C 1 Introduction to Calculus - Ms. Ma's...

Review of Prerequisite Skills, pp. 2–3

1. a.

b.

c.

d.

e.

f.

2. a. Substitute the given slope and y-intercept into

b. Substitute the given slope and y-intercept into

c. The slope of the line is

The equation of the line is in the formThe point is and

.The equation of the line is or

d.

e.f.3. a.

b.

c.

d.

4. a.

b.

c.

d.

5.

a.

b.c.

d.

6.

a.

b. s(21) 5 21

s(22) 5 21

2

s(t) 5 μ1

t, if 23 , t , 0

5, if t 5 0

t3, if t . 0

f(3) 5 "6

f(78) 5 9

f(0) 5 "3

f(233) 5 6

f(x) 5 •"3 2 x, if x , 0

"3 1 x, if x $ 0

55

52

f(10) 510

100 1 4

5 0

f(0) 50

0 1 4

5 23

13

f(23) 523

9 1 4

5 25

52

f(210) 5210

100 1 4

5 144

f(2) 5 (10 1 2)2

5 29

f(2) 5 23(4) 1 2(2) 2 1

5 0

f(2) 5 (8 2 2)(6 2 6)

5 21

f(2) 5 26 1 5

y 5 5

x 5 23

x 1 y 2 2 5 0

y 2 4 5 2x 2 2

y 2 4 5 21(x 2 (22))

5 21

m 58 2 4

26 2 (22)

y 5 65(x 1 1) 1 6.

y 2 6 5 65(x 1 1)

m 5 65

(21, 6)y 2 y1 5 m(x 2 x1).

56

5

m 512 2 6

4 2 (21)

y 5 22x 1 5

y 5 mx 1 b.

y 5 4x 2 2

y 5 mx 1 b.

5 21

2

5

22

4

1

m 521

4 2 14

74 2 3

4

5 24.1

m 54 2 4.41

22 2 (22.1)

5 24

m 54 2 0

21 2 0

5 4

m 54 2 0

1 2 0

5 22

m 54 2 (24)

21 2 3

5 23

m 527 2 5

6 2 2

CHAPTER 1Introduction to Calculus

1-1Calculus and Vectors Solutions Manual

c.d.e. or 7. a.b.c.

d.

e.

f.

8. a.

b.c.d.

e.f.

is a zero, so is a factor. Synthetic orlong division yields

9. a.b.c.d.e.

f.10. a.

m sb.

m s11. a. The average rate of change during the secondhour is the difference in the volume at and

(since t is measured in minutes), divided bythe difference in time.

b. To estimate the instantaneous rate of change in volume after exactly 60 minutes, calculate the averagerate of change in volume from minute 59 to minute 61.

c. The instantaneous rate of change in volume isnegative for because the volume ofwater in the hot tub is always decreasing during thattime period, a negative change.12. a., b.

The slope of the tangent line is c. The instantaneous rate of change in when

is

1.1 Radical Expressions:Rationalizating Denominators, p. 9

1. a.

b.

c.

d.

e.

f.

2. a.

b.

5 "6 2 3

5 2"6 2 6

2

2"3 2 3"2

"2?"2

"2

5 "6 1 "10

2

"3 1 "5

"2?"2

"2

2"5 2 2"2

"2 1 "5

3"3 2 "2

2"3 1 "2

"3 2 "2

2"3 1 4

28.x 5 5

f(x)

28.

4 60

4

–4

–8

8y

x

2–2

0 # t # 120

5 213.33 L>min

V(61) 2 V(59)

61 2 598

1186.56 2 1213.22

2

5 220 L>min

V(120) 2 V(60)

120 2 605

0 2 1200

60

t 5 60

t 5 120

>5 10.3

average rate of change 532.4 2 22.1

2 2 1

h(2) 5 32.4h(1) 5 22.1,

>5 20.1

average rate of change 522.1 2 2

1 2 0

h(1) 5 22.1h(0) 5 2,

5xPR 0 x 2 25, 22, 16exPR ` x 2 2

1

2, 3 f

2x2 2 5x 2 3 5 (2x 1 1)(x 2 3)

5xPR 0 x 2 065xPR 0 x 2 165xPR65xPR 0 x $ 2565 (x 2 1)(2x 2 3)(x 1 2)

2x3 2 x2 2 7x 1 6 5 (x 2 1)(2x2 1 x 2 6)

x 2 1x 5 1

2x3 2 x2 2 7x 1 6

27x3 2 64 5 (3x 2 4)(9x2 1 12x 1 16)

5 x(x 1 1)(x 1 1)

x3 1 2x2 1 x 5 x(x2 1 2x 1 1)

2x2 2 7x 1 6 5 (2x 2 3)(x 2 2)

x2 1 x 2 6 5 (x 1 3)(x 2 2)

5 x(x 1 1)(x 2 1)

x3 2 x 5 x(x2 2 1)

5 729a3 2 1215a2 1 675a 2 125

5 (81a2 2 90a 1 25)(9a 2 5)

(9a 2 5)3 5 (9a 2 5)(9a 2 5)(9a 2 5)

5 a3 1 6a2 1 12a 1 8

5 (a2 1 4a 1 4)(a 1 2)

(a 1 2)3 5 (a 1 2)(a 1 2)(a 1 2)

5 2x2 1 x 1 7

5 x2 1 2x 2 3 2 (2x2 1 x 2 10)

(x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)

5 2x2 2 7xx(5x 2 3) 2 2x(3x 1 2) 5 5x2 2 3x 2 6x2 2 4x(5 2 x)(3 1 4x) 5 15 1 17x 2 4x2

(x 2 6)(x 1 2) 5 x2 2 4x 2 12

106s(100) 5 1003

s(1) 5 1

s(0) 5 5

1-2 Chapter 1: Introduction to Calculus

c.

d.

3. a.

b.

c.

d.

e.

f.

4. a.

b.

c.

5. a.

b.

c. The expressions in the two parts are equivalent.The radicals in the denominator of part a. have beensimplified in part b.

5 8"10 1 24

5 16"10 1 48

2

5 16"10 1 48

20 2 18

8"2

2"5 2 3"2?

2"5 1 3"2

2"5 1 3"2

5 8"10 1 24

5 16"10 1 48

2

5 8"40 1 8"36

20 2 18

8"2

"20 2 "18?"20 1 "18

"20 1 "18

5 1

12 2 5!5

5 5 2 4

10 2 5"5 1 2

"5 1 2

2"5 2 1?"5 2 2

"5 2 2

5 27

2 1 3"2

5 4 2 18

2(2 1 3"2)

2 2 3"2

2?

2 1 3"2

2 1 3"2

5 1

!5 1 1

5 5 2 1

4("5 1 1)

"5 2 1

4?"5 1 1

"5 1 1

5 35 2 12"6

19

5 27 2 12"6 1 8

27 2 8

3"3 2 2"2

3"3 1 2"2?

3"3 2 2"2

3"3 2 2"2

5 11"6 2 16

47

5 10"6 2 6 2 10 1 "6

50 2 3

2"3 2 "2

5"2 1 "3?

5"2 2 "3

5"2 2 "3

5 4 2 2"5

5 44 2 22"5

11

5 20 2 22"5 1 24

20 2 9

2"5 2 8

2"5 1 3?

2"5 2 3

2"5 2 3

5 5 1 2"6

5 3 1 2"6 1 2

3 2 2

"3 2 "2

"3 1 "2?"3 2 "2

"3 2 "2

5 10 2 3"10

5 20 2 6"10

20 2 18

2"5

2"5 1 3"2?

2"5 2 3"2

2"5 2 3"2

5 "5 1 "2

5 3("5 1 "2)

3

3

"5 2 "2?"5 1 "2

"5 1 "2

5 3"10 2 2

4

3"5 2 "2

2"2?"2

"2

5 4 1 "6

2

5 12 1 3"6

6

4"3 1 3"2

2"3?"3

"3


6. a.

b.

c.

d.

e.

f.

7. a.

b.

c.

1.2 The Slope of a Tangent, pp. 18–21

1. a.

b.

c.

2. a. The slope of the given line is 3, so the slope of a line perpendicular to the given line is b.

The slope of the given line is so the slope of a lineperpendicular to the given line is 2 7

13.

137 ,

y 513

7x 1

11

7

27y 5 213x 2 11

13x 2 7y 2 11 5 0

213.

5 21

3

m 521 2 (22.6)

1.5 2 6.3

5 25

3

5210

262

m 527

2 2 32

72 2 1

2

5 3

m 528 2 7

23 2 2

51

!x 1 h 1 !x

5h

hA!x 1 h 1 !x B

5x 1 h 2 x

hA!x 1 h 1 !x B

!x 1 h 2 !xh

?!x 1 h 1 !x!x 1 h 1 !x

5 1

"x 1 4 2 2

5 x

x("x 1 4 1 2)

5 x 1 4 2 4

x("x 1 4 1 2)

"x 1 4 2 2

x?"x 1 4 1 2

"x 1 4 1 2

5 1

"a 2 2

5 a 2 4

(a 2 4)("a 2 2)

"a 2 2

a 2 4?"a 1 2

"a 1 2

5 5 1 2"6

5 30 1 12"6

6

5 18 1 2"216 1 12

18 2 12

"18 1 "12

"18 2 "12?"18 1 "12

"18 1 "12

5 212"15 1 15"10

2

5 12"15 1 15"10

48 2 50

3!5

4!3 2 5!2?

4!3 1 5!2

4!3 1 5!2

5 24 1 15"3

4

5 3"24 1 12 1 12 1 2"24

12 2 8

3"2 1 2"3

"12 2 "8?"12 1 "8

"12 1 "8

5 2"2 1 "6

5 8"2 1 4"6

16 2 12

5 2"2

4 2 2"3?

4 1 2"3

4 1 2"3

2"2

"16 2 "12

5 18"2 1 4"3

23

5 36"2 1 8"3

46

5 4"162 1 2"48

54 2 8

2"6

2"27 2 "8?

2"27 1 "8

2"27 1 "8

5 22"3 2 4

5 4"6 1 8

6 2 8

2"2

2"3 2 "8?

2"3 1 "8

2"3 1 "8


3. a.

b. The slope and y-intercept are given.

c. (5, 0)

d. The line is a vertical line because both pointshave the same x-coordinate.

4. a.

b.

c.

d.

53(1 1 2h 1 h2 2 1)

h

3(1 1 h)2 2 3

h5

3((1 1 h)2 2 1)

h

11 1 h 2 1

h5

1 2 1 2 hh(1 1 h)

5 21

1 1 h

5 108 1 54h 1 12h2 1 h3

5 (6 1 h)(18 1 6h 1 h2)

5(9 1 6h 1 h2 2 9)(9 1 6h 1 h2 1 9)

h

5((3 1 h)2 2 9)((3 1 h)2 1 9)

h

(3 1 h)4 2 81

h

5 75 1 15h 1 h2

5h(75 1 15h 1 h2)

h

5(5 1 h 2 5)((5 1 h)2 1 5(5 1 h) 1 25)

h

(5 1 h)3 2 125

h

4 60

2

–2

–4

4y

x

2–2

x 5 5

4 60

2

–2

–4

4y

x

2–2

3x 2 5y 2 15 5 0

y 2 0 53

5(x 2 5)

53

5

m 50 2 (23)

5 2 0

(0, 23),

40

4

–4

–8

8y

x

2–2–4

y 5 8x 1 6

4 60

2

–2

–4

4y

x2–2

7x 2 17y 2 40 5 0

17y 1 68 5 7x 1 28

y 2 (24) 57

17(x 2 (24))

57

17

573173

m 525

3 2 (24)53 2 (24)


e.

f.

5. a.

b.

c.

6. a.

b.

c.

7. a.

b. 12c.

d.

e. They are the same.f.

8. a.

b. at

5 5

5 limhS0

(5 1 h)

5 limhS0

9 1 6h 1 h2 2 3 2 h 2 6

h

m 5 limhS0

(3 1 h)2 2 (3 1 h) 2 6

h

y 5 6.x 5 3,y 5 x2 2 x 5 212

5 limhS0

(212 1 3h)

5 limhS0

12 2 12h 1 3h2 2 12

h

m 5 limhS0

3(22 1 h)2 2 12

h

(22, 12)y 5 3x2,

40

4

–4

8

12y

x

2–2–4

5 12

m 5 limhS0

(12 1 6h 1 h2)

5 12 1 6h 1 h2

58 1 12h 1 6h2 1 h3 2 8

h

m 5(2 1 h)3 2 8

2 1 h 2 2

(2, 8), ((2 1 h), (2 1 h)3)

51

"9 1 h 1 3

m 5"9 1 h 2 3

h?"9 1 h 1 3

"9 1 h 1 3

Q(9 1 h, "9 1 h)P(9, 3),

5 3 1 3h 1 h2

51 1 3h 1 3h2 1 h3 2 1

h

m 5(1 1 h)3 1 2 2 3

h

Q(1 1 h, (1 1 h)3 1 2)P(1, 3),

5 6 1 3h

m 53(1 1 h)2 2 3

h

f(x) 5 3x2Q(1 1 h, f(1 1 h)),P(1, 3),

51

"5 1 h 1 "5

"5 1 h 2 "5

h5

5 1 h 2 5

h("5 1 h 1 "5)

5h 1 5

"h2 1 5h 1 4 1 2

"h2 1 5h 1 4 2 2

h5

h2 1 5h 1 4 2 4

h("h2 1 5h 1 4 1 2)

51

"16 1 h 1 4

"16 1 h 2 4

h5

16 1 h 2 16

h("16 1 h 1 4)

51

4 1 2h

5h

2h(2 1 h)

212 1 h 1 1

2

h5

22 1 2 1 h2(2 1 h)

h

523

4(4 1 h)

34 1 h 2 3

4

h5

12 2 12 2 3h4(4 1 h)

h

5 6 1 3h

53(2h 1 h2)

h


P Q Slope of Line PQ

(2, 8) (3, 27) 19

(2, 8) (2.5, 15.625) 15.25

(2, 8) (2.1, 9.261) 12.61

(2, 8) (2.01, 8.120 601) 12.060 1

(2, 8) (1, 1) 7

(2, 8) (1.5, 3.375) 9.25

(2, 8) (1.9, 6.859) 11.41

(2, 8) (1.99, 7.880 599) 11.940 1

c. at

9. a.

b. at

c. at

10. a. at

b. at

c. at

11. a. Let

Using the limit of the difference quotient, the slopeof the tangent at is

Therefore, the slope of the tangent toat is 1.

b.


5 limhS0

c 2h22 1 h

?1

hd

5 limhS0

c4 2 4 1 2h22 1 h

?1

hd

5 limhS0

4

22 1 h1 2

h

5 limhS0

4

22 1 h2 (22)

h

m 5 limhS0

f(22 1 h) 2 f(22)

h

x 5 22

f(22 1 h) 54

22 1 h

f(22) 54

225 22

x 5 2y 5 f(x) 5 x2 2 3x

5 1

5 0 1 1

5 limhS0

(h 1 1)

5 limhS0

h2 1 h

h

5 limhS0

4 1 4h 1 h2 2 6 2 3h 1 2

h

5 limhS0

(2 1 h)2 2 3(2 1 h) 2 (22)

h

m 5 limhS0

f(2 1 h) 2 f(2)

h

x 5 2

f(2 1 h) 5 (2 1 h)2 2 3(2 1 h)

f(2) 5 (2)2 2 3(2) 5 4 2 6 5 22

y 5 f(x).

5 21

10

5 limhS0

21

5(5 1 h)

m 5 limhS0

15 1 h 2 1

5

h

y 51

5x 5 3;y 5

1

x 1 2

5 21

2

5 limhS0

22

4 1 h

m 5 limhS0

84 1 h 2 2

h

y 5 2x 5 1;y 58

3 1 x

5 22

5 limhS0

24

2 1 h

m 5 limhS0

82 1 h 2 4

h

(2, 4)y 58

x

55

6

5 limhS0

5

"9 1 5h 1 3

5 limhS0

£"9 1 5h 2 3

h3

"9 1 5h 1 3

"9 1 5h 1 3§

m 5 limhS0

"10 1 5h 2 1 2 3

h

y 5 3x 5 2,y 5 "5x 2 1

51

4

5 limhS0

1

"4 1 h 1 2

5 limhS0

£"4 1 h 2 2

h3

"4 1 h 1 2

"4 1 h 1 2§

m 5 limhS0

"9 1 h 2 5 2 2

h

y 5 2x 5 9,y 5 "x 2 5

51

2

5 limhS0

1

"1 1 h 1 1

5 limhS0

£"1 1 h 2 1

h3

"1 1 h 1 1

"1 1 h 1 1§

m 5 limhS0

"3 1 h 2 2 2 1

h

(3, 1)y 5 "x 2 2;

5 12

5 limhS0

(12 2 6h 1 h2)

5 limhS0

28 1 12h 2 6h2 1 h3 1 8

h

m 5 limhS0

(22 1 h)3 1 8

h

y 5 28.x 5 22,y 5 x3


Therefore, the slope of the tangent to at

is c. Let


Using the binomial formula to expand (orone could simply expand using algebra), the slope m is


d. Let


Therefore, the slope of the tangent toat is

e. Let



f. Let

f(8 1 h) 54 1 (8 1 h)

(8 1 h) 2 25

12 1 h6 1 h

f(8) 54 1 8

8 2 25

12

65 2

y 5 f(x).

234.x 5 3y 5 f(x) 5 "25 2 x2

5 23

4

526

8

526

!16 1 4

526 2 0

"16 2 6(0) 2 (0)2 1 4

5 limhS0

26 2 h

"16 2 6h 2 h2 1 4

5 limhS0

h(26 2 h)

h("16 2 6h 2 h2 1 4)

5 limhS0

16 2 6h 2 h2 2 16

h("16 2 6h 2 h2 1 4)

3"16 2 6h 2 h2 1 4

"16 2 6h 2 h2 1 4d

5 limhS0

c"16 2 6h 2 h2 2 4

h

5 limhS0

"16 2 6h 2 h2 2 4

h

m 5 limhS0

f(3 1 h) 2 f(3)

h

x 5 3

5 "16 2 6h 2 h2

5 "25 2 9 2 6h 2 h2

5 "25 2 (9 1 6h 1 h2)

f(3 1 h) 5 "25 2 (3 1 h)2

f(3) 5 "25 2 (3)2 5 !25 2 9 5 4

y 5 f(x).

16.x 5 16y 5 f(x) 5 !x 2 7

51

6

51

3 1 3

51

!0 1 9 1 3

5 limhS0

1

!h 1 9 1 3

5 limhS0

h

h(!h 1 9 1 3)

5 limhS0

(h 1 9) 2 9

h(!h 1 9 1 3)

5 limhS0

!h 1 9 2 3

h?!h 1 9 1 3

!h 1 9 1 3

5 limhS0

!h 1 9 2 3

h

m 5 limhS0

f(16 1 h) 2 f(16)

h

x 5 16

f(16 1 h) 5 !16 1 h 2 7 5 !h 1 9

f(16) 5 !16 2 7 5 !9 5 3

y 5 f(x).

x 5 1y 5 f(x) 5 3x3

5 9

5 3(0) 1 9(0) 1 9

5 limhS0

(3h2 1 9h 1 9)

5 limhS0

3h3 1 9h2 1 9h

h

5 limhS0

3h3 1 9h2 1 9h 1 3 2 3

h

5 limhS0

3(h3 1 3h2 1 3h 1 1) 2 (3)

h

(1 1 h)3

5 limhS0

3(1 1 h)3 2 3

h

m 5 limhS0

f(1 1 h) 2 f(1)

h

x 5 1

f(1 1 h) 5 3(1 1 h)3

f(1) 5 3(1)3 5 3

y 5 f(x).

21.x 5 22

f(x) 54

x

5 21

52

22 1 0

5 limhS0

2

22 1 h



Therefore, the slope of the tangent to

at is

12.

Semi-circle centre

rad 5,OA is a radius.The slope of OA is The slope of tangent is 13. Take values of x close to the point, then

determine 14.

Since the tangent is horizontal, the slope is 0.

15.

The slope of the tangent is 3.

16.

The slope of the tangent is .When

17. a. b.c. The slope of secant AB is

The equation of the secant is

d. Calculate the slope of the tangent.

When the slope is So the equation of the tangent at is

y 5 2x 2 8

y 1 2 5 2(x 2 3)

y 2 y1 5 m(x 2 x1)

A(3, 22)

2(3) 2 4 5 2.x 5 3,

5 2x 2 4

5 2x 1 0 2 4

5 limhS0

(2x 1 h 2 4)

5 limhS0

2xh 1 h2 2 4h

h

5 limhS0

x2 1 2xh 1 h2 2 4x 2 4h 1 1 2 x2 1 4x 2 1

h

5 limhS0

(x 1 h)2 2 4(x 1 h) 1 1 2 (x2 2 4x 1 1)

h

m 5 limhS0

f(x 1 h) 2 f(x)

h

y 5 4x 2 14

y 1 2 5 4(x 2 3)

y 2 y1 5 mAB(x 2 x1)

5 4

58

2

mAB 56 2 (22)

5 2 3

f(5) 5 25 2 20 1 1 5 6; (5, 6)

f(3) 5 9 2 12 1 1 5 22; (3, 22)

3x 1 y 2 8 5 0

y 2 2 5 23(x 2 2)

y 5 2.x 5 2,

23

5 23

5 limhS0

( 2 3 1 h)

5 limhS0

23h 1 h2

h

5 limhS0

4 1 4h 1 h2 2 14 2 7h 1 10

h

m 5 limhS0

(2 1 h)2 2 7(2 1 h) 1 12 2 2

h

3x 2 y 2 8 5 0

y 2 1 5 3(x 2 3)

5 3

5 limhS0

(3 1 h)

5 limhS0

3h 1 h2

h

5 limhS0

9 1 6h 1 h2 2 9 2 3h

h

m 5 limhS0

(3 1 h)2 2 3(3 1 h) 1 1 2 1

h

DyDx

.

234.

43.

y $ 0

(0, 0)y 5 "25 2 x2 S

80

4

–4

8

A

y

x

4–4

216.x 5 8y 5 f(x) 5

4 1 xx 2 2

5 21

6

521

6 1 0

5 limhS0

21

6 1 h

5 limhS0

2h

6 1 h?

1

h

5 limhS0

12 1 h 2 12 2 2h

6 1 h?

1

h

5 limhS0

12 1 h6 1 h

2 2

h

m 5 limhS0

f(8 1 h) 2 f(8)

h

x 5 8


e. When the slope of the tangent is

So the equation of the tangent at is

18. a.

The slope is undefined.b.

The slope is 0.c.

The slope is about –2.5.d.

The slope is about 1.e.

The slope is about f. There is no tangent at this point.

19. at

20. Slope at

Increasing at a rate of 1600 papers per month.21. Point on tangent parallel to

Therefore, tangent line has slope 8.

The point has coordinates

22.

Points on the graph for horizontal tangents are:

23. and

or

The points of intersection are

Tangent to

5 2a.

