
y 5 "25 2 x 2 S Semi-circle centre (0, 0)<br />
rad 5, y $ 0<br />
OA is a radius.<br />
4<br />
The slope of OA is 3.<br />
The slope of tangent is 2 3 4.<br />
13. Take values of x close to the point, then<br />
Dy<br />
determine<br />
Dx .<br />
14.<br />
Since the tangent is horizontal, the slope is 0.<br />
(3 1 h) 2 2 3(3 1 h) 1 1 2 1<br />
15. m 5 lim<br />
hS0<br />
h<br />
9 1 6h 1 h 2 2 9 2 3h<br />
5 lim<br />
hS0 h<br />
3h 1 h 2<br />
5 lim<br />
hS0 h<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
8<br />
x<br />
5 lim (3 1 h)<br />
hS0<br />
5 3<br />
The slope of the tangent is 3.<br />
y 2 1 5 3(x 2 3)<br />
3x 2 y 2 8 5 0<br />
(2 1 h) 2 2 7(2 1 h) 1 12 2 2<br />
16. m 5 lim<br />
hS0<br />
h<br />
4 1 4h 1 h 2 2 14 2 7h 1 10<br />
5 lim<br />
hS0<br />
h<br />
23h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim ( 2 3 1 h)<br />
hS0<br />
523<br />
The slope of the tangent is 23.<br />
When x 5 2, y 5 2.<br />
y 2 2 523(x 2 2)<br />
3x 1 y 2 8 5 0<br />
17. a. f(3) 5 9 2 12 1 1 522; (3, 22)<br />
b. f(5) 5 25 2 20 1 1 5 6; (5, 6)<br />
c. The slope of secant AB is<br />
m AB 5 6 2 (22)<br />
5 2 3<br />
5 8 2<br />
5 4<br />
The equation of the secant is<br />
y 2 y 1 5 m AB (x 2 x 1 )<br />
y 1 2 5 4(x 2 3)<br />
y 5 4x 2 14<br />
d. Calculate the slope of the tangent.<br />
f(x 1 h) 2 f(x)<br />
m 5 lim<br />
hS0 h<br />
(x 1 h) 2 2 4(x 1 h) 1 1 2 (x 2 2 4x 1 1)<br />
5 lim<br />
hS0<br />
h<br />
x 2 1 2xh 1 h 2 2 4x 2 4h 1 1 2 x 2 1 4x 2 1<br />
5 lim<br />
hS0<br />
h<br />
2xh 1 h 2 2 4h<br />
5 lim<br />
hS0 h<br />
5 lim (2x 1 h 2 4)<br />
hS0<br />
5 2x 1 0 2 4<br />
5 2x 2 4<br />
When x 5 3, the slope is 2(3) 2 4 5 2. So the<br />
equation of the tangent at A(3, 22) is<br />
y 2 y 1 5 m(x 2 x 1 )<br />
y 1 2 5 2(x 2 3)<br />
y 5 2x 2 8<br />
1-9
e. When x 5 5, the slope of the tangent is<br />
2(5) 2 4 5 6.<br />
So the equation of the tangent at B(5, 6) is<br />
y 2 y 1 5 m(x 2 x 1 )<br />
y 2 6 5 6(x 2 5)<br />
y 5 6x 2 24<br />
18. a.<br />
b.<br />
c.<br />
d.<br />
The slope is undefined.<br />
The slope is 0.<br />
P<br />
The slope is about –2.5.<br />
P<br />
P<br />
P<br />
20. C(t) 5 100t 2 1 400t 1 5000<br />
Slope at t 5 6<br />
Cr(t) 5 200t 1 400<br />
Cr(6) 5 1200 1 400 5 1600<br />
Increasing at a rate of 1600 papers per month.<br />
21. Point on f(x) 5 3x 2 2 4x tangent parallel to<br />
y 5 8x. Therefore, tangent line has slope 8.<br />
3(h 1 a) 2 2 4(h 1 a) 2 3(a 2 1 4a)<br />
m 5 lim<br />
5 8<br />
hS0<br />
h<br />
3h 2 1 6ah 2 4h<br />
lim<br />
5 8<br />
hS0 h<br />
6a 2 4 5 8<br />
a 5 2<br />
The point has coordinates (2, 4).<br />
22.<br />
y 5 1 3 x3 2 5x 2 4 x<br />
1<br />
3 (a 1 h)2 2 1 3 a3<br />
limaa 2 1 ah 1 1<br />
hS0<br />
3 h3 b 5 a 2<br />
5 lim 2<br />
hS0<br />
(a 1 h) 2 (2a)<br />
h<br />
2 4<br />
lim<br />
hS0<br />
525<br />
a 1 h 1 4 a<br />
4<br />
a(a 1 h) 5 4 a 2<br />
5 a 2 h 1 ah 2 1 1 3 h3<br />
524a<br />
1 4a 1 4h<br />
a(a 1 h)<br />
e.<br />
52 10 8<br />
52 5 4<br />
The slope is about 1.<br />
The slope is about 2 7 8.<br />
f. There is no tangent at this point.<br />
19. D(p) 5 20 p . 1 at (5, 10)<br />
"p 2 1 ,<br />
20<br />
!4 1 h 2 10<br />
m 5 lim<br />
hS0 h<br />
2 2 "4 1 h<br />
5 10 lim<br />
hS0<br />
5 10 lim<br />
hS0<br />
P<br />
h"4 1 h 3 2 1 "4 1 h<br />
2 1 "4 1 h<br />
4 2 4 2 h<br />
h"4 1 h(2 1 "4 1 h)<br />
a 4 2 5a 2 1 4 5 0<br />
( a 2 2 4)(a 2 2 1) 5 0<br />
a 562, a 561<br />
Points on the graph for horizontal tangents are:<br />
(22, 28 (21, 26 (1, 2 26 (2, 2 28 3 ), 3 ), 3 ), 3 ).<br />
23. y 5 x 2 and<br />
x 2 5 1 2 2 x2 y 5 1 2 2 x2<br />
x 2 5 1 4<br />
x 5 1 or x 52 1 2 2<br />
The points of intersection are<br />
P( 1 2, 1 4), Q(2 1 2, 1 4).<br />
Tangent to y 5 x 2 :<br />
(a 1 h) 2 2 a 2<br />
m 5 lim<br />
hS0 h<br />
2ah 1 h 2<br />
5 lim<br />
hS0 h<br />
5 2a.<br />
m 5 a 2 2 5 1 4 a 2 5 0<br />
1-10 <strong>Chapter</strong> 1: Introduction to Calculus
The slope of the tangent at a 5 1 2 is 1 5 m p ,<br />
at a 52 1 2 is 21 5 m q .<br />
Tangents to y 5 1 2 2 x 2 :<br />
S 1 2 2 (a 1 h) 2 T 2 S 1 2 2 a 2 T<br />
m 5 lim<br />
hS0<br />
h<br />
22ah 2 h 2<br />
5 lim<br />
hS0 h<br />
522a.<br />
The slope of the tangents at a 5 1 2 is 21 5 M p ;<br />
at a 52 1 2 is 1 5 M q<br />
m p M p 521 and m q M q 521<br />
Therefore, the tangents are perpendicular at the<br />
points of intersection.<br />
24. y 523x 3 2 2x, (21, 5)<br />
23(21 1 h) 3 2 2(21 1 h) 2 5<br />
m 5 lim<br />
hS0<br />
h<br />
23(21 1 3h 2 3h 2 1 h 3 ) 1 2 2 2h 2 5<br />
5 lim<br />
hS0<br />
h<br />
23(21 1 3h 2 3h 2 1 h 3 ) 1 2 2 2h 2 5<br />
5 lim<br />
hS0<br />
h<br />
3 2 9h 1 9h 2 2 3h 3 1 2 2 2h 2 5<br />
5 lim<br />
hS0<br />
h<br />
211h 1 9h 2 2 3h 3<br />
5 lim<br />
hS0 h<br />
5 lim(211 1 9h 2 3h 2 )<br />
hS0<br />
5211<br />
The slope of the tangent is 211.<br />
We want the line that is parallel to the tangent (i.e.<br />
has slope 211) and passes through (2, 2). Then,<br />
y 2 2 5211(x 2 2)<br />
y 5211x 1 24<br />
25. a. Let y 5 f(x).<br />
f(a) 5 4a 2 1 5a 2 2<br />
f(a 1 h) 5 4(a 1 h) 2 1 5(a 1 h) 2 2<br />
5 4(a 2 1 2ah 1 h 2 ) 1 5a 1 5h 2 2<br />
5 4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 a is<br />
f(a 1 h) 2 f(a)<br />
m 5 lim<br />
hS0 h<br />
5 lim c 4a2 1 8ah 1 4h 2 1 5a 1 5h 2 2<br />
hS0<br />
h<br />
2 (4a2 1 5a 2 2)<br />
d<br />
h<br />
5 lim c 4a2 1 8ah 1 4h 2 1 5a 1 5h 2 2<br />
hS0<br />
h<br />
1 24a2 2 5a 1 2<br />
d<br />
h<br />
8ah 1 4h 2 1 5h<br />
5 lim<br />
hS0 h<br />
5 lim (8a 1 4h 1 5)<br />
hS0<br />
5 8a 1 4(0) 1 5<br />
5 8a 1 5<br />
b. To be parallel, the point on the parabola and the<br />
line must have the same slope. So, first find the<br />
slope of the line. The line 10x 2 2y 2 18 5 0 can<br />
be rewritten as<br />
22y 5 18 2 10x<br />
18 2 10x<br />
y 5<br />
22<br />
y 529 1 5x<br />
y 5 5x 2 9<br />
So, the slope, m, of the line 10x 2 2y 2 18 5 0 is 5.<br />
To be parallel, the slope at a must equal 5. From<br />
part a., the slope of the tangent to the parabola at<br />
x 5 a is 8a 1 5.<br />
8a 1 5 5 5<br />
8a 5 0<br />
a 5 0<br />
Therefore, at the point (0, 22) the tangent line is<br />
parallel to the line 10x 2 2y 2 18 5 0.<br />
c. To be perpendicular, the point on the parabola<br />
and the line must have slopes that are negative<br />
reciprocals of each other. That is, their product must<br />
equal 21. So, first find the slope of the line. The<br />
line x 2 35y 1 7 5 0 can be rewritten as<br />
235y 52x 2 7<br />
y 5 2x 2 7<br />
235<br />
y 5 1<br />
35 x 1 7<br />
35<br />
1<br />
So, the slope, m, of the line x 2 35y 1 7 5 0 is 35.<br />
To be perpendicular, the slope at a must equal<br />
the negative reciprocal of the slope of the line<br />
x 2 35y 1 7 5 0. That is, the slope of a must equal<br />
235. From part a., the slope of the tangent to the<br />
parabola at x 5 a is 8a 1 5.<br />
8a 1 5 5235<br />
8a 5240<br />
a 525<br />
Therefore, at the point (25, 73) the tangent line is<br />
perpendicular to the line x 2 35y 1 7 5 0.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-11
1.3 Rates of Change, pp. 29–31<br />
1. v(t) 5 0 when t 5 0 or t 5 4.<br />
2. a. . Slope of the secant between the<br />
points (2, s(2)) and (9, s(9)).<br />
b. lim<br />
. Slope of the tangent at the<br />
hS0<br />
s(9) 2 s(2)<br />
7<br />
s(6 1 h) 2 s(6)<br />
h<br />
point (6, s(6)).<br />
"4 1 h 2 2<br />
3. lim . Slope of the tangent to the<br />
hS0 h<br />
function with equation y 5 !x at the point (4, 2).<br />
4. a. A and B<br />
b. greater; the secant line through these two points<br />
is steeper than the tangent line at B.<br />
c. y y = f(x)<br />
B C<br />
A D E<br />
x<br />
5. Speed is represented only by a number, not a<br />
direction.<br />
6. Yes, velocity needs to be described by a number<br />
and a direction. Only the speed of the school bus<br />
was given, not the direction, so it is not correct to<br />
use the word “velocity.”<br />
7. s(t) 5 320 2 5t 2 , 0 # t # 8<br />
a. Average velocity during the first second:<br />
s(1) 2 s(0)<br />
5 5 m>s;<br />
1<br />
third second:<br />
s(3) 2 s(2) 45 2 20<br />
5 5 25 m>s;<br />
1<br />
1<br />
eighth second:<br />
s(8) 2 s(7) 320 2 245<br />
5 5 75 m>s.<br />
1<br />
1<br />
b. Average velocity 3 # t # 8<br />
s(8) 2 s(3) 320 2 45<br />
5 5 275 5 55 m>s<br />
8 2 3 5 5<br />
c. s(t) 5 320 2 5t 2<br />
320 2 5(2 1 h) 2 2 (320 2 5(2) 2 )<br />
v(t) 5 lim<br />
hS0<br />
h<br />
24h 1 h 2<br />
5 5 lim<br />
hS0 h<br />
5220<br />
Velocity at t 5 2 is 20 m>s downward.<br />
8. s(t) 5 8t(t 1 2), 0 # t # 5<br />
a. i. from t 5 3 to t 5 4<br />
Average velocity<br />
s(4) 2 s(3)<br />
1<br />
5 32(6) 2 24(5)<br />
5 24(8 2 5)<br />
5 72 km>h<br />
ii. from t 5 3 to t 5 3.1<br />
s(3.1) 2 s(3)<br />
0.1<br />
126.48 2 120<br />
5<br />
0.1<br />
5 64.8 km>h<br />
iii. 3 # t # 3.01<br />
s(3.01) 2 s(3)<br />
0.01<br />
5 64.08 km>h<br />
b. Instantaneous velocity is approximately 64 km>h.<br />
c. At t 5 3<br />
s(t) 5 8t 2 1 16t<br />
v(t) 5 16t 1 16<br />
v(3) 5 48 1 16<br />
5 64 km>h<br />
N(t) 5 20t 2 t 2<br />
9. a.<br />
N(3) 2 N(2)<br />
1<br />
51 2 36<br />
5<br />
1<br />
5 15<br />
15 terms are learned between t 5 2 and t 5 3.<br />
20(2 1 h) 2 (2 1 h) 2 2 36<br />
b. lim<br />
hS0<br />
h<br />
40 1 20h 2 4 2 4h 2 h 2 2 36<br />
5 lim<br />
hS0<br />
h<br />
16h 2 h 2<br />
5 lim<br />
hS0 h<br />
5 lim (16 2 h)<br />
hS0<br />
5 16<br />
At t 5 2, the student is learning at a rate of 16 terms><br />
h.<br />
10. a. M in mg in 1 mL of blood t hours after the<br />
injection.<br />
M(t) 52 1 3 t2 1 t; 0 # t # 3<br />
Calculate the instantaneous rate of change when t 5 2.<br />
2 1<br />
lim<br />
3(2 1 h) 2 1 (2 1 h) 2 (2 4 3 1 2)<br />
hS0<br />
h<br />
2 4 3 2 4 3 h 2 1 3 h 2 1 2 1 h 1 4 3 2 2<br />
5 lim<br />
hS0<br />
h<br />
2 1 3<br />
5 lim<br />
h 2 1 3 h2<br />
hS0 h<br />
5 lim a2 1<br />
hS0 3 2 1 3 hb<br />
52 1 3<br />
1-12 <strong>Chapter</strong> 1: Introduction to Calculus
Rate of change is 2 1 3 mg><br />
h.<br />
b. Amount of medicine in 1 mL of blood is being<br />
dissipated throughout the system.<br />
s<br />
11. t 5 Å 5<br />
Calculate the instantaneous rate of change when<br />
s 5 125.<br />
125 1 h 125<br />
Ä 5 2 Ä 5<br />
lim<br />
hS0 h<br />
125 1 h<br />
2 5<br />
Ä 5<br />
5 lim<br />
hS0 h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0 ≥<br />
5 lim ≥<br />
hS0<br />
5 lim<br />
hS0 125 1 h<br />
5a Ä 5<br />
1<br />
5<br />
125<br />
5a Ä 5 1 5b<br />
1<br />
5<br />
5(5 1 5)<br />
5 1<br />
50<br />
At s 5 125, rate of change of time with respect to<br />
1<br />
height is 50 s>m.<br />
12. T(h) 5 60<br />
h 1 2<br />
Calculate the instantaneous rate of change when<br />
h 5 3.<br />
lim<br />
kS0<br />
60<br />
(3 1 k) 1 2 2 60<br />
(3 1 2)<br />
5 lim<br />
kS0<br />
125 1 h<br />
≥ Ä 5<br />
h<br />
125 1 h<br />
5<br />
ha Ä<br />
125 1 h<br />
5<br />
125 1 h 2 125<br />
5<br />
125 1 h<br />
ha Ä 5<br />
1<br />
k<br />
60<br />
5 1 k 2 12<br />
k<br />
2 5<br />
2 25<br />
¥<br />
1 5b<br />
¥<br />
1 5b<br />
1 5b<br />
125 1 h<br />
? Ä 5<br />
125 1 h<br />
Ä 5<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1 5<br />
¥<br />
1 5<br />
5 lim<br />
kS0<br />
60 60 1 12k<br />
2<br />
5 1 k 5 1 k<br />
k<br />
212k<br />
5 lim<br />
kS0 k(5 1 k)<br />
212<br />
5 lim<br />
kS0 (5 1 k)<br />
52 12 5<br />
12<br />
Temperature is decreasing at C km.<br />
13. h 5 25t 2 5 ° ><br />
2 100t 1 100<br />
When h 5 0, 25t 2 2 100t 1 100 5 0<br />
t 2 2 4t 1 4 5 0<br />
( t 2 2) 2 5 0<br />
t 5 2<br />
Calculate the instantaneous rate of change when t 5 2.<br />
25(2 1 h) 2 2 100(2 1 h) 1 100 2 0<br />
lim<br />
hS0<br />
h<br />
100 1 100h 1 25h 2 2 200 2 100h 1 100<br />
5 lim<br />
hS0<br />
h<br />
25h 2<br />
5 lim<br />
hS0 h<br />
5 lim 25h<br />
hS0<br />
5 0<br />
It hit the ground in 2 s at a speed of 0 m><br />
s.<br />
14. Sale of x balls per week:<br />
P(x) 5 160x 2 x 2 dollars.<br />
a. P(40) 5 160(40) 2 (40) 2<br />
5 4800<br />
Profit on the sale of 40 balls is $4800.<br />
b. Calculate the instantaneous rate of change when<br />
x 5 40.<br />
160(40 1 h) 2 (40 1 h) 2 2 4800<br />
lim<br />
hS0<br />
h<br />
6400 1 160h 2 1600 2 80h 2 h 2 2 4800<br />
5 lim<br />
hS0<br />
h<br />
80h 2 h 2<br />
5 lim<br />
hS0 h<br />
5 lim (80 2 h)<br />
hS0<br />
5 80<br />
Rate of change of profit is $80 per ball.<br />
c.<br />
Rate of change of profit is positive when the sales<br />
level is less than 80.<br />
1-13
15. a. f(x) 52x 2 1 2x 1 3; (22, 25) For the year 2005, x 5 2005 2 1982 5 23. Hence,<br />
f(x) 2 f(22)<br />
the rate at which the average annual salary is changing<br />
lim<br />
xS22 x 1 2<br />
in 2005 is<br />
2x 2 1 2x 1 3 1 5<br />
P r(23) 5 64 2 17.8(23) 1 2.85(23) 2 5<br />
5 lim<br />
xS22 x 1 2<br />
$1 162 250> years since 1982<br />
2 (x 2 2 2x 2 8)<br />
17. s(t) 5 3t 2<br />
5 lim<br />
xS22 x 1 2<br />
a. The distance travelled from 0 s to 5 s is<br />
(x 2 4)(x 1 2)<br />
s(5) 5 3(5) 2 5 75 m<br />
52lim<br />
b. s(10) 5 3(10)<br />
xS22 x 1 2<br />
2 5 300 m<br />
The rate at which the avalanche is moving from 0 s<br />
52lim (x 2 4)<br />
xS22<br />
to 10 s is<br />
5 6<br />
Ds<br />
b. f(x) 5<br />
x x 5 2<br />
x 2 1 ,<br />
Dt 5 300 2 0<br />
10 2 0<br />
5 30 ms ><br />
x<br />
x 2 1<br />
lim<br />
2 2<br />
c. Calculate the instantaneous rate of change when<br />
xS2 x 2 2<br />
t 5 10.<br />
x 2 2x 1 2<br />
3(10 1 h) 2 2 300<br />
5 lim<br />
lim<br />
hS0<br />
xS2 (x 2 1)(x 2 2)<br />
h<br />
2 (x 2 2)<br />
300 1 60h 1 3h 2 2 300<br />
5 lim<br />
5 lim<br />
hS0<br />
xS2 (x 2 1)(x 2 2)<br />
h<br />
521<br />
60h 1 3h 2<br />
5 lim<br />
c. f(x) 5 !x 1 1, x 5 24<br />
hS0 h<br />
f(x) 2 f(24)<br />
5 lim (60 1 3h)<br />
hS0<br />
5 lim<br />
xS24<br />
5 60<br />
x 2 24<br />
!x 1 1 2 5<br />
5 lim<br />
? !x 1 1 1 5<br />
xS24 x 2 24 !x 1 1 1 5<br />
x 2 24<br />
5 lim<br />
xS24 (x 2 24)( !x 1 1 1 5)<br />
5 1<br />
10<br />
16. S(x) 5 246 1 64x 2 8.9x 2 1 0.95x 3<br />
Calculate the instantaneous rate of change.<br />
S(x 1 h) 2 S(x)<br />
5 lim<br />
hS0 h<br />
5 64 2 17.8x 1 2.85x 2<br />
At 10 s the avalanche is moving at 60 m><br />
s.<br />
d. Set s(t) 5 600:<br />
3t 2 5 600<br />
t 2 5 200<br />
t 5610!2<br />
Since t $ 0, t 5 10!2 8 14 s.<br />
246 1 64(x 1 h) 2 8.9(x 1 h) 2 1 0.95(x 1 h) 3 2 (246 2 64x 2 8.9x 2 1 0.95x 3 )<br />
5 lim<br />
hS0<br />
h<br />
246 2 246 1 64(x 1 h 2 x) 2 8.9(x 2 1 2xh 1 h 2 2 x 2 ) 1 0.95(x 3 1 3x 2 h 1 3xh 2 1 h 3 2 x 3 )<br />
5 lim<br />
hS0<br />
h<br />
64h 2 8.9(2xh 1 h 2 ) 1 0.95(3x 2 h 1 3xh 2 1 h 3 )<br />
5 lim<br />
hS0<br />
h<br />
5 lim 364 2 8.9(2x 1 h) 1 0.95(3x 2 1 3xh 1 h 2 )4<br />
hS0<br />
5 64 2 8.9(2x 1 0) 1 0.95 33x 2 1 3x(0) 1 (0) 2 4<br />
1-14 <strong>Chapter</strong> 1: Introduction to Calculus
18. The coordinates of the point are . The slope<br />
of the tangent is<br />
. The equation of the tangent<br />
is y 2 1 or y 52 1 The<br />
a 2x 1 2 a 521 (x 2 a)<br />
a a .<br />
2<br />
intercepts are a0, 2 and (22a, 0). The tangent line<br />
a b<br />
and the axes form a right triangle with legs of length<br />
2<br />
1<br />
and 2a. The area of the triangle is<br />
2 a2 b (2a) 5 2.<br />
a<br />
a<br />
19. C(x) 5 F 1 V(x)<br />
C(x 1 h) 5 F 1 V(x 1 h)<br />
Rate of change of cost is<br />
C(x 1 h) 2 C(x)<br />
lim<br />
xSR h<br />
V(x 1 h) 2 V(x)<br />
5 lim<br />
h,<br />
xSh h<br />
which is independent of F (fixed costs).<br />
20. A(r) 5pr 2<br />
Rate of change of area is<br />
A(r 1 h) 2 A(r)<br />
lim<br />
hS0 h<br />
p(r 1 h) 2 2pr 2<br />
5 lim<br />
hS0 h<br />
(r 1 h 2 r)(r 1 h 1 r)<br />
5p lim<br />
hS0<br />
h<br />
5 2pr<br />
r 5 100 m<br />
Rate is 200p m 2 > m.<br />
21. Cube of dimensions x by x by x has volume<br />
V 5 x 3 . Surface area is 6x 2 .<br />
Vr(x) 5 3x 2 5 1 surface area.<br />
2<br />
22. a. The surface area of a sphere is given by<br />
A(r) 5 4pr 2 .<br />
The question asks for the instantaneous rate of<br />
change of the surface when r 5 10. This is<br />
A(10 1 h) 2 A(10)<br />
lim<br />
hS0 h<br />
4p(10 1 h) 2 2 4p(10) 2<br />
5 lim<br />
hS0<br />
h<br />
4p(100 1 20h 1 h 2 ) 2 4p(100)<br />
5 lim<br />
hS0<br />
h<br />
400p 180ph 1 4ph 2 2 400p<br />
5 lim<br />
hS0<br />
h<br />
80ph 1 4ph 2<br />
5 lim<br />
hS0 h<br />
5 lim (80p 14ph)<br />
hS0<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
5 80p 14p(0)<br />
5 80p<br />
Therefore, the instantaneous rate of change of<br />
the surface area of a spherical balloon as it is<br />
inflated when the radius reaches 10 cm is<br />
80p cm 2 > unit of time.<br />
b. The volume of a sphere is given by V(r) 5 4 3pr 3 .<br />
The question asks for the instantaneous rate of<br />
change of the volume when r 5 5.<br />
Note that the volume is deflating. So, find the rate<br />
of the change of the volume when r 5 5 and then<br />
make the answer negative to symbolize a deflating<br />
spherical balloon.<br />
V(5 1 h) 2 V(5)<br />
lim<br />
hS0 h<br />
4<br />
5 lim<br />
3p(5 1 h) 3 2 4 3p(5) 3<br />
hS0 h<br />
Using the binomial formula to expand<br />
(5 1 h) 3 (or one could simply expand using<br />
algebra), the limit is<br />
4<br />
5 lim<br />
3p(h 3 1 15h 2 1 75h 1 125) 2 4 3 p(5) 3<br />
hS0<br />
h<br />
4<br />
5 lim<br />
3ph 3 1 20ph 2 1 100ph 1 4 3p(125)<br />
hS0<br />
h<br />
2 4 3p(125)<br />
h<br />
4<br />
5 lim<br />
3ph 3 1 20ph 2 1 100ph<br />
hS0<br />
h<br />
5 lim a 4 3 ph2 1 20ph 1 100pb<br />
hS0<br />
5 4 3 p(0)2 1 20p(0) 1 100p<br />
5 100p<br />
Because the balloon is deflating, the instantaneous rate<br />
of change of the volume of the spherical balloon when<br />
the radius reaches 5 cm is 2100p cm 3 > unit of time.<br />
Mid-<strong>Chapter</strong> Review pp. 32–33<br />
1. a. Corresponding conjugate: !5 1 !2.<br />
( !5 2 !2)(!5 1 !2)<br />
5 ( !25 1 !10 2 !10 2 !4)<br />
5 5 2 2<br />
5 3<br />
b. Corresponding conjugate: 3!5 2 2!2.<br />
(3!5 1 2!2)(3!5 2 2!2)<br />
5 (9!25 2 6!10 1 6!10 2 4!4)<br />
5 9(5) 2 4(2)<br />
5 45 2 8<br />
5 37<br />
2 1 a 2 aa, 1 a b 1-15
c. Corresponding conjugate: 9 2 2!5.<br />
(9 1 2!5)(9 2 2!5)<br />
5 (81 2 18!5 1 18!5 2 4!25)<br />
5 81 2 4(5)<br />
5 81 2 20<br />
5 61<br />
d. Corresponding conjugate: 3!5 1 2!10.<br />
(3!5 2 2!10)(3!5 1 2!10)<br />
5 (9!25 1 6!50 2 6!50 2 4!100)<br />
5 9(5) 2 4(10)<br />
5 45 2 40<br />
5 5<br />
6 1 !2<br />
2. a. ? !3<br />
!3 !3<br />
6!3 1 !6<br />
5<br />
!9<br />
6!3 1 !6<br />
5<br />
3<br />
2!3 1 4<br />
b. ? !3<br />
!3 !3<br />
2!9 1 4!3<br />
5<br />
!9<br />
5 6 1 4!3<br />
3<br />
5<br />
c.<br />
!7 2 4 ? !7 1 4<br />
!7 1 4<br />
5(!7 1 4)<br />
5<br />
!49 1 4!7 2 4!7 2 16<br />
5(!7 1 4)<br />
5<br />
7 2 16<br />
5(!7 1 4)<br />
52<br />
9<br />
2!3<br />
d.<br />
!3 2 2 ? !3 1 2<br />
!3 1 2<br />
2!9 1 4!3<br />
5<br />
!9 1 2!3 2 2!3 2 4<br />
5 6 1 4!3<br />
3 2 4<br />
5 6 1 4!3<br />
21<br />
522(3 1 2!3)<br />
5!3<br />
e.<br />
2!3 1 4 ? 2!3 2 4<br />
2!3 2 4<br />
10!9 2 20!3<br />
5<br />
4!9 2 8!3 1 8!3 2 16<br />
30 2 20!3<br />
5<br />
12 2 16<br />
30 2 20!3<br />
5<br />
24<br />
10!3 2 15<br />
5<br />
2<br />
3!2<br />
f.<br />
2!3 2 5 ? 2!3 1 5<br />
2!3 1 5<br />
3!2(2!3 1 5)<br />
5<br />
4!9 1 10!3 2 10!3 2 25<br />
3!2(2!3 1 5)<br />
5<br />
4(3) 2 25<br />
3!2(2!3 1 5)<br />
5<br />
12 2 25<br />
3!2(2!3 1 5)<br />
5<br />
213<br />
3!2(2!3 1 5)<br />
52<br />
13<br />
!2<br />
3. a.<br />
5 ? !2<br />
!2<br />
5 !4<br />
5!2<br />
5 2<br />
5!2<br />
!3<br />
b.<br />
6 1 !2 ? !3<br />
!3<br />
!9<br />
5<br />
!3(6 1 !2)<br />
3<br />
5<br />
!3(6 1 !2)<br />
!7 2 4<br />
c. ? !7 1 4<br />
5 !7 1 4<br />
!49 1 4!7 2 4!7 2 16<br />
5<br />
5(!7 1 4)<br />
5 7 2 16<br />
5(!7 1 4)<br />
9<br />
52<br />
5(!7 1 4)<br />
2!3 2 5<br />
d. ? 2!3 1 5<br />
3!2 2!3 1 5<br />
4!9 1 10!3 2 10!3 2 25<br />
5<br />
3!2(2!3 1 5)<br />
4(3) 2 25<br />
5<br />
3!2(2!3 1 5)<br />
12 2 25<br />
13<br />
5<br />
52<br />
3!2(2!3 1 5) 3!2(2!3 1 5)<br />
1-16 <strong>Chapter</strong> 1: Introduction to Calculus
!3 2 !7 !3 1 !7<br />
e.<br />
?<br />
4 !3 1 !7<br />
!9 1 !21 2 !21 2 !49<br />
5<br />
4(!3 1 !7)<br />
3 2 7<br />
5<br />
4(!3 1 !7)<br />
4<br />
52<br />
4(!3 1 !7)<br />
1<br />
52<br />
( !3 1 !7)<br />
2!3 1 !7 2!3 2 !7<br />
f.<br />
?<br />
5 2!3 2 !7<br />
4!9 2 2!21 1 2!21 2 !49<br />
5<br />
5(2!3 2 !7)<br />
5 4(3) 2 7<br />
