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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

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