5 limhS0

2ah 1 h2

h

m 5 limhS0

(a 1 h)2 2 a2

h

y 5 x2:

Q(212,

14).P(1

2, 14),

x 5 21

2x 5

1

2

x2 51

4

x2 51

22 x2

y 51

22 x2y 5 x2

(2, 2283 ).(1, 226

3 ),(21, 263 ),(22, 28

3 ),

a 5 61a 5 62,

(a2 2 4)(a2 2 1) 5 0

a4 2 5a2 1 4 5 0

m 5 a2 2 5 14

a2 5 0

limhS0

4

a(a 1 h)5

4

a2

24

a 1 h1

4

a5 2

4a 1 4a 1 4ha(a 1 h)

5 limhS0

2(a 1 h) 2 (2a)

h5 25

limhS0

aa2 1 ah 11

3h3b 5 a2

5 a2h 1 ah2 11

3h31

3(a 1 h)2 2

1

3a3

y 51

3x3 2 5x 2

4

x

(2, 4).

a 5 2

6a 2 4 5 8

lim

hS0

3h2 1 6ah 2 4hh

5 8

m 5 lim hS0

3(h 1 a)2 2 4(h 1 a) 2 3(a2 1 4a)

h5 8

y 5 8x.

f(x) 5 3x2 2 4x

Cr(6) 5 1200 1 400 5 1600

Cr(t) 5 200t 1 400

t 5 6

C(t) 5 100t2 1 400t 1 5000

5 25

4

5 210

8

5 10 limhS0

4 2 4 2 h

h"4 1 h(2 1 "4 1 h)

5 10 limhS0

2 2 "4 1 h

h"4 1 h3

2 1 "4 1 h

2 1 "4 1 h

m 5 limhS0

20

!4 1 h2 10

h

(5, 10)p . 1D(p) 520

"p 2 1,

278.

P

P

P

P

P

y 5 6x 2 24

y 2 6 5 6(x 2 5)

y 2 y1 5 m(x 2 x1)

B(5, 6)

2(5) 2 4 5 6.

x 5 5,



The slope of the tangent at is at is Tangents to

The slope of the tangents at is

at is and

Therefore, the tangents are perpendicular at thepoints of intersection.24.

The slope of the tangent is We want the line that is parallel to the tangent (i.e.has slope ) and passes through (2, 2). Then,

25. a. Let


2(4a2 1 5a 2 2)

hd

5 limhS0

c 4a2 1 8ah 1 4h2 1 5a 1 5h 2 2

h

m 5 limhS0

f(a 1 h) 2 f(a)

h

x 5 a

5 4a2 1 8ah 1 4h2 1 5a 1 5h 2 2

5 4(a2 1 2ah 1 h2) 1 5a 1 5h 2 2

f(a 1 h) 5 4(a 1 h)2 1 5(a 1 h) 2 2

f(a) 5 4a2 1 5a 2 2

y 5 f(x).

y 5 211x 1 24

y 2 2 5 211(x 2 2)

211

211.

5 211

5 limhS0

(211 1 9h 2 3h2)

5 limhS0

211h 1 9h2 2 3h3

h

5 limhS0

3 2 9h 1 9h2 2 3h3 1 2 2 2h 2 5

h

5 limhS0

23(21 1 3h 2 3h2 1 h3) 1 2 2 2h 2 5

h

5 limhS0

23(21 1 3h 2 3h2 1 h3) 1 2 2 2h 2 5

h

m 5 limhS0

23(21 1 h)3 2 2(21 1 h) 2 5

h

(21, 5)y 5 23x3 2 2x,

mqMq 5 21mpMp 5 21

1 5 Mqa 5 212

21 5 Mp;a 5 12

5 22a.

5 limhS0

22ah 2 h2

h

m 5 limhS0

S12 2 (a 1 h)2T 2 S12 2 a2Th

y 5 12 2 x2:

21 5 mq.a 5 212

1 5 mp,a 5 12

b. To be parallel, the point on the parabola and theline must have the same slope. So, first find theslope of the line. The line canbe rewritten as

So, the slope, m, of the line is 5.To be parallel, the slope at a must equal 5. Frompart a., the slope of the tangent to the parabola at

is

Therefore, at the point the tangent line isparallel to the line c. To be perpendicular, the point on the parabolaand the line must have slopes that are negativereciprocals of each other. That is, their product mustequal So, first find the slope of the line. Theline can be rewritten as

So, the slope, m, of the line is To be perpendicular, the slope at a must equal the negative reciprocal of the slope of the line

That is, the slope of a must equalFrom part a., the slope of the tangent to the

parabola at is

Therefore, at the point the tangent line isperpendicular to the line x 2 35y 1 7 5 0.

(25, 73)

a 5 25

8a 5 240

8a 1 5 5 235

8a 1 5.x 5 a235.

x 2 35y 1 7 5 0.

135.x 2 35y 1 7 5 0

y 51

35x 1

7

35

y 52x 2 7

235

235y 5 2x 2 7

x 2 35y 1 7 5 0

21.

10x 2 2y 2 18 5 0.

(0, 22)

a 5 0

8a 5 0

8a 1 5 5 5

8a 1 5.x 5 a

10x 2 2y 2 18 5 0

y 5 5x 2 9

y 5 29 1 5x

y 518 2 10x

22

22y 5 18 2 10x

10x 2 2y 2 18 5 0

5 8a 1 5

5 8a 1 4(0) 1 5

5 limhS0

(8a 1 4h 1 5)

5 limhS0

8ah 1 4h2 1 5h

h

124a2 2 5a 1 2

hd

5 limhS0

c4a2 1 8ah 1 4h2 1 5a 1 5h 2 2

h

1.3 Rates of Change, pp. 29–311. when or

2. a. Slope of the secant between the

points and

b. Slope of the tangent at the

point

3. Slope of the tangent to the

function with equation at the point 4. a. A and Bb. greater; the secant line through these two pointsis steeper than the tangent line at B.c.

5. Speed is represented only by a number, not adirection.6. Yes, velocity needs to be described by a numberand a direction. Only the speed of the school buswas given, not the direction, so it is not correct touse the word “velocity.”7.a. Average velocity during the first second:

third second:

eighth second:

b. Average velocity

c.

Velocity at is downward.8.a. i. from to

Average velocity

ii. from to

iii.

b. Instantaneous velocity is approximately c. At

9. a.

15 terms are learned between and

b.

At the student is learning at a rate of 16 terms h.10. a. M in mg in 1 mL of blood t hours after theinjection.

Calculate the instantaneous rate of change when

5 21

3

5 limhS0

a21

32

1

3 hb

5 limhS0

21

3 h 2 13 h2

h

5 limhS0

24

3 2 43 h 2 1

3 h2 1 2 1 h 1 4

3 2 2

h

limhS0

21

3(2 1 h)2 1 (2 1 h) 2 (243 1 2)

h

t 5 2.

M(t) 5 21

3t2 1 t; 0 # t # 3

>t 5 2,

5 16

5 limhS0

(16 2 h)

5 limhS0

16h 2 h2

h

5 limhS0

40 1 20h 2 4 2 4h 2 h2 2 36

h

limhS0

20(2 1 h) 2 (2 1 h)2 2 36

h

t 5 3.t 5 2

5 15

551 2 36

1

N(3) 2 N(2)

1

N(t) 5 20t 2 t2

5 64 km>h v(3) 5 48 1 16

v(t) 5 16t 1 16

s(t) 5 8t2 1 16t t 5 3

64 km>h.

5 64.08 km>hs(3.01) 2 s(3)

0.01

3 # t # 3.01

5 64.8 km>h 5

126.48 2 120

0.1

s(3.1) 2 s(3)

0.1

t 5 3.1t 5 3

5 72 km>h 5 24(8 2 5)

5 32(6) 2 24(5)

s(4) 2 s(3)

1

t 5 4t 5 3

0 # t # 5s(t) 5 8t(t 1 2),

20 m>st 5 2

5 220

5 5 limhS0

24h 1 h2

h

v(t) 5 limhS0

320 2 5(2 1 h)2 2 (320 2 5(2)2)

h

s(t) 5 320 2 5t2

5 55 m>s5275

5 s(8) 2 s(3)

8 2 35

320 2 45

5

3 # t # 8

s(8) 2 s(7)

15

320 2 245

15 75 m>s.

s(3) 2 s(2)

15

45 2 20

15 25 m>s;

s(1) 2 s(0)

15 5 m>s;

0 # t # 8s(t) 5 320 2 5t2,

y

xED

CBA

y = f(x)

(4, 2).y 5 !x

limhS0

"4 1 h 2 2

h.

(6, s(6)).

limhS0

s(6 1 h) 2 s(6)

h.

(9, s(9)).(2, s(2))

s(9) 2 s(2)

7.

t 5 4.t 5 0v(t) 5 0



Rate of change is mg h.b. Amount of medicine in 1 mL of blood is beingdissipated throughout the system.

11.


At , rate of change of time with respect toheight is .

12.Calculate the instantaneous rate of change when

5 limkS0

60

5 1 k2 12

k

limkS0

60

(3 1 k) 1 22

60

(3 1 2)

k

h 5 3.

T(h) 560

h 1 2

s>m150

s 5 125

51

50

51

5(5 1 5)

51

5aÄ125

51 5b

5 limhS0

1

5aÄ125 1 h5

1 5b

5 limhS0

≥125 1 h 2 125

5

haÄ125 1 h5

1 5b¥

5 limhS0

≥125 1 h

52 25

haÄ125 1 h5

1 5b ¥

5 limhS0

≥ Ä125 1 h

52 5

h?Ä125 1 h

51 5

Ä125 1 h5

1 5

¥

5 limhS0

Ä125 1 h

52 5

h

limhS0

Ä125 1 h

5 2 Ä125

5

h

s 5 125.

t 5 Ås5

>213

Temperature is decreasing at C km.13.When


It hit the ground in 2 s at a speed of 0 m s.14. Sale of x balls per week:

dollars.a.

Profit on the sale of 40 balls is $4800.b. Calculate the instantaneous rate of change when

Rate of change of profit is $80 per ball.c.

Rate of change of profit is positive when the saleslevel is less than 80.

5 80

5 limhS0

(80 2 h)

5 limhS0

80h 2 h2

h

5 limhS0

6400 1 160h 2 1600 2 80h 2 h2 2 4800

h

limhS0

160(40 1 h) 2 (40 1 h)2 2 4800

h

x 5 40.

5 4800

P(40) 5 160(40) 2 (40)2

P(x) 5 160x 2 x2

>5 0

5 limhS0

25h

5 limhS0

25h2

h

5 limhS0

100 1 100h 1 25h2 2 200 2 100h 1 100

h

limhS0

25(2 1 h)2 2 100(2 1 h) 1 100 2 0

h

t 5 2.

t 5 2

(t 2 2)2 5 0

t2 2 4t 1 4 5 0

25t2 2 100t 1 100 5 0h 5 0,

h 5 25t2 2 100t 1 100

>°125

5 212

5

5 limkS0

212

(5 1 k)

5 limkS0

212k

k(5 1 k)

5 limkS0

60

5 1 k2

60 1 12k5 1 k

k


15. a.

b.

c.

16.

Calculate the instantaneous rate of change.

5 limhS0

S(x 1 h) 2 S(x)

h

S(x) 5 246 1 64x 2 8.9x2 1 0.95x3

51

10

5 limxS24

x 2 24

(x 2 24)(!x 1 1 1 5)

5 limxS24

!x 1 1 2 5

x 2 24?!x 1 1 1 5

!x 1 1 1 5

5 limxS24

f(x) 2 f(24)

x 2 24

x 5 24f(x) 5 !x 1 1,

5 21

5 limxS2

2 (x 2 2)

(x 2 1)(x 2 2)

5 limxS2

x 2 2x 1 2

(x 2 1)(x 2 2)

limxS2

xx 2 1

2 2

x 2 2

x 5 2f(x) 5x

x 2 1,

5 6

5 2 limxS22

(x 2 4)

5 2 limxS22

(x 2 4)(x 1 2)

x 1 2

5 limxS22

2 (x2 2 2x 2 8)

x 1 2

5 limxS22

2x2 1 2x 1 3 1 5

x 1 2

limxS22

f(x) 2 f(22)

x 1 2

f(x) 5 2x2 1 2x 1 3; (22, 25) For the year 2005, Hence,the rate at which the average annual salary is changingin 2005 is

years since 198217.a. The distance travelled from 0 s to 5 s is

mb. mThe rate at which the avalanche is moving from 0 sto 10 s is

m sc. Calculate the instantaneous rate of change when

At 10 s the avalanche is moving at 60 m s.d. Set

Since s.t 5 10!2 8 14t $ 0,

t 5 610!2

t2 5 200

3t2 5 600

s(t) 5 600:

>5 60

5 limhS0

(60 1 3h)

5 limhS0

60h 1 3h2

h

5 limhS0

300 1 60h 1 3h2 2 300

h

limhS0

3(10 1 h)2 2 300

h

t 5 10.

>5 30

DsDt

5300 2 0

10 2 0

s(10) 5 3(10)2 5 300

s(5) 5 3(5)2 5 75

s(t) 5 3t2

$1 162 250>P r(23) 5 64 2 17.8(23) 1 2.85(23)2 5

x 5 2005 2 1982 5 23.

5 64 2 17.8x 1 2.85x2

5 64 2 8.9(2x 1 0) 1 0.95 33x2 1 3x(0) 1 (0)245 lim

hS0 364 2 8.9(2x 1 h) 1 0.95(3x2 1 3xh 1 h2)4

5 limhS0

64h 2 8.9(2xh 1 h2) 1 0.95(3x2h 1 3xh2 1 h3)

h

5 limhS0

246 2 246 1 64(x 1 h 2 x) 2 8.9(x2 1 2xh 1 h2 2 x2) 1 0.95(x3 1 3x2h 1 3xh2 1 h3 2 x3)

h

5 limhS0

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 8.9x2 1 0.95x3)

h

18. The coordinates of the point are . The slope

of the tangent is . The equation of the tangent

is or The

intercepts are and The tangent line

and the axes form a right triangle with legs of length

and 2a. The area of the triangle is

19.

Rate of change of cost is

which is independent of F (fixed costs).20.Rate of change of area is

mRate is m2 m.21. Cube of dimensions x by x by x has volume

Surface area is

surface area.

22. a. The surface area of a sphere is given by

The question asks for the instantaneous rate ofchange of the surface when This is

Therefore, the instantaneous rate of change of the surface area of a spherical balloon as it isinflated when the radius reaches 10 cm is

cm2 unit of time.b. The volume of a sphere is given by The question asks for the instantaneous rate ofchange of the volume when Note that the volume is deflating. So, find the rateof the change of the volume when and thenmake the answer negative to symbolize a deflatingspherical balloon.

Using the binomial formula to expand(or one could simply expand using

algebra), the limit is

Because the balloon is deflating, the instantaneous rateof change of the volume of the spherical balloon whenthe radius reaches 5 cm is cm3 unit of time.

Mid-Chapter Review pp. 32–331. a. Corresponding conjugate:

b. Corresponding conjugate:

5 37

5 45 2 8

5 9(5) 2 4(2)

5 (9!25 2 6!10 1 6!10 2 4!4)

(3!5 1 2!2)(3!5 2 2!2)

3!5 2 2!2.

5 3

5 5 2 2

5 (!25 1 !10 2 !10 2 !4)

(!5 2 !2)(!5 1 !2)

!5 1 !2.

>2100p

5 100p

54

3p(0)2 1 20p(0) 1 100p

5 limhS0

a4

3ph2 1 20ph 1 100pb

5 limhS0

43 ph3 1 20ph2 1 100ph

h

2 43 p(125)

h

5 limhS0

43 ph3 1 20ph2 1 100ph 1 4

3 p(125)

h

5 limhS0

43 p(h3 1 15h2 1 75h 1 125) 2 4

3 p(5)3

h

(5 1 h)3

5 limhS0

43 p(5 1 h)3 2 4

3 p(5)3

h

limhS0

V(5 1 h) 2 V(5)

h

r 5 5

r 5 5.

V(r) 5 43pr3.

>80p

5 80p

5 80p 1 4p(0)

5 limhS0

(80p 1 4ph)

5 limhS0

80ph 1 4ph2

h

5 limhS0

400p 1 80ph 1 4ph2 2 400p

h

5 limhS0

4p(100 1 20h 1 h2) 2 4p(100)

h

5 limhS0

4p(10 1 h)2 2 4p(10)2

h

limhS0

A(10 1 h) 2 A(10)

h

r 5 10.

A(r) 5 4pr2.

Vr(x) 5 3x2 51

2

6x2.V 5 x3.

>200p

r 5 100

5 2pr

5 p limhS0

(r 1 h 2 r)(r 1 h 1 r)

h

5 limhS0

p(r 1 h)2 2 pr2

h

limhS0

A(r 1 h) 2 A(r)

h

A(r) 5 pr2

5 limxSh

V(x 1 h) 2 V(x)

h h,

limxSR

C(x 1 h) 2 C(x)

h

C(x 1 h) 5 F 1 V(x 1 h)

C(x) 5 F 1 V(x)

1

2a2

ab (2a) 5 2.

2

a

(22a, 0).a0, 2

ab

y 5 21

a2 x 12

a.y 2

1

a5 2

1

a2 (x 2 a)

21

a2

aa, 1

ab


c. Corresponding conjugate:

d. Corresponding conjugate:

2. a.

b.

c.

d.

e.

f.

3. a.

b.

c.

d.

5 213

3!2(2!3 1 5)5

12 2 25

3!2(2!3 1 5)

54(3) 2 25

3!2(2!3 1 5)

54!9 1 10!3 2 10!3 2 25

3!2(2!3 1 5)

2!3 2 5

3!2?

2!3 1 5

2!3 1 5

5 29

5(!7 1 4)

57 2 16

5(!7 1 4)

5!49 1 4!7 2 4!7 2 16

5(!7 1 4)

!7 2 4

5?!7 1 4

!7 1 4

53

!3(6 1 !2)

5!9

!3(6 1 !2)

!3

6 1 !2?!3

!3

52

5!2

5!4

5!2

!2

5?!2

!2

5 23!2(2!3 1 5)

13

53!2(2!3 1 5)

213

53!2(2!3 1 5)

12 2 25

53!2(2!3 1 5)

4(3) 2 25

53!2(2!3 1 5)

4!9 1 10!3 2 10!3 2 25

3!2

2!3 2 5?

2!3 1 5

2!3 1 5

510!3 2 15

2

530 2 20!3

24

530 2 20!3

12 2 16

510!9 2 20!3

4!9 2 8!3 1 8!3 2 16

5!3

2!3 1 4?

2!3 2 4

2!3 2 4

5 22(3 1 2!3)

56 1 4!3

21

56 1 4!3

3 2 4

52!9 1 4!3

!9 1 2!3 2 2!3 2 4

2!3

!3 2 2?!3 1 2

!3 1 2

5 25(!7 1 4)

9

55(!7 1 4)

7 2 16

55(!7 1 4)

!49 1 4!7 2 4!7 2 16

5

!7 2 4?!7 1 4

!7 1 4

56 1 4!3

3

52!9 1 4!3

!9

2!3 1 4

!3?!3

!3

56!3 1 !6

3

56!3 1 !6

!9

6 1 !2

!3?!3

!3

5 5

5 45 2 40

5 9(5) 2 4(10)

5 (9!25 1 6!50 2 6!50 2 4!100)

(3!5 2 2!10)(3!5 1 2!10)

3!5 1 2!10.

5 61

5 81 2 20

5 81 2 4(5)

5 (81 2 18!5 1 18!5 2 4!25)

(9 1 2!5)(9 2 2!5)

9 2 2!5.




(21, 1) (22, 6) 52

(21, 1) (21.5, 3.25) 4.52

(21, 1) (21.1, 1.41) 4.12

(21, 1) (21.01, 1.040 1) 4.012

(21, 1) (21.001, 1.004 001) 4.0012


(21, 1) (0, 22) 32

(21, 1) (20.5, 20.75) 3.52

(21, 1) (20.9, 0.61) 3.92

(21, 1) (20.99, 0.9601) 3.992

(21, 1) (20.999, 0.996 001) 3.9992

e.

f.

4. a.

b.

c.

d.

21

5x 1 y 1

10

52

1

55 0

y 1 2 51

5x 1

1

5

y 2 (22) 51

5(x 2 (21))

m 51

5

4x 2 y 2 2 5 0

24x 1 y 1 2 5 0

y 2 6 5 4x 2 8

y 2 6 5 4(x 2 2)

m 5 4

x 2 y 1 5 5 0

2x 1 y 2 5 5 0

y 2 7 5 x 2 2

y 2 7 5 1(x 2 2)

m 511 2 7

6 2 25

4

45 1

2

3x 1 y 2 6 5 0

y 2 6 5 22

3x

y 2 6 5 22

3(x 2 0)

m 5 22

3;

51

(2!3 2 !7)

512 2 7

5(2!3 2 !7)

54(3) 2 7

5(2!3 2 !7)

54!9 2 2!21 1 2!21 2 !49

5(2!3 2 !7)

2!3 1 !7

5?

2!3 2 !7

2!3 2 !7

5 21

(!3 1 !7)

5 24

4(!3 1 !7)

53 2 7

4(!3 1 !7)

5!9 1 !21 2 !21 2 !49

4(!3 1 !7)

!3 2 !7

4?!3 1 !7

!3 1 !7

5. The slope of PQ is

So, the slope of PQ with is 6. a. Unlisted y-coordinates for Q are found by substituting the x-coordinates into the given function.The slope of the line PQ with the given points isgiven by the following: Let and

Then, the

b. The slope from the right and from the left appearto approach The slope of the tangent to thegraph of at point P is about c. With the points and

the slope, m, of PQ isthe following:Q 5 (21 1 h, f(21 1 h)),

P 5 (21, 1)

24.f(x)

24.

slope 5 m 5y

22 y

1

x2 2 x1.Q 5 (y1, y2).

P 5 (x1, y1)

22.f(x) 5 2x2

5 22

5 22 2 (0)

5 limhS0

(22 2 h)

5 limhS0

22h 2 h2

h

5 limhS0

21 2 2h 2 h2 1 1

h

5 limhS0

2 (1 1 2h 1 h2) 1 1

h

5 limhS0

2 (1 1 h)2 1 1

h

m 5 limhS0

f(1 1 h) 2 (21)

(1 1 h) 2 1

x 2 5y 2 9 5 0

1

5x 2 y 2

9

55 0

21

5x 1 y 1

9

55 0


d. The slope of the tangent is

In this case, as h goes to zero, goes toThe slope of the tangent to

the graph of at the point P is e. The answers are equal.

24.f(x)

h 2 4 5 0 2 4 5 24.

h 2 4

limhS0

f(x).

5 h 2 4

5h2 2 4h

h

51 2 2h 1 h2 1 2 2 2h 2 2 2 1

21 1 h 1 1

53(21 1 h)2 2 2(21 1 h) 2 24 2 (1)

(21 1 h) 2 (21)

m 5y2 2 y1

x2 2 x1c.

5 limhS0

a4 2 (h 1 4)

h 1 4b 1

h

5 limhS0

4

h 1 42 1

h

5 limhS0

4

h 1 42

4

4

h

5 limhS0

4

6 1 h 2 22

4

6 2 2

h

m 5 limhS0

f(6 1 h) 2 f(6)

h

y 5 f(x) 54

x 2 2

7. a.