5(2!3 2 !7)<br />
12 2 7<br />
5<br />
5(2!3 2 !7)<br />
1<br />
5<br />
(2!3 2 !7)<br />
4. a.<br />
2<br />
3 x 1 y 2 6 5 0<br />
b.<br />
y 2 7 5 1(x 2 2)<br />
y 2 7 5 x 2 2<br />
2x 1 y 2 5 5 0<br />
x 2 y 1 5 5 0<br />
c.<br />
y 2 6 5 4x 2 8<br />
24x 1 y 1 2 5 0<br />
4x 2 y 2 2 5 0<br />
d.<br />
m 52 2 3 ;<br />
y 2 6 52 2 (x 2 0)<br />
3<br />
y 2 6 52 2 3 x<br />
m 5 11 2 7<br />
6 2 2 5 4 4 5 1<br />
m 5 4<br />
y 2 6 5 4(x 2 2)<br />
m 5 1 5<br />
y 2 (22) 5 1 (x 2 (21))<br />
5<br />
y 1 2 5 1 5 x 1 1 5<br />
2 1 5 x 1 y 1 10<br />
5 2 1 5 5 0<br />
2 1 5 x 1 y 1 9 5 5 0<br />
1<br />
5 x 2 y 2 9 5 5 0<br />
x 2 5y 2 9 5 0<br />
5. The slope of PQ is<br />
f(1 1 h) 2 (21)<br />
m 5 lim<br />
hS0 (1 1 h) 2 1<br />
2 (1 1 h) 2 1 1<br />
5 lim<br />
hS0 h<br />
2 (1 1 2h 1 h 2 ) 1 1<br />
5 lim<br />
hS0 h<br />
21 2 2h 2 h 2 1 1<br />
5 lim<br />
hS0 h<br />
22h 2 h 2<br />
5 lim<br />
hS0 h<br />
5 lim (22 2 h)<br />
hS0<br />
522 2 (0)<br />
522<br />
So, the slope of PQ with f(x) 52x 2 is 22.<br />
6. a. Unlisted y-coordinates for Q are found by<br />
substituting the x-coordinates into the given function.<br />
The slope of the line PQ with the given points is<br />
given by the following: Let P 5 (x 1 , y 1 ) and<br />
Then, the slope 5 m 5 y 2 y 2 1<br />
Q 5 (y 1 , y 2 ).<br />
.<br />
x 2 2 x 1<br />
P Q Slope of Line PQ<br />
(21, 1) (22, 6) 25<br />
(21, 1) (21.5, 3.25) 24.5<br />
(21, 1) (21.1, 1.41) 24.1<br />
(21, 1) (21.01, 1.040 1) 24.01<br />
(21, 1) (21.001, 1.004 001) 24.001<br />
P Q Slope of Line PQ<br />
(21, 1) (0, 22) 23<br />
(21, 1) (20.5, 20.75) 23.5<br />
(21, 1) (20.9, 0.61) 23.9<br />
(21, 1) (20.99, 0.9601) 23.99<br />
(21, 1) (20.999, 0.996 001) 23.999<br />
b. The slope from the right and from the left appear<br />
to approach 24. The slope of the tangent to the<br />
graph of f(x) at point P is about 24.<br />
c. With the points P 5 (21, 1) and<br />
Q 5 (21 1 h, f(21 1 h)), the slope, m, of PQ is<br />
the following:<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-17
m 5 y 2 2 y 1<br />
x 2 2 x 1<br />
c.<br />
5 3(21 1 h)2 2 2(21 1 h) 2 24 2 (1)<br />
(21 1 h) 2 (21)<br />
5 1 2 2h 1 h2 1 2 2 2h 2 2 2 1<br />
21 1 h 1 1<br />
5 h2 2 4h<br />
h<br />
5 h 2 4<br />
d. The slope of the tangent is lim f(x).<br />
hS0<br />
In this case, as h goes to zero, h 2 4 goes to<br />
h 2 4 5 0 2 4 524. The slope of the tangent to<br />
the graph of f(x) at the point P is 24.<br />
e. The answers are equal.<br />
7. a.<br />
f(23 1 h) 2 f(23)<br />
m 5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
h 2 2 3h<br />
5 lim<br />
hS0 h<br />
5 lim (h 2 3)<br />
hS0<br />
5 0 2 3<br />
523<br />
y 5 f(x) 5 4<br />
x 2 2<br />
f(6 1 h) 2 f(6)<br />
m 5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
h<br />
4<br />
6 1 h 2 2 2 4<br />
6 2 2<br />
h<br />
4<br />
h 1 4<br />
5 lim<br />
2 4 4<br />
hS0 h<br />
4<br />
h 1 4<br />
5 lim<br />
2 1<br />
hS0 h<br />
4 2 (h 1 4)<br />
5 lim a b 1<br />
hS0 h 1 4 h<br />
h<br />
3(23 1 h) 2 1 3(23 1 h) 2 54 2 3(23) 2 1 3(23) 2 54<br />
h<br />
9 2 6h 1 h 2 2 9 1 3h 2 5 2 (9 2 9 2 5)<br />
h<br />
h 2 2 3h 2 5 2 (25)<br />
h<br />
5 lim a 2h<br />
hS0<br />
5 lim<br />
hS0<br />
5 21<br />
0 1 4<br />
h 1 4 b 1 h<br />
21<br />
h 1 4<br />
b. y 5 f(x) 5 1 x<br />
f( 1 3 1 h) 2 f( 1<br />
m 5 lim<br />
3)<br />
hS0 h<br />
1<br />
1<br />
3 1 h 2 1 1<br />
3<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
a 2h<br />
1<br />
9 1 1 3h b 1 h<br />
21<br />
1<br />
9 1 1 3h<br />
5 21<br />
1<br />
9 1 1 3(0)<br />
529<br />
h<br />
( 1 3) 2 ( 1 3 1 h)<br />
1<br />
3( 1 3 1 h)<br />
h<br />
d.<br />
52 1 4<br />
f(5 1 h) 2 f(5)<br />
m 5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
h<br />
!5 1 h 1 4 2 !5 1 4<br />
h<br />
!9 1 h 2 !9<br />
h<br />
!9 1 h 2 3<br />
h<br />
!9 1 h 2 3<br />
? !9 1 h 1 3<br />
h !9 1 h 1 3<br />
9 1 h 1 3!9 1 h 2 3!9 1 h 2 9<br />
h( !9 1 h 1 3)<br />
h<br />
h( !9 1 h 1 3)<br />
1<br />
!9 1 h 1 3<br />
1-18 <strong>Chapter</strong> 1: Introduction to Calculus
5<br />
1<br />
!9 1 0 1 3<br />
5 1 6<br />
s(t) 5 6t(t 1 1) 5 6t 2 1 6t<br />
8.<br />
s(3) 2 s(2)<br />
a. i. average velocity 5<br />
3 2 2<br />
5 36(3) 2 1 6(3)42 36(2) 2 1 6(2)4<br />
5 6(9) 1 18 2 (24 1 12)<br />
5 54 1 18 2 36<br />
5 36 km><br />
h<br />
s(2.1) 2 s(2)<br />
ii. average velocity 5<br />
2.1 2 2<br />
5 36(2.1)2 1 6(2.1)4 2 36(2) 2 1 6(2)4<br />
0.1<br />
326.46 1 12.64 2 324 1 124<br />
5<br />
0.1<br />
39.06 2 36<br />
5<br />
0.1<br />
5 3.06<br />
0.1<br />
5 30.6 km><br />
h<br />
s(2.01) 2 s(2)<br />
iii. average velocity 5<br />
2.01 2 2<br />
5 36(2.01)2 1 6(2.01)4 2 36(2) 2 1 6(2)4<br />
0.01<br />
5 324.2406 1 12.064 2 36(2)2 1 6(2)4<br />
0.01<br />
36.3006 2 324 1 124<br />
5<br />
0.01<br />
36.3006 2 36<br />
5<br />
0.01<br />
5 0.3006<br />
0.01<br />
5 30.06 km><br />
h<br />
b. At the time t 5 2, the velocity of the car appears<br />
to approach 30 km><br />
h.<br />
f(2 1 h) 2 f(2)<br />
c. average velocity 5<br />
(2 1 h) 2 (2)<br />
5 36(2 1 h)2 1 6(2 1 h)4 2 36(2) 2 1 6(2)4<br />
h<br />
5 36(4 1 4h 1 h2 ) 1 12 1 6h4 2 324 1 124<br />
h<br />
5 324 1 24h 1 6h2 1 12 1 6h4 2 36<br />
h<br />
5 6h2 1 30h 1 36 2 36<br />
h<br />
5 6h2 1 30h<br />
h<br />
5 (6h 1 30) km><br />
h<br />
d. When t 5 2, the velocity is the limit as h<br />
approaches 0.<br />
velocity 5 lim (6h 1 30)<br />
hS0<br />
5 6(0) 1 30<br />
5 30<br />
Therefore, when t 5 2 the velocity is 30 km><br />
h.<br />
9. a. The instantaneous rate of change of f(x) with<br />
respect to x at x 5 2 is given by<br />
f(2 1 h) 2 f(2)<br />
lim<br />
hS0 h<br />
35 2 (2 1 h) 2 4 2 35 2 (2) 2 4<br />
5 lim<br />
hS0<br />
h<br />
5 2 (4 1 4h 1 h 2 ) 2 1<br />
5 lim<br />
hS0<br />
h<br />
5 2 4 2 4h 2 h 2 2 1<br />
5 lim<br />
hS0 h<br />
2h 2 2 4h<br />
5 lim<br />
hS0 h<br />
5 lim (2h 2 4)<br />
hS0<br />
52(0) 2 4<br />
524<br />
b. The instantaneous rate of change of f(x) with<br />
respect to x at x 5 1 2 is given by<br />
f( 1 2 1 h) 2 f( 1<br />
lim<br />
2)<br />
hS0 h<br />
3<br />
1<br />
2 1 h 2 3 1<br />
2<br />
5 lim<br />
hS0 h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
26h<br />
1<br />
5 26<br />
1<br />
2 1 0<br />
5212<br />
3<br />
1<br />
2 1 h 2 6<br />
h<br />
3 2 6( 1 2 1 h)<br />
1<br />
2 1 h<br />
3 2 3 2 6h<br />
1<br />
2 1 h<br />
2 1 h ? 1 h<br />
26<br />
1<br />
2 1 h<br />
? 1 h<br />
? 1 h<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-19
10. a. The average rate of change of V(t) with<br />
respect to t during the first 20 minutes is given by<br />
f(20) 2 f(0)<br />
20 2 0<br />
5 350(30 2 20)2 4 2 350(30 2 0) 2 4<br />
20<br />
5000 2 45 000<br />
5<br />
20<br />
40 000<br />
52<br />
20<br />
522000 Lmin ><br />
b. The rate of change of V(t) with respect to t at the<br />
time t 5 20 is given by<br />
f(20 1 h) 2 f(20)<br />
lim<br />
hS0 h<br />
350(30 2 (20 1 h)) 2 4 2 350(30 2 20) 2 4<br />
5 lim<br />
hS0<br />
h<br />
350(10 2 h) 2 4 2 350(10) 2 4<br />
5 lim<br />
hS0<br />
h<br />
350(100 2 20h 1 h 2 )4 2 350(100)4<br />
5 lim<br />
hS0<br />
h<br />
5000 2 1000h 1 50h 2 2 5000<br />
5 lim<br />
hS0<br />
h<br />
50h 2 2 1000h<br />
5 lim<br />
hS0 h<br />
5 lim 50h 2 1000<br />
hS0<br />
5 50(0) 2 1000<br />
521000 Lmin ><br />
11. a. Let y 5 f(x).<br />
f(4) 5 (4) 2 1 (4) 2 3 5 16 1 1 5 17<br />
f(4 1 h) 5 (4 1 h) 2 1 (4 1 h) 2 3<br />
5 16 1 8h 1 h 2 1 h 1 1<br />
5 h 2 1 9h 1 17<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 4 is<br />
f(4 1 h) 2 f(4)<br />
m 5 lim<br />
hS0 h<br />
h 2 1 9h 1 17 2 (17)<br />
5 lim<br />
hS0 h<br />
h 2 1 9h<br />
5 lim<br />
hS0 h<br />
5 lim (h 1 9)<br />
hS0<br />
5 0 1 9<br />
5 9<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 x 2 1 x 2 3 at x 5 4 is 9.<br />
So an equation of the tangent at x 5 4 is given by<br />
y 2 17 5 9(x 2 4)<br />
y 2 17 5 9x 2 36<br />
29x 1 y 2 17 1 36 5 0<br />
29x 1 y 1 19 5 0<br />
b. Let y 5 f(x).<br />
f(22) 5 2(22) 2 2 7 5 2(4) 2 7 5 1<br />
f(22 1 h) 5 2(22 1 h) 2 2 7<br />
5 2(4 2 4h 1 h 2 ) 2 7<br />
5 8 2 8h 1 2h 2 2 7<br />
5 2h 2 2 8h 1 1<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 4 is<br />
f(22 1 h) 2 f(22)<br />
m 5 lim<br />
hS0 h<br />
2h 2 2 8h 1 1 2 (1)<br />
5 lim<br />
hS0 h<br />
2h 2 2 8h<br />
5 lim<br />
hS0 h<br />
5 lim (2h 2 8)<br />
hS0<br />
5 2(0) 2 8<br />
528<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 2x 2 2 7 at x 522 is 28.<br />
So an equation of the tangent at x 522<br />
is given by<br />
y 2 1 528(x 2 (22))<br />
y 2 1 528x 2 16<br />
8x 1 y 2 1 1 16 5 0<br />
8x 1 y 1 15 5 0<br />
c. f(21) 5 3(21) 2 1 2(21) 2 5 5 3 2 2 2 5<br />
524<br />
f(21 1 h) 5 3(21 1 h) 2 1 2(21 1 h) 2 5<br />
5 3(1 2 2h 1 h 2 ) 2 2 1 2h 2 5<br />
5 3 2 6h 1 3h 2 2 7 1 2h<br />
5 3h 2 2 4h 2 4<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 4 is<br />
f(21 1 h) 2 f(21)<br />
m 5 lim<br />
hS0 h<br />
3h 2 2 4h 2 4 2 (24)<br />
5 lim<br />
hS0 h<br />
3h 2 2 4h<br />
5 lim<br />
hS0 h<br />
5 lim (3h 2 4)<br />
hS0<br />
5 3(0) 2 4<br />
524<br />
1-20 <strong>Chapter</strong> 1: Introduction to Calculus
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 3x 2 1 2x 2 5 at x 521 is 24.<br />
So an equation of the tangent at x 524 is given by<br />
y 2 (24) 524(x 2 (21))<br />
y 1 4 524(x 1 1)<br />
y 1 4 524x 2 4<br />
4x 1 y 1 4 1 4 5 0<br />
4x 1 y 1 8 5 0<br />
d. f(1) 5 5(1) 2 2 8(1) 1 3 5 5 2 8 1 3 5 0<br />
f(1 1 h) 5 5(1 1 h) 2 2 8(1 1 h) 1 3<br />
5 5(1 1 2h 1 h 2 ) 2 8 2 8h 1 3<br />
5 5 1 10h 1 5h 2 2 5 2 8h<br />
5 5h 2 1 2h<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 1 is<br />
f(1 1 h) 2 f(1)<br />
m 5 lim<br />
hS0 h<br />
5h 2 1 2h 2 (0)<br />
5 lim<br />
hS0 h<br />
5 lim (5h 1 2)<br />
hS0<br />
5 5(0) 1 2<br />
5 2<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 5x 2 2 8x 1 3 at x 5 1 is 2.<br />
So an equation of the tangent at x 5 1 is given by<br />
y 2 0 5 2(x 2 1)<br />
y 5 2x 2 2<br />
22x 1 y 1 2 5 0<br />
12. a. Using the limit of the difference quotient, the<br />
slope of the tangent at x 525 is<br />
f(25 1 h) 2 f(25)<br />
m 5 lim<br />
hS0 h<br />
25 1 h<br />
5 lim a<br />
hS0 25 1 h 1 3 2 25<br />
25 1 3 b ? 1 h<br />
5 lim a 25 1 h<br />
hS0 22 1 h 2 5 2 b ? 1 h<br />
210 1 2h 2 (210 1 5h)<br />
5 lim a b ? 1<br />
hS0 24 1 2h<br />
h<br />
210 1 2h 1 10 2 5h<br />
5 lim a b ? 1<br />
hS0 24 1 2h h<br />
23h<br />
5 lim a<br />
hS0 24 1 2h b ? 1 h<br />
23<br />
5 lim a<br />
hS0 24 1 2h b<br />
23<br />
5<br />
24 1 2(0)<br />
5 3 4<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
Therefore, the slope of the tangent to<br />
3<br />
f(x) 5<br />
x at x 525 is<br />
x 1 3<br />
4.<br />
So an equation of the tangent at x 5 3 4 is given by<br />
y 2 5 2 5 3 (x 2 (25))<br />
4<br />
y 2 5 2 5 3 4 x 1 15<br />
4<br />
2 3 4 x 1 y 2 10<br />
4 2 15<br />
4 5 0<br />
2 3 4 x 1 y 2 25<br />
4 5 0<br />
23x 1 4y 2 25 5 0<br />
b. Using the limit of the difference quotient, the<br />
slope of the tangent at x 521 is<br />
f(21 1 h) 2 f(21)<br />
m 5 lim<br />
hS0 h<br />
2(21 1 h) 1 5<br />
5 lim a<br />
hS0 5(21 1 h) 2 1 2 2(21) 1 5<br />
5(21) 2 1 b ? 1 h<br />
22 1 2h 1 5<br />
5 lim a<br />
hS0 25 1 5h 2 1 2 22 1 5<br />
25 2 1 b ? 1 h<br />
5 lim a 2h 1 3<br />
hS0 5h 2 6 2 3<br />
26 b ? 1 h<br />
5 lim a 2h 1 3<br />
hS0 5h 2 6 1 1 2 b ? 1 h<br />
4h 1 6 1 5h 2 6<br />
5 lim a b ? 1<br />
hS0 10h 2 12 h<br />
9h<br />
5 lim a<br />
hS0 10h 2 12 b ? 1 h<br />
9<br />
5 lim a<br />
hS0 10h 2 12 b<br />
9<br />
5<br />
10(0) 2 12<br />
52 9 12<br />
52 3 4<br />
Therefore, the slope of the tangent to<br />
f(x) 5 2x 1 5 at x 521 is 2 3 5x 2 1<br />
4.<br />
So an equation of the tangent at x 52 3 4 is given by<br />
y 2 a2 1 2 b 523 (x 2 (21))<br />
4<br />
y 1 1 2 523 4 x 2 3 4<br />
4y 1 2 523x 2 3<br />
3x 1 4y 1 2 1 3 5 0<br />
3x 1 4y 1 5 5 0<br />
1-21
1.4 The Limit of a Function,<br />
pp. 37–39<br />
27<br />
1. a.<br />
99<br />
b. p<br />
2. One way to find a limit is to evaluate the function<br />
for values of the independent variable that get<br />
progressively closer to the given value of the<br />
independent variable.<br />
3. a. A right-sided limit is the value that a<br />
function gets close to as the values of the<br />
independent variable decrease and get close<br />
to a given value.<br />
b. A left-sided limit is the value that a function<br />
gets close to as the values of the independent<br />
variable increase and get close to a given<br />
value.<br />
c. A (two-sided) limit is the value that a function<br />
gets close to as the values of the independent<br />
variable get close to a given value, regardless<br />
of whether the values increase or decrease<br />
toward the given value.<br />
4. a. 25<br />
b. 3 1 7 5 10<br />
c. 10 2 5 100<br />
d. 4 2 3(22) 2 528<br />
e. 4<br />
f. 2 3 5 8<br />
5. Even though f(4) 521, the limit is 1, since that<br />
is the value that the function approaches from the<br />
left and the right of x 5 4.<br />
6. a. 0<br />
b. 2<br />
c. 21<br />
d. 2<br />
7. a. 2<br />
b. 1<br />
c. does not exist<br />
8. a. 9 2 (21) 2 5 8<br />
b.<br />
c.<br />
0 1 20<br />
Å 0 1 5 5 "4<br />
5 2<br />
"5 2 1 5 "4<br />
5 2<br />
9. 2 2 1 1 5 5<br />
6<br />
4<br />
2<br />
y<br />
x<br />
–4 –2 0 2 4<br />
10. a. Since 0 is not a value for which the function is<br />
undefined, one may substitute 0 in for x to find that<br />
lim x 4 5 lim x 4<br />
xS0 1 xS0<br />
5 (0) 4<br />
5 0<br />
b. Since 2 is not a value for which the function is<br />
undefined, one may substitute 2 in for x to find that<br />
lim (x 2 2 4) 5 lim (x 2 2 4)<br />
xS2 2 xS2<br />
5 (2) 2 2 4<br />
5 4 2 4<br />
5 0<br />
c. Since 3 is not a value for which the function is<br />
undefined, one may substitute 3 in for x to find that<br />
lim (x 2 2 4) 5 lim (x 2 2 4)<br />
xS3 2 xS3<br />
5 (3) 2 2 4<br />
5 9 2 4<br />
5 5<br />
d. Since 1 is not a value for which the function is<br />
undefined, one may substitute 1 in for x to find that<br />
1<br />
lim<br />
xS1 1 x 2 3 5 lim 1<br />
xS1 x 2 3<br />
5 1<br />
1 2 3<br />
52 1 2<br />
e. Since 3 is not a value for which the function is<br />
undefined, one may substitute 3 in for x to find that<br />
1<br />
lim<br />
xS3 1 x 1 2 5 lim 1<br />
xS3 x 1 2<br />
5 1<br />
3 1 2<br />
5 1 5<br />
f. If 3 is substituted in the function for x, then the<br />
function is undefined because of division by zero.<br />
There does not exist a way to divide out the x 2 3 in<br />
1-22 <strong>Chapter</strong> 1: Introduction to Calculus
1<br />
the denominator. Also, lim approaches infinity,<br />
xS3 1 x 2 3<br />
1<br />
while lim approaches negative infinity.<br />
xS3 2 x 2 3<br />
1<br />
1<br />
Therefore, since lim<br />
lim<br />
xS3 1 x 2 3 2 lim 1<br />
xS3 2 x 2 3 ,<br />
xS3 x 2 3<br />
does not exist.<br />
11. a.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
–6<br />
–8<br />
lim f(x) 2 lim f(x). Therefore, lim f(x)<br />
xS21 1 xS21 2 xS21<br />
not exist.<br />
b.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
–6<br />
–8<br />
lim f(x) 5 lim f(x). Therefore, lim f(x)<br />
xS2 1 xS2 2 xS2<br />
is equal to 2.<br />
c.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
–6<br />
–8<br />
lim f(x) 5 lim f(x). Therefore, lim f(x)<br />
xS 1 xS 1 2 1 xS 1 2 2 2<br />
is equal to 2.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
does<br />
exists and<br />
exists and<br />
d.<br />
lim f(x) 2 lim f(x). Therefore, lim f(x)<br />
xS20.5 1 xS20.5 2 xS20.5<br />
does not exist.<br />
12. Answers may vary. For example:<br />
a.<br />
6<br />
y<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
b.<br />
c.<br />
d.<br />
13. f(x) 5 mx 1 b<br />
f(x) 522 m 1 b 522<br />
lim<br />
xS1<br />
lim<br />
xS21<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
–6<br />
–8<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
f(x) 5 4<br />
8<br />
6<br />
4<br />
2<br />
6<br />
4<br />
2<br />
6<br />
4<br />
2<br />
6<br />
4<br />
2<br />
y<br />
y<br />
y<br />
y<br />
2<br />
2<br />
2<br />
2<br />
4 6 8<br />
4 6 8<br />
4 6 8<br />
4 6 8<br />
2m 1 b 5 4<br />
2b 5 2<br />
b 5 1, m 523<br />
x<br />
x<br />
x<br />
x<br />
1-23
14. f(x) 5 ax 2 1 bx 1 c, a 2 0<br />
f(0) 5 0 c 5 0<br />
f(x) 5 5 a 1 b 5 5<br />
lim<br />
xS1<br />
lim<br />
xS22<br />
f(x) 5 8<br />
6a 5 18<br />
a 5 3, b 5 2<br />
Therefore, the values are a 5 3, b 5 2, and c 5 0.<br />
15. a.<br />
y<br />
10<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–4 –2 0 2 4 6 8 10 12<br />
–2<br />
b. lim p(t) 5 3 1 1<br />
tS6 2 12 (6)2<br />
5 3 1 36<br />
12<br />
5 3 1 3<br />
5 6<br />
lim p(t) 5 2 1 1<br />
tS6 1 18 (6)2<br />
5 2 1 36<br />
18<br />
5 2 1 2<br />
5 4<br />
c. Since p(t) is measured in thousands, right before<br />
the chemical spill there were 6000 fish in the lake.<br />
Right after the chemical spill there were 4000 fish<br />
in the lake. So, 6000 2 4000 5 2000 fish were<br />
killed by the spill.<br />
d. The question asks for the time, t, after the chemical<br />
spill when there are once again 6000 fish in the lake.<br />
Use the second equation to set up an equation that is<br />
modelled by<br />
6 5 2 1 1<br />
18 t2<br />
4 5 1<br />
18 t2<br />
4a 2 2b 5 8<br />
72 5 t 2<br />
!75 5 t<br />
(The question asks for time so the negative answer<br />
is disregarded.)<br />
So, at time t 5 !72 8 8.49 years the population<br />
has recovered to the level before the spill.<br />
1.5 Properties of Limits, pp. 45–47<br />
1. lim(3 1 x) and lim(x 1 3) have the same value,<br />
xS2<br />
xS2<br />
but 3 1 x does not. Since there are no brackets<br />
around the expression, the limit only applies to 3,<br />
and there is no value for the last term, x.<br />
2. Factor the numerator and denominator. Cancel<br />
any common factors. Substitute the given value of x.<br />
3. If the two one-sided limits have the same value,<br />
then the value of the limit is equal to the value of<br />
the one-sided limits. If the one-sided limits do not<br />
have the same value, then the limit does not exist.<br />
3(2)<br />
4. a.<br />
2 2 1 2 5 1<br />
b. (21) 4 1 (21) 3 1 (21) 2 5 1<br />
c.<br />
5 100<br />
9<br />
d. (2p) 3 1p 2 (2p) 2 5p 3 5 8p 3 1 2p 3 2 5p 3<br />
5 5p 3<br />
e. "3 1 "1 1 0 5 "3 1 1<br />
5 2<br />
23 2 3<br />
f.<br />
Å 2(23) 1 4 5 26<br />
Å 22<br />
5 "3<br />
(22) 3<br />
5. a.<br />
22 2 2 522<br />
2<br />
b.<br />
!1 1 1 5 2 !2<br />
5 "2<br />
6. Since substituting t 5 1 does not make the<br />
denominator 0, direct substitution works.<br />
1 2 1 2 5<br />
5 25<br />
6 2 1 5<br />
521<br />
4 2 x 2<br />
7. a. lim<br />
xS2 2 2 x 5 lim (2 2 x)(2 1 x)<br />
xS2 (2 2 x)<br />
5 lim(2 1 x)<br />
xS2<br />
5 4<br />
2x 2 1 5x 1 3 (x 1 1)(2x 1 3)<br />
b. lim<br />
5 lim<br />
xS21 x 1 1 xS21 x 1 1<br />
c.<br />
lim<br />
xS2<br />
c"9 1 1 2<br />
"9 d 5 a3 1 1 2<br />
3 b<br />
5 5<br />
x 3 2 27<br />
lim<br />
xS3 x 2 3<br />
5 lim (x 2 3)(x 2 1 3x 1 9)<br />
xS3 x 2 3<br />
5 9 1 9 1 9<br />
5 27<br />
1-24 <strong>Chapter</strong> 1: Introduction to Calculus
d.<br />
e.<br />
f.<br />
5 lim<br />
xS0<br />
52 1 4<br />
5 lim<br />
xS0<br />
52 1<br />
"7<br />
8. a.<br />
Let u 5 " 3 x. Therefore, u 3 5 x as x S 8, u S 2.<br />
u 2 2<br />
Here, lim<br />
xS2 u 3 2 8 5 lim 1<br />
xS2 u 2 1 2u 1 4<br />
5 1<br />
12<br />
27 2 x<br />
b. lim<br />
Let<br />
xS27 x 1 3 2 3 x 1 3 5 u<br />
x 5 u 3<br />
u 3 2 27<br />
x S 27, u S 3.<br />
5 lim<br />
xS3 u 2 3<br />
(u 2 3)(u 2 1 3u 1 9)<br />
52lim<br />
xS3 u 2 3<br />
52(9 1 9 1 9)<br />
5227<br />
c.<br />
5 lim<br />
xS1<br />
5 lim<br />
xS1<br />
5 1 6<br />
d.<br />
lim<br />
xS0<br />
"x 2 2<br />
lim<br />
xS4 x 2 4<br />
lim<br />
xS0<br />
x 1 6<br />
lim<br />
2 1<br />
xS1<br />
x 1 6<br />
lim<br />
2 1<br />
xS1<br />
5 lim<br />
xS1<br />
£ 2 2 "4 1 x<br />
x<br />
21<br />
2 1 "4 1 x<br />
"7 2 x 2 "7 1 x "7 2 x 1 "7 1 x<br />
£ 3<br />
x<br />
"7 2 x 1 "7 1 x §<br />
" 3 x 2 2<br />
lim<br />
xS8 x 2 8<br />
x 2 1<br />
u 2 1<br />
u 6 2 1<br />
(u 2 1)<br />
(u 2 1)(u 5 1 u 4 1 u 3 1 u 2 1 u 1 1)<br />
x 1 3<br />
u 2 2 1<br />
1<br />
u 2 2 1<br />
5 lim "x 2 2<br />
xS4 ("x 2 2)("x 1 2)<br />
5 1 4<br />
7 2 x 2 7 2 x<br />
3 2 1 "4 1 x<br />
x("7 2 x 1 "7 1 x)<br />
x 1 6 5 u, x 5 u 6<br />
x S 1, u S 1<br />
Let x 1 6 5 u<br />
u 6 5 x<br />
x 1 3 5 u2<br />
As x S 1, u S 1<br />
5 lim<br />
xS1<br />
5 1 2<br />
u 2 1<br />
(u 2 1)(u 1 1)<br />
"x 2 2<br />
e. lim<br />
Let x 1 2<br />
xS4 "x 3 2 8<br />
5 u<br />
x 3 2<br />
u 2 2<br />
5 u3<br />
5 lim<br />
x S 4, u S 2<br />
xS2 u 3 2 8<br />
u 2 2<br />
5 lim<br />
xS2 (u 2 2)(u 2 1 2u 1 4)<br />
5 1<br />
12<br />
(x 1 8) 1 3<br />
f. lim<br />
2 2<br />
Let (x 1 8) 1 3<br />
xS0 x<br />
5 u<br />
x 1 8 5 u 3<br />
u 2 2<br />
lim<br />
x 5 u 3 2 8<br />
xS2 u 3 2 8<br />
x S 0, u S 2<br />
5 1<br />
12<br />
16 2 16<br />
9. a.<br />
64 1 64 5 0<br />
16 2 16<br />
b.<br />
16 2 20 1 6 5 0<br />
x 2 1 x<br />
c. lim<br />
xS21 x 1 1 5 lim x(x 1 1)<br />
xS21 x 1 1<br />
521<br />
"x 1 1 2 1 "x 1 1 2 1<br />
d. lim<br />
5 lim<br />
xS0 x<br />
xS0 x 1 1 2 1<br />
"x 1 1 2 1<br />
5 lim<br />
xS0 ("x 1 1 2 1)("x 1 1 1 1)<br />
5 1 2<br />
(x 1 h) 2 2 x 2 2xh 1 h 2<br />
e. lim<br />
5 lim<br />
hS0 h<br />
hS0 h<br />
5 2x<br />
1<br />
f. lima<br />
xS1 x 2 1 ba 1<br />
x 1 3 2 2<br />
3x 1 5 b<br />
1 1 5 2 2x 2 6<br />
5 lima<br />
ba3x<br />
xS1 x 2 1 (x 1 3)(3x 1 5) b<br />
1<br />
5 lim<br />
xS1 (x 1 3)(3x 1 5)<br />
5 1<br />
4(8)<br />
5 1<br />
32<br />
2 1 "4 1 x § 1-25<br />
Calculus and Vectors <strong>Solutions</strong> Manual
0 x 2 5 0<br />
10. a. lim does not exist.<br />
xS5 x 2 5<br />
0 x 2 5 0<br />
lim<br />
xS5 1 x 2 5 5 lim x 2 5<br />
xS5 1 x 2 5<br />
5 1<br />
0 x 2 5 0<br />
lim<br />
xS5 2 x 2 5 5 lim 2 a x 2 5<br />
xS5 2 x 2 5 b<br />
0 2x 2 5 0 (x 1 1)<br />
b. lim<br />
does not exist.<br />
2x 2 5<br />
0 2x 2 5 0 5 2x 2 5, x $ 5 2<br />
(2x 2 5)(x 1 1)<br />
lim<br />
5 x 1 1<br />
xS 5 1<br />
2x 2 5<br />
2<br />
0 2x 2 5 0 52(2x 2 5), x , 5 2<br />
2 (2x 2 5)(x 1 1)<br />
lim<br />
52(x 1 1)<br />
xS 5 2<br />
2x 2 5<br />
2<br />
y<br />
4<br />
c.<br />
–8<br />
xS 5 2<br />
–4<br />
–4<br />
–2<br />
521<br />
2<br />
2<br />
–2<br />
–4<br />
0<br />
0<br />
x 2 2 x 2 2<br />
lim<br />
xS2 0 x 2 2 0<br />
1<br />
–1<br />
–2<br />
y<br />
4<br />
2<br />
(x 2 2)(x 1 1)<br />
5 lim<br />
xS2 0 x 2 2 0<br />
(x 2 2)(x 1 1) (x 2 2)(x 1 1)<br />
lim<br />
5 lim<br />
xS2 1 0 x 2 2 0<br />
xS2 1 x 2 2<br />
5 lim x 1 1<br />
xS2 1<br />
5 3<br />
8<br />
4<br />
x<br />
x<br />
(x 2 2)(x 1 1)<br />
lim<br />
5 lim<br />
xS2 2 0 x 2 2 0<br />
–4<br />
d. 0 x 1 2 0 5 x 1 2 if x .22<br />
52(x 1 2) if x ,22<br />
(x 1 2)(x 1 2) 2<br />
lim<br />
5 lim (x 1 2) 2 5 0<br />
xS22 1 x 1 2<br />
xS22 1<br />
(x 1 2)(x 1 2) 2<br />
lim<br />
5 0<br />
xS22 2 2 (x 1 2)<br />
–4<br />
11. a.<br />
–2<br />
–2<br />
4<br />
2<br />
–2<br />
–4<br />
0<br />
4<br />
2<br />
–2<br />
–4<br />
DT T V DV<br />
20<br />
20<br />
20<br />
20<br />
20<br />
20<br />
0<br />
y<br />
240 19.1482<br />
220 20.7908<br />
0 22.4334<br />
20 24.0760<br />
40 25.7186<br />
60 27.3612<br />
80 29.0038<br />
y<br />
5 lim 2 (x 1 1)<br />
xS2 2<br />
523<br />
2<br />
xS2 2 2<br />
DV is constant, therefore T and V form a linear<br />
relationship.<br />
b. V 5 DV<br />
DT ? T 1 K<br />
DV<br />
DT 5 1.6426 5 0.082 13<br />
20<br />
2<br />
1.6426<br />
1.6426<br />
1.6426<br />
1.6426<br />
1.6426<br />
1.6426<br />
4<br />
4<br />
x<br />
(x 2 2)(x 1 1)<br />
(x 2 2)<br />
x<br />
1-26 <strong>Chapter</strong> 1: Introduction to Calculus
V 5 0.082 13T 1 K<br />
T 5 0 V 5 22.4334<br />
Therefore, k 5 22.4334 and<br />
V 5 0.082 13T 1 22.4334.<br />
c. T 5 V 2 22.4334<br />
0.082 13<br />
d. limT 52273.145<br />
vS0<br />
e. V<br />
12<br />
12.<br />
5 21<br />
3<br />
5 7<br />
13. lim f(x) 5 3<br />
xS4<br />
a. 3f(x)4 3 5 3 3 5 27<br />
b.<br />
5<br />
lim<br />
xS4<br />
10<br />
8<br />
6<br />
4<br />
2<br />
0<br />
lim<br />
xS4<br />
0 2 4 6 8 10 12<br />
x 2 2 4<br />
lim<br />
xS5 f(x)<br />
lim<br />
xS5<br />
(x 2 2 4)<br />
lim<br />
xS5<br />
f(x)<br />
3f(x)4 2 2 x 2<br />
f(x) 1 x<br />
( f(x) 2 x)( f(x) 1 x)<br />
5 lim<br />
xS4 f(x) 1 x<br />
5 lim( f(x) 2 x)<br />
xS4<br />
5 3 2 4<br />
521<br />
c. lim"3f(x) 2 2x 5 "3 3 3 2 2 3 4<br />
xS4<br />
5 1<br />
f(x)<br />
14. lim<br />
xS0 x 5 1<br />
a. limf(x) 5 lim c f(x)<br />
xS0<br />
xS0 x 3 xd 5 0<br />
f(x)<br />
b. lim<br />
xS0 g(x) 5 lim x f(x)<br />
c<br />
xS0 g(x) x d 5 0<br />
T<br />
f(x)<br />
g(x)<br />
15. lim and lim 5 2<br />
xS0 x 5 1<br />
xS0 x<br />
a.<br />
f(x)<br />
b. lim<br />
xS0 g(x) 5 lim<br />
xS0<br />
16.<br />
lim<br />
xS0<br />
!x 1 1 2 !2x 1 1<br />
5 lim c<br />
xS0 !x 1 1 1 !2x 1 1<br />
5 lim<br />
xS0<br />
52 2 1 2<br />
1 1 1<br />
522<br />
x 2 1 0 x 2 1 021<br />
17. lim<br />
xS1 0 x 2 1 0<br />
x S 1 1 0 x 2 1 0 5 x 2 1<br />
x 2 1 x 2 2 (x 1 2)(x 2 1)<br />
5<br />
x 2 1 x 2 1<br />
x 2 1 0 x 2 1 021<br />
lim<br />
5 3<br />
xS1 1 0 x 2 1 0<br />
x S 1 2 0 x 2 1 0 52x 1 1<br />
x 2 2 x<br />
lim<br />
xS1 2 2x 1 1 5 lim x(x 2 1)<br />
xS1 2 2x 1 1<br />
521<br />
Therefore, this limit does not exist.<br />
y<br />
4<br />
–4<br />
lim<br />
xS0<br />
g(x) 5 limxa g(x)<br />
xS0<br />
!x 1 1 2 !2x 1 1<br />
!3x 1 4 2 !2x 1 4<br />
3<br />
3<br />
!x 1 1 1 !2x 1 1<br />
!3x 1 4 2 !2x 1 4<br />
!3x 1 4 1 !2x 1 4<br />
!3x 1 4 1 !2x 1 4 d<br />
(x 1 1 2 2x 2 1)<br />
c<br />
(3x 1 4 2 2x 2 4)<br />
–2<br />
2<br />
–2<br />
–4<br />
0<br />
f(x)<br />
x<br />
g(x)<br />
x<br />
x b 5 0 3 2<br />
2<br />
5 1 2<br />
5 0<br />
3<br />
!3x 1 4 1 !2x 1 4<br />
!x 1 1 1 !2x 1 1 d<br />
4<br />
x<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-27
1.6 Continuity, pp. 51–53<br />
1. Anywhere that you can see breaks or jumps is a<br />
place where the function is not continuous.<br />
2. It means that on that domain, you can trace the<br />
graph of the function without lifting your pencil.<br />
3. point discontinuity<br />
10<br />
8<br />
6<br />
4<br />
2<br />
–2 0<br />
–2<br />
y<br />
hole<br />
2 4<br />
jump discontinuity<br />
y<br />
10<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–2 0 2 4 6<br />
–2<br />
infinite discontinuity<br />
y<br />
10<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–1 0 1 2 3 4<br />
–2<br />
vertical<br />
–4 asymptote<br />
6<br />
x<br />
4. a. x 5 3 makes the denominator 0.<br />
b. x 5 0 makes the denominator 0.<br />
c. x 5 0 makes the denominator 0.<br />
d. x 5 3 and x 523 make the denominator 0.<br />
e. x 2 1 x 2 6 5 (x 1 3)(x 2 2)<br />
x 523 and x 5 2 make the denominator 0.<br />
f. The function has different one-sided limits at x 5 3.<br />
5. a. The function is a polynomial, so the function<br />
is continuous for all real numbers.<br />
b. The function is a polynomial, so the function is<br />
continuous for all real numbers.<br />
c. x 2 2 5x 5 x(x 2 5)<br />
The is continuous for all real numbers except<br />
0 and 5.<br />
d. The is continuous for all real numbers greater<br />
than or equal to 22.<br />
e. The is continuous for all real numbers.<br />
f. The is continuous for all real numbers.<br />
6. g(x) is a linear function (a polynomial),<br />
and so is continuous everywhere,<br />
including x 5 2.<br />
7.<br />
y<br />
8<br />
–8<br />
The function is continuous everywhere.<br />
8.<br />
y<br />
4<br />
–4<br />
The function is discontinuous at x 5 0.<br />
9. y<br />
4<br />
2<br />
0<br />
–4<br />
–2<br />
200<br />
4<br />
–4<br />
–8<br />
2<br />
–2<br />
–4<br />
0<br />
0<br />
4<br />
2<br />
400 600<br />
x<br />
8<br />
4<br />
x<br />
x<br />
1-28 <strong>Chapter</strong> 1: Introduction to Calculus
10.<br />
lim<br />
xS3<br />
–4<br />
x 1 3, if x 2 3<br />
12. g(x) 5 e<br />
2 1 !k, if x 5 3<br />
g(x) is continuous.<br />
–4<br />
x 2 2 x 2 6<br />
f(x) 5 lim<br />
xS3 x 2 3<br />
(x 2 3)(x 1 2)<br />
5 lim<br />
xS3 x 2 3<br />
5 5<br />
Function is discontinuous at x 5 3.<br />
11. Discontinuous at x 5 2<br />
y<br />
4<br />
–2<br />
2 1 "k 5 6<br />
"k 5 4, k 5 16<br />
13.<br />
21, if x , 0<br />
f(x) 5 • 0, if x 5 0<br />
1, if x . 0<br />
a.<br />
y<br />
4<br />
–2<br />
2<br />
–2<br />
–4<br />
2<br />
–2<br />
–4<br />
0<br />
0<br />
2<br />
b. i. From the graph, lim f(x) 521.<br />
xS0 2<br />
ii. From the graph, lim f(x) 5 1.<br />
xS0 1<br />
iii. Since the one-sided limits differ, limf(x)<br />
does<br />
xS0<br />
not exist.<br />
c. f is not continuous since limf(x)<br />
does not exist.<br />
xS0<br />
14. a. From the graph, f(3) 5 2.<br />
b. From the graph, lim f(x) 5 4.<br />
xS3 2<br />
c. lim f(x) 5 4 5 lim f(x)<br />
xS3 2<br />
xS3 2<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
2<br />
4<br />
4<br />
x<br />
x<br />
Thus, lim f(x) 5 4. But, f(3) 5 2. Hence f is not<br />
xS3<br />
continuous at x 5 2 (and also not continuous over<br />
23 , x , 8).<br />
15. The function is to be continuous at x 5 1 and<br />
discontinuous at x 5 2.<br />
f(x) 5 μ<br />
For f(x) to be continuous at x 5 1:<br />
A(1) 2 B<br />
5 3(1)<br />
1 2 2<br />
A(1) 2 B 523<br />
A 5 B 2 3<br />
For f(x) to be discontinuous at x 5 2:<br />
B(2) 2 2 A 2 3(2)<br />
4 B 2 A 2 6<br />
If 4B 2 A . 6, then if 4B 2 A , 6, then<br />
4B 2 (B 2 3) . 6 4B 2 B 1 3 , 6<br />
3B 1 3 . 6<br />
3B 1 3 , 6<br />
3B . 3<br />
3B , 3<br />
B . 1 and<br />
B , 1 and<br />
A .22<br />
A ,22<br />
This shows that A and B can be any set of real<br />
numbers such that<br />
(1) A 5 B 2 3<br />
(2) 4B 2 A 2 6 (if B . 1, then A .22 if B , 1,<br />
then A ,22)<br />
A 5 1 and B 522 is not a solution because then<br />
the graph would be continuous at x 5 2.<br />
2x, if 23 # x #22<br />
16. f(x) 5 • ax 2 1 b, if 22 , x , 0<br />
6, if x 5 0<br />
at x 522, 4a 1 b 5 2<br />
at x 5 0, b 5 6.<br />
a 521<br />
2x, if 23 # x #22<br />
f(x) 5 • 2x 2 1 b, if 22 , x , 0<br />
6, if x 5 0<br />
if a 521, b 5 6. f(x) is continuous.<br />
17.<br />
Ax 2 B<br />
x 2 2 ,if x # 1<br />
3x, if 1 , x , 2<br />
Bx 2 2 A, if x $ 2<br />
x0 x 2 1 0<br />
g(x) 5 • x 2 1 , if x 2 1<br />
0, if x 5 1<br />
lim g(x) 521<br />
a. xS1 2 limg(x)<br />
lim g(x) 5 1 xS1<br />
xS1 1<br />
limg(x)<br />
does not exist.<br />
xS1<br />
1-29
.<br />
g(x) is discontinuous at x 5 1.<br />
Review Exercise, pp. 56–59<br />
1. a. f(22) 5 36, f(3) 5 21<br />
21 2 36<br />
m 5<br />
3 2 (22)<br />
523<br />
b. f(21) 5 13, f(4) 5 48<br />
48 2 13<br />
m 5<br />
4 2 (21)<br />
5 7<br />
c. f(1) 523<br />
5(1 1 2h 1 h 2 ) 2 (23)<br />
m 5 lim<br />
hS0<br />
h<br />
2h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim 2 1 h<br />
hS0<br />
5 2<br />
y 2 (23) 5 2(x 2 1)<br />
2x 2 y 2 5 5 0<br />
2. a. f(x) 5 3 P(2, 1)<br />
x 1 1 ,<br />
3 1 h 2 1<br />
m 5<br />
h<br />
5 lim 2 1<br />
hS0 3 1 h<br />
b.<br />
–4<br />
52 1 3<br />
g(x) 5 "x 1 2,<br />
"21 1 h 1 2 2 1<br />
m 5 lim<br />
hS0 h<br />
5 lim c !h 1 1 2 1<br />
hS0<br />
5 lim<br />
hS0<br />
5 1 2<br />
3<br />
–2<br />
4<br />
2<br />
–2<br />
–4<br />
0<br />
y<br />
x<br />
1<br />
!h 1 1 1 1<br />
2<br />
P(21, 1)<br />
4<br />
x<br />
3 !h 1 1 1 1<br />
!h 1 1 1 1 d<br />
c. h(x) 5 2<br />
Pa4, 2 !x 1 5 , 3 b<br />
2<br />
!4 1 h 1 5 2 2 3<br />
m 5 lim<br />
hS0 h<br />
d.<br />
5 2 lim<br />
hS0<br />
5 2 lim c2<br />
hS0<br />
52 2<br />
9(6)<br />
52 1 27<br />
f(x) 5 5<br />
x 2 2 ,<br />
4 1 h 2 2 2 5 2<br />
m 5 lim<br />
hS0 h<br />
10 2 5(2 1 h)<br />
5 lim<br />
hS0 h(2 1 h)(2)<br />
25h<br />
5 lim 2<br />
hS0 h(2 1 h)(2)<br />
52 5 4<br />
3. f(x) 5 e 4 2 x2 , if x # 1<br />
2x 1 1, if x . 1<br />
a. Slope at P(21, 3) f(x) 5 4 2 x 2<br />
4 2 (21 1 h) 2 2 3<br />
m 5 lim<br />
hS0 h<br />
4 2 1 1 2h 2 h 2 2 3<br />
5 lim<br />
hS0 h<br />
5 lim(2 2 h)<br />
hS0<br />
5 2<br />
Slope of the graph at P(21, 3) is 2.<br />
b. Slope at P(2, 0.5)<br />
f(x) 5 2x 1 1<br />
f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5<br />
5 2h<br />
2h<br />
m 5 lim<br />
hS0 h 5 2<br />
Slope of the graph at P(2, 0.5) is 2.<br />
4. s(t) 525t 2 1 180<br />
a. s(0) 5 180, s(1) 5 175, s(2) 5 160<br />
Average velocity during the first second is<br />
s(1) 2 s(0)<br />
1<br />
c 3 2 !9 1 h<br />
3h!9 1 h 3 3 1 !9 1 h<br />
3 1 !9 1 h d<br />
1<br />
3!9 1 h(3 1 !9 1 h) d<br />
5<br />
525<br />
Pa4, 5 2 b<br />
ms. ><br />
1-30 <strong>Chapter</strong> 1: Introduction to Calculus
Average velocity during the second second is<br />
s(2) 2 s(1)<br />
5215 ms. ><br />
1<br />
b. At t 5 4:<br />
s(4 1 h) 2 s(4)<br />
525(4 1 h) 2 1 180 2 (25(16) 1 180)<br />
5280 2 40h 2 5h 2 1 180 1 80 2 180<br />
s(4 1 h) 2 s(4) 240h 2 5h2<br />
5<br />
h<br />
h<br />
v(4) 5 lim(240 2 5h) 5240<br />
hS0<br />
Velocity is 240 m><br />
s.<br />
c. Time to reach ground is when s(t) 5 0.<br />
Therefore, 25t 2 1 180 5 0<br />
t 2 5 36<br />
t 5 6, t . 0.<br />
Velocity at t 5 6:<br />
s(6 1 h) 525(36 1 12h 1 h 2 ) 1 180<br />
5260h 2 5h 2<br />
s(6) 5 0<br />
Therefore, v(6) 5 lim(260 2 5h) 5260.<br />
hS0<br />
5. M(t) 5 t 2 mass in grams<br />
a. Growth during 3 # t # 3.01<br />
M(3.01) 5 (3.01) 2 5 9.0601<br />
M(3) 5 3 2<br />
5 9<br />
Grew 0.0601 g during this time interval.<br />
b. Average rate of growth is<br />
0.0601<br />
g><br />
min.<br />
0.01 5 6.01<br />
c. s(3 1 h) 5 9 1 6h 1 h 2<br />
s(3) 5 9<br />
s(3 1 h) 2 s(3) 6h 1 h2<br />
5<br />
h<br />
h<br />
Rate of growth is (6 1 h) 5 6 g><br />
min.<br />
lim<br />
hS0<br />
6. Q(t) 5 10 4 (t 2 1 15t 1 70) tonnes of waste,<br />
0 # t # 10<br />
a. At t 5 0,<br />
Q(t) 5 70 3 10 4<br />
5 700 000.<br />
700 000 t have accumulated up to now.<br />
b. Over the next three years, the average rate of<br />
change:<br />
Q(3) 5 10 4 (9 1 45 1 70)<br />
5 124 3 10 4<br />
Q(0) 5 70 3 10 4<br />
Q(3) 2 Q(0) 54 3 104<br />
5<br />
3<br />
3<br />
5 18 3 10 4 t per year.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
c. Present rate of change:<br />
Q(h) 5 10 4 (h 2 1 15h 1 70)<br />
Q(0) 5 10 4 1 70<br />
Q(h) 2 Q(0)<br />
lim<br />
5 lim10 4 (h 1 15)<br />
hS0 h<br />
hS0<br />
5 15 3 10 4 t per year.<br />
d. Q(a 1 h)<br />
5 10 4 3a 2 1 2ah 1 h 2 1 15a 1 15h 1 704<br />
Q(a) 5 10 4 3a 2 1 15a 1 704<br />
Q(a 1 h) 2 Q(a)<br />
5 104 32ah 1 h 2 1 15h4<br />
h<br />
h<br />
Q(a 1 h) 2 Q(a)<br />
lim<br />
5 lim10 4 (2a 1 h 1 15)<br />
hS0 h<br />
hS0<br />
5 (2a 1 15)10 4<br />
Now,<br />
(2a 1 15)10 4 5 3 3 10 5<br />
2 a 1 15 5 30<br />
a 5 7.5<br />
3.0 3 10 5<br />
It will take 7.5 years to reach a rate of<br />
t per year.<br />
7. a. From the graph, the limit is 10.<br />
b. 7; 0<br />
c. p(t) is discontinuous for t 5 3 and t 5 4.<br />
8. a. Answers will vary. lim f(x) 5 0.5, f is<br />
xS21<br />
discontinuous at x 521<br />
y<br />
2<br />
–2<br />
b. f(x) 524 if x , 3; f is increasing for x . 3<br />
lim f(x) 5 1<br />
xS3 1 y<br />
4<br />
–4<br />
–1<br />
–2<br />
1<br />
–1<br />
–2<br />
0<br />
2<br />
–2<br />
–4<br />
0<br />
1<br />
2<br />
2<br />
4<br />
x<br />
x<br />
1-31
9. a.<br />
4<br />
y<br />
13. a.<br />
x 1.9 1.99 1.999 2.001 2.01 2.1<br />
–4<br />
–2<br />
2<br />
0<br />
2<br />
4<br />
x<br />
x 2 2<br />
0.344 83 0.334 45 0.333 44 0.333 22 0.332 23 0.322 58<br />
x 2 2 x 2 2<br />
1<br />
3<br />
–2<br />
–4<br />
x 1 1, if x ,21<br />
b. f(x) 5 • 2x 1 1, if 21 # x , 1<br />
x 2 2, if x . 1<br />
Discontinuous at x 521 and x 5 1.<br />
c. They do not exist.<br />
10. The function is not continuous at x 524<br />
because the function is not defined at x 524.<br />
( x 524 makes the denominator 0.)<br />
11. f(x) 5 2x 2 2<br />
x 2 1 x 2 2<br />
2(x 2 1)<br />
5<br />
(x 2 1)(x 1 2)<br />
a. f is discontinuous at x 5 1 and x 522.<br />
2<br />
b. limf(x) 5 lim<br />
xS1<br />
xS1 x 1 2<br />
2<br />
lim f(x): 5 lim<br />
xS22<br />
xS22 1 x 1 2 51`<br />
lim<br />
x 1 2 52`<br />
lim f(x) does not exist.<br />
xS22<br />
12. a. f(x) 5 1 limf(x)<br />
does not exist.<br />
x 2, xS0<br />
b. g(x) 5 x(x 2 5), limg(x) 5 0<br />
c. h(x) 5 x3 xS0<br />
2 27<br />
x 2 2 9 ,<br />
limh(x) 5 37<br />
xS4<br />
lim<br />
xS23<br />
5 2 3<br />
xS22 2 2<br />
7 5 5.2857<br />
h(x) does not exist.<br />
b.<br />
14.<br />
!x 1 3 2 !3<br />
lim c<br />
xS0<br />
5 lim<br />
xS0<br />
5 lim<br />
xS0<br />
5 lim<br />
xS0<br />
5 1<br />
2!3<br />
This agrees well with the values in the table.<br />
15. a.<br />
x 0.9 0.99 0.999 1.001 1.01 1.1<br />
x 2 1<br />
0.526 32 0.502 51 0.500 25 0.499 75 0.497 51 0.476 19<br />
x 2 2 1<br />
1<br />
2<br />
x 20.1 20.01 20.001 0.001 0.01 0.1<br />
"x 1 3 2 "3<br />
0.291 12 0.288 92 0.2887 0.288 65 0.288 43 0.286 31<br />
x<br />
?<br />
x<br />
x 1 3 2 3<br />
xA!x 1 3 1 !3B<br />
x<br />
xA!x 1 3 1 !3B<br />
1<br />
!x 1 3 1 !3<br />
f(x) 5 "x 1 2 2 2<br />
x 2 2<br />
!x 1 3 1 !3<br />
!x 1 3 1 !3 d<br />
x 2.1 2.01 2.001 2.0001<br />
f(x) 0.248 46 0.249 84 0.249 98 0.25<br />
x 5 2.0001<br />
f(x) 8 0.25<br />
1-32 <strong>Chapter</strong> 1: Introduction to Calculus
.<br />
5 lim<br />
xS0<br />
1<br />
A!x 1 5 1 !5 2 xB<br />
5 1 !5<br />
c.<br />
(5 1 h) 2 2 25<br />
16. a. lim<br />
hS0 h<br />
5 lim(10 1 h)<br />
hS0<br />
5 10<br />
Slope of the tangent to y 5 x 2 at x 5 5 is 10.<br />
b.<br />
52 1 16<br />
Slope of the tangent to y 5 1 at (x 5 4) is<br />
x<br />
(x 1 4)(x 1 8)<br />
17. a. lim<br />
5 lim (x 1 8)<br />
xS24 x 1 4<br />
xS24<br />
5 (24) 1 8<br />
5 4<br />
b.<br />
c.<br />
limf(x) 5 0.25<br />
xS2<br />
lim c !x 1 2 2 2<br />
xS2<br />
5 lim<br />
xS2<br />
Slope of the tangent to at x 5 4 is<br />
1<br />
4 1 h 2 1 4 4 2 4 2 h<br />
c. lim 5 lim<br />
hS0 h<br />
hS0 4(4 1 h)(h)<br />
1<br />
5 lim 2<br />
hS0 4(4 1 h)<br />
(x 1 4a) 2 2 25a 2<br />
lim<br />
xSa x 2 a<br />
lim c<br />
xS0<br />
5 lim<br />
xS0<br />
x 2 2<br />
1<br />
!x 1 2 1 2<br />
5 1 4 5 0.25<br />
"4 1 h 2 2<br />
lim<br />
hS0 h<br />
3 !x 1 2 1 2<br />
!x 1 2 1 2 d<br />
"4 1 h 2 2<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 1 4<br />
y 5 "x<br />
5 10a<br />
!x 1 5 2 !5 2 x<br />
3<br />
x<br />
x 1 5 2 5 1 x<br />
xA!x 1 5 1 !5 2 xB<br />
4 1 h 2 4<br />
1<br />
!4 1 h 1 2<br />
1<br />
4.<br />
2 1 16.<br />
(x 2 a)(x 1 9a)<br />
5 lim<br />
xSa x 2 a<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
!x 1 5 1 !5 2 x<br />
!x 1 5 1 !5 2 x d<br />
d.<br />
e.<br />
lim<br />
xS2<br />
x 1 2<br />
5 lim<br />
xS2 x 2 1 2x 1 4<br />
(2) 1 2<br />
5<br />
(2) 2 1 2(2) 1 4<br />
5 4 12<br />
5 1 3<br />
(x 2 2)(x 1 2)<br />
(x 2 2)(x 2 1 2x 1 4)<br />
lim c 4 2 !12 1 x ? 4 1 !12 1 x<br />
xS4 x 2 4 4 1 !12 1 x d<br />
16 2 (12 1 x)<br />
5 lim<br />
xS4 (x 2 4)(4 1 !12 1 x)<br />
4 2 x<br />
5 lim<br />
xS4 (x 2 4)(4 1 !12 1 x)<br />
2 (x 2 4)<br />
5 lim<br />
xS4 (x 2 4)(4 1 !12 1 x)<br />
21<br />
5 lim<br />
xS4 4 1 !12 1 x<br />
21<br />
5<br />
4 1 !12 1 (4)<br />
5 21<br />
4 1 4<br />
52 1 8<br />
1<br />
f. lim<br />
xS0 x a 1<br />
2 1 x 2 1 2 b<br />
5 lim c 1<br />
xS0 x 32 x<br />
2(2 1 x) d<br />
1<br />
5 lim c2<br />
xS0 2(2 1 x) d<br />
52 1 4<br />
18. a. The function is not defined for x , 3, so<br />
there is no left-side limit.<br />
b. Even after dividing out common factors from<br />
numerator and denominator, there is a factor of<br />
x 2 2 in the denominator; the graph has a vertical<br />
asymptote at x 5 2.<br />
25, if x , 1<br />
c. f(x) 5 e<br />
2, if x $ 1<br />
lim f(x) 525 2 lim f(x) 5 2<br />
xS1 2 xS1 1<br />
1-33
d. The function has a vertical asymptote at x 5 2.<br />
0 x 0<br />
e. lim<br />
xS0 x<br />
x S 0 2 0 x 0 52x<br />
0 x 0<br />
lim<br />
xS0 2 x 521<br />
0 x 0<br />
lim<br />
xS0 1 x 5 1<br />
0 x 0<br />
lim<br />
xS0 1 x 2 lim 0 x 0<br />
xS0 2 x<br />
5x 2 , if x ,21<br />
f. f(x) 5 e<br />
2x 1 1, if x $21<br />
lim f(x) 521<br />
xS21 1<br />
lim f(x) 5 5<br />
xS21 2<br />
lim f(x) 2 lim f(x)<br />
xS21 1 xS21 2<br />
Therefore, lim f(x) does not exist.<br />
xS21<br />
19. a.<br />
23(11 h) 2 1 6(11 h) 1 4 2 (23 1 6 1 4)<br />
m 5 lim<br />
hS0<br />
h<br />
23 2 6h 2 h 2 1 6 1 6h 1 4 2 7<br />
5 lim<br />
hS0<br />
h<br />
2h 2<br />
5 lim<br />
hS0 h<br />
5 lim 2h<br />
hS0<br />
5 0<br />
When x 5 1, y 5 7.<br />
The equation of the tangent is y 2 7 5 0(x 2 1)<br />
y 5 7<br />
b.<br />
(22 1 h) 2 2 (22 1 h) 2 1 2 (4 1 2 2 1)<br />
m 5 lim<br />
hS0<br />
h<br />
4 2 4h 1 h 2 1 2 2 h 2 1 2 5<br />
5 lim<br />
hS0<br />
h<br />
25h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim(25 1 h)<br />
hS0<br />
525<br />
When x 522, y 5 5.<br />
The equation of the tangent is y 2 5 525(x 1 2)<br />
y 525x 2 5<br />
6(21 1 h) 3 2 3 2 (26 2 3)<br />
c. m 5 lim<br />
hS0<br />
h<br />
6(21 1 3h 2 3h 2 1 h 3 ) 2 3 1 9<br />
5 lim<br />
hS0<br />
h<br />
18h 2 18h 2 1 6h 3<br />
5 lim<br />
hS0 h<br />
5 lim(18 2 18h 1 6h 2 )<br />
hS0<br />
5 18<br />