5 23

5 0 2 3

5 limhS0

(h 2 3)

5 limhS0

h2 2 3h

h

5 limhS0

h2 2 3h 2 5 2 (25)

h

5 limhS0

9 2 6h 1 h2 2 9 1 3h 2 5 2 (9 2 9 2 5)

h

5 limhS0

3(23 1 h)2 1 3(23 1 h) 2 54 2 3(23)2 1 3(23) 2 54

h

m 5 limhS0

f(23 1 h) 2 f(23)

h

b.

5 29

521

19 1 1

3(0)

5 limhS0

21

19 1 1

3h

5 limhS0

a 2h19 1 1

3hb 1

h

5 limhS0

(13) 2 (1

3 1 h)13(1

3 1 h)

h

5 limhS0

113 1 h

2113

h

m 5 limhS0

f(1

3 1 h) 2 f(13)

h

y 5 f(x) 51

x

d.

5 limhS0

1

!9 1 h 1 3

5 limhS0

h

h(!9 1 h 1 3)

5 limhS0

9 1 h 1 3!9 1 h 2 3!9 1 h 2 9

h(!9 1 h 1 3)

5 limhS0

!9 1 h 2 3

h?!9 1 h 1 3

!9 1 h 1 3

5 limhS0

!9 1 h 2 3

h

5 limhS0

!9 1 h 2 !9

h

5 limhS0

!5 1 h 1 4 2 !5 1 4

h

m 5 limhS0

f(5 1 h) 2 f(5)

h

5 21

4

521

0 1 4

5 limhS0

21

h 1 4

5 limhS0

a 2hh 1 4

b 1

h

8.

a. i.

km h

ii.

km h

iii.

km h

b. At the time the velocity of the car appearsto approach 30 km h.

c.

km hd. When the velocity is the limit as happroaches 0.

Therefore, when the velocity is 30 km h.9. a. The instantaneous rate of change of withrespect to x at is given by

b. The instantaneous rate of change of with

respect to x at is given by

5 212

526

12 1 0

5 limhS0

26

12 1 h

5 limhS0

26h

12 1 h

?1

h

5 limhS0

3 2 3 2 6h

12 1 h

?1

h

5 limhS0

3 2 6(1

2 1 h)12 1 h

?1

h

5 limhS0

312 1 h

2 6

h

5 limhS0

312 1 h

2312

h

limhS0

f(1

2 1 h) 2 f(12)

h

x 5 12

f(x)5 24

5 2 (0) 2 4

5 limhS0

(2h 2 4)

5 limhS0

2h2 2 4h

h

5 limhS0

5 2 4 2 4h 2 h2 2 1

h

5 limhS0

5 2 (4 1 4h 1 h2) 2 1

h

5 limhS0

35 2 (2 1 h)24 2 35 2 (2)24

h

limhS0

f(2 1 h) 2 f(2)

h

x 5 2

f(x)

>t 5 25 30

5 6(0) 1 30

velocity 5 limhS0

(6h 1 30)

t 5 2,

>5 (6h 1 30)

56h2 1 30h

h

56h2 1 30h 1 36 2 36

h

5324 1 24h 1 6h2 1 12 1 6h4 2 36

h

536(4 1 4h 1 h2) 1 12 1 6h4 2 324 1 124

h

536(2 1 h)2 1 6(2 1 h)4 2 36(2)2 1 6(2)4

h

average velocity 5f(2 1 h) 2 f(2)

(2 1 h) 2 (2)

>t 5 2,

>5 30.06

50.3006

0.01

536.3006 2 36

0.01

536.3006 2 324 1 124

0.01

5324.2406 1 12.064 2 36(2)2 1 6(2)4

0.01

536(2.01)2 1 6(2.01)4 2 36(2)2 1 6(2)4

0.01

average velocity 5s(2.01) 2 s(2)

2.01 2 2

>5 30.6

53.06

0.1

539.06 2 36

0.1

5326.46 1 12.64 2 324 1 124

0.1

536(2.1)2 1 6(2.1)4 2 36(2)2 1 6(2)4

0.1


2.1 2 2

>5 36

5 54 1 18 2 36

5 6(9) 1 18 2 (24 1 12)

5 36(3)2 1 6(3)42 36(2)2 1 6(2)4average velocity 5

s(3) 2 s(2)

3 2 2

s(t) 5 6t(t 1 1) 5 6t2 1 6t

51

6

51

!9 1 0 1 3


10. a. The average rate of change of withrespect to t during the first 20 minutes is given by

L minb. The rate of change of with respect to t at thetime is given by

L min11. a. Let



So an equation of the tangent at is given by

b. Let




c.


5 24

5 3(0) 2 4

5 limhS0

(3h 2 4)

5 limhS0

3h2 2 4h

h

5 limhS0

3h2 2 4h 2 4 2 (24)

h

m 5 limhS0

f(21 1 h) 2 f(21)

h

x 5 4

5 3h2 2 4h 2 45 3 2 6h 1 3h2 2 7 1 2h5 3(1 2 2h 1 h2) 2 2 1 2h 2 5

f(21 1 h) 5 3(21 1 h)2 1 2(21 1 h) 2 5

5 24f(21) 5 3(21)2 1 2(21) 2 5 5 3 2 2 2 5

8x 1 y 1 15 5 0

8x 1 y 2 1 1 16 5 0

y 2 1 5 28x 2 16

y 2 1 5 28(x 2 (22))

x 5 22

28.x 5 22y 5 f(x) 5 2x2 2 7

5 28

5 2(0) 2 8

5 limhS0

(2h 2 8)

5 limhS0

2h2 2 8h

h

5 limhS0

2h2 2 8h 1 1 2 (1)

h

m 5 limhS0

f(22 1 h) 2 f(22)

h

x 5 4

5 2h2 2 8h 1 1

5 8 2 8h 1 2h2 2 7

5 2(4 2 4h 1 h2) 2 7

f(22 1 h) 5 2(22 1 h)2 2 7

f(22) 5 2(22)2 2 7 5 2(4) 2 7 5 1

y 5 f(x).

29x 1 y 1 19 5 0

29x 1 y 2 17 1 36 5 0

y 2 17 5 9x 2 36

y 2 17 5 9(x 2 4)

x 5 4

x 5 4y 5 f(x) 5 x2 1 x 2 3

5 9

5 0 1 9

5 limhS0

(h 1 9)

5 limhS0

h2 1 9h

h

5 limhS0

h2 1 9h 1 17 2 (17)

h

m 5 limhS0

f(4 1 h) 2 f(4)

h

x 5 4

5 h2 1 9h 1 17

5 16 1 8h 1 h2 1 h 1 1

f(4 1 h) 5 (4 1 h)2 1 (4 1 h) 2 3

f(4) 5 (4)2 1 (4) 2 3 5 16 1 1 5 17

y 5 f(x).

>5 21000

5 50(0) 2 1000

5 limhS0

50h 2 1000

5 limhS0

50h2 2 1000h

h

5 limhS0

5000 2 1000h 1 50h2 2 5000

h

5 limhS0

350(100 2 20h 1 h2)4 2 350(100)4

h

5 limhS0

350(10 2 h)24 2 350(10)24

h

5 limhS0

350(30 2 (20 1 h))24 2 350(30 2 20)24

h

limhS0

f(20 1 h) 2 f(20)

h

t 5 20

V(t)>5 22000

5 240 000

20

55000 2 45 000

20

5350(30 2 20)24 2 350(30 2 0)24

20

f(20) 2 f(0)

20 2 0

V(t)


Therefore, the slope of the tangent toat is .


d.




12. a. Using the limit of the difference quotient, theslope of the tangent at is



b. Using the limit of the difference quotient, theslope of the tangent at is


at is


3x 1 4y 1 5 5 0

3x 1 4y 1 2 1 3 5 0

4y 1 2 5 23x 2 3

y 11

25 2

3

4x 2

3

4

y 2 a21

2b 5 2

3

4(x 2 (21))

x 5 234

234.x 5 21f(x) 5

2x 1 5

5x 2 1

5 23

4

5 29

12

59

10(0) 2 12

5 limhS0

a 9

10h 2 12b

5 limhS0

a 9h10h 2 12

b ?1

h

5 limhS0

a4h 1 6 1 5h 2 6

10h 2 12b ?

1

h

5 limhS0

a2h 1 3

5h 2 61

1

2b ?

1

h

5 limhS0

a2h 1 3

5h 2 62

3

26b ?

1

h

5 limhS0

a22 1 2h 1 5

25 1 5h 2 12

22 1 5

25 2 1b ?

1

h

5 limhS0

a2(21 1 h) 1 5

5(21 1 h) 2 12

2(21) 1 5

5(21) 2 1b ?

1

h

m 5 limhS0

f(21 1 h) 2 f(21)

h

x 5 21

23x 1 4y 2 25 5 0

23

4x 1 y 2

25

45 0

23

4x 1 y 2

10

42

15

45 0

y 25

25

3

4x 1

15

4

y 25

25

3

4(x 2 (25))

x 5 34

34.x 5 25f(x) 5

xx 1 3

53

4

523

24 1 2(0)

5 limhS0

a 23

24 1 2hb

5 limhS0

a 23h24 1 2h

b ?1

h

5 limhS0

a210 1 2h 1 10 2 5h24 1 2h

b ?1

h

5 limhS0

a210 1 2h 2 (210 1 5h)

24 1 2hb ?

1

h

5 limhS0

a25 1 h22 1 h

25

2b ?

1

h

5 limhS0

a 25 1 h25 1 h 1 3

225

25 1 3b ?

1

h

m 5 limhS0

f(25 1 h) 2 f(25)

h

x 5 25

22x 1 y 1 2 5 0

y 5 2x 2 2

y 2 0 5 2(x 2 1)

x 5 1

x 5 1y 5 f(x) 5 5x2 2 8x 1 3

5 2

5 5(0) 1 2

5 limhS0

(5h 1 2)

5 limhS0

5h2 1 2h 2 (0)

h

m 5 limhS0

f(1 1 h) 2 f(1)

h

x 5 1

5 5h2 1 2h5 5 1 10h 1 5h2 2 5 2 8h5 5(1 1 2h 1 h2) 2 8 2 8h 1 3

f(1 1 h) 5 5(1 1 h)2 2 8(1 1 h) 1 3

f(1) 5 5(1)2 2 8(1) 1 3 5 5 2 8 1 3 5 0

4x 1 y 1 8 5 0

4x 1 y 1 4 1 4 5 0

y 1 4 5 24x 2 4

y 1 4 5 24(x 1 1)y 2 (24) 5 24(x 2 (21))

x 5 24

24x 5 21y 5 f(x) 5 3x2 1 2x 2 5


1.4 The Limit of a Function, pp. 37–39

1. a.

b.2. One way to find a limit is to evaluate the functionfor values of the independent variable that get progressively closer to the given value of the independent variable.3. a. A right-sided limit is the value that a function gets close to as the values of the independent variable decrease and get close to a given value.b. A left-sided limit is the value that a function gets close to as the values of the independent variable increase and get close to a given value.c. A (two-sided) limit is the value that a functiongets close to as the values of the independent variable get close to a given value, regardless of whether the values increase or decrease toward the given value.4. a.b.c.d.e. 4f.5. Even though the limit is 1, since thatis the value that the function approaches from theleft and the right of 6. a. 0b. 2c.d. 27. a. 2b. 1c. does not exist8. a.

b.

c.

9.

10. a. Since 0 is not a value for which the function isundefined, one may substitute 0 in for x to find that

b. Since 2 is not a value for which the function isundefined, one may substitute 2 in for x to find that

c. Since 3 is not a value for which the function isundefined, one may substitute 3 in for x to find that

d. Since 1 is not a value for which the function isundefined, one may substitute 1 in for x to find that

e. Since 3 is not a value for which the function isundefined, one may substitute 3 in for x to find that

f. If 3 is substituted in the function for x, then thefunction is undefined because of division by zero.There does not exist a way to divide out the in x 2 3

51

5

51

3 1 2

limxS31

1

x 1 25 lim

xS3

1

x 1 2

5 21

2

51

1 2 3

limxS11

1

x 2 35 lim

xS1

1

x 2 3

5 55 9 2 4

5 (3)2 2 4

limxS32

(x2 2 4) 5 limxS3

(x2 2 4)

5 05 4 2 4

5 (2)2 2 4

limxS22

(x2 2 4) 5 limxS2

(x2 2 4)

5 0

5 (0)4

limxS01

x4 5 limxS0

x4

40

2

4

6y

x

2–2–4

22 1 1 5 5

5 2

"5 2 1 5 "4

5 2Å

0 1 20

0 1 55 "4

9 2 (21)2 5 8

21

x 5 4.

f(4) 5 21,

23 5 8

4 2 3(22)2 5 28

102 5 100

3 1 7 5 10

25

p

27

99


the denominator. Also, approaches infinity,

while approaches negative infinity.

Therefore, since

does not exist.

11. a.

Therefore, does

not exist.b.

Therefore, exists and

is equal to 2.c.


is equal to 2.

d.

Therefore,

does not exist.12. Answers may vary. For example:a.

b.

c.

d.

13.

m 5 23b 5 1,

2b 5 2

2m 1 b 5 4limxS21

f(x) 5 4

m 1 b 5 22limxS1

f(x) 5 22

f(x) 5 mx 1 b

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

limxS20.5

f(x)limxS20.51

f(x) 2 limxS20.52

f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS1

2

f(x)limxS1

21 f(x) 5 lim

xS12

2 f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS2

f(x)limxS21

f(x) 5 limxS22

f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS21

f(x)limxS211

f(x) 2 limxS212

f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS3

1

x 2 3lim

xS31

1

x 2 32 lim

xS32

1

x 2 3,

limxS32

1

x 2 3

limxS31

1

x 2 3


14.

Therefore, the values are , and 15. a.

b.

c. Since is measured in thousands, right beforethe chemical spill there were 6000 fish in the lake.Right after the chemical spill there were 4000 fishin the lake. So, fish werekilled by the spill.d. The question asks for the time, t, after the chemicalspill when there are once again 6000 fish in the lake.Use the second equation to set up an equation that ismodelled by

(The question asks for time so the negative answeris disregarded.)So, at time years the populationhas recovered to the level before the spill.

1.5 Properties of Limits, pp. 45–471. and have the same value,

but does not. Since there are no brackets

around the expression, the limit only applies to 3,and there is no value for the last term, x.2. Factor the numerator and denominator. Cancelany common factors. Substitute the given value of x.3. If the two one-sided limits have the same value,then the value of the limit is equal to the value ofthe one-sided limits. If the one-sided limits do nothave the same value, then the limit does not exist.

4. a.

b.

c.

d.

e.

f.

5. a.

b.

6. Since substituting does not make thedenominator 0, direct substitution works.

7. a.

b.

c.

5 27

5 9 1 9 1 9

limxS3

x3 2 27

x 2 35 lim

xS3

(x 2 3)(x2 1 3x 1 9)

x 2 3

5 5

limxS21

2x2 1 5x 1 3

x 1 15 lim

xS21

(x 1 1)(2x 1 3)

x 1 1

5 4

5 limxS2

(2 1 x)

limxS2

4 2 x2

2 2 x5 lim

xS2

(2 2 x)(2 1 x)

(2 2 x)

5 21

1 2 1 2 5

6 2 15

25

5

t 5 1

5 "2

2

!1 1 15

2

!2

(22)3

22 2 25 22

5 "3

Å23 2 3

2(23) 1 45 Å

26

22

5 2

"3 1 "1 1 0 5 "3 1 1

5 5p3

(2p)3 1 p2(2p) 2 5p3 5 8p3 1 2p3 2 5p3

5100

9

c"9 11

"9d 2

5 a3 11

3b2

(21)4 1 (21)3 1 (21)2 5 1

3(2)

22 1 25 1

limxS2

3 1 x

limxS2

(x 1 3)limxS2

(3 1 x)

t 5 !72 8 8.49

!75 5 t72 5 t2

4 51

18t2

6 5 2 11

18t2

6000 2 4000 5 2000

p(t)5 4

5 2 1 2

5 2 136

18

limtS61

p(t) 5 2 11

18(6)2

5 6

5 3 1 3

5 3 136

12

limtS62

p(t) 5 3 11

12(6)2

8 10 120

68

42

–2

10y

x

642–2–4

c 5 0.b 5 2a 5 3,

b 5 2a 5 3,

6a 5 18

4a 2 2b 5 8limxS22

f(x) 5 8

a 1 b 5 5limxS1

f(x) 5 5

c 5 0f(0) 5 0

a 2 0f(x) 5 ax2 1 bx 1 c,


d.

e.

f.

8. a.

Let Therefore, as

Here,

b.

c.

d.

e.

f.

9. a.

b.

c.

d.

e.

f.

51

32

51

4(8)

5 limxS1

1

(x 1 3)(3x 1 5)

5 limxS1

a 1

x 2 1b a3x 1 5 2 2x 2 6

(x 1 3)(3x 1 5)b

limxS1

a 1

x 2 1b a 1

x 1 32

2

3x 1 5b

5 2x

limhS0

(x 1 h)2 2 x2

h5 lim

hS0

2xh 1 h2

h

51

2

5 limxS0

"x 1 1 2 1

("x 1 1 2 1)("x 1 1 1 1)

limxS0

"x 1 1 2 1

x5 lim

xS0

"x 1 1 2 1

x 1 1 2 1

5 21

limxS21

x2 1 xx 1 1

5 limxS21

x(x 1 1)

x 1 1

16 2 16

16 2 20 1 65 0

16 2 16

64 1 645 0

51

12

lim

xS2

u 2 2

u3 2 8

limxS0

(x 1 8)13 2 2

x

51

12

5 limxS2

u 2 2

(u 2 2)(u2 1 2u 1 4)

5 lim

xS2

u 2 2

u3 2 8

limxS4

"x 2 2

"x3 2 8

51

2

5 limxS1

u 2 1

(u 2 1)(u 1 1)

5 limxS1

u 2 1

u2 2 1

limxS1

x

16 2 1

x13 2 1

51

6

5 limxS1

(u 2 1)

(u 2 1)(u5 1 u4 1 u3 1 u2 1 u 1 1)

5 limxS1

u 2 1

u6 2 1

limxS1

x

16 2 1

x 2 1

5 227

5 2 (9 1 9 1 9)

5 2limxS3

(u 2 3)(u2 1 3u 1 9)

u 2 3

5 limxS3

u3 2 27

u 2 3

limxS27

27 2 xx

13 2 3

51

12

limxS2

u 2 2

u3 2 85 lim

xS2

1

u2 1 2u 1 4

u S 2.x S 8,u3 5 xu 5 "3 x.

limxS8

"3 x 2 2

x 2 8

5 21

"7

5 limxS0

7 2 x 2 7 2 x

x("7 2 x 1 "7 1 x)

limxS0

£"7 2 x 2 "7 1 xx

3"7 2 x 1 "7 1 x

"7 2 x 1 "7 1 x§

51

4

limxS4

"x 2 2

x 2 45 lim

xS4

"x 2 2

("x 2 2)("x 1 2)

5 21

4

5 limxS0

21

2 1 "4 1 x

limxS0

£ 2 2 "4 1 xx

32 1 "4 1 x

2 1 "4 1 x§


Let

u S 3.x S 27,

x 5 u3x

13 5 u

u S 1x S 1,

x 5 u6x16 5 u,

Let

As u S 1x S 1,

x13 5 u2

u6 5 xx

16 5 u

Let

u S 2x S 4,

x32 5 u3

x12 5 u

Let

u S 2x S 0,

x 5 u3 2 8

x 1 8 5 u3

(x 1 8)13 5 u


10. a. does not exist.

b. does not exist.

c.

5 3

5 limxS21

x 1 1

limxS21

(x 2 2)(x 1 1)

0 x 2 2 0 5 limxS21

(x 2 2)(x 1 1)

x 2 2

limxS2

x2 2 x 2 2

0 x 2 2 0 5 limxS2

(x 2 2)(x 1 1)

0 x 2 2 0

40

2

–2

–4

4y

x

2–2–4

limxS5

2

2

2 (2x 2 5)(x 1 1)

2x 2 55 2 (x 1 1)

x ,5

20 2x 2 5 0 5 2 (2x 2 5),

limxS5

2

1

(2x 2 5)(x 1 1)

2x 2 55 x 1 1

x $5

20 2x 2 5 0 5 2x 2 5,

limxS5

2

0 2x 2 5 0 (x 1 1)

2x 2 5

80

1

–1

–2

2y

x

4–4–8

5 21

limxS52

0 x 2 5 0x 2 5

5 limxS52

2 ax 2 5

x 2 5b

5 1

limxS51

0 x 2 5 0x 2 5

5 limxS51

x 2 5

x 2 5

limxS5

0 x 2 5 0x 2 5

d. if if

11. a.

is constant, therefore T and V form a linear relationship.

b.

DVDT

51.6426

205 0.082 13

V 5DVDT

? T 1 K

DV

40

2

–2

–4

4y

x

2–2–4

limxS222

(x 1 2)(x 1 2)2

2 (x 1 2)5 0

limxS221

(x 1 2)(x 1 2)2

x 1 25 lim

xS221

(x 1 2)2 5 0

x , 225 2 (x 1 2)

x . 220 x 1 2 0 5 x 1 2

40

2

–2

–4

4y

x

2–2–4

5 23

5 limxS22

2 (x 1 1)

limxS22

(x 2 2)(x 1 1)

0 x 2 2 0 5 limxS22

2(x 2 2)(x 1 1)

(x 2 2)

DT T V DV

240 19.1482

220 20.7908

0 22.4334

20 24.0760

40 25.7186

60 27.3612

80 29.0038

20

20

20

20

20

20

1.6426

1.6426

1.6426

1.6426

1.6426

1.6426

Therefore, and

c.

d.

e.

12.

13.

a.

b.

c.

14.

a.

b.

15. and

a.

b.

16.

17.

Therefore, this limit does not exist.

40

2

–2

–4

4y

x

2–2–4

5 21

limxS12

x2 2 x2x 1 1

5 limxS12

x(x 2 1)

2x 1 1

x S 12 0 x 2 1 0 5 2x 1 1

limxS11

x2 1 0 x 2 1 021

0 x 2 1 0 5 3

x2 1 x 2 2

x 2 15

(x 1 2)(x 2 1)

x 2 1

x S 11 0 x 2 1 0 5 x 2 1

limxS1

x2 1 0 x 2 1 021

0 x 2 1 05 22

5 22 1 2

1 1 1

5 limxS0

c (x 1 1 2 2x 2 1)

(3x 1 4 2 2x 2 4)3

!3x 1 4 1 !2x 1 4

!x 1 1 1 !2x 1 1d

3!3x 1 4 1 !2x 1 4

!3x 1 4 1 !2x 1 4d

3!x 1 1 1 !2x 1 1

!3x 1 4 2 !2x 1 4

5 limxS0

c !x 1 1 2 !2x 1 1

!x 1 1 1 !2x 1 1

limxS0

!x 1 1 2 !2x 1 1

!3x 1 4 2 !2x 1 4

limxS0

f(x)

g(x)5 lim

xS0

f(x)x

g(x)x

51

2

5 0

limxS0

g(x) 5 limxS0

xag(x)

xb 5 0 3 2

limxS0

g(x)

x5 2lim

xS0

f(x)

x5 1

limxS0

f(x)

g(x)5 lim

xS0 c x

g(x)

f(x)

xd 5 0

limxS0

f(x) 5 limxS0

c f(x)

x3 x d 5 0

limxS0

f(x)

x5 1

5 1

limxS4

"3f(x) 2 2x 5 "3 3 3 2 2 3 4

5 21

5 3 2 4

5 limxS4

( f(x) 2 x)

limxS4

3 f(x)42 2 x2

f(x) 1 x5 lim

xS4

( f(x) 2 x)( f(x) 1 x)

f(x) 1 x

limxS4

3 f(x)43 5 33 5 27

limxS4

f(x) 5 3

5 7

521

3

5limxS5

(x2 2 4)

limxS5

f(x)

limxS5

x2 2 4

f(x)

12

10

8

6

4

2

020 4 6 8 10 12

T

V

limvS0

T 5 2273.145

T 5V 2 22.4334

0.082 13

V 5 0.082 13T 1 22.4334.

k 5 22.4334

V 5 22.4334 T 5 0

V 5 0.082 13T 1 K


1.6 Continuity, pp. 51–531. Anywhere that you can see breaks or jumps is aplace where the function is not continuous.2. It means that on that domain, you can trace thegraph of the function without lifting your pencil.3. point discontinuity

jump discontinuity

infinite discontinuity

4. a. makes the denominator 0.b. makes the denominator 0.c. makes the denominator 0.d. and make the denominator 0.e.

and make the denominator 0.f. The function has different one-sided limits at

5. a. The function is a polynomial, so the functionis continuous for all real numbers.b. The function is a polynomial, so the function iscontinuous for all real numbers.c.The is continuous for all real numbers except 0 and 5.d. The is continuous for all real numbers greaterthan or equal to e. The is continuous for all real numbers.f. The is continuous for all real numbers.6. is a linear function (a polynomial),and so is continuous everywhere,including 7.

The function is continuous everywhere.8.

The function is discontinuous at 9.

400 6000

2

4y

x

200

x 5 0.

40

2

–2

–4

4y

x

2–2–4

80

4

–4

–8

8y

x

4–4–8

x 5 2.

g(x)

22.

x2 2 5x 5 x(x 2 5)

x 5 3.

x 5 2x 5 23

x2 1 x 2 6 5 (x 1 3)(x 2 2)

x 5 23x 5 3

x 5 0

x 5 0

x 5 3

3 40

42

–2–4

6810

y

x

1 2–1verticalasymptote

60

42

–2

6810

y

x

2 4–2

60

42

–2

6810

y

x

2 4–2

hole


10.

Function is discontinuous at 11. Discontinuous at

12.

is continuous.

13.

a.

b. i. From the graph,

ii. From the graph,

iii. Since the one-sided limits differ, doesnot exist.c. f is not continuous since does not exist.

14. a. From the graph,b. From the graph,

c.

Thus, But, Hence f is not

continuous at (and also not continuous over).

15. The function is to be continuous at anddiscontinuous at

For to be continuous at

For to be discontinuous at

If then if then

and and

This shows that A and B can be any set of real numbers such that(1) (2) (if then if then )

and is not a solution because thenthe graph would be continuous at

16.

at at

if is continuous.

17.

a.

does not exist.limxS1

g(x)

limxS12

g(x) 5 21

limxS11

g(x) 5 1¶ lim

xS1 g(x)

g(x) 5 •x 0 x 2 1 0

x 2 1, if x 2 1

0, if x 5 1

f(x)b 5 6.a 5 21,

f(x) 5 • 2x, if 23 # x # 22

2x2 1 b, if 22 , x , 0

6, if x 5 0

a 5 21

b 5 6.x 5 0,

4a 1 b 5 2x 5 22,

f(x) 5 • 2x, if 23 # x # 22

ax2 1 b, if 22 , x , 0

6, if x 5 0

x 5 2.

B 5 22A 5 1

A , 22

B , 1,A . 22B . 1,4B 2 A 2 6

A 5 B 2 3

A , 22 A . 22

B , 1 B . 1

3B , 3 3B . 3

3B 1 3 , 6 3B 1 3 . 6

4B 2 B 1 3 , 6 4B 2 (B 2 3) . 6

4B 2 A , 6,4B 2 A . 6,

4B 2 A 2 6

B(2)2 2 A 2 3(2)

x 5 2:f(x)

A 5 B 2 3

A(1) 2 B 5 23

A(1) 2 B

1 2 25 3(1)

x 5 1:f(x)

f(x) 5 μ

Ax 2 Bx 2 2

, if x # 1

3x, if 1 , x , 2

Bx2 2 A, if x $ 2

x 5 2.

x 5 1

23 , x , 8

x 5 2

f(3) 5 2.limxS3

f(x) 5 4.

limxS32

f(x) 5 4 5 lim

xS32 f(x)

limxS32

f(x) 5 4.

f(3) 5 2.

limxS0

f(x)

limxS0

f(x)

limxS01

f(x) 5 1.

limxS02

f(x) 5 21.

40

2

–2

–4

4y

x

2–2–4

f(x) 5 •21, if x , 0

0, if x 5 0

1, if x . 0

k 5 16"k 5 4,

2 1 "k 5 6

g(x)

g(x) 5 e x 1 3, if x 2 3

2 1 !k, if x 5 3

40

2

–2

–4

4y

x

2–2–4

x 5 2

x 5 3.

5 5

5 limxS3

(x 2 3)(x 1 2)

x 2 3

limxS3

f(x) 5 limxS3

x2 2 x 2 6

x 2 3


b.

is discontinuous at .

Review Exercise, pp. 56–591. a.

b.

c.

2. a.

b.

c.

d.

3.

a. Slope at

Slope of the graph at is 2.b. Slope at

Slope of the graph at is 2.4.a.Average velocity during the first second is

m s.>s(1) 2 s(0)

15 25

s(2) 5 160s(1) 5 175,s(0) 5 180,

s(t) 5 25t2 1 180

P(2, 0.5)

m 5 limhS0

2hh

5 2

5 2h f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5

f(x) 5 2x 1 1

P(2, 0.5)

P(21, 3)

5 2

5 limhS0

(2 2 h)

5 limhS0

4 2 1 1 2h 2 h2 2 3

h

m 5 limhS0

4 2 (21 1 h)2 2 3

h

f(x) 5 4 2 x2P(21, 3)

f(x) 5 e4 2 x2 , if x # 1

2x 1 1, if x . 1

5 25

4

5 limhS0

225h

h(2 1 h)(2)

5 limhS0

10 2 5(2 1 h)

h(2 1 h)(2)

m 5 limhS0

5

4 1 h 2 22

5

2

h

Pa4, 5

2bf(x) 5

5

x 2 2,

5 21

27

5 22

9(6)

5 2 limhS0

c2 1

3!9 1 h(3 1 !9 1 h)d

5 2 limhS0

c 3 2 !9 1 h3h!9 1 h

33 1 !9 1 h3 1 !9 1 h

d m 5 lim

hS0

2

!4 1 h 1 52

2

3

h

Pa4, 2

3bh(x) 5

2

!x 1 5,

51

2

5 limhS0

1

!h 1 1 1 1

5 limhS0

c !h 1 1 2 1

x3

!h 1 1 1 1

!h 1 1 1 1d

m 5 limhS0

"21 1 h 1 2 2 1

h

P(21, 1)g(x) 5 "x 1 2,

5 21

3

5 limhS0

21

3 1 h

m 53

3 1 h 2 1

h

P(2, 1)f(x) 53

x 1 1,

2x 2 y 2 5 5 0

y 2 (23) 5 2(x 2 1)

5 2

5 limhS0

2 1 h

5 limhS0

2h 1 h2

h

m 5 limhS0

5(1 1 2h 1 h2) 2 (23)

h

f(1) 5 23

5 7

m 548 2 13

4 2 (21)

f(4) 5 48f(21) 5 13,

5 23

m 521 2 36

3 2 (22)

f(3) 5 21f(22) 5 36,

x 5 1g(x)

40

2

–2

–4

4y

x

2–2–4


Average velocity during the second second is

m s.b. At

Velocity is m s.c. Time to reach ground is when Therefore,

Velocity at

Therefore,

5. mass in gramsa. Growth during

Grew 0.0601 g during this time interval.b. Average rate of growth is

g min.

c.

Rate of growth is g min.

6. tonnes of waste,

a. At

700 000 t have accumulated up to now.b. Over the next three years, the average rate of change:

c. Present rate of change:

d.

Now,

It will take 7.5 years to reach a rate of t per year.

7. a. From the graph, the limit is 10.b. 7; 0c. is discontinuous for and 8. a. Answers will vary. f is

discontinuous at

b. if f is increasing for

40

2

–2

–4

4y

x

2–2–4

limxS31

f(x) 5 1

x . 3x , 3;f(x) 5 24

20

1

–1

–2

2y

x

1–1–2

x 5 21

limxS21

f(x) 5 0.5,

t 5 4.t 5 3p(t)

3.0 3 105

a 5 7.5

2a 1 15 5 30

(2a 1 15)104 5 3 3 105

5 (2a 1 15)104

5 limhS0

104(2a 1 h 1 15)limhS0

Q(a 1 h) 2 Q(a)

h

5104 32ah 1 h2 1 15h4

hQ(a 1 h) 2 Q(a)

h

Q(a) 5 104 3a2 1 15a 1 70415a 1 15h 1 7045 104 3a2 1 2ah 1 h2 1

Q(a 1 h)

5 15 3 104 t per year.

limhS0

Q(h) 2 Q(0)

h5 lim

hS0 104(h 1 15)

Q(0) 5 104 1 70

Q(h) 5 104(h2 1 15h 1 70)

5 18 3 104 t per year.

Q(3) 2 Q(0)

35

54 3 104

3

Q(0) 5 70 3 104

5 124 3 104

Q(3) 5 104(9 1 45 1 70)

5 700 000.

Q(t) 5 70 3 104

t 5 0,

0 # t # 10

Q(t) 5 104(t2 1 15t 1 70)

>limhS0

(6 1 h) 5 6

s(3 1 h) 2 s(3)

h5

6h 1 h2

h

s(3) 5 9

s(3 1 h) 5 9 1 6h 1 h2

>0.0601

0.015 6.01

5 9

M(3) 5 32

M(3.01) 5 (3.01)2 5 9.0601

3 # t # 3.01

M(t) 5 t2

v(6) 5 limhS0

(260 2 5h) 5 260.

s(6) 5 0

5 260h 2 5h2

s(6 1 h) 5 25(36 1 12h 1 h2) 1 180

t 5 6:

t 5 6, t . 0.

t2 5 36

25t2 1 180 5 0

s(t) 5 0.

>240

v(4) 5 limhS0

(240 2 5h) 5 240

s(4 1 h) 2 s(4)

h5

240h 2 5h2

h

5 280 2 40h 2 5h2 1 180 1 80 2 180

5 25(4 1 h)2 1 180 2 (25(16) 1 180)

s(4 1 h) 2 s(4)

t 5 4:

>s(2) 2 s(1)

15 215


9. a.

b.

Discontinuous at and c. They do not exist.10. The function is not continuous at because the function is not defined at ( makes the denominator 0.)

11.

a. f is discontinuous at and

b.

does not exist.


b.

c.

does not exist.

13. a.

b.

14.

This agrees well with the values in the table.

15. a.

f(x) 8 0.25

x 5 2.0001

f(x) 5"x 1 2 2 2

x 2 2

51

2!3

5 limxS0

1

!x 1 3 1 !3

5 limxS0

x

xA!x 1 3 1 !3 B

5 limxS0

x 1 3 2 3

xA!x 1 3 1 !3 B

limxS0

c !x 1 3 2 !3

x?!x 1 3 1 !3

!x 1 3 1 !3d

1

2

1

3

limxS23

h(x)

limxS4

h(x) 537

75 5.2857

h(x) 5x3 2 27

x2 2 9,

limxS0

g(x) 5 0g(x) 5 x(x 2 5),

limxS0

f(x)f(x) 51

x2,

limxS22

f(x)

limxS222

2

x 1 25 2`

limxS22

f(x): 5 limxS221

2

x 1 25 1`

52

3

limxS1

f(x) 5 limxS1

2

x 1 2

x 5 22.x 5 1

52(x 2 1)

(x 2 1)(x 1 2)

f(x) 52x 2 2

x2 1 x 2 2

x 5 24

x 5 24.

x 5 24

x 5 1.x 5 21

f(x) 5 • x 1 1, if x , 21

2x 1 1, if 21 # x , 1

x 2 2, if x . 1

40

2

–2

–4

4y

x

2–2–4


x 2.1 2.01 2.001 2.0001

f(x) 0.248 46 0.249 84 0.249 98 0.25

x 1.9 1.99 1.999 2.001 2.01 2.1

x 2 2x2 2 x 2 2

0.344 83 0.334 45 0.333 44 0.333 22 0.332 23 0.322 58

x 0.9 0.99 0.999 1.001 1.01 1.1

x 2 1x2 2 1

0.526 32 0.502 51 0.500 25 0.499 75 0.497 51 0.476 19

x 20.1 20.01 20.001 0.001 0.01 0.1

"x 1 3 2 "3x

0.291 12 0.288 92 0.2887 0.288 65 0.288 43 0.286 31

b.

c.

16. a.

Slope of the tangent to at is 10.

b.

Slope of the tangent to at is

c.


17. a.

b.

c.

d.

e.

f.

18. a. The function is not defined for sothere is no left-side limit.b. Even after dividing out common factors fromnumerator and denominator, there is a factor of

in the denominator; the graph has a verticalasymptote at

c.

limxS12

f(x) 5 25 2 limxS11

f(x) 5 2

f(x) 5 e25, if x , 1

2, if x $ 1

x 5 2.

x 2 2

x , 3,

5 21

4

5 limxS0

c2 1

2(2 1 x)d

5 limxS0

c1x

3 2x

2(2 1 x)d

limxS0

1

xa 1

2 1 x2

1

2b

5 21

8

521

4 1 4

521

4 1 !12 1 (4)

5 limxS4

21

4 1 !12 1 x

5 limxS4

2 (x 2 4)

(x 2 4)(4 1 !12 1 x)

5 limxS4

4 2 x

(x 2 4)(4 1 !12 1 x)

5 limxS4

16 2 (12 1 x)

(x 2 4)(4 1 !12 1 x)

limxS4

c4 2 !12 1 xx 2 4

?4 1 !12 1 x4 1 !12 1 x

d5

1

3

54

12

5(2) 1 2

(2)2 1 2(2) 1 4

5 limxS2

x 1 2

x2 1 2x 1 4

limxS2

(x 2 2)(x 1 2)

(x 2 2)(x2 1 2x 1 4)

51

!5

5 limxS0

1

A!x 1 5 1 !5 2 x B

5 limxS0

x 1 5 2 5 1 x

xA!x 1 5 1 !5 2 x B

limxS0

c !x 1 5 2 !5 2 xx

3!x 1 5 1 !5 2 x!x 1 5 1 !5 2 x

d5 10a

limxSa

(x 1 4a)2 2 25a2

x 2 a5 lim

xSa

(x 2 a)(x 1 9a)

x 2 a

5 4

5 (24) 1 8

5 limxS24

(x 1 8)limxS24

(x 1 4)(x 1 8)

x 1 4

2 116.(x 5 4)y 5

1

x

5 21

16

5 limhS0

21

4(4 1 h)

limhS0

1

4 1 h2

1

4

h5 lim

hS0

4 2 4 2 h4(4 1 h)(h)

14.x 5 4y 5 "x

51

4

5 limhS0

1

!4 1 h 1 2

limhS0

"4 1 h 2 2

h5 lim

hS0

"4 1 h 2 2

4 1 h 2 4

x 5 5y 5 x2

5 10

5 limhS0

(10 1 h)

limhS0

(5 1 h)2 2 25

h

51

45 0.25

5 limxS2

1

!x 1 2 1 2

limxS2

c !x 1 2 2 2

x 2 23

!x 1 2 1 2

!x 1 2 1 2d

limxS2

f(x) 5 0.25


d. The function has a vertical asymptote at

e.

f.

Therefore, does not exist.

19. a.

When The equation of the tangent is

b.


c.


d.


20.a.

b.

The population is changing at the rate of .

Chapter 1 Test, p. 60

1. does not exist since

2.

Slope of secant is

5 213

36 1 3

22 2 15 2

39

3

f(1) 5 5 2 8 5 23

f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36

f(x) 5 5x2 2 8x

limxS11

1

x 2 15 1` 2 lim

xS12

1

x 2 15 2`.

limxS1

1

x 2 1

109 000>h5 109

5 limhS0

(109 1 3h)

5 limhS0

109h 1 3h2

h

5 limhS0

20 1 488 1 61h 1 192 1 48h 1 3h2 2 700

h

5 limhS0

20 1 488 1 61h 1 3(64 1 16h 1 h2) 2 700

h

limhS0

20 1 61(8 1 h) 1 3(8 1 h)2 2 (20 1 488 1 192)

h

5 700 000

P(8) 5 20 1 61(8) 1 3(8)2

P(t) 5 20 1 61t 1 3t2

y 5 2216x 1 486

y 2 (2162) 5 2216(x 2 3)

x 5 3, y 5 2162.

5 2216

5 limhS0

( 2 216 2 108h 2 24h2 2 2h3)

5 limhS0

2216h 2 108h2 2 24h3 2 2h4

h

5 limhS0

22(81 1 108h 1 54h2 1 12h3 1 h4) 1 162

h

m 5 limhS0

22(3 1 h)4 2 (2162)

h

y 5 18x 1 9

y 2 (29) 5 18(x 2 (21))

x 5 21, y 5 29.

5 18

5 limhS0

(18 2 18h 1 6h2)

5 limhS0

18h 2 18h2 1 6h3

h

5 limhS0

6(21 1 3h 2 3h2 1 h3) 2 3 1 9

h

m 5 limhS0

6(21 1 h)3 2 3 2 (26 2 3)

h

y 5 25x 2 5

y 2 5 5 25(x 1 2)

x 5 22, y 5 5.

5 25

5 limhS0

(25 1 h)

5 limhS0

25h 1 h2

h

5 limhS0

4 2 4h 1 h2 1 2 2 h 2 1 2 5

h

m 5 limhS0

(22 1 h)2 2 (22 1 h) 2 1 2 (4 1 2 2 1)

h

y 5 7

y 2 7 5 0(x 2 1)

x 5 1, y 5 7.

5 0

5 limhS0

2h

5 limhS0

2h2

h

5 limhS0

23 2 6h 2 h2 1 6 1 6h 1 4 2 7

h

m 5 limhS0

23(1 1 h)2 1 6(1 1 h) 1 4 2 (23 1 6 1 4)

h

limxS21

f(x)

limxS211

f(x) 2 limxS212

f(x)

limxS212

f(x) 5 5

limxS211

f(x) 5 21

f(x) 5 e 5x2, if x , 21

2x 1 1, if x $ 21

limxS01

0 x 0x

2 limxS02

0 x 0x

limxS01

0 x 0x

5 1

limxS02

0 x 0x

5 21

x S 02 0 x 0 5 2x

limxS0

0 x 0x

x 5 2.



b.

c.

d. f is discontinuous at and 4. a. Average velocity from to

Average velocity from to is 1 km h.b.

Velocity at is 2 km h.5.Average rate of change from to

6.

Slope of the tangent at

Slope of the tangent at is

7. a.

b.

c.

d.

e.

f.

8.

is continuous.Therefore,

b 5 218

5

5b 5 218 25 1 5b 1 a 5 8

a 5 15a 1 3 5 8

f(x)

f(x) 5 • ax 1 3, if x . 5

8, if x 5 5

x2 1 bx 1 a, if x , 5

51

12

51

4 1 4 1 4

5 limxS0

(x 1 8)13 2 2

((x 1 8)13 2 2)((x 1 8)

23 1 2(x 1 8)

13 1 4)

limxS0

(x 1 8)13

2 2

x5 lim

xS0

(x 1 8)13

2 2

(x 1 8) 2 8

51

6

5 limxS3

1

x 1 3

limxS3

a 1

x 2 32

6

x2 2 9b 5 lim

xS3

(x 1 3) 2 6

(x 2 3)(x 1 3)

5 23

4

53

22(2)

limxS21

x3 1 1

x4 2 15 lim

xS21

(x 1 1)(x2 2 x 1 1)

(x 2 1)(x 1 1)(x2 1 1)

5 4

5 limxS5

A!x 2 1 2 2B A!x 2 1 1 2B!x 2 1 2 2

limxS5

x 2 5

!x 2 1 2 25 lim

xS5

(x 2 1) 2 4

!x 2 1 2 2

57

5

limxS2

2x2 2 x 2 6

3x2 2 7x 1 25 lim

xS2

(2x 1 3)(x 2 2)

(x 2 2)(3x 2 1)

5 12

limxS3

4x2 2 36

2x 2 65 lim

xS3

2(x 2 3)(x 1 3)

(x 2 3)

231.x 5 4

5 231

limhS0

f(4 1 h) 2 f(4)

h5 lim

hS0 (231 2 4h)

1 1 2h 1 h2

5 231h 2 4h2

(1 1 2h 1 h2)

54 1 h 2 4 2 32h 2 4h2

1 1 2h 1 h2

f(4 1 h) 2 f(4) 54 1 h

1 1 8h 1 h2 2 4

f(4) 54

1

54 1 h

1 1 8h 1 h2

f(4 1 h) 54 1 h

(4 1 h)2 2 15

x 5 4:

f(x) 5x

x2 2 15

5"16 1 h 2 "16

h

f(5 1 h) 2 f(5)

h

x 5 5 1 h:x 5 5

f(x) 5 "x 1 11

>t 5 3

v(3) 5 limhS0

2h 2 h2

h5 2

5 2h 2 h2

5 24 1 8h 2 9 2 6h 2 h2 2 15

5 8(3 1 h) 2 (3 1 h)2 2 (24 2 9)

s(3 1 h) 2 s(3)

>t 5 5t 5 2

5 1

515 2 12

3

s(5) 2 s(2)

35

(40 2 25) 2 (16 2 4)

3

t 5 5:t 5 2

x 5 2.x 5 1

limxS42

f(x) 5 1

limxS2

f(x) 5 1

limxS1

f(x)


Review of Prerequisite Skills, pp. 2–3

1. a.

b.

c.

d.

e.

f.

2. a. Substitute the given slope and y-intercept into

b. Substitute the given slope and y-intercept into

c. The slope of the line is

The equation of the line is in the formThe point is and

.The equation of the line is or

d.

e.f.3. a.

b.

c.

d.

4. a.

b.

c.

d.

5.

a.

b.c.

d.

6.

a.

b. s(21) 5 21

s(22) 5 21

2

s(t) 5 μ1

t, if 23 , t , 0

5, if t 5 0

t3, if t . 0

f(3) 5 "6

f(78) 5 9

f(0) 5 "3

f(233) 5 6

f(x) 5 •"3 2 x, if x , 0

"3 1 x, if x $ 0

55

52

f(10) 510

100 1 4

5 0

f(0) 50

0 1 4

5 23

13

f(23) 523

9 1 4

5 25

52

f(210) 5210

100 1 4

5 144

f(2) 5 (10 1 2)2

5 29

f(2) 5 23(4) 1 2(2) 2 1

5 0

f(2) 5 (8 2 2)(6 2 6)

5 21

f(2) 5 26 1 5

y 5 5

x 5 23

x 1 y 2 2 5 0

y 2 4 5 2x 2 2

y 2 4 5 21(x 2 (22))

5 21

m 58 2 4

26 2 (22)

y 5 65(x 1 1) 1 6.

y 2 6 5 65(x 1 1)

m 5 65

(21, 6)y 2 y1 5 m(x 2 x1).