When x 521, y 529.<br />
The equation of the tangent is<br />
y 2 (29) 5 18(x 2 (21))<br />
y 5 18x 1 9<br />
22(3 1 h) 4 2 (2162)<br />
d. m 5 lim<br />
hS0 h<br />
22(81 1 108h 1 54h 2 1 12h 3 1 h 4 ) 1 162<br />
5 lim<br />
hS0<br />
h<br />
2216h 2 108h 2 2 24h 3 2 2h 4<br />
5 lim<br />
hS0<br />
h<br />
5 lim( 2 216 2 108h 2 24h 2 2 2h 3 )<br />
hS0<br />
52216<br />
When x 5 3, y 52162.<br />
The equation of the tangent is<br />
y 2 (2162) 52216(x 2 3)<br />
y 52216x 1 486<br />
20. P(t) 5 20 1 61t 1 3t 2<br />
a. P(8) 5 20 1 61(8) 1 3(8) 2<br />
5 700000<br />
b.<br />
20 1 61(8 1 h) 1 3(8 1 h) 2 2 (20 1 488 1 192)<br />
lim<br />
hS0<br />
h<br />
20 1 488 1 61h 1 3(64 1 16h 1 h 2 ) 2 700<br />
5 lim<br />
hS0<br />
h<br />
20 1 488 1 61h 1 192 1 48h 1 3h 2 2 700<br />
5 lim<br />
hS0<br />
h<br />
109h 1 3h 2<br />
5 lim<br />
hS0 h<br />
5 lim(109 1 3h)<br />
hS0<br />
5 109<br />
The population is changing at the rate of<br />
109000>h.<br />
<strong>Chapter</strong> 1 Test, p. 60<br />
1. lim does not exist since<br />
xS1<br />
lim 51`2 lim<br />
x 2 1 x 2 1 52`.<br />
2. f(x) 5 5x 2 2 8x<br />
f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36<br />
f(1) 5 5 2 8 523<br />
36 1 3<br />
Slope of secant is<br />
22 2 1 5239 3<br />
5213<br />
xS1 1 1<br />
1<br />
x 2 1<br />
xS1 2 1<br />
1-34 <strong>Chapter</strong> 1: Introduction to Calculus
3. a. lim f(x) does not exist.<br />
xS1<br />
b. lim f(x) 5 1<br />
xS2<br />
c. lim f(x) 5 1<br />
xS4 2<br />
d. f is discontinuous at x 5 1 and x 5 2.<br />
4. a. Average velocity from t 5 2 to t 5 5:<br />
s(5) 2 s(2) (40 2 25) 2 (16 2 4)<br />
5<br />
3<br />
3<br />
15 2 12<br />
5<br />
3<br />
5 1<br />
Average velocity from t 5 2 to t 5 5 is 1 km><br />
h.<br />
b. s(3 1 h) 2 s(3)<br />
5 8(3 1 h) 2 (3 1 h) 2 2 (24 2 9)<br />
5 24 1 8h 2 9 2 6h 2 h 2 2 15<br />
5 2h 2 h 2<br />
2h 2 h 2<br />
v(3) 5 lim 5 2<br />
hS0 h<br />
Velocity at t 5 3 is 2 km><br />
h.<br />
5. f(x) 5 "x 1 11<br />
Average rate of change from x 5 5 to x 5 5 1 h:<br />
f(5 1 h) 2 f(5)<br />
h<br />
"16 1 h 2 "16<br />
5<br />
h<br />
x<br />
6. f(x) 5<br />
x 2 2 15<br />
Slope of the tangent at x 5 4:<br />
4 1 h<br />
f(4 1 h) 5<br />
(4 1 h) 2 2 15<br />
4 1 h<br />
5<br />
1 1 8h 1 h 2<br />
f(4) 5 4 1<br />
4 1 h<br />
f(4 1 h) 2 f(4) 5<br />
1 1 8h 1 h 2 2 4<br />
4 1 h 2 4 2 32h 2 4h2<br />
5<br />
1 1 2h 1 h 2<br />
31h 2 4h2<br />
52<br />
(1 1 2h 1 h 2 )<br />
f(4 1 h) 2 f(4) (231 2 4h)<br />
lim<br />
5 lim<br />
hS0 h<br />
hS0 1 1 2h 1 h 2<br />
5231<br />
Slope of the tangent at x 5 4 is 231.<br />
4x 2 2 36<br />
7. a. lim<br />
xS3 2x 2 6<br />
5 lim 2(x 2 3)(x 1 3)<br />
xS3 (x 2 3)<br />
5 12<br />
2x 2 2 x 2 6<br />
b. lim<br />
xS2 3x 2 2 7x 1 2 5 lim (2x 1 3)(x 2 2)<br />
xS2 (x 2 2)(3x 2 1)<br />
c.<br />
A!x 2 1 2 2BA!x 2 1 1 2B<br />
5 lim<br />
xS5 !x 2 1 2 2<br />
5 4<br />
x 3 1 1<br />
d. lim<br />
xS21 x 4 2 1 5 lim (x 1 1)(x 2 2 x 1 1)<br />
xS21 (x 2 1)(x 1 1)(x 2 1 1)<br />
5 3<br />
22(2)<br />
e.<br />
f.<br />
(x 1 8) 1 3<br />
2 2<br />
lim<br />
xS0 x<br />
(x 1 8) 1 3<br />
5 lim<br />
2 2<br />
xS0 ((x 1 8) 1 3 2 2)((x 1 8) 2 3 1 2(x 1 8) 1 3 1 4)<br />
5<br />
lim<br />
xS5<br />
lim<br />
xS3<br />
5 7 5<br />
x 2 5<br />
!x 2 1 2 2 5 lim (x 2 1) 2 4<br />
xS5 !x 2 1 2 2<br />
52 3 4<br />
1<br />
a<br />
x 2 3 2 6<br />
x 2 2 9 b 5 lim<br />
xS3<br />
1<br />
4 1 4 1 4<br />
5 lim<br />
xS3<br />
5 1 6<br />
(x 1 8) 1 3<br />
2 2<br />
5 lim<br />
xS0 (x 1 8) 2 8<br />
5 1<br />
12<br />
ax 1 3, if x . 5<br />
8. f(x) 5 • 8, if x 5 5<br />
x 2 1 bx 1 a, if x , 5<br />
f(x) is continuous.<br />
Therefore, 5a 1 3 5 8<br />
25 1 5b 1 a 5 8<br />
(x 1 3) 2 6<br />
(x 2 3)(x 1 3)<br />
1<br />
x 1 3<br />
a 5 1<br />
5 b 5218<br />
b 52 18 5<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-35
CHAPTER 1<br />
Introduction to Calculus<br />
Review of Prerequisite Skills, pp. 2–3<br />
1. a. m 5 27 2 5<br />
6 2 2<br />
523<br />
b. m 5 4 2 (24)<br />
21 2 3<br />
522<br />
c. m 5 4 2 0<br />
1 2 0<br />
5 4<br />
d. m 5 4 2 0<br />
21 2 0<br />
524<br />
e. m 5 4 2 4.41<br />
22 2 (22.1)<br />
524.1<br />
f.<br />
m 5 21 4 2 1 4<br />
7<br />
4 2 3 4<br />
2 2 4<br />
5<br />
1<br />
52 1 2<br />
2. a. Substitute the given slope and y-intercept into<br />
y 5 mx 1 b.<br />
y 5 4x 2 2<br />
b. Substitute the given slope and y-intercept into<br />
y 5 mx 1 b.<br />
y 522x 1 5<br />
c. The slope of the line is<br />
m 5 12 2 6<br />
4 2 (21)<br />
5 6 5<br />
The equation of the line is in the form<br />
y 2 y 1 5 m(x 2 x 1 ). The point is (21, 6) and<br />
m 5 6 5.<br />
The equation of the line is y 2 6 5 6 5(x 1 1) or<br />
y 5 6 5(x 1 1) 1 6.<br />
8 2 4<br />
d. m 5<br />
26 2 (22)<br />
521<br />
y 2 4 521(x 2 (22))<br />
y 2 4 52x 2 2<br />
x 1 y 2 2 5 0<br />
e.<br />
f.<br />
3. a.<br />
b.<br />
c.<br />
d.<br />
4. a.<br />
b.<br />
x 523<br />
y 5 5<br />
f(2) 526 1 5<br />
521<br />
f(2) 5 (8 2 2)(6 2 6)<br />
5 0<br />
f(2) 523(4) 1 2(2) 2 1<br />
529<br />
f(2) 5 (10 1 2) 2<br />
5 144<br />
f(210) 5 210<br />
100 1 4<br />
f(23) 5 23<br />
9 1 4<br />
52 3 13<br />
c. f(0) 5 0<br />
0 1 4<br />
5 0<br />
d. f(10) 5 10<br />
100 1 4<br />
5 5<br />
52<br />
"3 2 x, if x , 0<br />
5. f(x) 5 •<br />
"3 1 x, if x $ 0<br />
a. f(233) 5 6<br />
b. f(0) 5 "3<br />
c. f(78) 5 9<br />
d. f(3) 5 "6<br />
1<br />
, if 23 , t , 0<br />
t<br />
6. s(t) 5 μ<br />
5, if t 5 0<br />
t 3 , if t . 0<br />
a. s(22) 52 1 2<br />
b. s(21) 521<br />
52 5<br />
52<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-1
c. s(0) 5 5<br />
d. s(1) 5 1<br />
e. s(100) 5 100 3 or 10 6<br />
7. a. (x 2 6)(x 1 2) 5 x 2 2 4x 2 12<br />
b. (5 2 x)(3 1 4x) 5 15 1 17x 2 4x 2<br />
c. x(5x 2 3) 2 2x(3x 1 2) 5 5x 2 2 3x 2 6x 2 2 4x<br />
52x 2 2 7x<br />
d. (x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)<br />
5 x 2 1 2x 2 3 2 (2x 2 1 x 2 10)<br />
5 2x 2 1 x 1 7<br />
e. (a 1 2) 3 5 (a 1 2)(a 1 2)(a 1 2)<br />
5 (a 2 1 4a 1 4)(a 1 2)<br />
5 a 3 1 6a 2 1 12a 1 8<br />
f. (9a 2 5) 3 5 (9a 2 5)(9a 2 5)(9a 2 5)<br />
5 (81a 2 2 90a 1 25)(9a 2 5)<br />
5 729a 3 2 1215a 2 1 675a 2 125<br />
8. a. x 3 2 x 5 x(x 2 2 1)<br />
5 x(x 1 1)(x 2 1)<br />
b.<br />
c.<br />
d.<br />
x 2 1 x 2 6 5 (x 1 3)(x 2 2)<br />
2x 2 2 7x 1 6 5 (2x 2 3)(x 2 2)<br />
x 3 1 2x 2 1 x 5 x(x 2 1 2x 1 1)<br />
5 x(x 1 1)(x 1 1)<br />
e. 27x 3 2 64 5 (3x 2 4)(9x 2 1 12x 1 16)<br />
f. 2x 3 2 x 2 2 7x 1 6<br />
x 5 1 is a zero, so x 2 1 is a factor. Synthetic or<br />
long division yields<br />
2x 3 2 x 2 2 7x 1 6 5 (x 2 1)(2x 2 1 x 2 6)<br />
5 (x 2 1)(2x 2 3)(x 1 2)<br />
5xPR 0 x $256<br />
9. a.<br />
b. 5xPR6<br />
c. 5xPR 0 x 2 16<br />
d.<br />
e.<br />
5xPR 0 x 2 06<br />
2x 2 2 5x 2 3 5 (2x 1 1)(x 2 3)<br />
e xPR ` x 2 2 1 2 , 3 f<br />
f. 5xPR 0 x 2 25, 22, 16<br />
10. a. h(0) 5 2, h(1) 5 22.1<br />
average rate of change 5 22.1 2 2<br />
1 2 0<br />
5 20.1 ms ><br />
b. h(1) 5 22.1, h(2) 5 32.4<br />
32.4 2 22.1<br />
average rate of change 5<br />
2 2 1<br />
5 10.3 ms ><br />
11. a. The average rate of change during the second<br />
hour is the difference in the volume at t 5 120 and<br />
t 5 60 (since t is measured in minutes), divided by<br />
the difference in time.<br />
V(120) 2 V(60)<br />
120 2 60<br />
b. To estimate the instantaneous rate of change in<br />
volume after exactly 60 minutes, calculate the average<br />
rate of change in volume from minute 59 to minute 61.<br />
V(61) 2 V(59) 1186.56 2 1213.22<br />
8<br />
61 2 59<br />
2<br />
5213.33 L>min<br />
c. The instantaneous rate of change in volume is<br />
negative for 0 # t # 120 because the volume of<br />
water in the hot tub is always decreasing during that<br />
time period, a negative change.<br />
12. a., b.<br />
y<br />
8<br />
The slope of the tangent line is 28.<br />
c. The instantaneous rate of change in f(x) when<br />
x 5 5 is 28.<br />
1.1 Radical Expressions:<br />
Rationalizating Denominators, p. 9<br />
1. a. 2"3 1 4<br />
b. "3 2 "2<br />
c. 2"3 1 "2<br />
d. 3"3 2 "2<br />
e. "2 1 "5<br />
f. 2"5 2 2"2<br />
"3 1 "5<br />
2. a.<br />
? "2<br />
"2 "2<br />
5<br />
"6 1 "10<br />
2<br />
2"3 2 3"2<br />
b.<br />
"2<br />
5 2"6 2 6<br />
2<br />
5 "6 2 3<br />
–2<br />
4<br />
–4<br />
–8<br />
5 0 2 1200<br />
60<br />
5220 L>min<br />
0<br />
? "2<br />
"2<br />
2<br />
4 6<br />
x<br />
1-2 <strong>Chapter</strong> 1: Introduction to Calculus
c. s(0) 5 5<br />
d. s(1) 5 1<br />
e. s(100) 5 100 3 or 10 6<br />
7. a. (x 2 6)(x 1 2) 5 x 2 2 4x 2 12<br />
b. (5 2 x)(3 1 4x) 5 15 1 17x 2 4x 2<br />
c. x(5x 2 3) 2 2x(3x 1 2) 5 5x 2 2 3x 2 6x 2 2 4x<br />
52x 2 2 7x<br />
d. (x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)<br />
5 x 2 1 2x 2 3 2 (2x 2 1 x 2 10)<br />
5 2x 2 1 x 1 7<br />
e. (a 1 2) 3 5 (a 1 2)(a 1 2)(a 1 2)<br />
5 (a 2 1 4a 1 4)(a 1 2)<br />
5 a 3 1 6a 2 1 12a 1 8<br />
f. (9a 2 5) 3 5 (9a 2 5)(9a 2 5)(9a 2 5)<br />
5 (81a 2 2 90a 1 25)(9a 2 5)<br />
5 729a 3 2 1215a 2 1 675a 2 125<br />
8. a. x 3 2 x 5 x(x 2 2 1)<br />
5 x(x 1 1)(x 2 1)<br />
b.<br />
c.<br />
d.<br />
x 2 1 x 2 6 5 (x 1 3)(x 2 2)<br />
2x 2 2 7x 1 6 5 (2x 2 3)(x 2 2)<br />
x 3 1 2x 2 1 x 5 x(x 2 1 2x 1 1)<br />
5 x(x 1 1)(x 1 1)<br />
e. 27x 3 2 64 5 (3x 2 4)(9x 2 1 12x 1 16)<br />
f. 2x 3 2 x 2 2 7x 1 6<br />
x 5 1 is a zero, so x 2 1 is a factor. Synthetic or<br />
long division yields<br />
2x 3 2 x 2 2 7x 1 6 5 (x 2 1)(2x 2 1 x 2 6)<br />
5 (x 2 1)(2x 2 3)(x 1 2)<br />
5xPR 0 x $256<br />
9. a.<br />
b. 5xPR6<br />
c. 5xPR 0 x 2 16<br />
d.<br />
e.<br />
5xPR 0 x 2 06<br />
2x 2 2 5x 2 3 5 (2x 1 1)(x 2 3)<br />
e xPR ` x 2 2 1 2 , 3 f<br />
f. 5xPR 0 x 2 25, 22, 16<br />
10. a. h(0) 5 2, h(1) 5 22.1<br />
average rate of change 5 22.1 2 2<br />
1 2 0<br />
5 20.1 ms ><br />
b. h(1) 5 22.1, h(2) 5 32.4<br />
32.4 2 22.1<br />
average rate of change 5<br />
2 2 1<br />
5 10.3 ms ><br />
11. a. The average rate of change during the second<br />
hour is the difference in the volume at t 5 120 and<br />
t 5 60 (since t is measured in minutes), divided by<br />
the difference in time.<br />
V(120) 2 V(60)<br />
120 2 60<br />
b. To estimate the instantaneous rate of change in<br />
volume after exactly 60 minutes, calculate the average<br />
rate of change in volume from minute 59 to minute 61.<br />
V(61) 2 V(59) 1186.56 2 1213.22<br />
8<br />
61 2 59<br />
2<br />
5213.33 L>min<br />
c. The instantaneous rate of change in volume is<br />
negative for 0 # t # 120 because the volume of<br />
water in the hot tub is always decreasing during that<br />
time period, a negative change.<br />
12. a., b.<br />
y<br />
8<br />
The slope of the tangent line is 28.<br />
c. The instantaneous rate of change in f(x) when<br />
x 5 5 is 28.<br />
1.1 Radical Expressions:<br />
Rationalizating Denominators, p. 9<br />
1. a. 2"3 1 4<br />
b. "3 2 "2<br />
c. 2"3 1 "2<br />
d. 3"3 2 "2<br />
e. "2 1 "5<br />
f. 2"5 2 2"2<br />
"3 1 "5<br />
2. a.<br />
? "2<br />
"2 "2<br />
5<br />
"6 1 "10<br />
2<br />
2"3 2 3"2<br />
b.<br />
"2<br />
5 2"6 2 6<br />
2<br />
5 "6 2 3<br />
–2<br />
4<br />
–4<br />
–8<br />
5 0 2 1200<br />
60<br />
5220 L>min<br />
0<br />
? "2<br />
"2<br />
2<br />
4 6<br />
x<br />
1-2 <strong>Chapter</strong> 1: Introduction to Calculus
4"3 1 3"2<br />
c.<br />
2"3<br />
5<br />
5 4 1 "6<br />
2<br />
3"5 2 "2<br />
d.<br />
2"2<br />
2"5<br />
b.<br />
2"5 1 3"2<br />
5<br />
"3 2 "2<br />
c.<br />
"3 1 "2<br />
d.<br />
5<br />
5<br />
2"3 2 "2<br />
e.<br />
5"2 1 "3<br />
5<br />
5<br />
12 1 3"6<br />
6<br />
5 3"10 2 2<br />
4<br />
5 "5 1 "2<br />
20 2 6"10<br />
20 2 18<br />
5 10 2 3"10<br />
5 3 1 2"6 1 2<br />
3 2 2<br />
5 5 1 2"6<br />
44 2 22"5<br />
11<br />
5 4 2 2"5<br />
? "3<br />
"3<br />
? "2<br />
"2<br />
3 "5 1 "2<br />
3. a.<br />
?<br />
"5 2 "2 "5 1 "2<br />
3("5 1 "2)<br />
5<br />
3<br />
2"5 2 8<br />
2"5 1 3 ? 2"5 2 3<br />
2"5 2 3<br />
20 2 22"5 1 24<br />
20 2 9<br />
?<br />
2"5 2 3"2<br />
2"5 2 3"2<br />
?<br />
"3 2 "2<br />
"3 2 "2<br />
?<br />
5"2 2 "3<br />
5"2 2 "3<br />
10"6 2 6 2 10 1 "6<br />
50 2 3<br />
11"6 2 16<br />
47<br />
3"3 2 2"2<br />
f.<br />
3"3 1 2"2<br />
5<br />
5<br />
"5 2 1<br />
4. a. ? "5 1 1<br />
4 "5 1 1<br />
5 2 1<br />
5<br />
4("5 1 1)<br />
1<br />
5<br />
!5 1 1<br />
2 2 3"2<br />
b. ? 2 1 3"2<br />
2 2 1 3"2<br />
4 2 18<br />
5<br />
2(2 1 3"2)<br />
27<br />
5<br />
2 1 3"2<br />
"5 1 2<br />
c.<br />
2"5 2 1 ? "5 2 2<br />
"5 2 2<br />
5 2 4<br />
5<br />
10 2 5"5 1 2<br />
1<br />
5<br />
12 2 5!5<br />
8"2<br />
5. a.<br />
"20 2 "18<br />
5<br />
5<br />
5 8"10 1 24<br />
8"2<br />
b.<br />
2"5 2 3"2<br />
5<br />
5<br />
27 2 12"6 1 8<br />
27 2 8<br />
35 2 12"6<br />
19<br />
8"40 1 8"36<br />
20 2 18<br />
16"10 1 48<br />
2<br />
16"10 1 48<br />
20 2 18<br />
16"10 1 48<br />
2<br />
?<br />
3"3 2 2"2<br />
3"3 2 2"2<br />
?<br />
"20 1 "18<br />
"20 1 "18<br />
?<br />
2"5 1 3"2<br />
2"5 1 3"2<br />
5 8"10 1 24<br />
c. The expressions in the two parts are equivalent.<br />
The radicals in the denominator of part a. have been<br />
simplified in part b.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-3
2"2<br />
6. a.<br />
2"3 2 "8<br />
5 4"6 1 8<br />
6 2 8<br />
5 22"3 2 4<br />
2"6<br />
b.<br />
2"27 2 "8<br />
5<br />
5<br />
5<br />
2"2<br />
c.<br />
"16 2 "12<br />
5<br />
5<br />
3"2 1 2"3<br />
d.<br />
"12 2 "8<br />
5<br />
5<br />
e.<br />
5<br />
12"15 1 15"10<br />
5 2<br />
2<br />
"18 1 "12<br />
f.<br />
"18 2 "12<br />
5<br />
5<br />
4"162 1 2"48<br />
54 2 8<br />
36"2 1 8"3<br />
46<br />
18"2 1 4"3<br />
23<br />
2"2<br />
4 2 2"3 ? 4 1 2"3<br />
4 1 2"3<br />
8"2 1 4"6<br />
16 2 12<br />
5 2"2 1 "6<br />
3"24 1 12 1 12 1 2"24<br />
12 2 8<br />
24 1 15"3<br />
4<br />
3!5 4!3 1 5!2<br />
?<br />
4!3 2 5!2 4!3 1 5!2<br />
12"15 1 15"10<br />
48 2 50<br />
18 1 2"216 1 12<br />
18 2 12<br />
30 1 12"6<br />
6<br />
5 5 1 2"6<br />
?<br />
2"3 1 "8<br />
2"3 1 "8<br />
?<br />
2"27 1 "8<br />
2"27 1 "8<br />
?<br />
"12 1 "8<br />
"12 1 "8<br />
?<br />
"18 1 "12<br />
"18 1 "12<br />
"a 2 2<br />
7. a.<br />
a 2 4 ? "a 1 2<br />
"a 1 2<br />
a 2 4<br />
5<br />
(a 2 4)("a 2 2)<br />
1<br />
5<br />
"a 2 2<br />
"x 1 4 2 2<br />
b.<br />
? "x 1 4 1 2<br />
x "x 1 4 1 2<br />
x 1 4 2 4<br />
5<br />
x("x 1 4 1 2)<br />
x<br />
5<br />
x("x 1 4 1 2)<br />
1<br />
5<br />
"x 1 4 2 2<br />
!x 1 h 2 !x<br />
c.<br />
?<br />
h<br />
x 1 h 2 x<br />
5<br />
5<br />
5<br />
1.2 The Slope of a Tangent, pp. 18–21<br />
1. a. m 5 28 2 7<br />
23 2 2<br />
5 3<br />
b.<br />
hA!x 1 h 1 !xB<br />
h<br />
hA!x 1 h 1 !xB<br />
1<br />
!x 1 h 1 !x<br />
m 5 27 2 2 3 2<br />
7<br />
2 2 1 2<br />
5 210 2<br />
6<br />
2<br />
52 5 3<br />
21 2 (22.6)<br />
c. m 5<br />
1.5 2 6.3<br />
!x 1 h 1 !x<br />
!x 1 h 1 !x<br />
52 1 3<br />
2. a. The slope of the given line is 3, so the slope<br />
of a line perpendicular to the given line is 2 1 3.<br />
b. 13x 2 7y 2 11 5 0<br />
27y 5213x 2 11<br />
y 5 13<br />
7 x 1 11<br />
7<br />
13<br />
The slope of the given line is 7 , so the slope of a line<br />
perpendicular to the given line is 213.<br />
7<br />
1-4 <strong>Chapter</strong> 1: Introduction to Calculus
2"2<br />
6. a.<br />
2"3 2 "8<br />
5 4"6 1 8<br />
6 2 8<br />
5 22"3 2 4<br />
2"6<br />
b.<br />
2"27 2 "8<br />
5<br />
5<br />
5<br />
2"2<br />
c.<br />
"16 2 "12<br />
5<br />
5<br />
3"2 1 2"3<br />
d.<br />
"12 2 "8<br />
5<br />
5<br />
e.<br />
5<br />
12"15 1 15"10<br />
5 2<br />
2<br />
"18 1 "12<br />
f.<br />
"18 2 "12<br />
5<br />
5<br />
4"162 1 2"48<br />
54 2 8<br />
36"2 1 8"3<br />
46<br />
18"2 1 4"3<br />
23<br />
2"2<br />
4 2 2"3 ? 4 1 2"3<br />
4 1 2"3<br />
8"2 1 4"6<br />
16 2 12<br />
5 2"2 1 "6<br />
3"24 1 12 1 12 1 2"24<br />
12 2 8<br />
24 1 15"3<br />
4<br />
3!5 4!3 1 5!2<br />
?<br />
4!3 2 5!2 4!3 1 5!2<br />
12"15 1 15"10<br />
48 2 50<br />
18 1 2"216 1 12<br />
18 2 12<br />
30 1 12"6<br />
6<br />
5 5 1 2"6<br />
?<br />
2"3 1 "8<br />
2"3 1 "8<br />
?<br />
2"27 1 "8<br />
2"27 1 "8<br />
?<br />
"12 1 "8<br />
"12 1 "8<br />
?<br />
"18 1 "12<br />
"18 1 "12<br />
"a 2 2<br />
7. a.<br />
a 2 4 ? "a 1 2<br />
"a 1 2<br />
a 2 4<br />
5<br />
(a 2 4)("a 2 2)<br />
1<br />
5<br />
"a 2 2<br />
"x 1 4 2 2<br />
b.<br />
? "x 1 4 1 2<br />
x "x 1 4 1 2<br />
x 1 4 2 4<br />
5<br />
x("x 1 4 1 2)<br />
x<br />
5<br />
x("x 1 4 1 2)<br />
1<br />
5<br />
"x 1 4 2 2<br />
!x 1 h 2 !x<br />
c.<br />
?<br />
h<br />
x 1 h 2 x<br />
5<br />
5<br />
5<br />
1.2 The Slope of a Tangent, pp. 18–21<br />
1. a. m 5 28 2 7<br />
23 2 2<br />
5 3<br />
b.<br />
hA!x 1 h 1 !xB<br />
h<br />
hA!x 1 h 1 !xB<br />
1<br />
!x 1 h 1 !x<br />
m 5 27 2 2 3 2<br />
7<br />
2 2 1 2<br />
5 210 2<br />
6<br />
2<br />
52 5 3<br />
21 2 (22.6)<br />
c. m 5<br />
1.5 2 6.3<br />
!x 1 h 1 !x<br />
!x 1 h 1 !x<br />
52 1 3<br />
2. a. The slope of the given line is 3, so the slope<br />
of a line perpendicular to the given line is 2 1 3.<br />
b. 13x 2 7y 2 11 5 0<br />
27y 5213x 2 11<br />
y 5 13<br />
7 x 1 11<br />
7<br />
13<br />
The slope of the given line is 7 , so the slope of a line<br />
perpendicular to the given line is 213.<br />
7<br />
1-4 <strong>Chapter</strong> 1: Introduction to Calculus
3. a.<br />
y 2 (24) 5 7 (x 2 (24))<br />
17<br />
17y 1 68 5 7x 1 28<br />
7x 2 17y 2 40 5 0<br />
y<br />
4<br />
–2<br />
b. The slope and y-intercept are given.<br />
y 5 8x 1 6<br />
y<br />
8<br />
–4<br />
c. (0, 23), (5, 0)<br />
m 5 0 2 (23)<br />
5 2 0<br />
5 3 5<br />
m 5 25 3 2 (24)<br />
5<br />
3 2 (24)<br />
2<br />
–2<br />
–4<br />
5<br />
0<br />
–2<br />
7<br />
3<br />
17<br />
3<br />
5 7<br />
17<br />
y 2 0 5 3 (x 2 5)<br />
5<br />
3x 2 5y 2 15 5 0<br />
4<br />
–4<br />
–8<br />
0<br />
2<br />
4 6<br />
2<br />
4<br />
x<br />
x<br />
d. The line is a vertical line because both points<br />
have the same x-coordinate.<br />
x 5 5<br />
y<br />
4<br />
4. a.<br />
b.<br />
c.<br />
d.<br />
–2<br />
–2<br />
4<br />
2<br />
–2<br />
–4<br />
0<br />
2<br />
–2<br />
–4<br />
0<br />
y<br />
2<br />
2<br />
(5 1 h) 3 2 125<br />
h<br />
5 (5 1 h 2 5)((5 1 h)2 1 5(5 1 h) 1 25)<br />
h<br />
5 h(75 1 15h 1 h2 )<br />
h<br />
5 75 1 15h 1 h 2<br />
(3 1 h) 4 2 81<br />
h<br />
5 ((3 1 h)2 2 9)((3 1 h) 2 1 9)<br />
h<br />
5 (9 1 6h 1 h2 2 9)(9 1 6h 1 h 2 1 9)<br />
h<br />
5 (6 1 h)(18 1 6h 1 h 2 )<br />
5 108 1 54h 1 12h 2 1 h 3<br />
1<br />
1 1 h 2 1<br />
5 1 2 1 2 h<br />
h h(1 1 h) 52 1<br />
1 1 h<br />
3(1 1 h) 2 2 3<br />
h<br />
4 6<br />
4 6<br />
5 3((1 1 h)2 2 1)<br />
h<br />
5 3(1 1 2h 1 h2 2 1)<br />
h<br />
x<br />
x<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-5
e.<br />
f.<br />
3<br />
4 1 h 2 3 4<br />
5<br />
h<br />
21<br />
2 1 h 1 1 2<br />
5<br />
h<br />
5<br />
5 1<br />
4 1 2h<br />
"h 2 1 5h 1 4 2 2<br />
b.<br />
5 h2 1 5h 1 4 2 4<br />
h<br />
"5 1 h 2 "5 5 1 h 2 5<br />
c.<br />
5<br />
h h("5 1 h 1 "5)<br />
1<br />
5<br />
"5 1 h 1 "5<br />
6. a. P(1, 3), Q(1 1 h, f(1 1 h)), f(x) 5 3x 2<br />
m 5 3(1 1 h)2 2 3<br />
h<br />
5 6 1 3h<br />
b. P(1, 3), Q(1 1 h, (1 1 h) 3 1 2)<br />
m 5 (1 1 h)3 1 2 2 3<br />
h<br />
5 1 1 3h 1 3h2 1 h 3 2 1<br />
h<br />
5 3 1 3h 1 h 2<br />
c. P(9, 3), Q(9 1 h, "9 1 h)<br />
m 5 "9 1 h 2 3<br />
h<br />
1<br />
5<br />
"9 1 h 1 3<br />
5 3(2h 1 h2 )<br />
h<br />
5 6 1 3h<br />
12 2 12 2 3h<br />
4(4 1 h)<br />
h<br />
5 23<br />
4(4 1 h)<br />
22 1 2 1 h<br />
2(2 1 h)<br />
h<br />
h<br />
2h(2 1 h)<br />
"16 1 h 2 4 16 1 h 2 16<br />
5. a.<br />
5<br />
h h("16 1 h 1 4)<br />
1<br />
5<br />
"16 1 h 1 4<br />
h("h 2 1 5h 1 4 1 2)<br />
h 1 5<br />
5<br />
"h 2 1 5h 1 4 1 2<br />
? "9 1 h 1 3<br />
"9 1 h 1 3<br />
7. a.<br />
b. 12<br />
c. (2, 8), ((2 1 h), (2 1 h) 3 )<br />
m 5 (2 1 h)3 2 8<br />
2 1 h 2 2<br />
5 8 1 12h 1 6h2 1 h 3 2 8<br />
h<br />
5 12 1 6h 1 h 2<br />
d. m 5 lim(12 1 6h 1 h 2 )<br />
hS0<br />
5 12<br />
e. They are the same.<br />
f.<br />
y<br />
12<br />
–4<br />
P Q Slope of Line PQ<br />
(2, 8) (3, 27) 19<br />
(2, 8) (2.5, 15.625) 15.25<br />
(2, 8) (2.1, 9.261) 12.61<br />
(2, 8) (2.01, 8.120 601) 12.060 1<br />
(2, 8) (1, 1) 7<br />
(2, 8) (1.5, 3.375) 9.25<br />
(2, 8) (1.9, 6.859) 11.41<br />
(2, 8) (1.99, 7.880 599) 11.940 1<br />
–2<br />
8<br />
4<br />
–4<br />
0<br />
8. a. y 5 3x 2 , (22, 12)<br />
3(22 1 h) 2 2 12<br />
m 5 lim<br />
hS0 h<br />
12 2 12h 1 3h 2 2 12<br />
5 lim<br />
hS0 h<br />
5 lim(212 1 3h)<br />
hS0<br />
5212<br />
b. y 5 x 2 2 x at x 5 3, y 5 6.<br />
(3 1 h) 2 2 (3 1 h) 2 6<br />
m 5 lim<br />
hS0<br />
h<br />
9 1 6h 1 h 2 2 3 2 h 2 6<br />
5 lim<br />
hS0<br />
h<br />
5 lim(5 1 h)<br />
hS0<br />
5 5<br />
2<br />
4<br />
x<br />
1-6 <strong>Chapter</strong> 1: Introduction to Calculus
c. at x 522, y 528.<br />
(22 1 h) 3 1 8<br />
m 5 lim<br />
hS0 h<br />
28 1 12h 2 6h 2 1 h 3 1 8<br />
5 lim<br />
hS0<br />
h<br />
5 lim(12 2 6h 1 h 2 )<br />
5 12<br />
9. a. y 5 "x 2 2; (3, 1)<br />
"3 1 h 2 2 2 1<br />
m 5 lim<br />
hS0 h<br />
5 1 2<br />
b. y 5 "x 2 5 at x 5 9, y 5 2<br />
"9 1 h 2 5 2 2<br />
m 5 lim<br />
hS0 h<br />
5 lim £ "4 1 h 2 2 3 "4 1 h 1 2<br />
hS0 h "4 1 h 1 2 §<br />
1<br />
5 lim<br />
hS0 "4 1 h 1 2<br />
5 1 4<br />
c. y 5 "5x 2 1 at x 5 2, y 5 3<br />
"10 1 5h 2 1 2 3<br />
m 5 lim<br />
hS0 h<br />
"9 1 5h 2 3 "9 1 5h 1 3<br />
5 lim £ 3<br />
hS0 h "9 1 5h 1 3 §<br />
5<br />
5 lim<br />
hS0 "9 1 5h 1 3<br />
5 5 6<br />
hS0<br />
5 lim £ "1 1 h 2 1 3 "1 1 h 1 1<br />
hS0 h "1 1 h 1 1 §<br />
1<br />
5 lim<br />
hS0 "1 1 h 1 1<br />
10. a. y 5 8 at (2, 4)<br />
x<br />
8<br />
2 1 h 2 4<br />
m 5 lim<br />
hS0 h<br />
24<br />
5 lim<br />
hS0 2 1 h<br />
522<br />
b. y 5 8 at x 5 1; y 5 2<br />
3 1 x<br />
8<br />
4 1 h 2 2<br />
m 5 lim<br />
hS0 h<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
5 lim<br />
52 1 2<br />
c. y 5 1 at x 5 3; y 5 1 x 1 2<br />
5<br />
m 5 lim<br />
hS0 h<br />
21<br />
5 lim<br />
hS0 5(5 1 h)<br />
52 1<br />
10<br />
11. a. Let y 5 f(x).<br />
f(2) 5 (2) 2 2 3(2) 5 4 2 6 522<br />
f(2 1 h) 5 (2 1 h) 2 2 3(2 1 h)<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 2 is<br />
f(2 1 h) 2 f(2)<br />
m 5 lim<br />
hS0 h<br />
(2 1 h) 2 2 3(2 1 h) 2 (22)<br />
5 lim<br />
hS0<br />
h<br />
4 1 4h 1 h 2 2 6 2 3h 1 2<br />
5 lim<br />
hS0<br />
h<br />
h 2 1 h<br />
5 lim<br />
hS0 h<br />
5 lim (h 1 1)<br />
hS0<br />
hS0<br />
22<br />
4 1 h<br />
1<br />
5 1 h 2 1 5<br />
5 0 1 1<br />
5 1<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 x 2 2 3x at x 5 2 is 1.<br />
b. f(22) 5 4<br />
22 522 4<br />
f(22 1 h) 5<br />
22 1 h<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 522 is<br />
f(22 1 h) 2 f(22)<br />
m 5 lim<br />
hS0 h<br />
4<br />
22 1 h<br />
5 lim<br />
2 (22)<br />
hS0 h<br />
4<br />
22 1 h<br />
5 lim<br />
1 2<br />
hS0 h<br />
5 lim c 4 2 4 1 2h ? 1<br />
hS0 22 1 h h d<br />
2h<br />
5 lim c<br />
hS0 22 1 h ? 1 h d<br />
y 5 x 3 1-7
2<br />
5 lim<br />
hS0 22 1 h<br />
2<br />
5<br />
22 1 0<br />
521<br />
Therefore, the slope of the tangent to f(x) 5 4 at<br />
x<br />
x 522 is 21.<br />
c. Let y 5 f(x).<br />
f(1) 5 3(1) 3 5 3<br />
f(1 1 h) 5 3(1 1 h) 3<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 1 is<br />
f(1 1 h) 2 f(1)<br />
m 5 lim<br />
hS0 h<br />
3(1 1 h) 3 2 3<br />
5 lim<br />
hS0 h<br />
Using the binomial formula to expand (1 1 h) 3 (or<br />
one could simply expand using algebra), the slope m is<br />
3(h 3 1 3h 2 1 3h 1 1) 2 (3)<br />
5 lim<br />
hS0<br />
h<br />
3h 3 1 9h 2 1 9h 1 3 2 3<br />
5 lim<br />
hS0<br />
h<br />
3h 3 1 9h 2 1 9h<br />
5 lim<br />
hS0 h<br />
5 lim (3h 2 1 9h 1 9)<br />
hS0<br />
5 3(0) 1 9(0) 1 9<br />
5 9<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 3x 3 at x 5 1 is 9.<br />
d. Let y 5 f(x).<br />
f(16) 5 !16 2 7 5 !9 5 3<br />
f(16 1 h) 5 !16 1 h 2 7 5 !h 1 9<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 16 is<br />
f(16 1 h) 2 f(16)<br />
m 5 lim<br />
hS0 h<br />
!h 1 9 2 3<br />
5 lim<br />
hS0 h<br />
!h 1 9 2 3<br />
5 lim<br />
? !h 1 9 1 3<br />
hS0 h !h 1 9 1 3<br />
(h 1 9) 2 9<br />
5 lim<br />
hS0 h( !h 1 9 1 3)<br />
h<br />
5 lim<br />
hS0 h( !h 1 9 1 3)<br />
1<br />
5 lim<br />
hS0 !h 1 9 1 3<br />
1<br />
5<br />
!0 1 9 1 3<br />
5 1<br />
3 1 3<br />
5 1 6<br />
Therefore, the slope of the tangent to<br />
1<br />
y 5 f(x) 5 !x 2 7 at x 5 16 is 6.<br />
e. Let y 5 f(x).<br />
f(3) 5 "25 2 (3) 2 5 !25 2 9 5 4<br />
f(3 1 h) 5 "25 2 (3 1 h) 2<br />
5 "25 2 9 2 6h 2 h 2<br />
5 "16 2 6h 2 h 2<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 3 is<br />
f(3 1 h) 2 f(3)<br />
m 5 lim<br />
hS0 h<br />
"16 2 6h 2 h 2 2 4<br />
5 lim<br />
hS0 h<br />
5 lim c "16 2 6h 2 h2 2 4<br />
hS0 h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 "25 2 (9 1 6h 1 h 2 )<br />
3 "16 2 6h 2 h2 1 4<br />
"16 2 6h 2 h 2 1 4 d<br />
16 2 6h 2 h 2 2 16<br />
h("16 2 6h 2 h 2 1 4)<br />
h(26 2 h)<br />
h("16 2 6h 2 h 2 1 4)<br />
26 2 h<br />
5 lim<br />
hS0 "16 2 6h 2 h 2 1 4<br />
26 2 0<br />
5<br />
"16 2 6(0) 2 (0) 2 1 4<br />
5 26<br />
!16 1 4<br />
5 26<br />
8<br />
52 3 4<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 "25 2 x 2 at x 5 3 is 2 3 4.<br />
f. Let y 5 f(x).<br />
f(8) 5 4 1 8<br />
8 2 2 5 12<br />
6 5 2<br />
4 1 (8 1 h)<br />
f(8 1 h) 5<br />
(8 1 h) 2 2 5 12 1 h<br />
6 1 h<br />
1-8 <strong>Chapter</strong> 1: Introduction to Calculus
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 8 is<br />
f(8 1 h) 2 f(8)<br />
m 5 lim<br />
hS0 h<br />
12 1 h<br />
6 1 h<br />
5 lim<br />
2 2<br />
hS0 h<br />
12 1 h 2 12 2 2h<br />
5 lim<br />
? 1<br />
hS0 6 1 h h<br />
2h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 21<br />
6 1 0<br />
52 1 6<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 4 1 x at x 5 8 is 2 1 x 2 2<br />
6.<br />
12.<br />
y<br />
8<br />
–4<br />
6 1 h ? 1 h<br />
21<br />
6 1 h<br />
4<br />
–4<br />
0<br />
A<br />
4<br />
y 5 "25 2 x 2 S Semi-circle centre (0, 0)<br />
rad 5, y $ 0<br />
OA is a radius.<br />
4<br />
The slope of OA is 3.<br />
The slope of tangent is 2 3 4.<br />
13. Take values of x close to the point, then<br />
Dy<br />
determine<br />
Dx .<br />
14.<br />
Since the tangent is horizontal, the slope is 0.<br />
(3 1 h) 2 2 3(3 1 h) 1 1 2 1<br />
15. m 5 lim<br />
hS0<br />
h<br />
9 1 6h 1 h 2 2 9 2 3h<br />
5 lim<br />
hS0 h<br />
3h 1 h 2<br />
5 lim<br />
hS0 h<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
8<br />
x<br />
5 lim (3 1 h)<br />
hS0<br />
5 3<br />
The slope of the tangent is 3.<br />
y 2 1 5 3(x 2 3)<br />
3x 2 y 2 8 5 0<br />
(2 1 h) 2 2 7(2 1 h) 1 12 2 2<br />
16. m 5 lim<br />
hS0<br />
h<br />
4 1 4h 1 h 2 2 14 2 7h 1 10<br />
5 lim<br />
hS0<br />
h<br />
23h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim ( 2 3 1 h)<br />
hS0<br />
523<br />
The slope of the tangent is 23.<br />
When x 5 2, y 5 2.<br />
y 2 2 523(x 2 2)<br />
3x 1 y 2 8 5 0<br />
17. a. f(3) 5 9 2 12 1 1 522; (3, 22)<br />
b. f(5) 5 25 2 20 1 1 5 6; (5, 6)<br />
c. The slope of secant AB is<br />
m AB 5 6 2 (22)<br />
5 2 3<br />
5 8 2<br />
5 4<br />
The equation of the secant is<br />
y 2 y 1 5 m AB (x 2 x 1 )<br />
y 1 2 5 4(x 2 3)<br />
y 5 4x 2 14<br />
d. Calculate the slope of the tangent.<br />
f(x 1 h) 2 f(x)<br />
m 5 lim<br />
hS0 h<br />
(x 1 h) 2 2 4(x 1 h) 1 1 2 (x 2 2 4x 1 1)<br />
5 lim<br />
hS0<br />
h<br />
x 2 1 2xh 1 h 2 2 4x 2 4h 1 1 2 x 2 1 4x 2 1<br />
5 lim<br />
hS0<br />
h<br />
2xh 1 h 2 2 4h<br />
5 lim<br />
hS0 h<br />
5 lim (2x 1 h 2 4)<br />
hS0<br />
5 2x 1 0 2 4<br />
5 2x 2 4<br />
When x 5 3, the slope is 2(3) 2 4 5 2. So the<br />
equation of the tangent at A(3, 22) is<br />
y 2 y 1 5 m(x 2 x 1 )<br />
y 1 2 5 2(x 2 3)<br />
y 5 2x 2 8<br />
1-9
e. When x 5 5, the slope of the tangent is<br />
2(5) 2 4 5 6.<br />
So the equation of the tangent at B(5, 6) is<br />
y 2 y 1 5 m(x 2 x 1 )<br />
y 2 6 5 6(x 2 5)<br />
y 5 6x 2 24<br />
18. a.<br />
b.<br />
c.<br />
d.<br />
The slope is undefined.<br />
The slope is 0.<br />
P<br />
The slope is about –2.5.<br />
P<br />
P<br />
P<br />
20. C(t) 5 100t 2 1 400t 1 5000<br />
Slope at t 5 6<br />
Cr(t) 5 200t 1 400<br />
Cr(6) 5 1200 1 400 5 1600<br />
Increasing at a rate of 1600 papers per month.<br />
21. Point on f(x) 5 3x 2 2 4x tangent parallel to<br />
y 5 8x. Therefore, tangent line has slope 8.<br />
3(h 1 a) 2 2 4(h 1 a) 2 3(a 2 1 4a)<br />
m 5 lim<br />
5 8<br />
hS0<br />
h<br />
3h 2 1 6ah 2 4h<br />
lim<br />
5 8<br />
hS0 h<br />
6a 2 4 5 8<br />
a 5 2<br />
The point has coordinates (2, 4).<br />
22.<br />
y 5 1 3 x3 2 5x 2 4 x<br />
1<br />
3 (a 1 h)2 2 1 3 a3<br />
limaa 2 1 ah 1 1<br />
hS0<br />
3 h3 b 5 a 2<br />
5 lim 2<br />
hS0<br />
(a 1 h) 2 (2a)<br />
h<br />
2 4<br />
lim<br />
hS0<br />
525<br />
a 1 h 1 4 a<br />
4<br />
a(a 1 h) 5 4 a 2<br />
5 a 2 h 1 ah 2 1 1 3 h3<br />
524a<br />
1 4a 1 4h<br />
a(a 1 h)<br />
e.<br />
52 10 8<br />
52 5 4<br />
The slope is about 1.<br />
The slope is about 2 7 8.<br />
f. There is no tangent at this point.<br />
19. D(p) 5 20 p . 1 at (5, 10)<br />
"p 2 1 ,<br />
20<br />
!4 1 h 2 10<br />
m 5 lim<br />
hS0 h<br />
2 2 "4 1 h<br />
5 10 lim<br />
hS0<br />
5 10 lim<br />
hS0<br />
P<br />
h"4 1 h 3 2 1 "4 1 h<br />
2 1 "4 1 h<br />
4 2 4 2 h<br />
h"4 1 h(2 1 "4 1 h)<br />
a 4 2 5a 2 1 4 5 0<br />
( a 2 2 4)(a 2 2 1) 5 0<br />
a 562, a 561<br />
Points on the graph for horizontal tangents are:<br />
(22, 28 (21, 26 (1, 2 26 (2, 2 28 3 ), 3 ), 3 ), 3 ).<br />
23. y 5 x 2 and<br />
x 2 5 1 2 2 x2 y 5 1 2 2 x2<br />
x 2 5 1 4<br />
x 5 1 or x 52 1 2 2<br />
The points of intersection are<br />
P( 1 2, 1 4), Q(2 1 2, 1 4).<br />
Tangent to y 5 x 2 :<br />
(a 1 h) 2 2 a 2<br />
m 5 lim<br />
hS0 h<br />
2ah 1 h 2<br />
5 lim<br />
hS0 h<br />
5 2a.<br />
m 5 a 2 2 5 1 4 a 2 5 0<br />
1-10 <strong>Chapter</strong> 1: Introduction to Calculus
The slope of the tangent at a 5 1 2 is 1 5 m p ,<br />
at a 52 1 2 is 21 5 m q .<br />
Tangents to y 5 1 2 2 x 2 :<br />
S 1 2 2 (a 1 h) 2 T 2 S 1 2 2 a 2 T<br />
m 5 lim<br />
hS0<br />
h<br />
22ah 2 h 2<br />
5 lim<br />
hS0 h<br />
522a.<br />
The slope of the tangents at a 5 1 2 is 21 5 M p ;<br />
at a 52 1 2 is 1 5 M q<br />
m p M p 521 and m q M q 521<br />
Therefore, the tangents are perpendicular at the<br />
points of intersection.<br />
24. y 523x 3 2 2x, (21, 5)<br />
23(21 1 h) 3 2 2(21 1 h) 2 5<br />
m 5 lim<br />
hS0<br />
h<br />
23(21 1 3h 2 3h 2 1 h 3 ) 1 2 2 2h 2 5<br />
5 lim<br />
hS0<br />
h<br />
23(21 1 3h 2 3h 2 1 h 3 ) 1 2 2 2h 2 5<br />
5 lim<br />
hS0<br />
h<br />
3 2 9h 1 9h 2 2 3h 3 1 2 2 2h 2 5<br />
5 lim<br />
hS0<br />
h<br />
211h 1 9h 2 2 3h 3<br />
5 lim<br />
hS0 h<br />
5 lim(211 1 9h 2 3h 2 )<br />
hS0<br />
5211<br />
The slope of the tangent is 211.<br />
We want the line that is parallel to the tangent (i.e.<br />
has slope 211) and passes through (2, 2). Then,<br />
y 2 2 5211(x 2 2)<br />
y 5211x 1 24<br />
25. a. Let y 5 f(x).<br />
f(a) 5 4a 2 1 5a 2 2<br />
f(a 1 h) 5 4(a 1 h) 2 1 5(a 1 h) 2 2<br />
5 4(a 2 1 2ah 1 h 2 ) 1 5a 1 5h 2 2<br />
5 4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 a is<br />
f(a 1 h) 2 f(a)<br />
m 5 lim<br />
hS0 h<br />
5 lim c 4a2 1 8ah 1 4h 2 1 5a 1 5h 2 2<br />
hS0<br />
h<br />
2 (4a2 1 5a 2 2)<br />
d<br />
h<br />
5 lim c 4a2 1 8ah 1 4h 2 1 5a 1 5h 2 2<br />
hS0<br />
h<br />
1 24a2 2 5a 1 2<br />
d<br />
h<br />
8ah 1 4h 2 1 5h<br />
5 lim<br />
hS0 h<br />
5 lim (8a 1 4h 1 5)<br />
hS0<br />
5 8a 1 4(0) 1 5<br />
5 8a 1 5<br />
b. To be parallel, the point on the parabola and the<br />
line must have the same slope. So, first find the<br />
slope of the line. The line 10x 2 2y 2 18 5 0 can<br />
be rewritten as<br />
22y 5 18 2 10x<br />
18 2 10x<br />
y 5<br />
22<br />
y 529 1 5x<br />
y 5 5x 2 9<br />
So, the slope, m, of the line 10x 2 2y 2 18 5 0 is 5.<br />
To be parallel, the slope at a must equal 5. From<br />
part a., the slope of the tangent to the parabola at<br />
x 5 a is 8a 1 5.<br />
8a 1 5 5 5<br />
8a 5 0<br />
a 5 0<br />
Therefore, at the point (0, 22) the tangent line is<br />
parallel to the line 10x 2 2y 2 18 5 0.<br />
c. To be perpendicular, the point on the parabola<br />
and the line must have slopes that are negative<br />
reciprocals of each other. That is, their product must<br />
equal 21. So, first find the slope of the line. The<br />
line x 2 35y 1 7 5 0 can be rewritten as<br />
235y 52x 2 7<br />
y 5 2x 2 7<br />
235<br />
y 5 1<br />
35 x 1 7<br />
35<br />
1<br />
So, the slope, m, of the line x 2 35y 1 7 5 0 is 35.<br />
To be perpendicular, the slope at a must equal<br />
the negative reciprocal of the slope of the line<br />
x 2 35y 1 7 5 0. That is, the slope of a must equal<br />
235. From part a., the slope of the tangent to the<br />
parabola at x 5 a is 8a 1 5.<br />
8a 1 5 5235<br />
8a 5240<br />
a 525<br />
Therefore, at the point (25, 73) the tangent line is<br />
perpendicular to the line x 2 35y 1 7 5 0.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-11
1.3 Rates of Change, pp. 29–31<br />
1. v(t) 5 0 when t 5 0 or t 5 4.<br />
2. a. . Slope of the secant between the<br />
points (2, s(2)) and (9, s(9)).<br />
b. lim<br />
. Slope of the tangent at the<br />
hS0<br />
s(9) 2 s(2)<br />
7<br />
s(6 1 h) 2 s(6)<br />
h<br />
point (6, s(6)).<br />
"4 1 h 2 2<br />
3. lim . Slope of the tangent to the<br />
hS0 h<br />
function with equation y 5 !x at the point (4, 2).<br />
4. a. A and B<br />
b. greater; the secant line through these two points<br />
is steeper than the tangent line at B.<br />
c. y y = f(x)<br />
B C<br />
A D E<br />
x<br />
5. Speed is represented only by a number, not a<br />
direction.<br />
6. Yes, velocity needs to be described by a number<br />
and a direction. Only the speed of the school bus<br />
was given, not the direction, so it is not correct to<br />
use the word “velocity.”<br />
7. s(t) 5 320 2 5t 2 , 0 # t # 8<br />
a. Average velocity during the first second:<br />
s(1) 2 s(0)<br />
5 5 m>s;<br />
1<br />
third second:<br />
s(3) 2 s(2) 45 2 20<br />
5 5 25 m>s;<br />
1<br />
1<br />
eighth second:<br />
s(8) 2 s(7) 320 2 245<br />
5 5 75 m>s.<br />
1<br />
1<br />
b. Average velocity 3 # t # 8<br />
s(8) 2 s(3) 320 2 45<br />
5 5 275 5 55 m>s<br />
8 2 3 5 5<br />
c. s(t) 5 320 2 5t 2<br />
320 2 5(2 1 h) 2 2 (320 2 5(2) 2 )<br />
v(t) 5 lim<br />
hS0<br />
h<br />
24h 1 h 2<br />
5 5 lim<br />
hS0 h<br />
5220<br />
Velocity at t 5 2 is 20 m>s downward.<br />
8. s(t) 5 8t(t 1 2), 0 # t # 5<br />
a. i. from t 5 3 to t 5 4<br />
Average velocity<br />
s(4) 2 s(3)<br />
1<br />
5 32(6) 2 24(5)<br />
5 24(8 2 5)<br />
5 72 km>h<br />
ii. from t 5 3 to t 5 3.1<br />
s(3.1) 2 s(3)<br />
0.1<br />
126.48 2 120<br />
5<br />
0.1<br />
5 64.8 km>h<br />
iii. 3 # t # 3.01<br />
s(3.01) 2 s(3)<br />
0.01<br />
5 64.08 km>h<br />
b. Instantaneous velocity is approximately 64 km>h.<br />
c. At t 5 3<br />
s(t) 5 8t 2 1 16t<br />
v(t) 5 16t 1 16<br />
v(3) 5 48 1 16<br />
5 64 km>h<br />
N(t) 5 20t 2 t 2<br />
9. a.<br />
N(3) 2 N(2)<br />
1<br />
51 2 36<br />
5<br />
1<br />
5 15<br />
15 terms are learned between t 5 2 and t 5 3.<br />
20(2 1 h) 2 (2 1 h) 2 2 36<br />
b. lim<br />
hS0<br />
h<br />
40 1 20h 2 4 2 4h 2 h 2 2 36<br />
5 lim<br />
hS0<br />
h<br />
16h 2 h 2<br />
5 lim<br />
hS0 h<br />
5 lim (16 2 h)<br />
hS0<br />
5 16<br />
At t 5 2, the student is learning at a rate of 16 terms><br />
h.<br />
10. a. M in mg in 1 mL of blood t hours after the<br />
injection.<br />
M(t) 52 1 3 t2 1 t; 0 # t # 3<br />
Calculate the instantaneous rate of change when t 5 2.<br />
2 1<br />
lim<br />
3(2 1 h) 2 1 (2 1 h) 2 (2 4 3 1 2)<br />
hS0<br />
h<br />
2 4 3 2 4 3 h 2 1 3 h 2 1 2 1 h 1 4 3 2 2<br />
5 lim<br />
hS0<br />
h<br />
2 1 3<br />
5 lim<br />
h 2 1 3 h2<br />
hS0 h<br />
5 lim a2 1<br />
hS0 3 2 1 3 hb<br />
52 1 3<br />
1-12 <strong>Chapter</strong> 1: Introduction to Calculus
Rate of change is 2 1 3 mg><br />
h.<br />
b. Amount of medicine in 1 mL of blood is being<br />
dissipated throughout the system.<br />
s<br />
11. t 5 Å 5<br />
Calculate the instantaneous rate of change when<br />
s 5 125.<br />
125 1 h 125<br />
Ä 5 2 Ä 5<br />
lim<br />
hS0 h<br />
125 1 h<br />
2 5<br />
Ä 5<br />
5 lim<br />
hS0 h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0 ≥<br />
5 lim ≥<br />
hS0<br />
5 lim<br />
hS0 125 1 h<br />
5a Ä 5<br />
1<br />
5<br />
125<br />
5a Ä 5 1 5b<br />
1<br />
5<br />
5(5 1 5)<br />
5 1<br />
50<br />
At s 5 125, rate of change of time with respect to<br />
1<br />
height is 50 s>m.<br />
12. T(h) 5 60<br />
h 1 2<br />
Calculate the instantaneous rate of change when<br />
h 5 3.<br />
lim<br />
kS0<br />
60<br />
(3 1 k) 1 2 2 60<br />
(3 1 2)<br />
5 lim<br />
kS0<br />
125 1 h<br />
≥ Ä 5<br />
h<br />
125 1 h<br />
5<br />
ha Ä<br />
125 1 h<br />
5<br />
125 1 h 2 125<br />
5<br />
125 1 h<br />
ha Ä 5<br />
1<br />
k<br />
60<br />
5 1 k 2 12<br />
k<br />
2 5<br />
2 25<br />
¥<br />
1 5b<br />
¥<br />
1 5b<br />
1 5b<br />
125 1 h<br />
? Ä 5<br />
125 1 h<br />
Ä 5<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1 5<br />
¥<br />
1 5<br />
5 lim<br />
kS0<br />
60 60 1 12k<br />
2<br />
5 1 k 5 1 k<br />
k<br />
212k<br />
5 lim<br />
kS0 k(5 1 k)<br />
212<br />
5 lim<br />
kS0 (5 1 k)<br />
52 12 5<br />
12<br />
Temperature is decreasing at C km.<br />
13. h 5 25t 2 5 ° ><br />
2 100t 1 100<br />
When h 5 0, 25t 2 2 100t 1 100 5 0<br />
t 2 2 4t 1 4 5 0<br />
( t 2 2) 2 5 0<br />
t 5 2<br />
Calculate the instantaneous rate of change when t 5 2.<br />
25(2 1 h) 2 2 100(2 1 h) 1 100 2 0<br />
lim<br />
hS0<br />
h<br />
100 1 100h 1 25h 2 2 200 2 100h 1 100<br />
5 lim<br />
hS0<br />
h<br />
25h 2<br />
5 lim<br />
hS0 h<br />
5 lim 25h<br />
hS0<br />
5 0<br />
It hit the ground in 2 s at a speed of 0 m><br />
s.<br />
14. Sale of x balls per week:<br />
P(x) 5 160x 2 x 2 dollars.<br />
a. P(40) 5 160(40) 2 (40) 2<br />
5 4800<br />
Profit on the sale of 40 balls is $4800.<br />
b. Calculate the instantaneous rate of change when<br />
x 5 40.<br />
160(40 1 h) 2 (40 1 h) 2 2 4800<br />
lim<br />
hS0<br />
h<br />
6400 1 160h 2 1600 2 80h 2 h 2 2 4800<br />
5 lim<br />
hS0<br />
h<br />
80h 2 h 2<br />
5 lim<br />
hS0 h<br />
5 lim (80 2 h)<br />
hS0<br />
5 80<br />
Rate of change of profit is $80 per ball.<br />
c.<br />
Rate of change of profit is positive when the sales<br />
level is less than 80.<br />
1-13
15. a. f(x) 52x 2 1 2x 1 3; (22, 25) For the year 2005, x 5 2005 2 1982 5 23. Hence,<br />
f(x) 2 f(22)<br />
the rate at which the average annual salary is changing<br />
lim<br />
xS22 x 1 2<br />
in 2005 is<br />
2x 2 1 2x 1 3 1 5<br />
P r(23) 5 64 2 17.8(23) 1 2.85(23) 2 5<br />
5 lim<br />
xS22 x 1 2<br />
$1 162 250> years since 1982<br />
2 (x 2 2 2x 2 8)<br />
17. s(t) 5 3t 2<br />
5 lim<br />
xS22 x 1 2<br />
a. The distance travelled from 0 s to 5 s is<br />
(x 2 4)(x 1 2)<br />
s(5) 5 3(5) 2 5 75 m<br />
52lim<br />
b. s(10) 5 3(10)<br />
xS22 x 1 2<br />
2 5 300 m<br />
The rate at which the avalanche is moving from 0 s<br />
52lim (x 2 4)<br />
xS22<br />
to 10 s is<br />
5 6<br />
Ds<br />
b. f(x) 5<br />
x x 5 2<br />
x 2 1 ,<br />
Dt 5 300 2 0<br />
10 2 0<br />
5 30 ms ><br />
x<br />
x 2 1<br />
lim<br />
2 2<br />
c. Calculate the instantaneous rate of change when<br />
xS2 x 2 2<br />
t 5 10.<br />
x 2 2x 1 2<br />
3(10 1 h) 2 2 300<br />
5 lim<br />
lim<br />
hS0<br />
xS2 (x 2 1)(x 2 2)<br />
h<br />
2 (x 2 2)<br />
300 1 60h 1 3h 2 2 300<br />
5 lim<br />
5 lim<br />
hS0<br />
xS2 (x 2 1)(x 2 2)<br />
h<br />
521<br />
60h 1 3h 2<br />
5 lim<br />
c. f(x) 5 !x 1 1, x 5 24<br />
hS0 h<br />
f(x) 2 f(24)<br />
5 lim (60 1 3h)<br />
hS0<br />
5 lim<br />
xS24<br />
5 60<br />
x 2 24<br />
!x 1 1 2 5<br />
5 lim<br />
? !x 1 1 1 5<br />
xS24 x 2 24 !x 1 1 1 5<br />
x 2 24<br />
5 lim<br />
xS24 (x 2 24)( !x 1 1 1 5)<br />
5 1<br />
10<br />
16. S(x) 5 246 1 64x 2 8.9x 2 1 0.95x 3<br />
Calculate the instantaneous rate of change.<br />
S(x 1 h) 2 S(x)<br />
5 lim<br />
hS0 h<br />
5 64 2 17.8x 1 2.85x 2<br />
At 10 s the avalanche is moving at 60 m><br />
s.<br />
d. Set s(t) 5 600:<br />
3t 2 5 600<br />
t 2 5 200<br />
t 5610!2<br />
Since t $ 0, t 5 10!2 8 14 s.<br />
246 1 64(x 1 h) 2 8.9(x 1 h) 2 1 0.95(x 1 h) 3 2 (246 2 64x 2 8.9x 2 1 0.95x 3 )<br />
5 lim<br />
hS0<br />
h<br />
246 2 246 1 64(x 1 h 2 x) 2 8.9(x 2 1 2xh 1 h 2 2 x 2 ) 1 0.95(x 3 1 3x 2 h 1 3xh 2 1 h 3 2 x 3 )<br />
5 lim<br />
hS0<br />
h<br />
64h 2 8.9(2xh 1 h 2 ) 1 0.95(3x 2 h 1 3xh 2 1 h 3 )<br />
5 lim<br />
hS0<br />
h<br />
5 lim 364 2 8.9(2x 1 h) 1 0.95(3x 2 1 3xh 1 h 2 )4<br />
hS0<br />
5 64 2 8.9(2x 1 0) 1 0.95 33x 2 1 3x(0) 1 (0) 2 4<br />
1-14 <strong>Chapter</strong> 1: Introduction to Calculus
18. The coordinates of the point are . The slope<br />
of the tangent is<br />
. The equation of the tangent<br />