56

5

m 512 2 6

4 2 (21)

y 5 22x 1 5

y 5 mx 1 b.

y 5 4x 2 2

y 5 mx 1 b.

5 21

2

5

22

4

1

m 521

4 2 14

74 2 3

4

5 24.1

m 54 2 4.41

22 2 (22.1)

5 24

m 54 2 0

21 2 0

5 4

m 54 2 0

1 2 0

5 22

m 54 2 (24)

21 2 3

5 23

m 527 2 5

6 2 2

CHAPTER 1Introduction to Calculus


c.d.e. or 7. a.b.c.

d.

e.

f.

8. a.

b.c.d.

e.f.


9. a.b.c.d.e.

f.10. a.

m sb.






is


1. a.

b.

c.

d.

e.

f.

2. a.

b.

5 "6 2 3

5 2"6 2 6

2

2"3 2 3"2

"2?"2

"2

5 "6 1 "10

2

"3 1 "5

"2?"2

"2

2"5 2 2"2

"2 1 "5

3"3 2 "2

2"3 1 "2

"3 2 "2

2"3 1 4

28.x 5 5

f(x)

28.

4 60

4

–4

–8

8y

x

2–2

0 # t # 120

5 213.33 L>min

V(61) 2 V(59)

61 2 598

1186.56 2 1213.22

2

5 220 L>min

V(120) 2 V(60)

120 2 605

0 2 1200

60

t 5 60

t 5 120

>5 10.3


2 2 1

h(2) 5 32.4h(1) 5 22.1,

>5 20.1


1 2 0

h(1) 5 22.1h(0) 5 2,

5xPR 0 x 2 25, 22, 16exPR ` x 2 2

1

2, 3 f

2x2 2 5x 2 3 5 (2x 1 1)(x 2 3)


2x3 2 x2 2 7x 1 6 5 (x 2 1)(2x2 1 x 2 6)

x 2 1x 5 1

2x3 2 x2 2 7x 1 6

27x3 2 64 5 (3x 2 4)(9x2 1 12x 1 16)

5 x(x 1 1)(x 1 1)

x3 1 2x2 1 x 5 x(x2 1 2x 1 1)

2x2 2 7x 1 6 5 (2x 2 3)(x 2 2)

x2 1 x 2 6 5 (x 1 3)(x 2 2)

5 x(x 1 1)(x 2 1)

x3 2 x 5 x(x2 2 1)

5 729a3 2 1215a2 1 675a 2 125

5 (81a2 2 90a 1 25)(9a 2 5)

(9a 2 5)3 5 (9a 2 5)(9a 2 5)(9a 2 5)

5 a3 1 6a2 1 12a 1 8

5 (a2 1 4a 1 4)(a 1 2)

(a 1 2)3 5 (a 1 2)(a 1 2)(a 1 2)

5 2x2 1 x 1 7

5 x2 1 2x 2 3 2 (2x2 1 x 2 10)

(x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)

5 2x2 2 7xx(5x 2 3) 2 2x(3x 1 2) 5 5x2 2 3x 2 6x2 2 4x(5 2 x)(3 1 4x) 5 15 1 17x 2 4x2

(x 2 6)(x 1 2) 5 x2 2 4x 2 12

106s(100) 5 1003

s(1) 5 1

s(0) 5 5


c.d.e. or 7. a.b.c.

d.

e.

f.

8. a.

b.c.d.

e.f.


9. a.b.c.d.e.

f.10. a.

m sb.






is


1. a.

b.

c.

d.

e.

f.

2. a.

b.

5 "6 2 3

5 2"6 2 6

2

2"3 2 3"2

"2?"2

"2

5 "6 1 "10

2

"3 1 "5

"2?"2

"2

2"5 2 2"2

"2 1 "5

3"3 2 "2

2"3 1 "2

"3 2 "2

2"3 1 4

28.x 5 5

f(x)

28.

4 60

4

–4

–8

8y

x

2–2

0 # t # 120

5 213.33 L>min

V(61) 2 V(59)

61 2 598

1186.56 2 1213.22

2

5 220 L>min

V(120) 2 V(60)

120 2 605

0 2 1200

60

t 5 60

t 5 120

>5 10.3


2 2 1

h(2) 5 32.4h(1) 5 22.1,

>5 20.1


1 2 0

h(1) 5 22.1h(0) 5 2,

5xPR 0 x 2 25, 22, 16exPR ` x 2 2

1

2, 3 f

2x2 2 5x 2 3 5 (2x 1 1)(x 2 3)


2x3 2 x2 2 7x 1 6 5 (x 2 1)(2x2 1 x 2 6)

x 2 1x 5 1

2x3 2 x2 2 7x 1 6

27x3 2 64 5 (3x 2 4)(9x2 1 12x 1 16)

5 x(x 1 1)(x 1 1)

x3 1 2x2 1 x 5 x(x2 1 2x 1 1)

2x2 2 7x 1 6 5 (2x 2 3)(x 2 2)

x2 1 x 2 6 5 (x 1 3)(x 2 2)

5 x(x 1 1)(x 2 1)

x3 2 x 5 x(x2 2 1)

5 729a3 2 1215a2 1 675a 2 125

5 (81a2 2 90a 1 25)(9a 2 5)

(9a 2 5)3 5 (9a 2 5)(9a 2 5)(9a 2 5)

5 a3 1 6a2 1 12a 1 8

5 (a2 1 4a 1 4)(a 1 2)

(a 1 2)3 5 (a 1 2)(a 1 2)(a 1 2)

5 2x2 1 x 1 7

5 x2 1 2x 2 3 2 (2x2 1 x 2 10)

(x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)

5 2x2 2 7xx(5x 2 3) 2 2x(3x 1 2) 5 5x2 2 3x 2 6x2 2 4x(5 2 x)(3 1 4x) 5 15 1 17x 2 4x2

(x 2 6)(x 1 2) 5 x2 2 4x 2 12

106s(100) 5 1003

s(1) 5 1

s(0) 5 5


c.

d.

3. a.

b.

c.

d.

e.

f.

4. a.

b.

c.

5. a.

b.

c. The expressions in the two parts are equivalent.The radicals in the denominator of part a. have beensimplified in part b.

5 8"10 1 24

5 16"10 1 48

2

5 16"10 1 48

20 2 18

8"2

2"5 2 3"2?

2"5 1 3"2

2"5 1 3"2

5 8"10 1 24

5 16"10 1 48

2

5 8"40 1 8"36

20 2 18

8"2

"20 2 "18?"20 1 "18

"20 1 "18

5 1

12 2 5!5

5 5 2 4

10 2 5"5 1 2

"5 1 2

2"5 2 1?"5 2 2

"5 2 2

5 27

2 1 3"2

5 4 2 18

2(2 1 3"2)

2 2 3"2

2?

2 1 3"2

2 1 3"2

5 1

!5 1 1

5 5 2 1

4("5 1 1)

"5 2 1

4?"5 1 1

"5 1 1

5 35 2 12"6

19

5 27 2 12"6 1 8

27 2 8

3"3 2 2"2

3"3 1 2"2?

3"3 2 2"2

3"3 2 2"2

5 11"6 2 16

47

5 10"6 2 6 2 10 1 "6

50 2 3

2"3 2 "2

5"2 1 "3?

5"2 2 "3

5"2 2 "3

5 4 2 2"5

5 44 2 22"5

11

5 20 2 22"5 1 24

20 2 9

2"5 2 8

2"5 1 3?

2"5 2 3

2"5 2 3

5 5 1 2"6

5 3 1 2"6 1 2

3 2 2

"3 2 "2

"3 1 "2?"3 2 "2

"3 2 "2

5 10 2 3"10

5 20 2 6"10

20 2 18

2"5

2"5 1 3"2?

2"5 2 3"2

2"5 2 3"2

5 "5 1 "2

5 3("5 1 "2)

3

3

"5 2 "2?"5 1 "2

"5 1 "2

5 3"10 2 2

4

3"5 2 "2

2"2?"2

"2

5 4 1 "6

2

5 12 1 3"6

6

4"3 1 3"2

2"3?"3

"3


6. a.

b.

c.

d.

e.

f.

7. a.

b.

c.


1. a.

b.

c.



13.

137 ,

y 513

7x 1

11

7

27y 5 213x 2 11

13x 2 7y 2 11 5 0

213.

5 21

3

m 521 2 (22.6)

1.5 2 6.3

5 25

3

5210

262

m 527

2 2 32

72 2 1

2

5 3

m 528 2 7

23 2 2

51

!x 1 h 1 !x

5h

hA!x 1 h 1 !x B

5x 1 h 2 x

hA!x 1 h 1 !x B

!x 1 h 2 !xh

?!x 1 h 1 !x!x 1 h 1 !x

5 1

"x 1 4 2 2

5 x

x("x 1 4 1 2)

5 x 1 4 2 4

x("x 1 4 1 2)

"x 1 4 2 2

x?"x 1 4 1 2

"x 1 4 1 2

5 1

"a 2 2

5 a 2 4

(a 2 4)("a 2 2)

"a 2 2

a 2 4?"a 1 2

"a 1 2

5 5 1 2"6

5 30 1 12"6

6

5 18 1 2"216 1 12

18 2 12

"18 1 "12

"18 2 "12?"18 1 "12

"18 1 "12

5 212"15 1 15"10

2

5 12"15 1 15"10

48 2 50

3!5

4!3 2 5!2?

4!3 1 5!2

4!3 1 5!2

5 24 1 15"3

4

5 3"24 1 12 1 12 1 2"24

12 2 8

3"2 1 2"3

"12 2 "8?"12 1 "8

"12 1 "8

5 2"2 1 "6

5 8"2 1 4"6

16 2 12

5 2"2

4 2 2"3?

4 1 2"3

4 1 2"3

2"2

"16 2 "12

5 18"2 1 4"3

23

5 36"2 1 8"3

46

5 4"162 1 2"48

54 2 8

2"6

2"27 2 "8?

2"27 1 "8

2"27 1 "8

5 22"3 2 4

5 4"6 1 8

6 2 8

2"2

2"3 2 "8?

2"3 1 "8

2"3 1 "8


6. a.

b.

c.

d.

e.

f.

7. a.

b.

c.


1. a.

b.

c.



13.

137 ,

y 513

7x 1

11

7

27y 5 213x 2 11

13x 2 7y 2 11 5 0

213.

5 21

3

m 521 2 (22.6)

1.5 2 6.3

5 25

3

5210

262

m 527

2 2 32

72 2 1

2

5 3

m 528 2 7

23 2 2

51

!x 1 h 1 !x

5h

hA!x 1 h 1 !x B

5x 1 h 2 x

hA!x 1 h 1 !x B

!x 1 h 2 !xh

?!x 1 h 1 !x!x 1 h 1 !x

5 1

"x 1 4 2 2

5 x

x("x 1 4 1 2)

5 x 1 4 2 4

x("x 1 4 1 2)

"x 1 4 2 2

x?"x 1 4 1 2

"x 1 4 1 2

5 1

"a 2 2

5 a 2 4

(a 2 4)("a 2 2)

"a 2 2

a 2 4?"a 1 2

"a 1 2

5 5 1 2"6

5 30 1 12"6

6

5 18 1 2"216 1 12

18 2 12

"18 1 "12

"18 2 "12?"18 1 "12

"18 1 "12

5 212"15 1 15"10

2

5 12"15 1 15"10

48 2 50

3!5

4!3 2 5!2?

4!3 1 5!2

4!3 1 5!2

5 24 1 15"3

4

5 3"24 1 12 1 12 1 2"24

12 2 8

3"2 1 2"3

"12 2 "8?"12 1 "8

"12 1 "8

5 2"2 1 "6

5 8"2 1 4"6

16 2 12

5 2"2

4 2 2"3?

4 1 2"3

4 1 2"3

2"2

"16 2 "12

5 18"2 1 4"3

23

5 36"2 1 8"3

46

5 4"162 1 2"48

54 2 8

2"6

2"27 2 "8?

2"27 1 "8

2"27 1 "8

5 22"3 2 4

5 4"6 1 8

6 2 8

2"2

2"3 2 "8?

2"3 1 "8

2"3 1 "8


3. a.

b. The slope and y-intercept are given.

c. (5, 0)

d. The line is a vertical line because both pointshave the same x-coordinate.

4. a.

b.

c.

d.

53(1 1 2h 1 h2 2 1)

h

3(1 1 h)2 2 3

h5

3((1 1 h)2 2 1)

h

11 1 h 2 1

h5

1 2 1 2 hh(1 1 h)

5 21

1 1 h

5 108 1 54h 1 12h2 1 h3

5 (6 1 h)(18 1 6h 1 h2)

5(9 1 6h 1 h2 2 9)(9 1 6h 1 h2 1 9)

h

5((3 1 h)2 2 9)((3 1 h)2 1 9)

h

(3 1 h)4 2 81

h

5 75 1 15h 1 h2

5h(75 1 15h 1 h2)

h

5(5 1 h 2 5)((5 1 h)2 1 5(5 1 h) 1 25)

h

(5 1 h)3 2 125

h

4 60

2

–2

–4

4y

x

2–2

x 5 5

4 60

2

–2

–4

4y

x

2–2

3x 2 5y 2 15 5 0

y 2 0 53

5(x 2 5)

53

5

m 50 2 (23)

5 2 0

(0, 23),

40

4

–4

–8

8y

x

2–2–4

y 5 8x 1 6

4 60

2

–2

–4

4y

x2–2

7x 2 17y 2 40 5 0

17y 1 68 5 7x 1 28

y 2 (24) 57

17(x 2 (24))

57

17

573173

m 525

3 2 (24)53 2 (24)


e.

f.

5. a.

b.

c.

6. a.

b.

c.

7. a.

b. 12c.

d.

e. They are the same.f.

8. a.

b. at

5 5

5 limhS0

(5 1 h)

5 limhS0

9 1 6h 1 h2 2 3 2 h 2 6

h

m 5 limhS0

(3 1 h)2 2 (3 1 h) 2 6

h

y 5 6.x 5 3,y 5 x2 2 x 5 212

5 limhS0

(212 1 3h)

5 limhS0

12 2 12h 1 3h2 2 12

h

m 5 limhS0

3(22 1 h)2 2 12

h

(22, 12)y 5 3x2,

40

4

–4

8

12y

x

2–2–4

5 12

m 5 limhS0

(12 1 6h 1 h2)

5 12 1 6h 1 h2

58 1 12h 1 6h2 1 h3 2 8

h

m 5(2 1 h)3 2 8

2 1 h 2 2

(2, 8), ((2 1 h), (2 1 h)3)

51

"9 1 h 1 3

m 5"9 1 h 2 3

h?"9 1 h 1 3

"9 1 h 1 3

Q(9 1 h, "9 1 h)P(9, 3),

5 3 1 3h 1 h2

51 1 3h 1 3h2 1 h3 2 1

h

m 5(1 1 h)3 1 2 2 3

h

Q(1 1 h, (1 1 h)3 1 2)P(1, 3),

5 6 1 3h

m 53(1 1 h)2 2 3

h

f(x) 5 3x2Q(1 1 h, f(1 1 h)),P(1, 3),

51

"5 1 h 1 "5

"5 1 h 2 "5

h5

5 1 h 2 5

h("5 1 h 1 "5)

5h 1 5

"h2 1 5h 1 4 1 2

"h2 1 5h 1 4 2 2

h5

h2 1 5h 1 4 2 4

h("h2 1 5h 1 4 1 2)

51

"16 1 h 1 4

"16 1 h 2 4

h5

16 1 h 2 16

h("16 1 h 1 4)

51

4 1 2h

5h

2h(2 1 h)

212 1 h 1 1

2

h5

22 1 2 1 h2(2 1 h)

h

523

4(4 1 h)

34 1 h 2 3

4

h5

12 2 12 2 3h4(4 1 h)

h

5 6 1 3h

53(2h 1 h2)

h



(2, 8) (3, 27) 19

(2, 8) (2.5, 15.625) 15.25

(2, 8) (2.1, 9.261) 12.61

(2, 8) (2.01, 8.120 601) 12.060 1

(2, 8) (1, 1) 7

(2, 8) (1.5, 3.375) 9.25

(2, 8) (1.9, 6.859) 11.41

(2, 8) (1.99, 7.880 599) 11.940 1

c. at

9. a.

b. at

c. at

10. a. at

b. at

c. at

11. a. Let



b.


5 limhS0

c 2h22 1 h

?1

hd

5 limhS0

c4 2 4 1 2h22 1 h

?1

hd

5 limhS0

4

22 1 h1 2

h

5 limhS0

4

22 1 h2 (22)

h

m 5 limhS0

f(22 1 h) 2 f(22)

h

x 5 22

f(22 1 h) 54

22 1 h

f(22) 54

225 22

x 5 2y 5 f(x) 5 x2 2 3x

5 1

5 0 1 1

5 limhS0

(h 1 1)

5 limhS0

h2 1 h

h

5 limhS0

4 1 4h 1 h2 2 6 2 3h 1 2

h

5 limhS0

(2 1 h)2 2 3(2 1 h) 2 (22)

h

m 5 limhS0

f(2 1 h) 2 f(2)

h

x 5 2

f(2 1 h) 5 (2 1 h)2 2 3(2 1 h)

f(2) 5 (2)2 2 3(2) 5 4 2 6 5 22

y 5 f(x).

5 21

10

5 limhS0

21

5(5 1 h)

m 5 limhS0

15 1 h 2 1

5

h

y 51

5x 5 3;y 5

1

x 1 2

5 21

2

5 limhS0

22

4 1 h

m 5 limhS0

84 1 h 2 2

h

y 5 2x 5 1;y 58

3 1 x

5 22

5 limhS0

24

2 1 h

m 5 limhS0

82 1 h 2 4

h

(2, 4)y 58

x

55

6

5 limhS0

5

"9 1 5h 1 3

5 limhS0

£"9 1 5h 2 3

h3

"9 1 5h 1 3

"9 1 5h 1 3§

m 5 limhS0

"10 1 5h 2 1 2 3

h

y 5 3x 5 2,y 5 "5x 2 1

51

4

5 limhS0

1

"4 1 h 1 2

5 limhS0

£"4 1 h 2 2

h3

"4 1 h 1 2

"4 1 h 1 2§

m 5 limhS0

"9 1 h 2 5 2 2

h

y 5 2x 5 9,y 5 "x 2 5

51

2

5 limhS0

1

"1 1 h 1 1

5 limhS0

£"1 1 h 2 1

h3

"1 1 h 1 1

"1 1 h 1 1§

m 5 limhS0

"3 1 h 2 2 2 1

h

(3, 1)y 5 "x 2 2;

5 12

5 limhS0

(12 2 6h 1 h2)

5 limhS0

28 1 12h 2 6h2 1 h3 1 8

h

m 5 limhS0

(22 1 h)3 1 8

h

y 5 28.x 5 22,y 5 x3


Therefore, the slope of the tangent to at

is c. Let


Using the binomial formula to expand (orone could simply expand using algebra), the slope m is


d. Let



e. Let



f. Let

f(8 1 h) 54 1 (8 1 h)

(8 1 h) 2 25

12 1 h6 1 h

f(8) 54 1 8

8 2 25

12

65 2

y 5 f(x).

234.x 5 3y 5 f(x) 5 "25 2 x2

5 23

4

526

8

526

!16 1 4

526 2 0

"16 2 6(0) 2 (0)2 1 4

5 limhS0

26 2 h

"16 2 6h 2 h2 1 4

5 limhS0

h(26 2 h)

h("16 2 6h 2 h2 1 4)

5 limhS0

16 2 6h 2 h2 2 16

h("16 2 6h 2 h2 1 4)

3"16 2 6h 2 h2 1 4

"16 2 6h 2 h2 1 4d

5 limhS0

c"16 2 6h 2 h2 2 4

h

5 limhS0

"16 2 6h 2 h2 2 4

h

m 5 limhS0

f(3 1 h) 2 f(3)

h

x 5 3

5 "16 2 6h 2 h2

5 "25 2 9 2 6h 2 h2

5 "25 2 (9 1 6h 1 h2)

f(3 1 h) 5 "25 2 (3 1 h)2

f(3) 5 "25 2 (3)2 5 !25 2 9 5 4

y 5 f(x).

16.x 5 16y 5 f(x) 5 !x 2 7

51

6

51

3 1 3

51

!0 1 9 1 3

5 limhS0

1

!h 1 9 1 3

5 limhS0

h

h(!h 1 9 1 3)

5 limhS0

(h 1 9) 2 9

h(!h 1 9 1 3)

5 limhS0

!h 1 9 2 3

h?!h 1 9 1 3

!h 1 9 1 3

5 limhS0

!h 1 9 2 3

h

m 5 limhS0

f(16 1 h) 2 f(16)

h

x 5 16

f(16 1 h) 5 !16 1 h 2 7 5 !h 1 9

f(16) 5 !16 2 7 5 !9 5 3

y 5 f(x).

x 5 1y 5 f(x) 5 3x3

5 9

5 3(0) 1 9(0) 1 9

5 limhS0

(3h2 1 9h 1 9)

5 limhS0

3h3 1 9h2 1 9h

h

5 limhS0

3h3 1 9h2 1 9h 1 3 2 3

h

5 limhS0

3(h3 1 3h2 1 3h 1 1) 2 (3)

h

(1 1 h)3

5 limhS0

3(1 1 h)3 2 3

h

m 5 limhS0

f(1 1 h) 2 f(1)

h

x 5 1

f(1 1 h) 5 3(1 1 h)3

f(1) 5 3(1)3 5 3

y 5 f(x).

21.x 5 22

f(x) 54

x

5 21

52

22 1 0

5 limhS0

2

22 1 h




at is

12.

Semi-circle centre

rad 5,OA is a radius.The slope of OA is The slope of tangent is 13. Take values of x close to the point, then

determine 14.

Since the tangent is horizontal, the slope is 0.

15.

The slope of the tangent is 3.

16.

The slope of the tangent is .When

17. a. b.c. The slope of secant AB is

The equation of the secant is

d. Calculate the slope of the tangent.

When the slope is So the equation of the tangent at is

y 5 2x 2 8

y 1 2 5 2(x 2 3)

y 2 y1 5 m(x 2 x1)

A(3, 22)

2(3) 2 4 5 2.x 5 3,

5 2x 2 4

5 2x 1 0 2 4

5 limhS0

(2x 1 h 2 4)

5 limhS0

2xh 1 h2 2 4h

h

5 limhS0

x2 1 2xh 1 h2 2 4x 2 4h 1 1 2 x2 1 4x 2 1

h

5 limhS0

(x 1 h)2 2 4(x 1 h) 1 1 2 (x2 2 4x 1 1)

h

m 5 limhS0

f(x 1 h) 2 f(x)

h

y 5 4x 2 14

y 1 2 5 4(x 2 3)

y 2 y1 5 mAB(x 2 x1)

5 4

58

2

mAB 56 2 (22)

5 2 3

f(5) 5 25 2 20 1 1 5 6; (5, 6)

f(3) 5 9 2 12 1 1 5 22; (3, 22)

3x 1 y 2 8 5 0

y 2 2 5 23(x 2 2)

y 5 2.x 5 2,

23

5 23

5 limhS0

( 2 3 1 h)

5 limhS0

23h 1 h2

h

5 limhS0

4 1 4h 1 h2 2 14 2 7h 1 10

h

m 5 limhS0

(2 1 h)2 2 7(2 1 h) 1 12 2 2

h

3x 2 y 2 8 5 0

y 2 1 5 3(x 2 3)

5 3

5 limhS0

(3 1 h)

5 limhS0

3h 1 h2

h

5 limhS0

9 1 6h 1 h2 2 9 2 3h

h

m 5 limhS0

(3 1 h)2 2 3(3 1 h) 1 1 2 1

h

DyDx

.

234.

43.

y $ 0

(0, 0)y 5 "25 2 x2 S

80

4

–4

8

A

y

x

4–4

216.x 5 8y 5 f(x) 5

4 1 xx 2 2

5 21

6

521

6 1 0

5 limhS0

21

6 1 h

5 limhS0

2h

6 1 h?

1

h

5 limhS0

12 1 h 2 12 2 2h

6 1 h?

1

h

5 limhS0

12 1 h6 1 h

2 2

h

m 5 limhS0

f(8 1 h) 2 f(8)

h

x 5 8


e. When the slope of the tangent is

So the equation of the tangent at is

18. a.

The slope is undefined.b.

The slope is 0.c.

The slope is about –2.5.d.

The slope is about 1.e.

The slope is about f. There is no tangent at this point.

19. at

20. Slope at

Increasing at a rate of 1600 papers per month.21. Point on tangent parallel to

Therefore, tangent line has slope 8.

The point has coordinates

22.

Points on the graph for horizontal tangents are:

23. and

or

The points of intersection are

Tangent to

5 2a.

5 limhS0

2ah 1 h2

h

m 5 limhS0

(a 1 h)2 2 a2

h

y 5 x2:

Q(212,

14).P(1

2, 14),

x 5 21

2x 5

1

2

x2 51

4

x2 51

22 x2

y 51

22 x2y 5 x2

(2, 2283 ).(1, 226

3 ),(21, 263 ),(22, 28

3 ),

a 5 61a 5 62,

(a2 2 4)(a2 2 1) 5 0

a4 2 5a2 1 4 5 0

m 5 a2 2 5 14

a2 5 0

limhS0

4

a(a 1 h)5

4

a2

24

a 1 h1

4

a5 2

4a 1 4a 1 4ha(a 1 h)

5 limhS0

2(a 1 h) 2 (2a)

h5 25

limhS0

aa2 1 ah 11

3h3b 5 a2

5 a2h 1 ah2 11

3h31

3(a 1 h)2 2

1

3a3

y 51

3x3 2 5x 2

4

x

(2, 4).

a 5 2

6a 2 4 5 8

lim

hS0

3h2 1 6ah 2 4hh

5 8

m 5 lim hS0

3(h 1 a)2 2 4(h 1 a) 2 3(a2 1 4a)

h5 8

y 5 8x.

f(x) 5 3x2 2 4x

Cr(6) 5 1200 1 400 5 1600

Cr(t) 5 200t 1 400

t 5 6

C(t) 5 100t2 1 400t 1 5000

5 25

4

5 210

8

5 10 limhS0

4 2 4 2 h

h"4 1 h(2 1 "4 1 h)

5 10 limhS0

2 2 "4 1 h

h"4 1 h3

2 1 "4 1 h

2 1 "4 1 h

m 5 limhS0

20

!4 1 h2 10

h

(5, 10)p . 1D(p) 520

"p 2 1,

278.

P

P

P

P

P

y 5 6x 2 24

y 2 6 5 6(x 2 5)

y 2 y1 5 m(x 2 x1)

B(5, 6)

2(5) 2 4 5 6.

x 5 5,



The slope of the tangent at is at is Tangents to

The slope of the tangents at is

at is and

Therefore, the tangents are perpendicular at thepoints of intersection.24.

The slope of the tangent is We want the line that is parallel to the tangent (i.e.has slope ) and passes through (2, 2). Then,

25. a. Let


2(4a2 1 5a 2 2)

hd

5 limhS0

c 4a2 1 8ah 1 4h2 1 5a 1 5h 2 2

h

m 5 limhS0

f(a 1 h) 2 f(a)

h

x 5 a

5 4a2 1 8ah 1 4h2 1 5a 1 5h 2 2

5 4(a2 1 2ah 1 h2) 1 5a 1 5h 2 2

f(a 1 h) 5 4(a 1 h)2 1 5(a 1 h) 2 2

f(a) 5 4a2 1 5a 2 2

y 5 f(x).

y 5 211x 1 24

y 2 2 5 211(x 2 2)

211

211.

5 211

5 limhS0

(211 1 9h 2 3h2)

5 limhS0

211h 1 9h2 2 3h3

h

5 limhS0

3 2 9h 1 9h2 2 3h3 1 2 2 2h 2 5

h

5 limhS0

23(21 1 3h 2 3h2 1 h3) 1 2 2 2h 2 5

h

5 limhS0

23(21 1 3h 2 3h2 1 h3) 1 2 2 2h 2 5

h

m 5 limhS0

23(21 1 h)3 2 2(21 1 h) 2 5

h

(21, 5)y 5 23x3 2 2x,

mqMq 5 21mpMp 5 21

1 5 Mqa 5 212

21 5 Mp;a 5 12

5 22a.

5 limhS0

22ah 2 h2

h

m 5 limhS0

S12 2 (a 1 h)2T 2 S12 2 a2Th

y 5 12 2 x2:

21 5 mq.a 5 212

1 5 mp,a 5 12

b. To be parallel, the point on the parabola and theline must have the same slope. So, first find theslope of the line. The line canbe rewritten as

So, the slope, m, of the line is 5.To be parallel, the slope at a must equal 5. Frompart a., the slope of the tangent to the parabola at

is

Therefore, at the point the tangent line isparallel to the line c. To be perpendicular, the point on the parabolaand the line must have slopes that are negativereciprocals of each other. That is, their product mustequal So, first find the slope of the line. Theline can be rewritten as

So, the slope, m, of the line is To be perpendicular, the slope at a must equal the negative reciprocal of the slope of the line

That is, the slope of a must equalFrom part a., the slope of the tangent to the

parabola at is

Therefore, at the point the tangent line isperpendicular to the line x 2 35y 1 7 5 0.

(25, 73)

a 5 25

8a 5 240

8a 1 5 5 235

8a 1 5.x 5 a235.

x 2 35y 1 7 5 0.

135.x 2 35y 1 7 5 0

y 51

35x 1

7

35

y 52x 2 7

235

235y 5 2x 2 7

x 2 35y 1 7 5 0

21.

10x 2 2y 2 18 5 0.

(0, 22)

a 5 0

8a 5 0

8a 1 5 5 5

8a 1 5.x 5 a

10x 2 2y 2 18 5 0

y 5 5x 2 9

y 5 29 1 5x

y 518 2 10x

22

22y 5 18 2 10x

10x 2 2y 2 18 5 0

5 8a 1 5

5 8a 1 4(0) 1 5

5 limhS0

(8a 1 4h 1 5)

5 limhS0

8ah 1 4h2 1 5h

h

124a2 2 5a 1 2

hd

5 limhS0

c4a2 1 8ah 1 4h2 1 5a 1 5h 2 2

h

1.3 Rates of Change, pp. 29–311. when or

2. a. Slope of the secant between the

points and

b. Slope of the tangent at the

point

3. Slope of the tangent to the

function with equation at the point 4. a. A and Bb. greater; the secant line through these two pointsis steeper than the tangent line at B.c.

5. Speed is represented only by a number, not adirection.6. Yes, velocity needs to be described by a numberand a direction. Only the speed of the school buswas given, not the direction, so it is not correct touse the word “velocity.”7.a. Average velocity during the first second:

third second:

eighth second:

b. Average velocity

c.

Velocity at is downward.8.a. i. from to

Average velocity

ii. from to

iii.

b. Instantaneous velocity is approximately c. At

9. a.

15 terms are learned between and

b.

At the student is learning at a rate of 16 terms h.10. a. M in mg in 1 mL of blood t hours after theinjection.


5 21

3

5 limhS0

a21

32

1

3 hb

5 limhS0

21

3 h 2 13 h2

h

5 limhS0

24

3 2 43 h 2 1

3 h2 1 2 1 h 1 4

3 2 2

h

limhS0

21

3(2 1 h)2 1 (2 1 h) 2 (243 1 2)

h

t 5 2.

M(t) 5 21

3t2 1 t; 0 # t # 3

>t 5 2,

5 16

5 limhS0

(16 2 h)

5 limhS0

16h 2 h2

h

5 limhS0

40 1 20h 2 4 2 4h 2 h2 2 36

h

limhS0

20(2 1 h) 2 (2 1 h)2 2 36

h

t 5 3.t 5 2

5 15

551 2 36

1

N(3) 2 N(2)

1

N(t) 5 20t 2 t2

5 64 km>h v(3) 5 48 1 16

v(t) 5 16t 1 16

s(t) 5 8t2 1 16t t 5 3

64 km>h.

5 64.08 km>hs(3.01) 2 s(3)

0.01

3 # t # 3.01

5 64.8 km>h 5

126.48 2 120

0.1

s(3.1) 2 s(3)

0.1

t 5 3.1t 5 3

5 72 km>h 5 24(8 2 5)

5 32(6) 2 24(5)

s(4) 2 s(3)

1

t 5 4t 5 3

0 # t # 5s(t) 5 8t(t 1 2),

20 m>st 5 2

5 220

5 5 limhS0

24h 1 h2

h

v(t) 5 limhS0

320 2 5(2 1 h)2 2 (320 2 5(2)2)

h

s(t) 5 320 2 5t2

5 55 m>s5275

5 s(8) 2 s(3)

8 2 35

320 2 45

5

3 # t # 8

s(8) 2 s(7)

15

320 2 245

15 75 m>s.

s(3) 2 s(2)

15

45 2 20

15 25 m>s;

s(1) 2 s(0)

15 5 m>s;

0 # t # 8s(t) 5 320 2 5t2,

y

xED

CBA

y = f(x)

(4, 2).y 5 !x

limhS0

"4 1 h 2 2

h.

(6, s(6)).

limhS0

s(6 1 h) 2 s(6)

h.

(9, s(9)).(2, s(2))

s(9) 2 s(2)

7.

t 5 4.t 5 0v(t) 5 0



Rate of change is mg h.b. Amount of medicine in 1 mL of blood is beingdissipated throughout the system.

11.


At , rate of change of time with respect toheight is .

12.Calculate the instantaneous rate of change when

5 limkS0

60

5 1 k2 12

k

limkS0

60

(3 1 k) 1 22

60

(3 1 2)

k

h 5 3.

T(h) 560

h 1 2

s>m150

s 5 125

51

50

51

5(5 1 5)

51

5aÄ125

51 5b

5 limhS0

1

5aÄ125 1 h5

1 5b

5 limhS0

≥125 1 h 2 125

5

haÄ125 1 h5

1 5b¥

5 limhS0

≥125 1 h

52 25

haÄ125 1 h5

1 5b ¥

5 limhS0

≥ Ä125 1 h

52 5

h?Ä125 1 h

51 5

Ä125 1 h5

1 5

¥

5 limhS0

Ä125 1 h

52 5

h

limhS0

Ä125 1 h

5 2 Ä125

5

h

s 5 125.

t 5 Ås5

>213

Temperature is decreasing at C km.13.When


It hit the ground in 2 s at a speed of 0 m s.14. Sale of x balls per week:

dollars.a.

Profit on the sale of 40 balls is $4800.b. Calculate the instantaneous rate of change when

Rate of change of profit is $80 per ball.c.

Rate of change of profit is positive when the saleslevel is less than 80.

5 80

5 limhS0

(80 2 h)

5 limhS0

80h 2 h2

h

5 limhS0

6400 1 160h 2 1600 2 80h 2 h2 2 4800

h

limhS0

160(40 1 h) 2 (40 1 h)2 2 4800

h

x 5 40.

5 4800

P(40) 5 160(40) 2 (40)2

P(x) 5 160x 2 x2

>5 0

5 limhS0

25h

5 limhS0

25h2

h

5 limhS0

100 1 100h 1 25h2 2 200 2 100h 1 100

h

limhS0

25(2 1 h)2 2 100(2 1 h) 1 100 2 0

h

t 5 2.

t 5 2

(t 2 2)2 5 0

t2 2 4t 1 4 5 0

25t2 2 100t 1 100 5 0h 5 0,

h 5 25t2 2 100t 1 100

>°125

5 212

5

5 limkS0

212

(5 1 k)

5 limkS0

212k

k(5 1 k)

5 limkS0

60

5 1 k2

60 1 12k5 1 k

k


15. a.

b.

c.

16.

Calculate the instantaneous rate of change.

5 limhS0

S(x 1 h) 2 S(x)

h

S(x) 5 246 1 64x 2 8.9x2 1 0.95x3

51

10

5 limxS24

x 2 24

(x 2 24)(!x 1 1 1 5)

5 limxS24

!x 1 1 2 5

x 2 24?!x 1 1 1 5

!x 1 1 1 5

5 limxS24

f(x) 2 f(24)

x 2 24

x 5 24f(x) 5 !x 1 1,

5 21

5 limxS2

2 (x 2 2)

(x 2 1)(x 2 2)

5 limxS2

x 2 2x 1 2

(x 2 1)(x 2 2)

limxS2

xx 2 1

2 2

x 2 2

x 5 2f(x) 5x

x 2 1,

5 6

5 2 limxS22

(x 2 4)

5 2 limxS22

(x 2 4)(x 1 2)

x 1 2

5 limxS22

2 (x2 2 2x 2 8)

x 1 2

5 limxS22

2x2 1 2x 1 3 1 5

x 1 2

limxS22

f(x) 2 f(22)

x 1 2

f(x) 5 2x2 1 2x 1 3; (22, 25) For the year 2005, Hence,the rate at which the average annual salary is changingin 2005 is

years since 198217.a. The distance travelled from 0 s to 5 s is

mb. mThe rate at which the avalanche is moving from 0 sto 10 s is

m sc. Calculate the instantaneous rate of change when

At 10 s the avalanche is moving at 60 m s.d. Set

Since s.t 5 10!2 8 14t $ 0,

t 5 610!2

t2 5 200

3t2 5 600

s(t) 5 600:

>5 60

5 limhS0

(60 1 3h)

5 limhS0

60h 1 3h2

h

5 limhS0

300 1 60h 1 3h2 2 300

h

limhS0

3(10 1 h)2 2 300

h

t 5 10.

>5 30

DsDt

5300 2 0

10 2 0

s(10) 5 3(10)2 5 300

s(5) 5 3(5)2 5 75

s(t) 5 3t2

$1 162 250>P r(23) 5 64 2 17.8(23) 1 2.85(23)2 5

x 5 2005 2 1982 5 23.

5 64 2 17.8x 1 2.85x2

5 64 2 8.9(2x 1 0) 1 0.95 33x2 1 3x(0) 1 (0)245 lim

hS0 364 2 8.9(2x 1 h) 1 0.95(3x2 1 3xh 1 h2)4

5 limhS0

64h 2 8.9(2xh 1 h2) 1 0.95(3x2h 1 3xh2 1 h3)

h

5 limhS0

246 2 246 1 64(x 1 h 2 x) 2 8.9(x2 1 2xh 1 h2 2 x2) 1 0.95(x3 1 3x2h 1 3xh2 1 h3 2 x3)

h

5 limhS0

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 8.9x2 1 0.95x3)

h



is or The




19.




Surface area is

surface area.










5 37

5 45 2 8

5 9(5) 2 4(2)

5 (9!25 2 6!10 1 6!10 2 4!4)

(3!5 1 2!2)(3!5 2 2!2)

3!5 2 2!2.

5 3

5 5 2 2

5 (!25 1 !10 2 !10 2 !4)

(!5 2 !2)(!5 1 !2)

!5 1 !2.

>2100p

5 100p

54

3p(0)2 1 20p(0) 1 100p

5 limhS0

a4

3ph2 1 20ph 1 100pb

5 limhS0

43 ph3 1 20ph2 1 100ph

h

2 43 p(125)

h

5 limhS0

43 ph3 1 20ph2 1 100ph 1 4

3 p(125)

h

5 limhS0

43 p(h3 1 15h2 1 75h 1 125) 2 4

3 p(5)3

h

(5 1 h)3

5 limhS0

43 p(5 1 h)3 2 4

3 p(5)3

h

limhS0

V(5 1 h) 2 V(5)

h

r 5 5

r 5 5.

V(r) 5 43pr3.

>80p

5 80p

5 80p 1 4p(0)

5 limhS0

(80p 1 4ph)

5 limhS0

80ph 1 4ph2

h

5 limhS0

400p 1 80ph 1 4ph2 2 400p

h

5 limhS0

4p(100 1 20h 1 h2) 2 4p(100)

h

5 limhS0

4p(10 1 h)2 2 4p(10)2

h

limhS0

A(10 1 h) 2 A(10)

h

r 5 10.

A(r) 5 4pr2.

Vr(x) 5 3x2 51

2

6x2.V 5 x3.

>200p

r 5 100

5 2pr

5 p limhS0

(r 1 h 2 r)(r 1 h 1 r)

h

5 limhS0

p(r 1 h)2 2 pr2

h

limhS0

A(r 1 h) 2 A(r)

h

A(r) 5 pr2

5 limxSh

V(x 1 h) 2 V(x)

h h,

limxSR

C(x 1 h) 2 C(x)

h

C(x 1 h) 5 F 1 V(x 1 h)

C(x) 5 F 1 V(x)

1

2a2

ab (2a) 5 2.

2

a

(22a, 0).a0, 2

ab

y 5 21

a2 x 12

a.y 2

1

a5 2

1

a2 (x 2 a)

21

a2

aa, 1

ab




is or The




19.




Surface area is

surface area.










5 37

5 45 2 8

5 9(5) 2 4(2)

5 (9!25 2 6!10 1 6!10 2 4!4)

(3!5 1 2!2)(3!5 2 2!2)

3!5 2 2!2.

5 3

5 5 2 2

5 (!25 1 !10 2 !10 2 !4)

(!5 2 !2)(!5 1 !2)

!5 1 !2.

>2100p

5 100p

54

3p(0)2 1 20p(0) 1 100p

5 limhS0

a4

3ph2 1 20ph 1 100pb

5 limhS0

43 ph3 1 20ph2 1 100ph

h

2 43 p(125)

h

5 limhS0

43 ph3 1 20ph2 1 100ph 1 4

3 p(125)

h

5 limhS0

43 p(h3 1 15h2 1 75h 1 125) 2 4

3 p(5)3

h

(5 1 h)3

5 limhS0

43 p(5 1 h)3 2 4

3 p(5)3

h

limhS0

V(5 1 h) 2 V(5)

h

r 5 5

r 5 5.

V(r) 5 43pr3.

>80p

5 80p

5 80p 1 4p(0)

5 limhS0

(80p 1 4ph)

5 limhS0

80ph 1 4ph2

h

5 limhS0

400p 1 80ph 1 4ph2 2 400p

h

5 limhS0

4p(100 1 20h 1 h2) 2 4p(100)

h

5 limhS0

4p(10 1 h)2 2 4p(10)2

h

limhS0

A(10 1 h) 2 A(10)

h

r 5 10.

A(r) 5 4pr2.

Vr(x) 5 3x2 51

2

6x2.V 5 x3.

>200p

r 5 100

5 2pr

5 p limhS0

(r 1 h 2 r)(r 1 h 1 r)

h

5 limhS0

p(r 1 h)2 2 pr2

h

limhS0

A(r 1 h) 2 A(r)

h

A(r) 5 pr2

5 limxSh

V(x 1 h) 2 V(x)

h h,

limxSR

C(x 1 h) 2 C(x)

h

C(x 1 h) 5 F 1 V(x 1 h)

C(x) 5 F 1 V(x)

1

2a2

ab (2a) 5 2.

2

a

(22a, 0).a0, 2

ab

y 5 21

a2 x 12

a.y 2

1

a5 2

1

a2 (x 2 a)

21

a2

aa, 1

ab


c. Corresponding conjugate:

d. Corresponding conjugate:

2. a.

b.

c.

d.

e.

f.

3. a.

b.

c.

d.

5 213

3!2(2!3 1 5)5

12 2 25

3!2(2!3 1 5)

54(3) 2 25

3!2(2!3 1 5)

54!9 1 10!3 2 10!3 2 25

3!2(2!3 1 5)

2!3 2 5

3!2?

2!3 1 5

2!3 1 5

5 29

5(!7 1 4)

57 2 16

5(!7 1 4)

5!49 1 4!7 2 4!7 2 16

5(!7 1 4)

!7 2 4

5?!7 1 4

!7 1 4

53

!3(6 1 !2)

5!9

!3(6 1 !2)

!3

6 1 !2?!3

!3

52

5!2

5!4

5!2

!2

5?!2

!2

5 23!2(2!3 1 5)

13

53!2(2!3 1 5)

213

53!2(2!3 1 5)

12 2 25

53!2(2!3 1 5)

4(3) 2 25

53!2(2!3 1 5)

4!9 1 10!3 2 10!3 2 25

3!2

2!3 2 5?

2!3 1 5

2!3 1 5

510!3 2 15

2

530 2 20!3

24

530 2 20!3

12 2 16

510!9 2 20!3

4!9 2 8!3 1 8!3 2 16

5!3

2!3 1 4?

2!3 2 4

2!3 2 4

5 22(3 1 2!3)

56 1 4!3

21

56 1 4!3

3 2 4

52!9 1 4!3

!9 1 2!3 2 2!3 2 4

2!3

!3 2 2?!3 1 2

!3 1 2

5 25(!7 1 4)

9

55(!7 1 4)

7 2 16

55(!7 1 4)

!49 1 4!7 2 4!7 2 16

5

!7 2 4?!7 1 4

!7 1 4

56 1 4!3

3

52!9 1 4!3

!9

2!3 1 4

!3?!3

!3

56!3 1 !6

3

56!3 1 !6

!9

6 1 !2

!3?!3

!3

5 5

5 45 2 40

5 9(5) 2 4(10)

5 (9!25 1 6!50 2 6!50 2 4!100)

(3!5 2 2!10)(3!5 1 2!10)

3!5 1 2!10.

5 61

5 81 2 20

5 81 2 4(5)

5 (81 2 18!5 1 18!5 2 4!25)

(9 1 2!5)(9 2 2!5)

9 2 2!5.




(21, 1) (22, 6) 52

(21, 1) (21.5, 3.25) 4.52

(21, 1) (21.1, 1.41) 4.12

(21, 1) (21.01, 1.040 1) 4.012

(21, 1) (21.001, 1.004 001) 4.0012


(21, 1) (0, 22) 32

(21, 1) (20.5, 20.75) 3.52

(21, 1) (20.9, 0.61) 3.92

(21, 1) (20.99, 0.9601) 3.992

(21, 1) (20.999, 0.996 001) 3.9992

e.

f.