is y 2 1 or y 52 1 The<br />
a 2x 1 2 a 521 (x 2 a)<br />
a a .<br />
2<br />
intercepts are a0, 2 and (22a, 0). The tangent line<br />
a b<br />
and the axes form a right triangle with legs of length<br />
2<br />
1<br />
and 2a. The area of the triangle is<br />
2 a2 b (2a) 5 2.<br />
a<br />
a<br />
19. C(x) 5 F 1 V(x)<br />
C(x 1 h) 5 F 1 V(x 1 h)<br />
Rate of change of cost is<br />
C(x 1 h) 2 C(x)<br />
lim<br />
xSR h<br />
V(x 1 h) 2 V(x)<br />
5 lim<br />
h,<br />
xSh h<br />
which is independent of F (fixed costs).<br />
20. A(r) 5pr 2<br />
Rate of change of area is<br />
A(r 1 h) 2 A(r)<br />
lim<br />
hS0 h<br />
p(r 1 h) 2 2pr 2<br />
5 lim<br />
hS0 h<br />
(r 1 h 2 r)(r 1 h 1 r)<br />
5p lim<br />
hS0<br />
h<br />
5 2pr<br />
r 5 100 m<br />
Rate is 200p m 2 > m.<br />
21. Cube of dimensions x by x by x has volume<br />
V 5 x 3 . Surface area is 6x 2 .<br />
Vr(x) 5 3x 2 5 1 surface area.<br />
2<br />
22. a. The surface area of a sphere is given by<br />
A(r) 5 4pr 2 .<br />
The question asks for the instantaneous rate of<br />
change of the surface when r 5 10. This is<br />
A(10 1 h) 2 A(10)<br />
lim<br />
hS0 h<br />
4p(10 1 h) 2 2 4p(10) 2<br />
5 lim<br />
hS0<br />
h<br />
4p(100 1 20h 1 h 2 ) 2 4p(100)<br />
5 lim<br />
hS0<br />
h<br />
400p 180ph 1 4ph 2 2 400p<br />
5 lim<br />
hS0<br />
h<br />
80ph 1 4ph 2<br />
5 lim<br />
hS0 h<br />
5 lim (80p 14ph)<br />
hS0<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
5 80p 14p(0)<br />
5 80p<br />
Therefore, the instantaneous rate of change of<br />
the surface area of a spherical balloon as it is<br />
inflated when the radius reaches 10 cm is<br />
80p cm 2 > unit of time.<br />
b. The volume of a sphere is given by V(r) 5 4 3pr 3 .<br />
The question asks for the instantaneous rate of<br />
change of the volume when r 5 5.<br />
Note that the volume is deflating. So, find the rate<br />
of the change of the volume when r 5 5 and then<br />
make the answer negative to symbolize a deflating<br />
spherical balloon.<br />
V(5 1 h) 2 V(5)<br />
lim<br />
hS0 h<br />
4<br />
5 lim<br />
3p(5 1 h) 3 2 4 3p(5) 3<br />
hS0 h<br />
Using the binomial formula to expand<br />
(5 1 h) 3 (or one could simply expand using<br />
algebra), the limit is<br />
4<br />
5 lim<br />
3p(h 3 1 15h 2 1 75h 1 125) 2 4 3 p(5) 3<br />
hS0<br />
h<br />
4<br />
5 lim<br />
3ph 3 1 20ph 2 1 100ph 1 4 3p(125)<br />
hS0<br />
h<br />
2 4 3p(125)<br />
h<br />
4<br />
5 lim<br />
3ph 3 1 20ph 2 1 100ph<br />
hS0<br />
h<br />
5 lim a 4 3 ph2 1 20ph 1 100pb<br />
hS0<br />
5 4 3 p(0)2 1 20p(0) 1 100p<br />
5 100p<br />
Because the balloon is deflating, the instantaneous rate<br />
of change of the volume of the spherical balloon when<br />
the radius reaches 5 cm is 2100p cm 3 > unit of time.<br />
Mid-<strong>Chapter</strong> Review pp. 32–33<br />
1. a. Corresponding conjugate: !5 1 !2.<br />
( !5 2 !2)(!5 1 !2)<br />
5 ( !25 1 !10 2 !10 2 !4)<br />
5 5 2 2<br />
5 3<br />
b. Corresponding conjugate: 3!5 2 2!2.<br />
(3!5 1 2!2)(3!5 2 2!2)<br />
5 (9!25 2 6!10 1 6!10 2 4!4)<br />
5 9(5) 2 4(2)<br />
5 45 2 8<br />
5 37<br />
2 1 a 2 aa, 1 a b 1-15
18. The coordinates of the point are . The slope<br />
of the tangent is<br />
. The equation of the tangent<br />
is y 2 1 or y 52 1 The<br />
a 2x 1 2 a 521 (x 2 a)<br />
a a .<br />
2<br />
intercepts are a0, 2 and (22a, 0). The tangent line<br />
a b<br />
and the axes form a right triangle with legs of length<br />
2<br />
1<br />
and 2a. The area of the triangle is<br />
2 a2 b (2a) 5 2.<br />
a<br />
a<br />
19. C(x) 5 F 1 V(x)<br />
C(x 1 h) 5 F 1 V(x 1 h)<br />
Rate of change of cost is<br />
C(x 1 h) 2 C(x)<br />
lim<br />
xSR h<br />
V(x 1 h) 2 V(x)<br />
5 lim<br />
h,<br />
xSh h<br />
which is independent of F (fixed costs).<br />
20. A(r) 5pr 2<br />
Rate of change of area is<br />
A(r 1 h) 2 A(r)<br />
lim<br />
hS0 h<br />
p(r 1 h) 2 2pr 2<br />
5 lim<br />
hS0 h<br />
(r 1 h 2 r)(r 1 h 1 r)<br />
5p lim<br />
hS0<br />
h<br />
5 2pr<br />
r 5 100 m<br />
Rate is 200p m 2 > m.<br />
21. Cube of dimensions x by x by x has volume<br />
V 5 x 3 . Surface area is 6x 2 .<br />
Vr(x) 5 3x 2 5 1 surface area.<br />
2<br />
22. a. The surface area of a sphere is given by<br />
A(r) 5 4pr 2 .<br />
The question asks for the instantaneous rate of<br />
change of the surface when r 5 10. This is<br />
A(10 1 h) 2 A(10)<br />
lim<br />
hS0 h<br />
4p(10 1 h) 2 2 4p(10) 2<br />
5 lim<br />
hS0<br />
h<br />
4p(100 1 20h 1 h 2 ) 2 4p(100)<br />
5 lim<br />
hS0<br />
h<br />
400p 180ph 1 4ph 2 2 400p<br />
5 lim<br />
hS0<br />
h<br />
80ph 1 4ph 2<br />
5 lim<br />
hS0 h<br />
5 lim (80p 14ph)<br />
hS0<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
5 80p 14p(0)<br />
5 80p<br />
Therefore, the instantaneous rate of change of<br />
the surface area of a spherical balloon as it is<br />
inflated when the radius reaches 10 cm is<br />
80p cm 2 > unit of time.<br />
b. The volume of a sphere is given by V(r) 5 4 3pr 3 .<br />
The question asks for the instantaneous rate of<br />
change of the volume when r 5 5.<br />
Note that the volume is deflating. So, find the rate<br />
of the change of the volume when r 5 5 and then<br />
make the answer negative to symbolize a deflating<br />
spherical balloon.<br />
V(5 1 h) 2 V(5)<br />
lim<br />
hS0 h<br />
4<br />
5 lim<br />
3p(5 1 h) 3 2 4 3p(5) 3<br />
hS0 h<br />
Using the binomial formula to expand<br />
(5 1 h) 3 (or one could simply expand using<br />
algebra), the limit is<br />
4<br />
5 lim<br />
3p(h 3 1 15h 2 1 75h 1 125) 2 4 3 p(5) 3<br />
hS0<br />
h<br />
4<br />
5 lim<br />
3ph 3 1 20ph 2 1 100ph 1 4 3p(125)<br />
hS0<br />
h<br />
2 4 3p(125)<br />
h<br />
4<br />
5 lim<br />
3ph 3 1 20ph 2 1 100ph<br />
hS0<br />
h<br />
5 lim a 4 3 ph2 1 20ph 1 100pb<br />
hS0<br />
5 4 3 p(0)2 1 20p(0) 1 100p<br />
5 100p<br />
Because the balloon is deflating, the instantaneous rate<br />
of change of the volume of the spherical balloon when<br />
the radius reaches 5 cm is 2100p cm 3 > unit of time.<br />
Mid-<strong>Chapter</strong> Review pp. 32–33<br />
1. a. Corresponding conjugate: !5 1 !2.<br />
( !5 2 !2)(!5 1 !2)<br />
5 ( !25 1 !10 2 !10 2 !4)<br />
5 5 2 2<br />
5 3<br />
b. Corresponding conjugate: 3!5 2 2!2.<br />
(3!5 1 2!2)(3!5 2 2!2)<br />
5 (9!25 2 6!10 1 6!10 2 4!4)<br />
5 9(5) 2 4(2)<br />
5 45 2 8<br />
5 37<br />
2 1 a 2 aa, 1 a b 1-15
c. Corresponding conjugate: 9 2 2!5.<br />
(9 1 2!5)(9 2 2!5)<br />
5 (81 2 18!5 1 18!5 2 4!25)<br />
5 81 2 4(5)<br />
5 81 2 20<br />
5 61<br />
d. Corresponding conjugate: 3!5 1 2!10.<br />
(3!5 2 2!10)(3!5 1 2!10)<br />
5 (9!25 1 6!50 2 6!50 2 4!100)<br />
5 9(5) 2 4(10)<br />
5 45 2 40<br />
5 5<br />
6 1 !2<br />
2. a. ? !3<br />
!3 !3<br />
6!3 1 !6<br />
5<br />
!9<br />
6!3 1 !6<br />
5<br />
3<br />
2!3 1 4<br />
b. ? !3<br />
!3 !3<br />
2!9 1 4!3<br />
5<br />
!9<br />
5 6 1 4!3<br />
3<br />
5<br />
c.<br />
!7 2 4 ? !7 1 4<br />
!7 1 4<br />
5(!7 1 4)<br />
5<br />
!49 1 4!7 2 4!7 2 16<br />
5(!7 1 4)<br />
5<br />
7 2 16<br />
5(!7 1 4)<br />
52<br />
9<br />
2!3<br />
d.<br />
!3 2 2 ? !3 1 2<br />
!3 1 2<br />
2!9 1 4!3<br />
5<br />
!9 1 2!3 2 2!3 2 4<br />
5 6 1 4!3<br />
3 2 4<br />
5 6 1 4!3<br />
21<br />
522(3 1 2!3)<br />
5!3<br />
e.<br />
2!3 1 4 ? 2!3 2 4<br />
2!3 2 4<br />
10!9 2 20!3<br />
5<br />
4!9 2 8!3 1 8!3 2 16<br />
30 2 20!3<br />
5<br />
12 2 16<br />
30 2 20!3<br />
5<br />
24<br />
10!3 2 15<br />
5<br />
2<br />
3!2<br />
f.<br />
2!3 2 5 ? 2!3 1 5<br />
2!3 1 5<br />
3!2(2!3 1 5)<br />
5<br />
4!9 1 10!3 2 10!3 2 25<br />
3!2(2!3 1 5)<br />
5<br />
4(3) 2 25<br />
3!2(2!3 1 5)<br />
5<br />
12 2 25<br />
3!2(2!3 1 5)<br />
5<br />
213<br />
3!2(2!3 1 5)<br />
52<br />
13<br />
!2<br />
3. a.<br />
5 ? !2<br />
!2<br />
5 !4<br />
5!2<br />
5 2<br />
5!2<br />
!3<br />
b.<br />
6 1 !2 ? !3<br />
!3<br />
!9<br />
5<br />
!3(6 1 !2)<br />
3<br />
5<br />
!3(6 1 !2)<br />
!7 2 4<br />
c. ? !7 1 4<br />
5 !7 1 4<br />
!49 1 4!7 2 4!7 2 16<br />
5<br />
5(!7 1 4)<br />
5 7 2 16<br />
5(!7 1 4)<br />
9<br />
52<br />
5(!7 1 4)<br />
2!3 2 5<br />
d. ? 2!3 1 5<br />
3!2 2!3 1 5<br />
4!9 1 10!3 2 10!3 2 25<br />
5<br />
3!2(2!3 1 5)<br />
4(3) 2 25<br />
5<br />
3!2(2!3 1 5)<br />
12 2 25<br />
13<br />
5<br />
52<br />
3!2(2!3 1 5) 3!2(2!3 1 5)<br />
1-16 <strong>Chapter</strong> 1: Introduction to Calculus
!3 2 !7 !3 1 !7<br />
e.<br />
?<br />
4 !3 1 !7<br />
!9 1 !21 2 !21 2 !49<br />
5<br />
4(!3 1 !7)<br />
3 2 7<br />
5<br />
4(!3 1 !7)<br />
4<br />
52<br />
4(!3 1 !7)<br />
1<br />
52<br />
( !3 1 !7)<br />
2!3 1 !7 2!3 2 !7<br />
f.<br />
?<br />
5 2!3 2 !7<br />
4!9 2 2!21 1 2!21 2 !49<br />
5<br />
5(2!3 2 !7)<br />
5 4(3) 2 7<br />
5(2!3 2 !7)<br />
12 2 7<br />
5<br />
5(2!3 2 !7)<br />
1<br />
5<br />
(2!3 2 !7)<br />
4. a.<br />
2<br />
3 x 1 y 2 6 5 0<br />
b.<br />
y 2 7 5 1(x 2 2)<br />
y 2 7 5 x 2 2<br />
2x 1 y 2 5 5 0<br />
x 2 y 1 5 5 0<br />
c.<br />
y 2 6 5 4x 2 8<br />
24x 1 y 1 2 5 0<br />
4x 2 y 2 2 5 0<br />
d.<br />
m 52 2 3 ;<br />
y 2 6 52 2 (x 2 0)<br />
3<br />
y 2 6 52 2 3 x<br />
m 5 11 2 7<br />
6 2 2 5 4 4 5 1<br />
m 5 4<br />
y 2 6 5 4(x 2 2)<br />
m 5 1 5<br />
y 2 (22) 5 1 (x 2 (21))<br />
5<br />
y 1 2 5 1 5 x 1 1 5<br />
2 1 5 x 1 y 1 10<br />
5 2 1 5 5 0<br />
2 1 5 x 1 y 1 9 5 5 0<br />
1<br />
5 x 2 y 2 9 5 5 0<br />
x 2 5y 2 9 5 0<br />
5. The slope of PQ is<br />
f(1 1 h) 2 (21)<br />
m 5 lim<br />
hS0 (1 1 h) 2 1<br />
2 (1 1 h) 2 1 1<br />
5 lim<br />
hS0 h<br />
2 (1 1 2h 1 h 2 ) 1 1<br />
5 lim<br />
hS0 h<br />
21 2 2h 2 h 2 1 1<br />
5 lim<br />
hS0 h<br />
22h 2 h 2<br />
5 lim<br />
hS0 h<br />
5 lim (22 2 h)<br />
hS0<br />
522 2 (0)<br />
522<br />
So, the slope of PQ with f(x) 52x 2 is 22.<br />
6. a. Unlisted y-coordinates for Q are found by<br />
substituting the x-coordinates into the given function.<br />
The slope of the line PQ with the given points is<br />
given by the following: Let P 5 (x 1 , y 1 ) and<br />
Then, the slope 5 m 5 y 2 y 2 1<br />
Q 5 (y 1 , y 2 ).<br />
.<br />
x 2 2 x 1<br />
P Q Slope of Line PQ<br />
(21, 1) (22, 6) 25<br />
(21, 1) (21.5, 3.25) 24.5<br />
(21, 1) (21.1, 1.41) 24.1<br />
(21, 1) (21.01, 1.040 1) 24.01<br />
(21, 1) (21.001, 1.004 001) 24.001<br />
P Q Slope of Line PQ<br />
(21, 1) (0, 22) 23<br />
(21, 1) (20.5, 20.75) 23.5<br />
(21, 1) (20.9, 0.61) 23.9<br />
(21, 1) (20.99, 0.9601) 23.99<br />
(21, 1) (20.999, 0.996 001) 23.999<br />
b. The slope from the right and from the left appear<br />
to approach 24. The slope of the tangent to the<br />
graph of f(x) at point P is about 24.<br />
c. With the points P 5 (21, 1) and<br />
Q 5 (21 1 h, f(21 1 h)), the slope, m, of PQ is<br />
the following:<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-17
m 5 y 2 2 y 1<br />
x 2 2 x 1<br />
c.<br />
5 3(21 1 h)2 2 2(21 1 h) 2 24 2 (1)<br />
(21 1 h) 2 (21)<br />
5 1 2 2h 1 h2 1 2 2 2h 2 2 2 1<br />
21 1 h 1 1<br />
5 h2 2 4h<br />
h<br />
5 h 2 4<br />
d. The slope of the tangent is lim f(x).<br />
hS0<br />
In this case, as h goes to zero, h 2 4 goes to<br />
h 2 4 5 0 2 4 524. The slope of the tangent to<br />
the graph of f(x) at the point P is 24.<br />
e. The answers are equal.<br />
7. a.<br />
f(23 1 h) 2 f(23)<br />
m 5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
h 2 2 3h<br />
5 lim<br />
hS0 h<br />
5 lim (h 2 3)<br />
hS0<br />
5 0 2 3<br />
523<br />
y 5 f(x) 5 4<br />
x 2 2<br />
f(6 1 h) 2 f(6)<br />
m 5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
h<br />
4<br />
6 1 h 2 2 2 4<br />
6 2 2<br />
h<br />
4<br />
h 1 4<br />
5 lim<br />
2 4 4<br />
hS0 h<br />
4<br />
h 1 4<br />
5 lim<br />
2 1<br />
hS0 h<br />
4 2 (h 1 4)<br />
5 lim a b 1<br />
hS0 h 1 4 h<br />
h<br />
3(23 1 h) 2 1 3(23 1 h) 2 54 2 3(23) 2 1 3(23) 2 54<br />
h<br />
9 2 6h 1 h 2 2 9 1 3h 2 5 2 (9 2 9 2 5)<br />
h<br />
h 2 2 3h 2 5 2 (25)<br />
h<br />
5 lim a 2h<br />
hS0<br />
5 lim<br />
hS0<br />
5 21<br />
0 1 4<br />
h 1 4 b 1 h<br />
21<br />
h 1 4<br />
b. y 5 f(x) 5 1 x<br />
f( 1 3 1 h) 2 f( 1<br />
m 5 lim<br />
3)<br />
hS0 h<br />
1<br />
1<br />
3 1 h 2 1 1<br />
3<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
a 2h<br />
1<br />
9 1 1 3h b 1 h<br />
21<br />
1<br />
9 1 1 3h<br />
5 21<br />
1<br />
9 1 1 3(0)<br />
529<br />
h<br />
( 1 3) 2 ( 1 3 1 h)<br />
1<br />
3( 1 3 1 h)<br />
h<br />
d.<br />
52 1 4<br />
f(5 1 h) 2 f(5)<br />
m 5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
h<br />
!5 1 h 1 4 2 !5 1 4<br />
h<br />
!9 1 h 2 !9<br />
h<br />
!9 1 h 2 3<br />
h<br />
!9 1 h 2 3<br />
? !9 1 h 1 3<br />
h !9 1 h 1 3<br />
9 1 h 1 3!9 1 h 2 3!9 1 h 2 9<br />
h( !9 1 h 1 3)<br />
h<br />
h( !9 1 h 1 3)<br />
1<br />
!9 1 h 1 3<br />
1-18 <strong>Chapter</strong> 1: Introduction to Calculus
5<br />
1<br />
!9 1 0 1 3<br />
5 1 6<br />
s(t) 5 6t(t 1 1) 5 6t 2 1 6t<br />
8.<br />
s(3) 2 s(2)<br />
a. i. average velocity 5<br />
3 2 2<br />
5 36(3) 2 1 6(3)42 36(2) 2 1 6(2)4<br />
5 6(9) 1 18 2 (24 1 12)<br />
5 54 1 18 2 36<br />
5 36 km><br />
h<br />
s(2.1) 2 s(2)<br />
ii. average velocity 5<br />
2.1 2 2<br />
5 36(2.1)2 1 6(2.1)4 2 36(2) 2 1 6(2)4<br />
0.1<br />
326.46 1 12.64 2 324 1 124<br />
5<br />
0.1<br />
39.06 2 36<br />
5<br />
0.1<br />
5 3.06<br />
0.1<br />
5 30.6 km><br />
h<br />
s(2.01) 2 s(2)<br />
iii. average velocity 5<br />
2.01 2 2<br />
5 36(2.01)2 1 6(2.01)4 2 36(2) 2 1 6(2)4<br />
0.01<br />
5 324.2406 1 12.064 2 36(2)2 1 6(2)4<br />
0.01<br />
36.3006 2 324 1 124<br />
5<br />
0.01<br />
36.3006 2 36<br />
5<br />
0.01<br />
5 0.3006<br />
0.01<br />
5 30.06 km><br />
h<br />
b. At the time t 5 2, the velocity of the car appears<br />
to approach 30 km><br />
h.<br />
f(2 1 h) 2 f(2)<br />
c. average velocity 5<br />
(2 1 h) 2 (2)<br />
5 36(2 1 h)2 1 6(2 1 h)4 2 36(2) 2 1 6(2)4<br />
h<br />
5 36(4 1 4h 1 h2 ) 1 12 1 6h4 2 324 1 124<br />
h<br />
5 324 1 24h 1 6h2 1 12 1 6h4 2 36<br />
h<br />
5 6h2 1 30h 1 36 2 36<br />
h<br />
5 6h2 1 30h<br />
h<br />
5 (6h 1 30) km><br />
h<br />
d. When t 5 2, the velocity is the limit as h<br />
approaches 0.<br />
velocity 5 lim (6h 1 30)<br />
hS0<br />
5 6(0) 1 30<br />
5 30<br />
Therefore, when t 5 2 the velocity is 30 km><br />
h.<br />
9. a. The instantaneous rate of change of f(x) with<br />
respect to x at x 5 2 is given by<br />
f(2 1 h) 2 f(2)<br />
lim<br />
hS0 h<br />
35 2 (2 1 h) 2 4 2 35 2 (2) 2 4<br />
5 lim<br />
hS0<br />
h<br />
5 2 (4 1 4h 1 h 2 ) 2 1<br />
5 lim<br />
hS0<br />
h<br />
5 2 4 2 4h 2 h 2 2 1<br />
5 lim<br />
hS0 h<br />
2h 2 2 4h<br />
5 lim<br />
hS0 h<br />
5 lim (2h 2 4)<br />
hS0<br />
52(0) 2 4<br />
524<br />
b. The instantaneous rate of change of f(x) with<br />
respect to x at x 5 1 2 is given by<br />
f( 1 2 1 h) 2 f( 1<br />
lim<br />
2)<br />
hS0 h<br />
3<br />
1<br />
2 1 h 2 3 1<br />
2<br />
5 lim<br />
hS0 h<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
26h<br />
1<br />
5 26<br />
1<br />
2 1 0<br />
5212<br />
3<br />
1<br />
2 1 h 2 6<br />
h<br />
3 2 6( 1 2 1 h)<br />
1<br />
2 1 h<br />
3 2 3 2 6h<br />
1<br />
2 1 h<br />
2 1 h ? 1 h<br />
26<br />
1<br />
2 1 h<br />
? 1 h<br />
? 1 h<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-19
10. a. The average rate of change of V(t) with<br />
respect to t during the first 20 minutes is given by<br />
f(20) 2 f(0)<br />
20 2 0<br />
5 350(30 2 20)2 4 2 350(30 2 0) 2 4<br />
20<br />
5000 2 45 000<br />
5<br />
20<br />
40 000<br />
52<br />
20<br />
522000 Lmin ><br />
b. The rate of change of V(t) with respect to t at the<br />
time t 5 20 is given by<br />
f(20 1 h) 2 f(20)<br />
lim<br />
hS0 h<br />
350(30 2 (20 1 h)) 2 4 2 350(30 2 20) 2 4<br />
5 lim<br />
hS0<br />
h<br />
350(10 2 h) 2 4 2 350(10) 2 4<br />
5 lim<br />
hS0<br />
h<br />
350(100 2 20h 1 h 2 )4 2 350(100)4<br />
5 lim<br />
hS0<br />
h<br />
5000 2 1000h 1 50h 2 2 5000<br />
5 lim<br />
hS0<br />
h<br />
50h 2 2 1000h<br />
5 lim<br />
hS0 h<br />
5 lim 50h 2 1000<br />
hS0<br />
5 50(0) 2 1000<br />
521000 Lmin ><br />
11. a. Let y 5 f(x).<br />
f(4) 5 (4) 2 1 (4) 2 3 5 16 1 1 5 17<br />
f(4 1 h) 5 (4 1 h) 2 1 (4 1 h) 2 3<br />
5 16 1 8h 1 h 2 1 h 1 1<br />
5 h 2 1 9h 1 17<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 4 is<br />
f(4 1 h) 2 f(4)<br />
m 5 lim<br />
hS0 h<br />
h 2 1 9h 1 17 2 (17)<br />
5 lim<br />
hS0 h<br />
h 2 1 9h<br />
5 lim<br />
hS0 h<br />
5 lim (h 1 9)<br />
hS0<br />
5 0 1 9<br />
5 9<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 x 2 1 x 2 3 at x 5 4 is 9.<br />
So an equation of the tangent at x 5 4 is given by<br />
y 2 17 5 9(x 2 4)<br />
y 2 17 5 9x 2 36<br />
29x 1 y 2 17 1 36 5 0<br />
29x 1 y 1 19 5 0<br />
b. Let y 5 f(x).<br />
f(22) 5 2(22) 2 2 7 5 2(4) 2 7 5 1<br />
f(22 1 h) 5 2(22 1 h) 2 2 7<br />
5 2(4 2 4h 1 h 2 ) 2 7<br />
5 8 2 8h 1 2h 2 2 7<br />
5 2h 2 2 8h 1 1<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 4 is<br />
f(22 1 h) 2 f(22)<br />
m 5 lim<br />
hS0 h<br />
2h 2 2 8h 1 1 2 (1)<br />
5 lim<br />
hS0 h<br />
2h 2 2 8h<br />
5 lim<br />
hS0 h<br />
5 lim (2h 2 8)<br />
hS0<br />
5 2(0) 2 8<br />
528<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 2x 2 2 7 at x 522 is 28.<br />
So an equation of the tangent at x 522<br />
is given by<br />
y 2 1 528(x 2 (22))<br />
y 2 1 528x 2 16<br />
8x 1 y 2 1 1 16 5 0<br />
8x 1 y 1 15 5 0<br />
c. f(21) 5 3(21) 2 1 2(21) 2 5 5 3 2 2 2 5<br />
524<br />
f(21 1 h) 5 3(21 1 h) 2 1 2(21 1 h) 2 5<br />
5 3(1 2 2h 1 h 2 ) 2 2 1 2h 2 5<br />
5 3 2 6h 1 3h 2 2 7 1 2h<br />
5 3h 2 2 4h 2 4<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 4 is<br />
f(21 1 h) 2 f(21)<br />
m 5 lim<br />
hS0 h<br />
3h 2 2 4h 2 4 2 (24)<br />
5 lim<br />
hS0 h<br />
3h 2 2 4h<br />
5 lim<br />
hS0 h<br />
5 lim (3h 2 4)<br />
hS0<br />
5 3(0) 2 4<br />
524<br />
1-20 <strong>Chapter</strong> 1: Introduction to Calculus
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 3x 2 1 2x 2 5 at x 521 is 24.<br />
So an equation of the tangent at x 524 is given by<br />
y 2 (24) 524(x 2 (21))<br />
y 1 4 524(x 1 1)<br />
y 1 4 524x 2 4<br />
4x 1 y 1 4 1 4 5 0<br />
4x 1 y 1 8 5 0<br />
d. f(1) 5 5(1) 2 2 8(1) 1 3 5 5 2 8 1 3 5 0<br />
f(1 1 h) 5 5(1 1 h) 2 2 8(1 1 h) 1 3<br />
5 5(1 1 2h 1 h 2 ) 2 8 2 8h 1 3<br />
5 5 1 10h 1 5h 2 2 5 2 8h<br />
5 5h 2 1 2h<br />
Using the limit of the difference quotient, the slope<br />
of the tangent at x 5 1 is<br />
f(1 1 h) 2 f(1)<br />
m 5 lim<br />
hS0 h<br />
5h 2 1 2h 2 (0)<br />
5 lim<br />
hS0 h<br />
5 lim (5h 1 2)<br />
hS0<br />
5 5(0) 1 2<br />
5 2<br />
Therefore, the slope of the tangent to<br />
y 5 f(x) 5 5x 2 2 8x 1 3 at x 5 1 is 2.<br />
So an equation of the tangent at x 5 1 is given by<br />
y 2 0 5 2(x 2 1)<br />
y 5 2x 2 2<br />
22x 1 y 1 2 5 0<br />
12. a. Using the limit of the difference quotient, the<br />
slope of the tangent at x 525 is<br />
f(25 1 h) 2 f(25)<br />
m 5 lim<br />
hS0 h<br />
25 1 h<br />
5 lim a<br />
hS0 25 1 h 1 3 2 25<br />
25 1 3 b ? 1 h<br />
5 lim a 25 1 h<br />
hS0 22 1 h 2 5 2 b ? 1 h<br />
210 1 2h 2 (210 1 5h)<br />
5 lim a b ? 1<br />
hS0 24 1 2h<br />
h<br />
210 1 2h 1 10 2 5h<br />
5 lim a b ? 1<br />
hS0 24 1 2h h<br />
23h<br />
5 lim a<br />
hS0 24 1 2h b ? 1 h<br />
23<br />
5 lim a<br />
hS0 24 1 2h b<br />
23<br />
5<br />
24 1 2(0)<br />
5 3 4<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
Therefore, the slope of the tangent to<br />
3<br />
f(x) 5<br />
x at x 525 is<br />
x 1 3<br />
4.<br />
So an equation of the tangent at x 5 3 4 is given by<br />
y 2 5 2 5 3 (x 2 (25))<br />
4<br />
y 2 5 2 5 3 4 x 1 15<br />
4<br />
2 3 4 x 1 y 2 10<br />
4 2 15<br />
4 5 0<br />
2 3 4 x 1 y 2 25<br />
4 5 0<br />
23x 1 4y 2 25 5 0<br />
b. Using the limit of the difference quotient, the<br />
slope of the tangent at x 521 is<br />
f(21 1 h) 2 f(21)<br />
m 5 lim<br />
hS0 h<br />
2(21 1 h) 1 5<br />
5 lim a<br />
hS0 5(21 1 h) 2 1 2 2(21) 1 5<br />
5(21) 2 1 b ? 1 h<br />
22 1 2h 1 5<br />
5 lim a<br />
hS0 25 1 5h 2 1 2 22 1 5<br />
25 2 1 b ? 1 h<br />
5 lim a 2h 1 3<br />
hS0 5h 2 6 2 3<br />
26 b ? 1 h<br />
5 lim a 2h 1 3<br />
hS0 5h 2 6 1 1 2 b ? 1 h<br />
4h 1 6 1 5h 2 6<br />
5 lim a b ? 1<br />
hS0 10h 2 12 h<br />
9h<br />
5 lim a<br />
hS0 10h 2 12 b ? 1 h<br />
9<br />
5 lim a<br />
hS0 10h 2 12 b<br />
9<br />
5<br />
10(0) 2 12<br />
52 9 12<br />
52 3 4<br />
Therefore, the slope of the tangent to<br />
f(x) 5 2x 1 5 at x 521 is 2 3 5x 2 1<br />
4.<br />