4. a.

b.

c.

d.

21

5x 1 y 1

10

52

1

55 0

y 1 2 51

5x 1

1

5

y 2 (22) 51

5(x 2 (21))

m 51

5

4x 2 y 2 2 5 0

24x 1 y 1 2 5 0

y 2 6 5 4x 2 8

y 2 6 5 4(x 2 2)

m 5 4

x 2 y 1 5 5 0

2x 1 y 2 5 5 0

y 2 7 5 x 2 2

y 2 7 5 1(x 2 2)

m 511 2 7

6 2 25

4

45 1

2

3x 1 y 2 6 5 0

y 2 6 5 22

3x

y 2 6 5 22

3(x 2 0)

m 5 22

3;

51

(2!3 2 !7)

512 2 7

5(2!3 2 !7)

54(3) 2 7

5(2!3 2 !7)

54!9 2 2!21 1 2!21 2 !49

5(2!3 2 !7)

2!3 1 !7

5?

2!3 2 !7

2!3 2 !7

5 21

(!3 1 !7)

5 24

4(!3 1 !7)

53 2 7

4(!3 1 !7)

5!9 1 !21 2 !21 2 !49

4(!3 1 !7)

!3 2 !7

4?!3 1 !7

!3 1 !7

5. The slope of PQ is

So, the slope of PQ with is 6. a. Unlisted y-coordinates for Q are found by substituting the x-coordinates into the given function.The slope of the line PQ with the given points isgiven by the following: Let and

Then, the

b. The slope from the right and from the left appearto approach The slope of the tangent to thegraph of at point P is about c. With the points and

the slope, m, of PQ isthe following:Q 5 (21 1 h, f(21 1 h)),

P 5 (21, 1)

24.f(x)

24.

slope 5 m 5y

22 y

1

x2 2 x1.Q 5 (y1, y2).

P 5 (x1, y1)

22.f(x) 5 2x2

5 22

5 22 2 (0)

5 limhS0

(22 2 h)

5 limhS0

22h 2 h2

h

5 limhS0

21 2 2h 2 h2 1 1

h

5 limhS0

2 (1 1 2h 1 h2) 1 1

h

5 limhS0

2 (1 1 h)2 1 1

h

m 5 limhS0

f(1 1 h) 2 (21)

(1 1 h) 2 1

x 2 5y 2 9 5 0

1

5x 2 y 2

9

55 0

21

5x 1 y 1

9

55 0


d. The slope of the tangent is

In this case, as h goes to zero, goes toThe slope of the tangent to

the graph of at the point P is e. The answers are equal.

24.f(x)

h 2 4 5 0 2 4 5 24.

h 2 4

limhS0

f(x).

5 h 2 4

5h2 2 4h

h

51 2 2h 1 h2 1 2 2 2h 2 2 2 1

21 1 h 1 1

53(21 1 h)2 2 2(21 1 h) 2 24 2 (1)

(21 1 h) 2 (21)

m 5y2 2 y1

x2 2 x1c.

5 limhS0

a4 2 (h 1 4)

h 1 4b 1

h

5 limhS0

4

h 1 42 1

h

5 limhS0

4

h 1 42

4

4

h

5 limhS0

4

6 1 h 2 22

4

6 2 2

h

m 5 limhS0

f(6 1 h) 2 f(6)

h

y 5 f(x) 54

x 2 2

7. a.

5 23

5 0 2 3

5 limhS0

(h 2 3)

5 limhS0

h2 2 3h

h

5 limhS0

h2 2 3h 2 5 2 (25)

h

5 limhS0

9 2 6h 1 h2 2 9 1 3h 2 5 2 (9 2 9 2 5)

h

5 limhS0

3(23 1 h)2 1 3(23 1 h) 2 54 2 3(23)2 1 3(23) 2 54

h

m 5 limhS0

f(23 1 h) 2 f(23)

h

b.

5 29

521

19 1 1

3(0)

5 limhS0

21

19 1 1

3h

5 limhS0

a 2h19 1 1

3hb 1

h

5 limhS0

(13) 2 (1

3 1 h)13(1

3 1 h)

h

5 limhS0

113 1 h

2113

h

m 5 limhS0

f(1

3 1 h) 2 f(13)

h

y 5 f(x) 51

x

d.

5 limhS0

1

!9 1 h 1 3

5 limhS0

h

h(!9 1 h 1 3)

5 limhS0

9 1 h 1 3!9 1 h 2 3!9 1 h 2 9

h(!9 1 h 1 3)

5 limhS0

!9 1 h 2 3

h?!9 1 h 1 3

!9 1 h 1 3

5 limhS0

!9 1 h 2 3

h

5 limhS0

!9 1 h 2 !9

h

5 limhS0

!5 1 h 1 4 2 !5 1 4

h

m 5 limhS0

f(5 1 h) 2 f(5)

h

5 21

4

521

0 1 4

5 limhS0

21

h 1 4

5 limhS0

a 2hh 1 4

b 1

h

8.

a. i.

km h

ii.

km h

iii.

km h

b. At the time the velocity of the car appearsto approach 30 km h.

c.

km hd. When the velocity is the limit as happroaches 0.

Therefore, when the velocity is 30 km h.9. a. The instantaneous rate of change of withrespect to x at is given by

b. The instantaneous rate of change of with

respect to x at is given by

5 212

526

12 1 0

5 limhS0

26

12 1 h

5 limhS0

26h

12 1 h

?1

h

5 limhS0

3 2 3 2 6h

12 1 h

?1

h

5 limhS0

3 2 6(1

2 1 h)12 1 h

?1

h

5 limhS0

312 1 h

2 6

h

5 limhS0

312 1 h

2312

h

limhS0

f(1

2 1 h) 2 f(12)

h

x 5 12

f(x)5 24

5 2 (0) 2 4

5 limhS0

(2h 2 4)

5 limhS0

2h2 2 4h

h

5 limhS0

5 2 4 2 4h 2 h2 2 1

h

5 limhS0

5 2 (4 1 4h 1 h2) 2 1

h

5 limhS0

35 2 (2 1 h)24 2 35 2 (2)24

h

limhS0

f(2 1 h) 2 f(2)

h

x 5 2

f(x)

>t 5 25 30

5 6(0) 1 30

velocity 5 limhS0

(6h 1 30)

t 5 2,

>5 (6h 1 30)

56h2 1 30h

h

56h2 1 30h 1 36 2 36

h

5324 1 24h 1 6h2 1 12 1 6h4 2 36

h

536(4 1 4h 1 h2) 1 12 1 6h4 2 324 1 124

h

536(2 1 h)2 1 6(2 1 h)4 2 36(2)2 1 6(2)4

h

average velocity 5f(2 1 h) 2 f(2)

(2 1 h) 2 (2)

>t 5 2,

>5 30.06

50.3006

0.01

536.3006 2 36

0.01

536.3006 2 324 1 124

0.01

5324.2406 1 12.064 2 36(2)2 1 6(2)4

0.01

536(2.01)2 1 6(2.01)4 2 36(2)2 1 6(2)4

0.01


2.01 2 2

>5 30.6

53.06

0.1

539.06 2 36

0.1

5326.46 1 12.64 2 324 1 124

0.1

536(2.1)2 1 6(2.1)4 2 36(2)2 1 6(2)4

0.1


2.1 2 2

>5 36

5 54 1 18 2 36

5 6(9) 1 18 2 (24 1 12)

5 36(3)2 1 6(3)42 36(2)2 1 6(2)4average velocity 5

s(3) 2 s(2)

3 2 2

s(t) 5 6t(t 1 1) 5 6t2 1 6t

51

6

51

!9 1 0 1 3


10. a. The average rate of change of withrespect to t during the first 20 minutes is given by

L minb. The rate of change of with respect to t at thetime is given by

L min11. a. Let




b. Let




c.


5 24

5 3(0) 2 4

5 limhS0

(3h 2 4)

5 limhS0

3h2 2 4h

h

5 limhS0

3h2 2 4h 2 4 2 (24)

h

m 5 limhS0

f(21 1 h) 2 f(21)

h

x 5 4

5 3h2 2 4h 2 45 3 2 6h 1 3h2 2 7 1 2h5 3(1 2 2h 1 h2) 2 2 1 2h 2 5

f(21 1 h) 5 3(21 1 h)2 1 2(21 1 h) 2 5

5 24f(21) 5 3(21)2 1 2(21) 2 5 5 3 2 2 2 5

8x 1 y 1 15 5 0

8x 1 y 2 1 1 16 5 0

y 2 1 5 28x 2 16

y 2 1 5 28(x 2 (22))

x 5 22

28.x 5 22y 5 f(x) 5 2x2 2 7

5 28

5 2(0) 2 8

5 limhS0

(2h 2 8)

5 limhS0

2h2 2 8h

h

5 limhS0

2h2 2 8h 1 1 2 (1)

h

m 5 limhS0

f(22 1 h) 2 f(22)

h

x 5 4

5 2h2 2 8h 1 1

5 8 2 8h 1 2h2 2 7

5 2(4 2 4h 1 h2) 2 7

f(22 1 h) 5 2(22 1 h)2 2 7

f(22) 5 2(22)2 2 7 5 2(4) 2 7 5 1

y 5 f(x).

29x 1 y 1 19 5 0

29x 1 y 2 17 1 36 5 0

y 2 17 5 9x 2 36

y 2 17 5 9(x 2 4)

x 5 4

x 5 4y 5 f(x) 5 x2 1 x 2 3

5 9

5 0 1 9

5 limhS0

(h 1 9)

5 limhS0

h2 1 9h

h

5 limhS0

h2 1 9h 1 17 2 (17)

h

m 5 limhS0

f(4 1 h) 2 f(4)

h

x 5 4

5 h2 1 9h 1 17

5 16 1 8h 1 h2 1 h 1 1

f(4 1 h) 5 (4 1 h)2 1 (4 1 h) 2 3

f(4) 5 (4)2 1 (4) 2 3 5 16 1 1 5 17

y 5 f(x).

>5 21000

5 50(0) 2 1000

5 limhS0

50h 2 1000

5 limhS0

50h2 2 1000h

h

5 limhS0

5000 2 1000h 1 50h2 2 5000

h

5 limhS0

350(100 2 20h 1 h2)4 2 350(100)4

h

5 limhS0

350(10 2 h)24 2 350(10)24

h

5 limhS0

350(30 2 (20 1 h))24 2 350(30 2 20)24

h

limhS0

f(20 1 h) 2 f(20)

h

t 5 20

V(t)>5 22000

5 240 000

20

55000 2 45 000

20

5350(30 2 20)24 2 350(30 2 0)24

20

f(20) 2 f(0)

20 2 0

V(t)


Therefore, the slope of the tangent toat is .


d.




12. a. Using the limit of the difference quotient, theslope of the tangent at is



b. Using the limit of the difference quotient, theslope of the tangent at is


at is


3x 1 4y 1 5 5 0

3x 1 4y 1 2 1 3 5 0

4y 1 2 5 23x 2 3

y 11

25 2

3

4x 2

3

4

y 2 a21

2b 5 2

3

4(x 2 (21))

x 5 234

234.x 5 21f(x) 5

2x 1 5

5x 2 1

5 23

4

5 29

12

59

10(0) 2 12

5 limhS0

a 9

10h 2 12b

5 limhS0

a 9h10h 2 12

b ?1

h

5 limhS0

a4h 1 6 1 5h 2 6

10h 2 12b ?

1

h

5 limhS0

a2h 1 3

5h 2 61

1

2b ?

1

h

5 limhS0

a2h 1 3

5h 2 62

3

26b ?

1

h

5 limhS0

a22 1 2h 1 5

25 1 5h 2 12

22 1 5

25 2 1b ?

1

h

5 limhS0

a2(21 1 h) 1 5

5(21 1 h) 2 12

2(21) 1 5

5(21) 2 1b ?

1

h

m 5 limhS0

f(21 1 h) 2 f(21)

h

x 5 21

23x 1 4y 2 25 5 0

23

4x 1 y 2

25

45 0

23

4x 1 y 2

10

42

15

45 0

y 25

25

3

4x 1

15

4

y 25

25

3

4(x 2 (25))

x 5 34

34.x 5 25f(x) 5

xx 1 3

53

4

523

24 1 2(0)

5 limhS0

a 23

24 1 2hb

5 limhS0

a 23h24 1 2h

b ?1

h

5 limhS0

a210 1 2h 1 10 2 5h24 1 2h

b ?1

h

5 limhS0

a210 1 2h 2 (210 1 5h)

24 1 2hb ?

1

h

5 limhS0

a25 1 h22 1 h

25

2b ?

1

h

5 limhS0

a 25 1 h25 1 h 1 3

225

25 1 3b ?

1

h

m 5 limhS0

f(25 1 h) 2 f(25)

h

x 5 25

22x 1 y 1 2 5 0

y 5 2x 2 2

y 2 0 5 2(x 2 1)

x 5 1

x 5 1y 5 f(x) 5 5x2 2 8x 1 3

5 2

5 5(0) 1 2

5 limhS0

(5h 1 2)

5 limhS0

5h2 1 2h 2 (0)

h

m 5 limhS0

f(1 1 h) 2 f(1)

h

x 5 1

5 5h2 1 2h5 5 1 10h 1 5h2 2 5 2 8h5 5(1 1 2h 1 h2) 2 8 2 8h 1 3

f(1 1 h) 5 5(1 1 h)2 2 8(1 1 h) 1 3

f(1) 5 5(1)2 2 8(1) 1 3 5 5 2 8 1 3 5 0

4x 1 y 1 8 5 0

4x 1 y 1 4 1 4 5 0

y 1 4 5 24x 2 4

y 1 4 5 24(x 1 1)y 2 (24) 5 24(x 2 (21))

x 5 24

24x 5 21y 5 f(x) 5 3x2 1 2x 2 5


1.4 The Limit of a Function, pp. 37–39

1. a.

b.2. One way to find a limit is to evaluate the functionfor values of the independent variable that get progressively closer to the given value of the independent variable.3. a. A right-sided limit is the value that a function gets close to as the values of the independent variable decrease and get close to a given value.b. A left-sided limit is the value that a function gets close to as the values of the independent variable increase and get close to a given value.c. A (two-sided) limit is the value that a functiongets close to as the values of the independent variable get close to a given value, regardless of whether the values increase or decrease toward the given value.4. a.b.c.d.e. 4f.5. Even though the limit is 1, since thatis the value that the function approaches from theleft and the right of 6. a. 0b. 2c.d. 27. a. 2b. 1c. does not exist8. a.

b.

c.

9.

10. a. Since 0 is not a value for which the function isundefined, one may substitute 0 in for x to find that

b. Since 2 is not a value for which the function isundefined, one may substitute 2 in for x to find that

c. Since 3 is not a value for which the function isundefined, one may substitute 3 in for x to find that

d. Since 1 is not a value for which the function isundefined, one may substitute 1 in for x to find that

e. Since 3 is not a value for which the function isundefined, one may substitute 3 in for x to find that

f. If 3 is substituted in the function for x, then thefunction is undefined because of division by zero.There does not exist a way to divide out the in x 2 3

51

5

51

3 1 2

limxS31

1

x 1 25 lim

xS3

1

x 1 2

5 21

2

51

1 2 3

limxS11

1

x 2 35 lim

xS1

1

x 2 3

5 55 9 2 4

5 (3)2 2 4

limxS32

(x2 2 4) 5 limxS3

(x2 2 4)

5 05 4 2 4

5 (2)2 2 4

limxS22

(x2 2 4) 5 limxS2

(x2 2 4)

5 0

5 (0)4

limxS01

x4 5 limxS0

x4

40

2

4

6y

x

2–2–4

22 1 1 5 5

5 2

"5 2 1 5 "4

5 2Å

0 1 20

0 1 55 "4

9 2 (21)2 5 8

21

x 5 4.

f(4) 5 21,

23 5 8

4 2 3(22)2 5 28

102 5 100

3 1 7 5 10

25

p

27

99


the denominator. Also, approaches infinity,

while approaches negative infinity.

Therefore, since

does not exist.

11. a.

Therefore, does

not exist.b.


is equal to 2.c.


is equal to 2.

d.

Therefore,

does not exist.12. Answers may vary. For example:a.

b.

c.

d.

13.

m 5 23b 5 1,

2b 5 2

2m 1 b 5 4limxS21

f(x) 5 4

m 1 b 5 22limxS1

f(x) 5 22

f(x) 5 mx 1 b

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

4 6 80

46

2

–4–2

y

x

2–2–4–6–8

limxS20.5

f(x)limxS20.51

f(x) 2 limxS20.52

f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS1

2

f(x)limxS1

21 f(x) 5 lim

xS12

2 f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS2

f(x)limxS21

f(x) 5 limxS22

f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS21

f(x)limxS211

f(x) 2 limxS212

f(x).

4 6 80

46

2

–4–6

–2

–8

8y

x

2–2–4–6–8

limxS3

1

x 2 3lim

xS31

1

x 2 32 lim

xS32

1

x 2 3,

limxS32

1

x 2 3

limxS31

1

x 2 3


14.


b.






4. a.

b.

c.

d.

e.

f.

5. a.

b.


7. a.

b.

c.

5 27

5 9 1 9 1 9

limxS3

x3 2 27

x 2 35 lim

xS3

(x 2 3)(x2 1 3x 1 9)

x 2 3

5 5

limxS21

2x2 1 5x 1 3

x 1 15 lim

xS21

(x 1 1)(2x 1 3)

x 1 1

5 4

5 limxS2

(2 1 x)

limxS2

4 2 x2

2 2 x5 lim

xS2

(2 2 x)(2 1 x)

(2 2 x)

5 21

1 2 1 2 5

6 2 15

25

5

t 5 1

5 "2

2

!1 1 15

2

!2

(22)3

22 2 25 22

5 "3

Å23 2 3

2(23) 1 45 Å

26

22

5 2

"3 1 "1 1 0 5 "3 1 1

5 5p3

(2p)3 1 p2(2p) 2 5p3 5 8p3 1 2p3 2 5p3

5100

9

c"9 11

"9d 2

5 a3 11

3b2

(21)4 1 (21)3 1 (21)2 5 1

3(2)

22 1 25 1

limxS2

3 1 x

limxS2

(x 1 3)limxS2

(3 1 x)

t 5 !72 8 8.49

!75 5 t72 5 t2

4 51

18t2

6 5 2 11

18t2

6000 2 4000 5 2000

p(t)5 4

5 2 1 2

5 2 136

18

limtS61

p(t) 5 2 11

18(6)2

5 6

5 3 1 3

5 3 136

12

limtS62

p(t) 5 3 11

12(6)2

8 10 120

68

42

–2

10y

x

642–2–4

c 5 0.b 5 2a 5 3,

b 5 2a 5 3,

6a 5 18

4a 2 2b 5 8limxS22

f(x) 5 8

a 1 b 5 5limxS1

f(x) 5 5

c 5 0f(0) 5 0

a 2 0f(x) 5 ax2 1 bx 1 c,


14.


b.






4. a.

b.

c.

d.

e.

f.

5. a.

b.


7. a.

b.

c.

5 27

5 9 1 9 1 9

limxS3

x3 2 27

x 2 35 lim

xS3

(x 2 3)(x2 1 3x 1 9)

x 2 3

5 5

limxS21

2x2 1 5x 1 3

x 1 15 lim

xS21

(x 1 1)(2x 1 3)

x 1 1

5 4

5 limxS2

(2 1 x)

limxS2

4 2 x2

2 2 x5 lim

xS2

(2 2 x)(2 1 x)

(2 2 x)

5 21

1 2 1 2 5

6 2 15

25

5

t 5 1

5 "2

2

!1 1 15

2

!2

(22)3

22 2 25 22

5 "3

Å23 2 3

2(23) 1 45 Å

26

22

5 2

"3 1 "1 1 0 5 "3 1 1

5 5p3

(2p)3 1 p2(2p) 2 5p3 5 8p3 1 2p3 2 5p3

5100

9

c"9 11

"9d 2

5 a3 11

3b2

(21)4 1 (21)3 1 (21)2 5 1

3(2)

22 1 25 1

limxS2

3 1 x

limxS2

(x 1 3)limxS2

(3 1 x)

t 5 !72 8 8.49

!75 5 t72 5 t2

4 51

18t2

6 5 2 11

18t2

6000 2 4000 5 2000

p(t)5 4

5 2 1 2

5 2 136

18

limtS61

p(t) 5 2 11

18(6)2

5 6

5 3 1 3

5 3 136

12

limtS62

p(t) 5 3 11

12(6)2

8 10 120

68

42

–2

10y

x

642–2–4

c 5 0.b 5 2a 5 3,

b 5 2a 5 3,

6a 5 18

4a 2 2b 5 8limxS22

f(x) 5 8

a 1 b 5 5limxS1

f(x) 5 5

c 5 0f(0) 5 0

a 2 0f(x) 5 ax2 1 bx 1 c,


d.

e.

f.

8. a.

Let Therefore, as

Here,

b.

c.

d.

e.

f.

9. a.

b.

c.

d.

e.

f.

51

32

51

4(8)

5 limxS1

1

(x 1 3)(3x 1 5)

5 limxS1

a 1

x 2 1b a3x 1 5 2 2x 2 6

(x 1 3)(3x 1 5)b

limxS1

a 1

x 2 1b a 1

x 1 32

2

3x 1 5b

5 2x

limhS0

(x 1 h)2 2 x2

h5 lim

hS0

2xh 1 h2

h

51

2

5 limxS0

"x 1 1 2 1

("x 1 1 2 1)("x 1 1 1 1)

limxS0

"x 1 1 2 1

x5 lim

xS0

"x 1 1 2 1

x 1 1 2 1

5 21

limxS21

x2 1 xx 1 1

5 limxS21

x(x 1 1)

x 1 1

16 2 16

16 2 20 1 65 0

16 2 16

64 1 645 0

51

12

lim

xS2

u 2 2

u3 2 8

limxS0

(x 1 8)13 2 2

x

51

12

5 limxS2

u 2 2

(u 2 2)(u2 1 2u 1 4)

5 lim

xS2

u 2 2

u3 2 8

limxS4

"x 2 2

"x3 2 8

51

2

5 limxS1

u 2 1

(u 2 1)(u 1 1)

5 limxS1

u 2 1

u2 2 1

limxS1

x

16 2 1

x13 2 1

51

6

5 limxS1

(u 2 1)

(u 2 1)(u5 1 u4 1 u3 1 u2 1 u 1 1)

5 limxS1

u 2 1

u6 2 1

limxS1

x

16 2 1

x 2 1

5 227

5 2 (9 1 9 1 9)

5 2limxS3

(u 2 3)(u2 1 3u 1 9)

u 2 3

5 limxS3

u3 2 27

u 2 3

limxS27

27 2 xx

13 2 3

51

12

limxS2

u 2 2

u3 2 85 lim

xS2

1

u2 1 2u 1 4

u S 2.x S 8,u3 5 xu 5 "3 x.

limxS8

"3 x 2 2

x 2 8

5 21

"7

5 limxS0

7 2 x 2 7 2 x

x("7 2 x 1 "7 1 x)

limxS0

£"7 2 x 2 "7 1 xx

3"7 2 x 1 "7 1 x

"7 2 x 1 "7 1 x§

51

4

limxS4

"x 2 2

x 2 45 lim

xS4

"x 2 2

("x 2 2)("x 1 2)

5 21

4

5 limxS0

21

2 1 "4 1 x

limxS0

£ 2 2 "4 1 xx

32 1 "4 1 x

2 1 "4 1 x§


Let

u S 3.x S 27,

x 5 u3x

13 5 u

u S 1x S 1,

x 5 u6x16 5 u,

Let

As u S 1x S 1,

x13 5 u2

u6 5 xx

16 5 u

Let

u S 2x S 4,

x32 5 u3

x12 5 u

Let

u S 2x S 0,

x 5 u3 2 8

x 1 8 5 u3

(x 1 8)13 5 u



b. does not exist.

c.