So an equation of the tangent at x 52 3 4 is given by<br />
y 2 a2 1 2 b 523 (x 2 (21))<br />
4<br />
y 1 1 2 523 4 x 2 3 4<br />
4y 1 2 523x 2 3<br />
3x 1 4y 1 2 1 3 5 0<br />
3x 1 4y 1 5 5 0<br />
1-21
1.4 The Limit of a Function,<br />
pp. 37–39<br />
27<br />
1. a.<br />
99<br />
b. p<br />
2. One way to find a limit is to evaluate the function<br />
for values of the independent variable that get<br />
progressively closer to the given value of the<br />
independent variable.<br />
3. a. A right-sided limit is the value that a<br />
function gets close to as the values of the<br />
independent variable decrease and get close<br />
to a given value.<br />
b. A left-sided limit is the value that a function<br />
gets close to as the values of the independent<br />
variable increase and get close to a given<br />
value.<br />
c. A (two-sided) limit is the value that a function<br />
gets close to as the values of the independent<br />
variable get close to a given value, regardless<br />
of whether the values increase or decrease<br />
toward the given value.<br />
4. a. 25<br />
b. 3 1 7 5 10<br />
c. 10 2 5 100<br />
d. 4 2 3(22) 2 528<br />
e. 4<br />
f. 2 3 5 8<br />
5. Even though f(4) 521, the limit is 1, since that<br />
is the value that the function approaches from the<br />
left and the right of x 5 4.<br />
6. a. 0<br />
b. 2<br />
c. 21<br />
d. 2<br />
7. a. 2<br />
b. 1<br />
c. does not exist<br />
8. a. 9 2 (21) 2 5 8<br />
b.<br />
c.<br />
0 1 20<br />
Å 0 1 5 5 "4<br />
5 2<br />
"5 2 1 5 "4<br />
5 2<br />
9. 2 2 1 1 5 5<br />
6<br />
4<br />
2<br />
y<br />
x<br />
–4 –2 0 2 4<br />
10. a. Since 0 is not a value for which the function is<br />
undefined, one may substitute 0 in for x to find that<br />
lim x 4 5 lim x 4<br />
xS0 1 xS0<br />
5 (0) 4<br />
5 0<br />
b. Since 2 is not a value for which the function is<br />
undefined, one may substitute 2 in for x to find that<br />
lim (x 2 2 4) 5 lim (x 2 2 4)<br />
xS2 2 xS2<br />
5 (2) 2 2 4<br />
5 4 2 4<br />
5 0<br />
c. Since 3 is not a value for which the function is<br />
undefined, one may substitute 3 in for x to find that<br />
lim (x 2 2 4) 5 lim (x 2 2 4)<br />
xS3 2 xS3<br />
5 (3) 2 2 4<br />
5 9 2 4<br />
5 5<br />
d. Since 1 is not a value for which the function is<br />
undefined, one may substitute 1 in for x to find that<br />
1<br />
lim<br />
xS1 1 x 2 3 5 lim 1<br />
xS1 x 2 3<br />
5 1<br />
1 2 3<br />
52 1 2<br />
e. Since 3 is not a value for which the function is<br />
undefined, one may substitute 3 in for x to find that<br />
1<br />
lim<br />
xS3 1 x 1 2 5 lim 1<br />
xS3 x 1 2<br />
5 1<br />
3 1 2<br />
5 1 5<br />
f. If 3 is substituted in the function for x, then the<br />
function is undefined because of division by zero.<br />
There does not exist a way to divide out the x 2 3 in<br />
1-22 <strong>Chapter</strong> 1: Introduction to Calculus
1<br />
the denominator. Also, lim approaches infinity,<br />
xS3 1 x 2 3<br />
1<br />
while lim approaches negative infinity.<br />
xS3 2 x 2 3<br />
1<br />
1<br />
Therefore, since lim<br />
lim<br />
xS3 1 x 2 3 2 lim 1<br />
xS3 2 x 2 3 ,<br />
xS3 x 2 3<br />
does not exist.<br />
11. a.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
–6<br />
–8<br />
lim f(x) 2 lim f(x). Therefore, lim f(x)<br />
xS21 1 xS21 2 xS21<br />
not exist.<br />
b.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
–6<br />
–8<br />
lim f(x) 5 lim f(x). Therefore, lim f(x)<br />
xS2 1 xS2 2 xS2<br />
is equal to 2.<br />
c.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
–6<br />
–8<br />
lim f(x) 5 lim f(x). Therefore, lim f(x)<br />
xS 1 xS 1 2 1 xS 1 2 2 2<br />
is equal to 2.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
does<br />
exists and<br />
exists and<br />
d.<br />
lim f(x) 2 lim f(x). Therefore, lim f(x)<br />
xS20.5 1 xS20.5 2 xS20.5<br />
does not exist.<br />
12. Answers may vary. For example:<br />
a.<br />
6<br />
y<br />
4<br />
2<br />
x<br />
–8 –6 –4 –2 0<br />
–2<br />
2 4 6 8<br />
–4<br />
b.<br />
c.<br />
d.<br />
13. f(x) 5 mx 1 b<br />
f(x) 522 m 1 b 522<br />
lim<br />
xS1<br />
lim<br />
xS21<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
–6<br />
–8<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
–8 –6 –4 –2 0<br />
–2<br />
–4<br />
f(x) 5 4<br />
8<br />
6<br />
4<br />
2<br />
6<br />
4<br />
2<br />
6<br />
4<br />
2<br />
6<br />
4<br />
2<br />
y<br />
y<br />
y<br />
y<br />
2<br />
2<br />
2<br />
2<br />
4 6 8<br />
4 6 8<br />
4 6 8<br />
4 6 8<br />
2m 1 b 5 4<br />
2b 5 2<br />
b 5 1, m 523<br />
x<br />
x<br />
x<br />
x<br />
1-23
14. f(x) 5 ax 2 1 bx 1 c, a 2 0<br />
f(0) 5 0 c 5 0<br />
f(x) 5 5 a 1 b 5 5<br />
lim<br />
xS1<br />
lim<br />
xS22<br />
f(x) 5 8<br />
6a 5 18<br />
a 5 3, b 5 2<br />
Therefore, the values are a 5 3, b 5 2, and c 5 0.<br />
15. a.<br />
y<br />
10<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–4 –2 0 2 4 6 8 10 12<br />
–2<br />
b. lim p(t) 5 3 1 1<br />
tS6 2 12 (6)2<br />
5 3 1 36<br />
12<br />
5 3 1 3<br />
5 6<br />
lim p(t) 5 2 1 1<br />
tS6 1 18 (6)2<br />
5 2 1 36<br />
18<br />
5 2 1 2<br />
5 4<br />
c. Since p(t) is measured in thousands, right before<br />
the chemical spill there were 6000 fish in the lake.<br />
Right after the chemical spill there were 4000 fish<br />
in the lake. So, 6000 2 4000 5 2000 fish were<br />
killed by the spill.<br />
d. The question asks for the time, t, after the chemical<br />
spill when there are once again 6000 fish in the lake.<br />
Use the second equation to set up an equation that is<br />
modelled by<br />
6 5 2 1 1<br />
18 t2<br />
4 5 1<br />
18 t2<br />
4a 2 2b 5 8<br />
72 5 t 2<br />
!75 5 t<br />
(The question asks for time so the negative answer<br />
is disregarded.)<br />
So, at time t 5 !72 8 8.49 years the population<br />
has recovered to the level before the spill.<br />
1.5 Properties of Limits, pp. 45–47<br />
1. lim(3 1 x) and lim(x 1 3) have the same value,<br />
xS2<br />
xS2<br />
but 3 1 x does not. Since there are no brackets<br />
around the expression, the limit only applies to 3,<br />
and there is no value for the last term, x.<br />
2. Factor the numerator and denominator. Cancel<br />
any common factors. Substitute the given value of x.<br />
3. If the two one-sided limits have the same value,<br />
then the value of the limit is equal to the value of<br />
the one-sided limits. If the one-sided limits do not<br />
have the same value, then the limit does not exist.<br />
3(2)<br />
4. a.<br />
2 2 1 2 5 1<br />
b. (21) 4 1 (21) 3 1 (21) 2 5 1<br />
c.<br />
5 100<br />
9<br />
d. (2p) 3 1p 2 (2p) 2 5p 3 5 8p 3 1 2p 3 2 5p 3<br />
5 5p 3<br />
e. "3 1 "1 1 0 5 "3 1 1<br />
5 2<br />
23 2 3<br />
f.<br />
Å 2(23) 1 4 5 26<br />
Å 22<br />
5 "3<br />
(22) 3<br />
5. a.<br />
22 2 2 522<br />
2<br />
b.<br />
!1 1 1 5 2 !2<br />
5 "2<br />
6. Since substituting t 5 1 does not make the<br />
denominator 0, direct substitution works.<br />
1 2 1 2 5<br />
5 25<br />
6 2 1 5<br />
521<br />
4 2 x 2<br />
7. a. lim<br />
xS2 2 2 x 5 lim (2 2 x)(2 1 x)<br />
xS2 (2 2 x)<br />
5 lim(2 1 x)<br />
xS2<br />
5 4<br />
2x 2 1 5x 1 3 (x 1 1)(2x 1 3)<br />
b. lim<br />
5 lim<br />
xS21 x 1 1 xS21 x 1 1<br />
c.<br />
lim<br />
xS2<br />
c"9 1 1 2<br />
"9 d 5 a3 1 1 2<br />
3 b<br />
5 5<br />
x 3 2 27<br />
lim<br />
xS3 x 2 3<br />
5 lim (x 2 3)(x 2 1 3x 1 9)<br />
xS3 x 2 3<br />
5 9 1 9 1 9<br />
5 27<br />
1-24 <strong>Chapter</strong> 1: Introduction to Calculus
14. f(x) 5 ax 2 1 bx 1 c, a 2 0<br />
f(0) 5 0 c 5 0<br />
f(x) 5 5 a 1 b 5 5<br />
lim<br />
xS1<br />
lim<br />
xS22<br />
f(x) 5 8<br />
6a 5 18<br />
a 5 3, b 5 2<br />
Therefore, the values are a 5 3, b 5 2, and c 5 0.<br />
15. a.<br />
y<br />
10<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–4 –2 0 2 4 6 8 10 12<br />
–2<br />
b. lim p(t) 5 3 1 1<br />
tS6 2 12 (6)2<br />
5 3 1 36<br />
12<br />
5 3 1 3<br />
5 6<br />
lim p(t) 5 2 1 1<br />
tS6 1 18 (6)2<br />
5 2 1 36<br />
18<br />
5 2 1 2<br />
5 4<br />
c. Since p(t) is measured in thousands, right before<br />
the chemical spill there were 6000 fish in the lake.<br />
Right after the chemical spill there were 4000 fish<br />
in the lake. So, 6000 2 4000 5 2000 fish were<br />
killed by the spill.<br />
d. The question asks for the time, t, after the chemical<br />
spill when there are once again 6000 fish in the lake.<br />
Use the second equation to set up an equation that is<br />
modelled by<br />
6 5 2 1 1<br />
18 t2<br />
4 5 1<br />
18 t2<br />
4a 2 2b 5 8<br />
72 5 t 2<br />
!75 5 t<br />
(The question asks for time so the negative answer<br />
is disregarded.)<br />
So, at time t 5 !72 8 8.49 years the population<br />
has recovered to the level before the spill.<br />
1.5 Properties of Limits, pp. 45–47<br />
1. lim(3 1 x) and lim(x 1 3) have the same value,<br />
xS2<br />
xS2<br />
but 3 1 x does not. Since there are no brackets<br />
around the expression, the limit only applies to 3,<br />
and there is no value for the last term, x.<br />
2. Factor the numerator and denominator. Cancel<br />
any common factors. Substitute the given value of x.<br />
3. If the two one-sided limits have the same value,<br />
then the value of the limit is equal to the value of<br />
the one-sided limits. If the one-sided limits do not<br />
have the same value, then the limit does not exist.<br />
3(2)<br />
4. a.<br />
2 2 1 2 5 1<br />
b. (21) 4 1 (21) 3 1 (21) 2 5 1<br />
c.<br />
5 100<br />
9<br />
d. (2p) 3 1p 2 (2p) 2 5p 3 5 8p 3 1 2p 3 2 5p 3<br />
5 5p 3<br />
e. "3 1 "1 1 0 5 "3 1 1<br />
5 2<br />
23 2 3<br />
f.<br />
Å 2(23) 1 4 5 26<br />
Å 22<br />
5 "3<br />
(22) 3<br />
5. a.<br />
22 2 2 522<br />
2<br />
b.<br />
!1 1 1 5 2 !2<br />
5 "2<br />
6. Since substituting t 5 1 does not make the<br />
denominator 0, direct substitution works.<br />
1 2 1 2 5<br />
5 25<br />
6 2 1 5<br />
521<br />
4 2 x 2<br />
7. a. lim<br />
xS2 2 2 x 5 lim (2 2 x)(2 1 x)<br />
xS2 (2 2 x)<br />
5 lim(2 1 x)<br />
xS2<br />
5 4<br />
2x 2 1 5x 1 3 (x 1 1)(2x 1 3)<br />
b. lim<br />
5 lim<br />
xS21 x 1 1 xS21 x 1 1<br />
c.<br />
lim<br />
xS2<br />
c"9 1 1 2<br />
"9 d 5 a3 1 1 2<br />
3 b<br />
5 5<br />
x 3 2 27<br />
lim<br />
xS3 x 2 3<br />
5 lim (x 2 3)(x 2 1 3x 1 9)<br />
xS3 x 2 3<br />
5 9 1 9 1 9<br />
5 27<br />
1-24 <strong>Chapter</strong> 1: Introduction to Calculus
d.<br />
e.<br />
f.<br />
5 lim<br />
xS0<br />
52 1 4<br />
5 lim<br />
xS0<br />
52 1<br />
"7<br />
8. a.<br />
Let u 5 " 3 x. Therefore, u 3 5 x as x S 8, u S 2.<br />
u 2 2<br />
Here, lim<br />
xS2 u 3 2 8 5 lim 1<br />
xS2 u 2 1 2u 1 4<br />
5 1<br />
12<br />
27 2 x<br />
b. lim<br />
Let<br />
xS27 x 1 3 2 3 x 1 3 5 u<br />
x 5 u 3<br />
u 3 2 27<br />
x S 27, u S 3.<br />
5 lim<br />
xS3 u 2 3<br />
(u 2 3)(u 2 1 3u 1 9)<br />
52lim<br />
xS3 u 2 3<br />
52(9 1 9 1 9)<br />
5227<br />
c.<br />
5 lim<br />
xS1<br />
5 lim<br />
xS1<br />
5 1 6<br />
d.<br />
lim<br />
xS0<br />
"x 2 2<br />
lim<br />
xS4 x 2 4<br />
lim<br />
xS0<br />
x 1 6<br />
lim<br />
2 1<br />
xS1<br />
x 1 6<br />
lim<br />
2 1<br />
xS1<br />
5 lim<br />
xS1<br />
£ 2 2 "4 1 x<br />
x<br />
21<br />
2 1 "4 1 x<br />
"7 2 x 2 "7 1 x "7 2 x 1 "7 1 x<br />
£ 3<br />
x<br />
"7 2 x 1 "7 1 x §<br />
" 3 x 2 2<br />
lim<br />
xS8 x 2 8<br />
x 2 1<br />
u 2 1<br />
u 6 2 1<br />
(u 2 1)<br />
(u 2 1)(u 5 1 u 4 1 u 3 1 u 2 1 u 1 1)<br />
x 1 3<br />
u 2 2 1<br />
1<br />
u 2 2 1<br />
5 lim "x 2 2<br />
xS4 ("x 2 2)("x 1 2)<br />
5 1 4<br />
7 2 x 2 7 2 x<br />
3 2 1 "4 1 x<br />
x("7 2 x 1 "7 1 x)<br />
x 1 6 5 u, x 5 u 6<br />
x S 1, u S 1<br />
Let x 1 6 5 u<br />
u 6 5 x<br />
x 1 3 5 u2<br />
As x S 1, u S 1<br />
5 lim<br />
xS1<br />
5 1 2<br />
u 2 1<br />
(u 2 1)(u 1 1)<br />
"x 2 2<br />
e. lim<br />
Let x 1 2<br />
xS4 "x 3 2 8<br />
5 u<br />
x 3 2<br />
u 2 2<br />
5 u3<br />
5 lim<br />
x S 4, u S 2<br />
xS2 u 3 2 8<br />
u 2 2<br />
5 lim<br />
xS2 (u 2 2)(u 2 1 2u 1 4)<br />
5 1<br />
12<br />
(x 1 8) 1 3<br />
f. lim<br />
2 2<br />
Let (x 1 8) 1 3<br />
xS0 x<br />
5 u<br />
x 1 8 5 u 3<br />
u 2 2<br />
lim<br />
x 5 u 3 2 8<br />
xS2 u 3 2 8<br />
x S 0, u S 2<br />
5 1<br />
12<br />
16 2 16<br />
9. a.<br />
64 1 64 5 0<br />
16 2 16<br />
b.<br />
16 2 20 1 6 5 0<br />
x 2 1 x<br />
c. lim<br />
xS21 x 1 1 5 lim x(x 1 1)<br />
xS21 x 1 1<br />
521<br />
"x 1 1 2 1 "x 1 1 2 1<br />
d. lim<br />
5 lim<br />
xS0 x<br />
xS0 x 1 1 2 1<br />
"x 1 1 2 1<br />
5 lim<br />
xS0 ("x 1 1 2 1)("x 1 1 1 1)<br />
5 1 2<br />
(x 1 h) 2 2 x 2 2xh 1 h 2<br />
e. lim<br />
5 lim<br />
hS0 h<br />
hS0 h<br />
5 2x<br />
1<br />
f. lima<br />
xS1 x 2 1 ba 1<br />
x 1 3 2 2<br />
3x 1 5 b<br />
1 1 5 2 2x 2 6<br />
5 lima<br />
ba3x<br />
xS1 x 2 1 (x 1 3)(3x 1 5) b<br />
1<br />
5 lim<br />
xS1 (x 1 3)(3x 1 5)<br />
5 1<br />
4(8)<br />
5 1<br />
32<br />
2 1 "4 1 x § 1-25<br />
Calculus and Vectors <strong>Solutions</strong> Manual
0 x 2 5 0<br />
10. a. lim does not exist.<br />
xS5 x 2 5<br />
0 x 2 5 0<br />
lim<br />
xS5 1 x 2 5 5 lim x 2 5<br />
xS5 1 x 2 5<br />
5 1<br />
0 x 2 5 0<br />
lim<br />
xS5 2 x 2 5 5 lim 2 a x 2 5<br />
xS5 2 x 2 5 b<br />
0 2x 2 5 0 (x 1 1)<br />
b. lim<br />
does not exist.<br />
2x 2 5<br />
0 2x 2 5 0 5 2x 2 5, x $ 5 2<br />
(2x 2 5)(x 1 1)<br />
lim<br />
5 x 1 1<br />
xS 5 1<br />
2x 2 5<br />
2<br />
0 2x 2 5 0 52(2x 2 5), x , 5 2<br />
2 (2x 2 5)(x 1 1)<br />
lim<br />
52(x 1 1)<br />
xS 5 2<br />
2x 2 5<br />
2<br />
y<br />
4<br />
c.<br />
–8<br />
xS 5 2<br />
–4<br />
–4<br />
–2<br />
521<br />
2<br />
2<br />
–2<br />
–4<br />
0<br />
0<br />
x 2 2 x 2 2<br />
lim<br />
xS2 0 x 2 2 0<br />
1<br />
–1<br />
–2<br />
y<br />
4<br />
2<br />
(x 2 2)(x 1 1)<br />
5 lim<br />
xS2 0 x 2 2 0<br />
(x 2 2)(x 1 1) (x 2 2)(x 1 1)<br />
lim<br />
5 lim<br />
xS2 1 0 x 2 2 0<br />
xS2 1 x 2 2<br />
5 lim x 1 1<br />
xS2 1<br />
5 3<br />
8<br />
4<br />
x<br />
x<br />
(x 2 2)(x 1 1)<br />
lim<br />
5 lim<br />
xS2 2 0 x 2 2 0<br />
–4<br />
d. 0 x 1 2 0 5 x 1 2 if x .22<br />
52(x 1 2) if x ,22<br />
(x 1 2)(x 1 2) 2<br />
lim<br />
5 lim (x 1 2) 2 5 0<br />
xS22 1 x 1 2<br />
xS22 1<br />
(x 1 2)(x 1 2) 2<br />
lim<br />
5 0<br />
xS22 2 2 (x 1 2)<br />
–4<br />
11. a.<br />
–2<br />
–2<br />
4<br />
2<br />
–2<br />
–4<br />
0<br />
4<br />
2<br />
–2<br />
–4<br />
DT T V DV<br />
20<br />
20<br />
20<br />
20<br />
20<br />
20<br />
0<br />
y<br />
240 19.1482<br />
220 20.7908<br />
0 22.4334<br />
20 24.0760<br />
40 25.7186<br />
60 27.3612<br />
80 29.0038<br />
y<br />
5 lim 2 (x 1 1)<br />
xS2 2<br />
523<br />
2<br />
xS2 2 2<br />
DV is constant, therefore T and V form a linear<br />
relationship.<br />
b. V 5 DV<br />
DT ? T 1 K<br />
DV<br />
DT 5 1.6426 5 0.082 13<br />
20<br />
2<br />
1.6426<br />
1.6426<br />
1.6426<br />
1.6426<br />
1.6426<br />
1.6426<br />
4<br />
4<br />
x<br />
(x 2 2)(x 1 1)<br />
(x 2 2)<br />
x<br />
1-26 <strong>Chapter</strong> 1: Introduction to Calculus
V 5 0.082 13T 1 K<br />
T 5 0 V 5 22.4334<br />
Therefore, k 5 22.4334 and<br />
V 5 0.082 13T 1 22.4334.<br />
c. T 5 V 2 22.4334<br />
0.082 13<br />
d. limT 52273.145<br />
vS0<br />
e. V<br />
12<br />
12.<br />
5 21<br />
3<br />
5 7<br />
13. lim f(x) 5 3<br />
xS4<br />
a. 3f(x)4 3 5 3 3 5 27<br />
b.<br />
5<br />
lim<br />
xS4<br />
10<br />
8<br />
6<br />
4<br />
2<br />
0<br />
lim<br />
xS4<br />
0 2 4 6 8 10 12<br />
x 2 2 4<br />
lim<br />
xS5 f(x)<br />
lim<br />
xS5<br />
(x 2 2 4)<br />
lim<br />
xS5<br />
f(x)<br />
3f(x)4 2 2 x 2<br />
f(x) 1 x<br />
( f(x) 2 x)( f(x) 1 x)<br />
5 lim<br />
xS4 f(x) 1 x<br />
5 lim( f(x) 2 x)<br />
xS4<br />
5 3 2 4<br />
521<br />
c. lim"3f(x) 2 2x 5 "3 3 3 2 2 3 4<br />
xS4<br />
5 1<br />
f(x)<br />
14. lim<br />
xS0 x 5 1<br />
a. limf(x) 5 lim c f(x)<br />
xS0<br />
xS0 x 3 xd 5 0<br />
f(x)<br />
b. lim<br />
xS0 g(x) 5 lim x f(x)<br />
c<br />
xS0 g(x) x d 5 0<br />
T<br />
f(x)<br />
g(x)<br />
15. lim and lim 5 2<br />
xS0 x 5 1<br />
xS0 x<br />
a.<br />
f(x)<br />
b. lim<br />
xS0 g(x) 5 lim<br />
xS0<br />
16.<br />
lim<br />
xS0<br />
!x 1 1 2 !2x 1 1<br />
5 lim c<br />
xS0 !x 1 1 1 !2x 1 1<br />
5 lim<br />
xS0<br />
52 2 1 2<br />
1 1 1<br />
522<br />
x 2 1 0 x 2 1 021<br />
17. lim<br />
xS1 0 x 2 1 0<br />
x S 1 1 0 x 2 1 0 5 x 2 1<br />
x 2 1 x 2 2 (x 1 2)(x 2 1)<br />
5<br />
x 2 1 x 2 1<br />
x 2 1 0 x 2 1 021<br />
lim<br />
5 3<br />
xS1 1 0 x 2 1 0<br />
x S 1 2 0 x 2 1 0 52x 1 1<br />
x 2 2 x<br />
lim<br />
xS1 2 2x 1 1 5 lim x(x 2 1)<br />
xS1 2 2x 1 1<br />
521<br />
Therefore, this limit does not exist.<br />
y<br />
4<br />
–4<br />
lim<br />
xS0<br />
g(x) 5 limxa g(x)<br />
xS0<br />
!x 1 1 2 !2x 1 1<br />
!3x 1 4 2 !2x 1 4<br />
3<br />
3<br />
!x 1 1 1 !2x 1 1<br />
!3x 1 4 2 !2x 1 4<br />
!3x 1 4 1 !2x 1 4<br />
!3x 1 4 1 !2x 1 4 d<br />
(x 1 1 2 2x 2 1)<br />
c<br />
(3x 1 4 2 2x 2 4)<br />
–2<br />
2<br />
–2<br />
–4<br />
0<br />
f(x)<br />
x<br />
g(x)<br />
x<br />
x b 5 0 3 2<br />
2<br />
5 1 2<br />
5 0<br />
3<br />
!3x 1 4 1 !2x 1 4<br />
!x 1 1 1 !2x 1 1 d<br />
4<br />
x<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-27
1.6 Continuity, pp. 51–53<br />
1. Anywhere that you can see breaks or jumps is a<br />
place where the function is not continuous.<br />
2. It means that on that domain, you can trace the<br />
graph of the function without lifting your pencil.<br />
3. point discontinuity<br />
10<br />
8<br />
6<br />
4<br />
2<br />
–2 0<br />
–2<br />
y<br />
hole<br />
2 4<br />
jump discontinuity<br />
y<br />
10<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–2 0 2 4 6<br />
–2<br />
infinite discontinuity<br />
y<br />
10<br />
8<br />
6<br />
4<br />
2<br />
x<br />
–1 0 1 2 3 4<br />
–2<br />
vertical<br />
–4 asymptote<br />
6<br />
x<br />
4. a. x 5 3 makes the denominator 0.<br />
b. x 5 0 makes the denominator 0.<br />
c. x 5 0 makes the denominator 0.<br />
d. x 5 3 and x 523 make the denominator 0.<br />
e. x 2 1 x 2 6 5 (x 1 3)(x 2 2)<br />
x 523 and x 5 2 make the denominator 0.<br />
f. The function has different one-sided limits at x 5 3.<br />
5. a. The function is a polynomial, so the function<br />
is continuous for all real numbers.<br />
b. The function is a polynomial, so the function is<br />
continuous for all real numbers.<br />
c. x 2 2 5x 5 x(x 2 5)<br />
The is continuous for all real numbers except<br />
0 and 5.<br />
d. The is continuous for all real numbers greater<br />
than or equal to 22.<br />
e. The is continuous for all real numbers.<br />
f. The is continuous for all real numbers.<br />
6. g(x) is a linear function (a polynomial),<br />
and so is continuous everywhere,<br />
including x 5 2.<br />
7.<br />
y<br />
8<br />
–8<br />
The function is continuous everywhere.<br />
8.<br />
y<br />
4<br />
–4<br />
The function is discontinuous at x 5 0.<br />
9. y<br />
4<br />
2<br />
0<br />
–4<br />
–2<br />
200<br />
4<br />
–4<br />
–8<br />
2<br />
–2<br />
–4<br />
0<br />
0<br />
4<br />
2<br />
400 600<br />
x<br />
8<br />
4<br />
x<br />
x<br />
1-28 <strong>Chapter</strong> 1: Introduction to Calculus
10.<br />
lim<br />
xS3<br />
–4<br />
x 1 3, if x 2 3<br />
12. g(x) 5 e<br />
2 1 !k, if x 5 3<br />
g(x) is continuous.<br />
–4<br />
x 2 2 x 2 6<br />
f(x) 5 lim<br />
xS3 x 2 3<br />
(x 2 3)(x 1 2)<br />
5 lim<br />
xS3 x 2 3<br />
5 5<br />
Function is discontinuous at x 5 3.<br />
11. Discontinuous at x 5 2<br />
y<br />
4<br />
–2<br />
2 1 "k 5 6<br />
"k 5 4, k 5 16<br />
13.<br />
21, if x , 0<br />
f(x) 5 • 0, if x 5 0<br />
1, if x . 0<br />
a.<br />
y<br />
4<br />
–2<br />
2<br />
–2<br />
–4<br />
2<br />
–2<br />
–4<br />
0<br />
0<br />
2<br />
b. i. From the graph, lim f(x) 521.<br />
xS0 2<br />
ii. From the graph, lim f(x) 5 1.<br />
xS0 1<br />
iii. Since the one-sided limits differ, limf(x)<br />
does<br />
xS0<br />
not exist.<br />
c. f is not continuous since limf(x)<br />
does not exist.<br />
xS0<br />
14. a. From the graph, f(3) 5 2.<br />
b. From the graph, lim f(x) 5 4.<br />
xS3 2<br />
c. lim f(x) 5 4 5 lim f(x)<br />
xS3 2<br />
xS3 2<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
2<br />
4<br />
4<br />
x<br />
x<br />
Thus, lim f(x) 5 4. But, f(3) 5 2. Hence f is not<br />
xS3<br />
continuous at x 5 2 (and also not continuous over<br />
23 , x , 8).<br />
15. The function is to be continuous at x 5 1 and<br />
discontinuous at x 5 2.<br />
f(x) 5 μ<br />
For f(x) to be continuous at x 5 1:<br />
A(1) 2 B<br />
5 3(1)<br />
1 2 2<br />
A(1) 2 B 523<br />
A 5 B 2 3<br />
For f(x) to be discontinuous at x 5 2:<br />
B(2) 2 2 A 2 3(2)<br />
4 B 2 A 2 6<br />
If 4B 2 A . 6, then if 4B 2 A , 6, then<br />
4B 2 (B 2 3) . 6 4B 2 B 1 3 , 6<br />
3B 1 3 . 6<br />
3B 1 3 , 6<br />
3B . 3<br />
3B , 3<br />
B . 1 and<br />
B , 1 and<br />
A .22<br />
A ,22<br />
This shows that A and B can be any set of real<br />
numbers such that<br />
(1) A 5 B 2 3<br />
(2) 4B 2 A 2 6 (if B . 1, then A .22 if B , 1,<br />
then A ,22)<br />
A 5 1 and B 522 is not a solution because then<br />
the graph would be continuous at x 5 2.<br />
2x, if 23 # x #22<br />
16. f(x) 5 • ax 2 1 b, if 22 , x , 0<br />
6, if x 5 0<br />
at x 522, 4a 1 b 5 2<br />
at x 5 0, b 5 6.<br />
a 521<br />
2x, if 23 # x #22<br />
f(x) 5 • 2x 2 1 b, if 22 , x , 0<br />
6, if x 5 0<br />
if a 521, b 5 6. f(x) is continuous.<br />
17.<br />
Ax 2 B<br />
x 2 2 ,if x # 1<br />
3x, if 1 , x , 2<br />
Bx 2 2 A, if x $ 2<br />
x0 x 2 1 0<br />
g(x) 5 • x 2 1 , if x 2 1<br />
0, if x 5 1<br />
lim g(x) 521<br />
a. xS1 2 limg(x)<br />
lim g(x) 5 1 xS1<br />
xS1 1<br />
limg(x)<br />
does not exist.<br />
xS1<br />
1-29
.<br />
g(x) is discontinuous at x 5 1.<br />
Review Exercise, pp. 56–59<br />
1. a. f(22) 5 36, f(3) 5 21<br />
21 2 36<br />
m 5<br />
3 2 (22)<br />
523<br />
b. f(21) 5 13, f(4) 5 48<br />
48 2 13<br />
m 5<br />
4 2 (21)<br />
5 7<br />
c. f(1) 523<br />
5(1 1 2h 1 h 2 ) 2 (23)<br />
m 5 lim<br />
hS0<br />
h<br />
2h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim 2 1 h<br />
hS0<br />
5 2<br />
y 2 (23) 5 2(x 2 1)<br />
2x 2 y 2 5 5 0<br />
2. a. f(x) 5 3 P(2, 1)<br />
x 1 1 ,<br />
3 1 h 2 1<br />
m 5<br />
h<br />
5 lim 2 1<br />
hS0 3 1 h<br />
b.<br />
–4<br />
52 1 3<br />
g(x) 5 "x 1 2,<br />
"21 1 h 1 2 2 1<br />
m 5 lim<br />
hS0 h<br />
5 lim c !h 1 1 2 1<br />
hS0<br />
5 lim<br />
hS0<br />
5 1 2<br />
3<br />
–2<br />
4<br />
2<br />
–2<br />
–4<br />
0<br />
y<br />
x<br />
1<br />
!h 1 1 1 1<br />
2<br />
P(21, 1)<br />
4<br />
x<br />
3 !h 1 1 1 1<br />
!h 1 1 1 1 d<br />
c. h(x) 5 2<br />
Pa4, 2 !x 1 5 , 3 b<br />
2<br />
!4 1 h 1 5 2 2 3<br />
m 5 lim<br />
hS0 h<br />
d.<br />
5 2 lim<br />
hS0<br />
5 2 lim c2<br />
hS0<br />
52 2<br />
9(6)<br />
52 1 27<br />
f(x) 5 5<br />
x 2 2 ,<br />
4 1 h 2 2 2 5 2<br />
m 5 lim<br />
hS0 h<br />
10 2 5(2 1 h)<br />
5 lim<br />
hS0 h(2 1 h)(2)<br />
25h<br />
5 lim 2<br />
hS0 h(2 1 h)(2)<br />
52 5 4<br />
3. f(x) 5 e 4 2 x2 , if x # 1<br />
2x 1 1, if x . 1<br />
a. Slope at P(21, 3) f(x) 5 4 2 x 2<br />
4 2 (21 1 h) 2 2 3<br />
m 5 lim<br />
hS0 h<br />
4 2 1 1 2h 2 h 2 2 3<br />
5 lim<br />
hS0 h<br />
5 lim(2 2 h)<br />
hS0<br />
5 2<br />
Slope of the graph at P(21, 3) is 2.<br />
b. Slope at P(2, 0.5)<br />
f(x) 5 2x 1 1<br />
f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5<br />
5 2h<br />
2h<br />
m 5 lim<br />
hS0 h 5 2<br />
Slope of the graph at P(2, 0.5) is 2.<br />
4. s(t) 525t 2 1 180<br />
a. s(0) 5 180, s(1) 5 175, s(2) 5 160<br />
Average velocity during the first second is<br />
s(1) 2 s(0)<br />
1<br />
c 3 2 !9 1 h<br />
3h!9 1 h 3 3 1 !9 1 h<br />
3 1 !9 1 h d<br />
1<br />
3!9 1 h(3 1 !9 1 h) d<br />
5<br />
525<br />
Pa4, 5 2 b<br />
ms. ><br />
1-30 <strong>Chapter</strong> 1: Introduction to Calculus
.<br />
g(x) is discontinuous at x 5 1.<br />
Review Exercise, pp. 56–59<br />
1. a. f(22) 5 36, f(3) 5 21<br />
21 2 36<br />
m 5<br />
3 2 (22)<br />
523<br />
b. f(21) 5 13, f(4) 5 48<br />
48 2 13<br />
m 5<br />
4 2 (21)<br />
5 7<br />
c. f(1) 523<br />
5(1 1 2h 1 h 2 ) 2 (23)<br />
m 5 lim<br />
hS0<br />
h<br />
2h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim 2 1 h<br />
hS0<br />
5 2<br />
y 2 (23) 5 2(x 2 1)<br />
2x 2 y 2 5 5 0<br />
2. a. f(x) 5 3 P(2, 1)<br />
x 1 1 ,<br />
3 1 h 2 1<br />
m 5<br />
h<br />
5 lim 2 1<br />
hS0 3 1 h<br />
b.<br />
–4<br />
52 1 3<br />
g(x) 5 "x 1 2,<br />
"21 1 h 1 2 2 1<br />
m 5 lim<br />
hS0 h<br />
5 lim c !h 1 1 2 1<br />
hS0<br />
5 lim<br />
hS0<br />
5 1 2<br />
3<br />
–2<br />
4<br />
2<br />
–2<br />
–4<br />
0<br />
y<br />
x<br />
1<br />
!h 1 1 1 1<br />
2<br />
P(21, 1)<br />
4<br />
x<br />
3 !h 1 1 1 1<br />
!h 1 1 1 1 d<br />
c. h(x) 5 2<br />
Pa4, 2 !x 1 5 , 3 b<br />
2<br />
!4 1 h 1 5 2 2 3<br />
m 5 lim<br />
hS0 h<br />
d.<br />
5 2 lim<br />
hS0<br />
5 2 lim c2<br />
hS0<br />
52 2<br />
9(6)<br />
52 1 27<br />
f(x) 5 5<br />
x 2 2 ,<br />
4 1 h 2 2 2 5 2<br />
m 5 lim<br />
hS0 h<br />
10 2 5(2 1 h)<br />
5 lim<br />
hS0 h(2 1 h)(2)<br />
25h<br />
5 lim 2<br />
hS0 h(2 1 h)(2)<br />
52 5 4<br />
3. f(x) 5 e 4 2 x2 , if x # 1<br />
2x 1 1, if x . 1<br />
a. Slope at P(21, 3) f(x) 5 4 2 x 2<br />
4 2 (21 1 h) 2 2 3<br />
m 5 lim<br />
hS0 h<br />
4 2 1 1 2h 2 h 2 2 3<br />
5 lim<br />
hS0 h<br />
5 lim(2 2 h)<br />
hS0<br />
5 2<br />
Slope of the graph at P(21, 3) is 2.<br />
b. Slope at P(2, 0.5)<br />
f(x) 5 2x 1 1<br />
f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5<br />
5 2h<br />
2h<br />
m 5 lim<br />
hS0 h 5 2<br />
Slope of the graph at P(2, 0.5) is 2.<br />
4. s(t) 525t 2 1 180<br />
a. s(0) 5 180, s(1) 5 175, s(2) 5 160<br />
Average velocity during the first second is<br />
s(1) 2 s(0)<br />
1<br />
c 3 2 !9 1 h<br />
3h!9 1 h 3 3 1 !9 1 h<br />
3 1 !9 1 h d<br />
1<br />
3!9 1 h(3 1 !9 1 h) d<br />
5<br />
525<br />
Pa4, 5 2 b<br />
ms. ><br />
1-30 <strong>Chapter</strong> 1: Introduction to Calculus
Average velocity during the second second is<br />
s(2) 2 s(1)<br />
5215 ms. ><br />
1<br />
b. At t 5 4:<br />
s(4 1 h) 2 s(4)<br />
525(4 1 h) 2 1 180 2 (25(16) 1 180)<br />
5280 2 40h 2 5h 2 1 180 1 80 2 180<br />
s(4 1 h) 2 s(4) 240h 2 5h2<br />
5<br />
h<br />
h<br />
v(4) 5 lim(240 2 5h) 5240<br />
hS0<br />
Velocity is 240 m><br />
s.<br />
c. Time to reach ground is when s(t) 5 0.<br />
Therefore, 25t 2 1 180 5 0<br />
t 2 5 36<br />
t 5 6, t . 0.<br />
Velocity at t 5 6:<br />
s(6 1 h) 525(36 1 12h 1 h 2 ) 1 180<br />
5260h 2 5h 2<br />
s(6) 5 0<br />
Therefore, v(6) 5 lim(260 2 5h) 5260.<br />
hS0<br />
5. M(t) 5 t 2 mass in grams<br />
a. Growth during 3 # t # 3.01<br />
M(3.01) 5 (3.01) 2 5 9.0601<br />
M(3) 5 3 2<br />
5 9<br />
Grew 0.0601 g during this time interval.<br />
b. Average rate of growth is<br />
0.0601<br />
g><br />
min.<br />
0.01 5 6.01<br />
c. s(3 1 h) 5 9 1 6h 1 h 2<br />
s(3) 5 9<br />
s(3 1 h) 2 s(3) 6h 1 h2<br />
5<br />
h<br />
h<br />
Rate of growth is (6 1 h) 5 6 g><br />
min.<br />
lim<br />
hS0<br />
6. Q(t) 5 10 4 (t 2 1 15t 1 70) tonnes of waste,<br />
0 # t # 10<br />
a. At t 5 0,<br />
Q(t) 5 70 3 10 4<br />
5 700 000.<br />
700 000 t have accumulated up to now.<br />
b. Over the next three years, the average rate of<br />
change:<br />
Q(3) 5 10 4 (9 1 45 1 70)<br />
5 124 3 10 4<br />
Q(0) 5 70 3 10 4<br />
Q(3) 2 Q(0) 54 3 104<br />
5<br />
3<br />
3<br />
5 18 3 10 4 t per year.<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
c. Present rate of change:<br />
Q(h) 5 10 4 (h 2 1 15h 1 70)<br />
Q(0) 5 10 4 1 70<br />
Q(h) 2 Q(0)<br />
lim<br />
5 lim10 4 (h 1 15)<br />
hS0 h<br />
hS0<br />
5 15 3 10 4 t per year.<br />
d. Q(a 1 h)<br />
5 10 4 3a 2 1 2ah 1 h 2 1 15a 1 15h 1 704<br />
Q(a) 5 10 4 3a 2 1 15a 1 704<br />
Q(a 1 h) 2 Q(a)<br />
5 104 32ah 1 h 2 1 15h4<br />
h<br />
h<br />
Q(a 1 h) 2 Q(a)<br />
lim<br />
5 lim10 4 (2a 1 h 1 15)<br />
hS0 h<br />
hS0<br />
5 (2a 1 15)10 4<br />
Now,<br />
(2a 1 15)10 4 5 3 3 10 5<br />
2 a 1 15 5 30<br />
a 5 7.5<br />
3.0 3 10 5<br />
It will take 7.5 years to reach a rate of<br />
t per year.<br />
7. a. From the graph, the limit is 10.<br />
b. 7; 0<br />
c. p(t) is discontinuous for t 5 3 and t 5 4.<br />
8. a. Answers will vary. lim f(x) 5 0.5, f is<br />
xS21<br />
discontinuous at x 521<br />
y<br />
2<br />
–2<br />
b. f(x) 524 if x , 3; f is increasing for x . 3<br />
lim f(x) 5 1<br />
xS3 1 y<br />
4<br />
–4<br />
–1<br />
–2<br />
1<br />
–1<br />
–2<br />
0<br />
2<br />
–2<br />
–4<br />
0<br />
1<br />
2<br />
2<br />
4<br />
x<br />
x<br />
1-31
9. a.<br />
4<br />
y<br />
13. a.<br />
x 1.9 1.99 1.999 2.001 2.01 2.1<br />
–4<br />
–2<br />
2<br />
0<br />
2<br />
4<br />
x<br />
x 2 2<br />
0.344 83 0.334 45 0.333 44 0.333 22 0.332 23 0.322 58<br />
x 2 2 x 2 2<br />
1<br />
3<br />
–2<br />
–4<br />
x 1 1, if x ,21<br />
b. f(x) 5 • 2x 1 1, if 21 # x , 1<br />
x 2 2, if x . 1<br />
Discontinuous at x 521 and x 5 1.<br />
c. They do not exist.<br />
10. The function is not continuous at x 524<br />
because the function is not defined at x 524.<br />
( x 524 makes the denominator 0.)<br />
11. f(x) 5 2x 2 2<br />
x 2 1 x 2 2<br />
2(x 2 1)<br />
5<br />
(x 2 1)(x 1 2)<br />
a. f is discontinuous at x 5 1 and x 522.<br />
2<br />
b. limf(x) 5 lim<br />
xS1<br />
xS1 x 1 2<br />
2<br />
lim f(x): 5 lim<br />
xS22<br />
xS22 1 x 1 2 51`<br />
lim<br />
x 1 2 52`<br />
lim f(x) does not exist.<br />
xS22<br />
12. a. f(x) 5 1 limf(x)<br />
does not exist.<br />
x 2, xS0<br />
b. g(x) 5 x(x 2 5), limg(x) 5 0<br />
c. h(x) 5 x3 xS0<br />
2 27<br />
x 2 2 9 ,<br />
limh(x) 5 37<br />
xS4<br />
lim<br />
xS23<br />
5 2 3<br />
xS22 2 2<br />
7 5 5.2857<br />
h(x) does not exist.<br />
b.<br />
14.<br />
!x 1 3 2 !3<br />
lim c<br />
xS0<br />
5 lim<br />
xS0<br />
5 lim<br />
xS0<br />
5 lim<br />
xS0<br />
5 1<br />
2!3<br />
This agrees well with the values in the table.<br />
15. a.<br />
x 0.9 0.99 0.999 1.001 1.01 1.1<br />
x 2 1<br />
0.526 32 0.502 51 0.500 25 0.499 75 0.497 51 0.476 19<br />
x 2 2 1<br />
1<br />
2<br />
x 20.1 20.01 20.001 0.001 0.01 0.1<br />
"x 1 3 2 "3<br />
0.291 12 0.288 92 0.2887 0.288 65 0.288 43 0.286 31<br />
x<br />
?<br />
x<br />
x 1 3 2 3<br />
xA!x 1 3 1 !3B<br />
x<br />
xA!x 1 3 1 !3B<br />
1<br />
!x 1 3 1 !3<br />
f(x) 5 "x 1 2 2 2<br />
x 2 2<br />
!x 1 3 1 !3<br />
!x 1 3 1 !3 d<br />
x 2.1 2.01 2.001 2.0001<br />
f(x) 0.248 46 0.249 84 0.249 98 0.25<br />
x 5 2.0001<br />
f(x) 8 0.25<br />
1-32 <strong>Chapter</strong> 1: Introduction to Calculus
.<br />
5 lim<br />
xS0<br />
1<br />
A!x 1 5 1 !5 2 xB<br />
5 1 !5<br />
c.<br />
(5 1 h) 2 2 25<br />
16. a. lim<br />
hS0 h<br />
5 lim(10 1 h)<br />
hS0<br />
5 10<br />
Slope of the tangent to y 5 x 2 at x 5 5 is 10.<br />
b.<br />
52 1 16<br />
Slope of the tangent to y 5 1 at (x 5 4) is<br />
x<br />
(x 1 4)(x 1 8)<br />
17. a. lim<br />
5 lim (x 1 8)<br />
xS24 x 1 4<br />
xS24<br />
5 (24) 1 8<br />
5 4<br />
b.<br />
c.<br />
limf(x) 5 0.25<br />
xS2<br />
lim c !x 1 2 2 2<br />
xS2<br />
5 lim<br />
xS2<br />
Slope of the tangent to at x 5 4 is<br />
1<br />
4 1 h 2 1 4 4 2 4 2 h<br />
c. lim 5 lim<br />
hS0 h<br />
hS0 4(4 1 h)(h)<br />
1<br />
5 lim 2<br />
hS0 4(4 1 h)<br />
(x 1 4a) 2 2 25a 2<br />
lim<br />
xSa x 2 a<br />
lim c<br />
xS0<br />
5 lim<br />
xS0<br />
x 2 2<br />
1<br />
!x 1 2 1 2<br />
5 1 4 5 0.25<br />
"4 1 h 2 2<br />
lim<br />
hS0 h<br />
3 !x 1 2 1 2<br />
!x 1 2 1 2 d<br />
"4 1 h 2 2<br />
5 lim<br />
hS0<br />
5 lim<br />
hS0<br />
5 1 4<br />
y 5 "x<br />
5 10a<br />
!x 1 5 2 !5 2 x<br />
3<br />
x<br />
x 1 5 2 5 1 x<br />
xA!x 1 5 1 !5 2 xB<br />
4 1 h 2 4<br />
1<br />
!4 1 h 1 2<br />
1<br />
4.<br />
2 1 16.<br />
(x 2 a)(x 1 9a)<br />
5 lim<br />
xSa x 2 a<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
!x 1 5 1 !5 2 x<br />
!x 1 5 1 !5 2 x d<br />
d.<br />
e.<br />
lim<br />
xS2<br />
x 1 2<br />
5 lim<br />
xS2 x 2 1 2x 1 4<br />
(2) 1 2<br />
5<br />
(2) 2 1 2(2) 1 4<br />
5 4 12<br />
5 1 3<br />
(x 2 2)(x 1 2)<br />
(x 2 2)(x 2 1 2x 1 4)<br />
lim c 4 2 !12 1 x ? 4 1 !12 1 x<br />
xS4 x 2 4 4 1 !12 1 x d<br />
16 2 (12 1 x)<br />
5 lim<br />
xS4 (x 2 4)(4 1 !12 1 x)<br />
4 2 x<br />
5 lim<br />
xS4 (x 2 4)(4 1 !12 1 x)<br />
2 (x 2 4)<br />
5 lim<br />
xS4 (x 2 4)(4 1 !12 1 x)<br />
21<br />
5 lim<br />
xS4 4 1 !12 1 x<br />
21<br />
5<br />
4 1 !12 1 (4)<br />
5 21<br />
4 1 4<br />
52 1 8<br />
1<br />
f. lim<br />
xS0 x a 1<br />
2 1 x 2 1 2 b<br />
5 lim c 1<br />
xS0 x 32 x<br />
2(2 1 x) d<br />
1<br />
5 lim c2<br />
xS0 2(2 1 x) d<br />
52 1 4<br />
18. a. The function is not defined for x , 3, so<br />
there is no left-side limit.<br />
b. Even after dividing out common factors from<br />
numerator and denominator, there is a factor of<br />
x 2 2 in the denominator; the graph has a vertical<br />
asymptote at x 5 2.<br />
25, if x , 1<br />
c. f(x) 5 e<br />
2, if x $ 1<br />
lim f(x) 525 2 lim f(x) 5 2<br />
xS1 2 xS1 1<br />
1-33
d. The function has a vertical asymptote at x 5 2.<br />
0 x 0<br />
e. lim<br />
xS0 x<br />
x S 0 2 0 x 0 52x<br />
0 x 0<br />
lim<br />
xS0 2 x 521<br />
0 x 0<br />
lim<br />
xS0 1 x 5 1<br />
0 x 0<br />
lim<br />
xS0 1 x 2 lim 0 x 0<br />
xS0 2 x<br />
5x 2 , if x ,21<br />
f. f(x) 5 e<br />
2x 1 1, if x $21<br />
lim f(x) 521<br />
xS21 1<br />
lim f(x) 5 5<br />
xS21 2<br />
lim f(x) 2 lim f(x)<br />
xS21 1 xS21 2<br />
Therefore, lim f(x) does not exist.<br />
xS21<br />
19. a.<br />
23(11 h) 2 1 6(11 h) 1 4 2 (23 1 6 1 4)<br />
m 5 lim<br />
hS0<br />
h<br />
23 2 6h 2 h 2 1 6 1 6h 1 4 2 7<br />
5 lim<br />
hS0<br />
h<br />
2h 2<br />
5 lim<br />
hS0 h<br />
5 lim 2h<br />
hS0<br />
5 0<br />
When x 5 1, y 5 7.<br />
The equation of the tangent is y 2 7 5 0(x 2 1)<br />
y 5 7<br />
b.<br />
(22 1 h) 2 2 (22 1 h) 2 1 2 (4 1 2 2 1)<br />
m 5 lim<br />
hS0<br />
h<br />
4 2 4h 1 h 2 1 2 2 h 2 1 2 5<br />
5 lim<br />
hS0<br />
h<br />
25h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim(25 1 h)<br />
hS0<br />
525<br />
When x 522, y 5 5.<br />
The equation of the tangent is y 2 5 525(x 1 2)<br />
y 525x 2 5<br />
6(21 1 h) 3 2 3 2 (26 2 3)<br />
c. m 5 lim<br />
hS0<br />
h<br />
6(21 1 3h 2 3h 2 1 h 3 ) 2 3 1 9<br />
5 lim<br />
hS0<br />
h<br />
18h 2 18h 2 1 6h 3<br />
5 lim<br />
hS0 h<br />
5 lim(18 2 18h 1 6h 2 )<br />
hS0<br />
5 18<br />
When x 521, y 529.<br />
The equation of the tangent is<br />
y 2 (29) 5 18(x 2 (21))<br />
y 5 18x 1 9<br />
22(3 1 h) 4 2 (2162)<br />
d. m 5 lim<br />
hS0 h<br />
22(81 1 108h 1 54h 2 1 12h 3 1 h 4 ) 1 162<br />
5 lim<br />
hS0<br />
h<br />
2216h 2 108h 2 2 24h 3 2 2h 4<br />
5 lim<br />
hS0<br />
h<br />
5 lim( 2 216 2 108h 2 24h 2 2 2h 3 )<br />
hS0<br />
52216<br />
When x 5 3, y 52162.<br />
The equation of the tangent is<br />
y 2 (2162) 52216(x 2 3)<br />
y 52216x 1 486<br />
20. P(t) 5 20 1 61t 1 3t 2<br />
a. P(8) 5 20 1 61(8) 1 3(8) 2<br />
5 700000<br />
b.<br />
20 1 61(8 1 h) 1 3(8 1 h) 2 2 (20 1 488 1 192)<br />
lim<br />
hS0<br />
h<br />
20 1 488 1 61h 1 3(64 1 16h 1 h 2 ) 2 700<br />
5 lim<br />
hS0<br />
h<br />
20 1 488 1 61h 1 192 1 48h 1 3h 2 2 700<br />
5 lim<br />
hS0<br />
h<br />
109h 1 3h 2<br />
5 lim<br />
hS0 h<br />
5 lim(109 1 3h)<br />
hS0<br />
5 109<br />
The population is changing at the rate of<br />
109000>h.<br />
<strong>Chapter</strong> 1 Test, p. 60<br />
1. lim does not exist since<br />
xS1<br />
lim 51`2 lim<br />
x 2 1 x 2 1 52`.<br />
2. f(x) 5 5x 2 2 8x<br />
f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36<br />
f(1) 5 5 2 8 523<br />
36 1 3<br />
Slope of secant is<br />
22 2 1 5239 3<br />
5213<br />
xS1 1 1<br />
1<br />
x 2 1<br />
xS1 2 1<br />
1-34 <strong>Chapter</strong> 1: Introduction to Calculus
d. The function has a vertical asymptote at x 5 2.<br />
0 x 0<br />
e. lim<br />
xS0 x<br />
x S 0 2 0 x 0 52x<br />
0 x 0<br />
lim<br />
xS0 2 x 521<br />
0 x 0<br />
lim<br />
xS0 1 x 5 1<br />
0 x 0<br />
lim<br />
xS0 1 x 2 lim 0 x 0<br />
xS0 2 x<br />
5x 2 , if x ,21<br />
f. f(x) 5 e<br />
2x 1 1, if x $21<br />
lim f(x) 521<br />
xS21 1<br />
lim f(x) 5 5<br />
xS21 2<br />
lim f(x) 2 lim f(x)<br />
xS21 1 xS21 2<br />
Therefore, lim f(x) does not exist.<br />
xS21<br />
19. a.<br />
23(11 h) 2 1 6(11 h) 1 4 2 (23 1 6 1 4)<br />
m 5 lim<br />
hS0<br />
h<br />
23 2 6h 2 h 2 1 6 1 6h 1 4 2 7<br />
5 lim<br />
hS0<br />
h<br />
2h 2<br />
5 lim<br />
hS0 h<br />
5 lim 2h<br />
hS0<br />
5 0<br />
When x 5 1, y 5 7.<br />
The equation of the tangent is y 2 7 5 0(x 2 1)<br />
y 5 7<br />
b.<br />
(22 1 h) 2 2 (22 1 h) 2 1 2 (4 1 2 2 1)<br />
m 5 lim<br />
hS0<br />
h<br />
4 2 4h 1 h 2 1 2 2 h 2 1 2 5<br />
5 lim<br />
hS0<br />
h<br />
25h 1 h 2<br />
5 lim<br />
hS0 h<br />
5 lim(25 1 h)<br />
hS0<br />
525<br />
When x 522, y 5 5.<br />
The equation of the tangent is y 2 5 525(x 1 2)<br />
y 525x 2 5<br />
6(21 1 h) 3 2 3 2 (26 2 3)<br />
c. m 5 lim<br />
hS0<br />
h<br />
6(21 1 3h 2 3h 2 1 h 3 ) 2 3 1 9<br />
5 lim<br />
hS0<br />
h<br />
18h 2 18h 2 1 6h 3<br />
5 lim<br />
hS0 h<br />
5 lim(18 2 18h 1 6h 2 )<br />
hS0<br />
5 18<br />
When x 521, y 529.<br />
The equation of the tangent is<br />
y 2 (29) 5 18(x 2 (21))<br />
y 5 18x 1 9<br />
22(3 1 h) 4 2 (2162)<br />
d. m 5 lim<br />
hS0 h<br />
22(81 1 108h 1 54h 2 1 12h 3 1 h 4 ) 1 162<br />
5 lim<br />
hS0<br />
h<br />
2216h 2 108h 2 2 24h 3 2 2h 4<br />
5 lim<br />
hS0<br />
h<br />
5 lim( 2 216 2 108h 2 24h 2 2 2h 3 )<br />
hS0<br />
52216<br />
When x 5 3, y 52162.<br />
The equation of the tangent is<br />
y 2 (2162) 52216(x 2 3)<br />
y 52216x 1 486<br />
20. P(t) 5 20 1 61t 1 3t 2<br />
a. P(8) 5 20 1 61(8) 1 3(8) 2<br />
5 700000<br />
b.<br />
20 1 61(8 1 h) 1 3(8 1 h) 2 2 (20 1 488 1 192)<br />
lim<br />
hS0<br />
h<br />
20 1 488 1 61h 1 3(64 1 16h 1 h 2 ) 2 700<br />
5 lim<br />
hS0<br />
h<br />
20 1 488 1 61h 1 192 1 48h 1 3h 2 2 700<br />
5 lim<br />
hS0<br />
h<br />
109h 1 3h 2<br />
5 lim<br />
hS0 h<br />
5 lim(109 1 3h)<br />
hS0<br />
5 109<br />
The population is changing at the rate of<br />
109000>h.<br />
<strong>Chapter</strong> 1 Test, p. 60<br />
1. lim does not exist since<br />
xS1<br />
lim 51`2 lim<br />
x 2 1 x 2 1 52`.<br />
2. f(x) 5 5x 2 2 8x<br />
f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36<br />
f(1) 5 5 2 8 523<br />
36 1 3<br />
Slope of secant is<br />
22 2 1 5239 3<br />
5213<br />
xS1 1 1<br />
1<br />
x 2 1<br />
xS1 2 1<br />
1-34 <strong>Chapter</strong> 1: Introduction to Calculus
3. a. lim f(x) does not exist.<br />
xS1<br />
b. lim f(x) 5 1<br />
xS2<br />
c. lim f(x) 5 1<br />
xS4 2<br />
d. f is discontinuous at x 5 1 and x 5 2.<br />
4. a. Average velocity from t 5 2 to t 5 5:<br />
s(5) 2 s(2) (40 2 25) 2 (16 2 4)<br />
5<br />
3<br />
3<br />
15 2 12<br />
5<br />
3<br />
5 1<br />
Average velocity from t 5 2 to t 5 5 is 1 km><br />
h.<br />
b. s(3 1 h) 2 s(3)<br />
5 8(3 1 h) 2 (3 1 h) 2 2 (24 2 9)<br />
5 24 1 8h 2 9 2 6h 2 h 2 2 15<br />
5 2h 2 h 2<br />
2h 2 h 2<br />
v(3) 5 lim 5 2<br />
hS0 h<br />
Velocity at t 5 3 is 2 km><br />
h.<br />
5. f(x) 5 "x 1 11<br />
Average rate of change from x 5 5 to x 5 5 1 h:<br />
f(5 1 h) 2 f(5)<br />
h<br />
"16 1 h 2 "16<br />
5<br />
h<br />
x<br />
6. f(x) 5<br />
x 2 2 15<br />
Slope of the tangent at x 5 4:<br />
4 1 h<br />
f(4 1 h) 5<br />
(4 1 h) 2 2 15<br />
4 1 h<br />
5<br />
1 1 8h 1 h 2<br />
f(4) 5 4 1<br />
4 1 h<br />
f(4 1 h) 2 f(4) 5<br />
1 1 8h 1 h 2 2 4<br />
4 1 h 2 4 2 32h 2 4h2<br />
5<br />
1 1 2h 1 h 2<br />
31h 2 4h2<br />
52<br />
(1 1 2h 1 h 2 )<br />
f(4 1 h) 2 f(4) (231 2 4h)<br />
lim<br />
5 lim<br />
hS0 h<br />
hS0 1 1 2h 1 h 2<br />
5231<br />
Slope of the tangent at x 5 4 is 231.<br />
4x 2 2 36<br />
7. a. lim<br />
xS3 2x 2 6<br />
5 lim 2(x 2 3)(x 1 3)<br />
xS3 (x 2 3)<br />
5 12<br />
2x 2 2 x 2 6<br />
b. lim<br />
xS2 3x 2 2 7x 1 2 5 lim (2x 1 3)(x 2 2)<br />
xS2 (x 2 2)(3x 2 1)<br />
c.<br />
A!x 2 1 2 2BA!x 2 1 1 2B<br />
5 lim<br />
xS5 !x 2 1 2 2<br />
5 4<br />
x 3 1 1<br />
d. lim<br />
xS21 x 4 2 1 5 lim (x 1 1)(x 2 2 x 1 1)<br />
xS21 (x 2 1)(x 1 1)(x 2 1 1)<br />
5 3<br />
22(2)<br />
e.<br />
f.<br />
(x 1 8) 1 3<br />
2 2<br />
lim<br />
xS0 x<br />
(x 1 8) 1 3<br />
5 lim<br />
2 2<br />
xS0 ((x 1 8) 1 3 2 2)((x 1 8) 2 3 1 2(x 1 8) 1 3 1 4)<br />
5<br />
lim<br />
xS5<br />
lim<br />
xS3<br />
5 7 5<br />
x 2 5<br />
!x 2 1 2 2 5 lim (x 2 1) 2 4<br />
xS5 !x 2 1 2 2<br />
52 3 4<br />
1<br />
a<br />
x 2 3 2 6<br />
x 2 2 9 b 5 lim<br />
xS3<br />
1<br />
4 1 4 1 4<br />
5 lim<br />
xS3<br />
5 1 6<br />
(x 1 8) 1 3<br />
2 2<br />
5 lim<br />
xS0 (x 1 8) 2 8<br />
5 1<br />
12<br />
ax 1 3, if x . 5<br />
8. f(x) 5 • 8, if x 5 5<br />
x 2 1 bx 1 a, if x , 5<br />
f(x) is continuous.<br />
Therefore, 5a 1 3 5 8<br />
25 1 5b 1 a 5 8<br />
(x 1 3) 2 6<br />
(x 2 3)(x 1 3)<br />
1<br />
x 1 3<br />
a 5 1<br />
5 b 5218<br />
b 52 18 5<br />
Calculus and Vectors <strong>Solutions</strong> Manual<br />
1-35