5 3

5 limxS21

x 1 1

limxS21

(x 2 2)(x 1 1)

0 x 2 2 0 5 limxS21

(x 2 2)(x 1 1)

x 2 2

limxS2

x2 2 x 2 2

0 x 2 2 0 5 limxS2

(x 2 2)(x 1 1)

0 x 2 2 0

40

2

–2

–4

4y

x

2–2–4

limxS5

2

2

2 (2x 2 5)(x 1 1)

2x 2 55 2 (x 1 1)

x ,5

20 2x 2 5 0 5 2 (2x 2 5),

limxS5

2

1

(2x 2 5)(x 1 1)

2x 2 55 x 1 1

x $5

20 2x 2 5 0 5 2x 2 5,

limxS5

2

0 2x 2 5 0 (x 1 1)

2x 2 5

80

1

–1

–2

2y

x

4–4–8

5 21

limxS52

0 x 2 5 0x 2 5

5 limxS52

2 ax 2 5

x 2 5b

5 1

limxS51

0 x 2 5 0x 2 5

5 limxS51

x 2 5

x 2 5

limxS5

0 x 2 5 0x 2 5

d. if if

11. a.

is constant, therefore T and V form a linear relationship.

b.

DVDT

51.6426

205 0.082 13

V 5DVDT

? T 1 K

DV

40

2

–2

–4

4y

x

2–2–4

limxS222

(x 1 2)(x 1 2)2

2 (x 1 2)5 0

limxS221

(x 1 2)(x 1 2)2

x 1 25 lim

xS221

(x 1 2)2 5 0

x , 225 2 (x 1 2)

x . 220 x 1 2 0 5 x 1 2

40

2

–2

–4

4y

x

2–2–4

5 23

5 limxS22

2 (x 1 1)

limxS22

(x 2 2)(x 1 1)

0 x 2 2 0 5 limxS22

2(x 2 2)(x 1 1)

(x 2 2)

DT T V DV

240 19.1482

220 20.7908

0 22.4334

20 24.0760

40 25.7186

60 27.3612

80 29.0038

20

20

20

20

20

20

1.6426

1.6426

1.6426

1.6426

1.6426

1.6426

Therefore, and

c.

d.

e.

12.

13.

a.

b.

c.

14.

a.

b.

15. and

a.

b.

16.

17.

Therefore, this limit does not exist.

40

2

–2

–4

4y

x

2–2–4

5 21

limxS12

x2 2 x2x 1 1

5 limxS12

x(x 2 1)

2x 1 1

x S 12 0 x 2 1 0 5 2x 1 1

limxS11

x2 1 0 x 2 1 021

0 x 2 1 0 5 3

x2 1 x 2 2

x 2 15

(x 1 2)(x 2 1)

x 2 1

x S 11 0 x 2 1 0 5 x 2 1

limxS1

x2 1 0 x 2 1 021

0 x 2 1 05 22

5 22 1 2

1 1 1

5 limxS0

c (x 1 1 2 2x 2 1)

(3x 1 4 2 2x 2 4)3

!3x 1 4 1 !2x 1 4

!x 1 1 1 !2x 1 1d

3!3x 1 4 1 !2x 1 4

!3x 1 4 1 !2x 1 4d

3!x 1 1 1 !2x 1 1

!3x 1 4 2 !2x 1 4

5 limxS0

c !x 1 1 2 !2x 1 1

!x 1 1 1 !2x 1 1

limxS0

!x 1 1 2 !2x 1 1

!3x 1 4 2 !2x 1 4

limxS0

f(x)

g(x)5 lim

xS0

f(x)x

g(x)x

51

2

5 0

limxS0

g(x) 5 limxS0

xag(x)

xb 5 0 3 2

limxS0

g(x)

x5 2lim

xS0

f(x)

x5 1

limxS0

f(x)

g(x)5 lim

xS0 c x

g(x)

f(x)

xd 5 0

limxS0

f(x) 5 limxS0

c f(x)

x3 x d 5 0

limxS0

f(x)

x5 1

5 1

limxS4

"3f(x) 2 2x 5 "3 3 3 2 2 3 4

5 21

5 3 2 4

5 limxS4

( f(x) 2 x)

limxS4

3 f(x)42 2 x2

f(x) 1 x5 lim

xS4

( f(x) 2 x)( f(x) 1 x)

f(x) 1 x

limxS4

3 f(x)43 5 33 5 27

limxS4

f(x) 5 3

5 7

521

3

5limxS5

(x2 2 4)

limxS5

f(x)

limxS5

x2 2 4

f(x)

12

10

8

6

4

2

020 4 6 8 10 12

T

V

limvS0

T 5 2273.145

T 5V 2 22.4334

0.082 13

V 5 0.082 13T 1 22.4334.

k 5 22.4334

V 5 22.4334 T 5 0

V 5 0.082 13T 1 K


1.6 Continuity, pp. 51–531. Anywhere that you can see breaks or jumps is aplace where the function is not continuous.2. It means that on that domain, you can trace thegraph of the function without lifting your pencil.3. point discontinuity

jump discontinuity

infinite discontinuity

4. a. makes the denominator 0.b. makes the denominator 0.c. makes the denominator 0.d. and make the denominator 0.e.

and make the denominator 0.f. The function has different one-sided limits at

5. a. The function is a polynomial, so the functionis continuous for all real numbers.b. The function is a polynomial, so the function iscontinuous for all real numbers.c.The is continuous for all real numbers except 0 and 5.d. The is continuous for all real numbers greaterthan or equal to e. The is continuous for all real numbers.f. The is continuous for all real numbers.6. is a linear function (a polynomial),and so is continuous everywhere,including 7.

The function is continuous everywhere.8.

The function is discontinuous at 9.

400 6000

2

4y

x

200

x 5 0.

40

2

–2

–4

4y

x

2–2–4

80

4

–4

–8

8y

x

4–4–8

x 5 2.

g(x)

22.

x2 2 5x 5 x(x 2 5)

x 5 3.

x 5 2x 5 23

x2 1 x 2 6 5 (x 1 3)(x 2 2)

x 5 23x 5 3

x 5 0

x 5 0

x 5 3

3 40

42

–2–4

6810

y

x

1 2–1verticalasymptote

60

42

–2

6810

y

x

2 4–2

60

42

–2

6810

y

x

2 4–2

hole


10.

Function is discontinuous at 11. Discontinuous at

12.

is continuous.

13.

a.

b. i. From the graph,

ii. From the graph,

iii. Since the one-sided limits differ, doesnot exist.c. f is not continuous since does not exist.

14. a. From the graph,b. From the graph,

c.

Thus, But, Hence f is not

continuous at (and also not continuous over).

15. The function is to be continuous at anddiscontinuous at

For to be continuous at

For to be discontinuous at

If then if then

and and

This shows that A and B can be any set of real numbers such that(1) (2) (if then if then )

and is not a solution because thenthe graph would be continuous at

16.

at at

if is continuous.

17.

a.

does not exist.limxS1

g(x)

limxS12

g(x) 5 21

limxS11

g(x) 5 1¶ lim

xS1 g(x)

g(x) 5 •x 0 x 2 1 0

x 2 1, if x 2 1

0, if x 5 1

f(x)b 5 6.a 5 21,

f(x) 5 • 2x, if 23 # x # 22

2x2 1 b, if 22 , x , 0

6, if x 5 0

a 5 21

b 5 6.x 5 0,

4a 1 b 5 2x 5 22,

f(x) 5 • 2x, if 23 # x # 22

ax2 1 b, if 22 , x , 0

6, if x 5 0

x 5 2.

B 5 22A 5 1

A , 22

B , 1,A . 22B . 1,4B 2 A 2 6

A 5 B 2 3

A , 22 A . 22

B , 1 B . 1

3B , 3 3B . 3

3B 1 3 , 6 3B 1 3 . 6

4B 2 B 1 3 , 6 4B 2 (B 2 3) . 6

4B 2 A , 6,4B 2 A . 6,

4B 2 A 2 6

B(2)2 2 A 2 3(2)

x 5 2:f(x)

A 5 B 2 3

A(1) 2 B 5 23

A(1) 2 B

1 2 25 3(1)

x 5 1:f(x)

f(x) 5 μ

Ax 2 Bx 2 2

, if x # 1

3x, if 1 , x , 2

Bx2 2 A, if x $ 2

x 5 2.

x 5 1

23 , x , 8

x 5 2

f(3) 5 2.limxS3

f(x) 5 4.

limxS32

f(x) 5 4 5 lim

xS32 f(x)

limxS32

f(x) 5 4.

f(3) 5 2.

limxS0

f(x)

limxS0

f(x)

limxS01

f(x) 5 1.

limxS02

f(x) 5 21.

40

2

–2

–4

4y

x

2–2–4

f(x) 5 •21, if x , 0

0, if x 5 0

1, if x . 0

k 5 16"k 5 4,

2 1 "k 5 6

g(x)

g(x) 5 e x 1 3, if x 2 3

2 1 !k, if x 5 3

40

2

–2

–4

4y

x

2–2–4

x 5 2

x 5 3.

5 5

5 limxS3

(x 2 3)(x 1 2)

x 2 3

limxS3

f(x) 5 limxS3

x2 2 x 2 6

x 2 3


b.



b.

c.

2. a.

b.

c.

d.

3.

a. Slope at



m s.>s(1) 2 s(0)

15 25

s(2) 5 160s(1) 5 175,s(0) 5 180,

s(t) 5 25t2 1 180

P(2, 0.5)

m 5 limhS0

2hh

5 2

5 2h f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5

f(x) 5 2x 1 1

P(2, 0.5)

P(21, 3)

5 2

5 limhS0

(2 2 h)

5 limhS0

4 2 1 1 2h 2 h2 2 3

h

m 5 limhS0

4 2 (21 1 h)2 2 3

h

f(x) 5 4 2 x2P(21, 3)

f(x) 5 e4 2 x2 , if x # 1

2x 1 1, if x . 1

5 25

4

5 limhS0

225h

h(2 1 h)(2)

5 limhS0

10 2 5(2 1 h)

h(2 1 h)(2)

m 5 limhS0

5

4 1 h 2 22

5

2

h

Pa4, 5

2bf(x) 5

5

x 2 2,

5 21

27

5 22

9(6)

5 2 limhS0

c2 1

3!9 1 h(3 1 !9 1 h)d

5 2 limhS0

c 3 2 !9 1 h3h!9 1 h

33 1 !9 1 h3 1 !9 1 h

d m 5 lim

hS0

2

!4 1 h 1 52

2

3

h

Pa4, 2

3bh(x) 5

2

!x 1 5,

51

2

5 limhS0

1

!h 1 1 1 1

5 limhS0

c !h 1 1 2 1

x3

!h 1 1 1 1

!h 1 1 1 1d

m 5 limhS0

"21 1 h 1 2 2 1

h

P(21, 1)g(x) 5 "x 1 2,

5 21

3

5 limhS0

21

3 1 h

m 53

3 1 h 2 1

h

P(2, 1)f(x) 53

x 1 1,

2x 2 y 2 5 5 0

y 2 (23) 5 2(x 2 1)

5 2

5 limhS0

2 1 h

5 limhS0

2h 1 h2

h

m 5 limhS0

5(1 1 2h 1 h2) 2 (23)

h

f(1) 5 23

5 7

m 548 2 13

4 2 (21)

f(4) 5 48f(21) 5 13,

5 23

m 521 2 36

3 2 (22)

f(3) 5 21f(22) 5 36,

x 5 1g(x)

40

2

–2

–4

4y

x

2–2–4


b.



b.

c.

2. a.

b.

c.

d.

3.

a. Slope at



m s.>s(1) 2 s(0)

15 25

s(2) 5 160s(1) 5 175,s(0) 5 180,

s(t) 5 25t2 1 180

P(2, 0.5)

m 5 limhS0

2hh

5 2

5 2h f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5

f(x) 5 2x 1 1

P(2, 0.5)

P(21, 3)

5 2

5 limhS0

(2 2 h)

5 limhS0

4 2 1 1 2h 2 h2 2 3

h

m 5 limhS0

4 2 (21 1 h)2 2 3

h

f(x) 5 4 2 x2P(21, 3)

f(x) 5 e4 2 x2 , if x # 1

2x 1 1, if x . 1

5 25

4

5 limhS0

225h

h(2 1 h)(2)

5 limhS0

10 2 5(2 1 h)

h(2 1 h)(2)

m 5 limhS0

5

4 1 h 2 22

5

2

h

Pa4, 5

2bf(x) 5

5

x 2 2,

5 21

27

5 22

9(6)

5 2 limhS0

c2 1

3!9 1 h(3 1 !9 1 h)d

5 2 limhS0

c 3 2 !9 1 h3h!9 1 h

33 1 !9 1 h3 1 !9 1 h

d m 5 lim

hS0

2

!4 1 h 1 52

2

3

h

Pa4, 2

3bh(x) 5

2

!x 1 5,

51

2

5 limhS0

1

!h 1 1 1 1

5 limhS0

c !h 1 1 2 1

x3

!h 1 1 1 1

!h 1 1 1 1d

m 5 limhS0

"21 1 h 1 2 2 1

h

P(21, 1)g(x) 5 "x 1 2,

5 21

3

5 limhS0

21

3 1 h

m 53

3 1 h 2 1

h

P(2, 1)f(x) 53

x 1 1,

2x 2 y 2 5 5 0

y 2 (23) 5 2(x 2 1)

5 2

5 limhS0

2 1 h

5 limhS0

2h 1 h2

h

m 5 limhS0

5(1 1 2h 1 h2) 2 (23)

h

f(1) 5 23

5 7

m 548 2 13

4 2 (21)

f(4) 5 48f(21) 5 13,

5 23

m 521 2 36

3 2 (22)

f(3) 5 21f(22) 5 36,

x 5 1g(x)

40

2

–2

–4

4y

x

2–2–4


Average velocity during the second second is

m s.b. At

Velocity is m s.c. Time to reach ground is when Therefore,

Velocity at

Therefore,

5. mass in gramsa. Growth during

Grew 0.0601 g during this time interval.b. Average rate of growth is

g min.

c.

Rate of growth is g min.

6. tonnes of waste,

a. At

700 000 t have accumulated up to now.b. Over the next three years, the average rate of change:

c. Present rate of change:

d.

Now,

It will take 7.5 years to reach a rate of t per year.

7. a. From the graph, the limit is 10.b. 7; 0c. is discontinuous for and 8. a. Answers will vary. f is

discontinuous at

b. if f is increasing for

40

2

–2

–4

4y

x

2–2–4

limxS31

f(x) 5 1

x . 3x , 3;f(x) 5 24

20

1

–1

–2

2y

x

1–1–2

x 5 21

limxS21

f(x) 5 0.5,

t 5 4.t 5 3p(t)

3.0 3 105

a 5 7.5

2a 1 15 5 30

(2a 1 15)104 5 3 3 105

5 (2a 1 15)104

5 limhS0

104(2a 1 h 1 15)limhS0

Q(a 1 h) 2 Q(a)

h

5104 32ah 1 h2 1 15h4

hQ(a 1 h) 2 Q(a)

h

Q(a) 5 104 3a2 1 15a 1 70415a 1 15h 1 7045 104 3a2 1 2ah 1 h2 1

Q(a 1 h)

5 15 3 104 t per year.

limhS0

Q(h) 2 Q(0)

h5 lim

hS0 104(h 1 15)

Q(0) 5 104 1 70

Q(h) 5 104(h2 1 15h 1 70)

5 18 3 104 t per year.

Q(3) 2 Q(0)

35

54 3 104

3

Q(0) 5 70 3 104

5 124 3 104

Q(3) 5 104(9 1 45 1 70)

5 700 000.

Q(t) 5 70 3 104

t 5 0,

0 # t # 10

Q(t) 5 104(t2 1 15t 1 70)

>limhS0

(6 1 h) 5 6

s(3 1 h) 2 s(3)

h5

6h 1 h2

h

s(3) 5 9

s(3 1 h) 5 9 1 6h 1 h2

>0.0601

0.015 6.01

5 9

M(3) 5 32

M(3.01) 5 (3.01)2 5 9.0601

3 # t # 3.01

M(t) 5 t2

v(6) 5 limhS0

(260 2 5h) 5 260.

s(6) 5 0

5 260h 2 5h2

s(6 1 h) 5 25(36 1 12h 1 h2) 1 180

t 5 6:

t 5 6, t . 0.

t2 5 36

25t2 1 180 5 0

s(t) 5 0.

>240

v(4) 5 limhS0

(240 2 5h) 5 240

s(4 1 h) 2 s(4)

h5

240h 2 5h2

h

5 280 2 40h 2 5h2 1 180 1 80 2 180

5 25(4 1 h)2 1 180 2 (25(16) 1 180)

s(4 1 h) 2 s(4)

t 5 4:

>s(2) 2 s(1)

15 215


9. a.

b.

Discontinuous at and c. They do not exist.10. The function is not continuous at because the function is not defined at ( makes the denominator 0.)

11.

a. f is discontinuous at and

b.

does not exist.


b.

c.

does not exist.

13. a.

b.

14.

This agrees well with the values in the table.

15. a.

f(x) 8 0.25

x 5 2.0001

f(x) 5"x 1 2 2 2

x 2 2

51

2!3

5 limxS0

1

!x 1 3 1 !3

5 limxS0

x

xA!x 1 3 1 !3 B

5 limxS0

x 1 3 2 3

xA!x 1 3 1 !3 B

limxS0

c !x 1 3 2 !3

x?!x 1 3 1 !3

!x 1 3 1 !3d

1

2

1

3

limxS23

h(x)

limxS4

h(x) 537

75 5.2857

h(x) 5x3 2 27

x2 2 9,

limxS0

g(x) 5 0g(x) 5 x(x 2 5),

limxS0

f(x)f(x) 51

x2,

limxS22

f(x)

limxS222

2

x 1 25 2`

limxS22

f(x): 5 limxS221

2

x 1 25 1`

52

3

limxS1

f(x) 5 limxS1

2

x 1 2

x 5 22.x 5 1

52(x 2 1)

(x 2 1)(x 1 2)

f(x) 52x 2 2

x2 1 x 2 2

x 5 24

x 5 24.

x 5 24

x 5 1.x 5 21

f(x) 5 • x 1 1, if x , 21

2x 1 1, if 21 # x , 1

x 2 2, if x . 1

40

2

–2

–4

4y

x

2–2–4


x 2.1 2.01 2.001 2.0001

f(x) 0.248 46 0.249 84 0.249 98 0.25

x 1.9 1.99 1.999 2.001 2.01 2.1

x 2 2x2 2 x 2 2

0.344 83 0.334 45 0.333 44 0.333 22 0.332 23 0.322 58

x 0.9 0.99 0.999 1.001 1.01 1.1

x 2 1x2 2 1

0.526 32 0.502 51 0.500 25 0.499 75 0.497 51 0.476 19

x 20.1 20.01 20.001 0.001 0.01 0.1

"x 1 3 2 "3x

0.291 12 0.288 92 0.2887 0.288 65 0.288 43 0.286 31

b.

c.

16. a.

Slope of the tangent to at is 10.

b.


c.


17. a.

b.

c.

d.

e.

f.

18. a. The function is not defined for sothere is no left-side limit.b. Even after dividing out common factors fromnumerator and denominator, there is a factor of

in the denominator; the graph has a verticalasymptote at

c.

limxS12

f(x) 5 25 2 limxS11

f(x) 5 2

f(x) 5 e25, if x , 1

2, if x $ 1

x 5 2.

x 2 2

x , 3,

5 21

4

5 limxS0

c2 1

2(2 1 x)d

5 limxS0

c1x

3 2x

2(2 1 x)d

limxS0

1

xa 1

2 1 x2

1

2b

5 21

8

521

4 1 4

521

4 1 !12 1 (4)

5 limxS4

21

4 1 !12 1 x

5 limxS4

2 (x 2 4)

(x 2 4)(4 1 !12 1 x)

5 limxS4

4 2 x

(x 2 4)(4 1 !12 1 x)

5 limxS4

16 2 (12 1 x)

(x 2 4)(4 1 !12 1 x)

limxS4

c4 2 !12 1 xx 2 4

?4 1 !12 1 x4 1 !12 1 x

d5

1

3

54

12

5(2) 1 2

(2)2 1 2(2) 1 4

5 limxS2

x 1 2

x2 1 2x 1 4

limxS2

(x 2 2)(x 1 2)

(x 2 2)(x2 1 2x 1 4)

51

!5

5 limxS0

1

A!x 1 5 1 !5 2 x B

5 limxS0

x 1 5 2 5 1 x

xA!x 1 5 1 !5 2 x B

limxS0

c !x 1 5 2 !5 2 xx

3!x 1 5 1 !5 2 x!x 1 5 1 !5 2 x

d5 10a

limxSa

(x 1 4a)2 2 25a2

x 2 a5 lim

xSa

(x 2 a)(x 1 9a)

x 2 a

5 4

5 (24) 1 8

5 limxS24

(x 1 8)limxS24

(x 1 4)(x 1 8)

x 1 4

2 116.(x 5 4)y 5

1

x

5 21

16

5 limhS0

21

4(4 1 h)

limhS0

1

4 1 h2

1

4

h5 lim

hS0

4 2 4 2 h4(4 1 h)(h)

14.x 5 4y 5 "x

51

4

5 limhS0

1

!4 1 h 1 2

limhS0

"4 1 h 2 2

h5 lim

hS0

"4 1 h 2 2

4 1 h 2 4

x 5 5y 5 x2

5 10

5 limhS0

(10 1 h)

limhS0

(5 1 h)2 2 25

h

51

45 0.25

5 limxS2

1

!x 1 2 1 2

limxS2

c !x 1 2 2 2

x 2 23

!x 1 2 1 2

!x 1 2 1 2d

limxS2

f(x) 5 0.25



e.

f.


19. a.


b.


c.


d.


20.a.

b.




2.

Slope of secant is

5 213

36 1 3

22 2 15 2

39

3

f(1) 5 5 2 8 5 23

f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36

f(x) 5 5x2 2 8x

limxS11

1

x 2 15 1` 2 lim

xS12

1

x 2 15 2`.

limxS1

1

x 2 1

109 000>h5 109

5 limhS0

(109 1 3h)

5 limhS0

109h 1 3h2

h

5 limhS0

20 1 488 1 61h 1 192 1 48h 1 3h2 2 700

h

5 limhS0

20 1 488 1 61h 1 3(64 1 16h 1 h2) 2 700

h

limhS0

20 1 61(8 1 h) 1 3(8 1 h)2 2 (20 1 488 1 192)

h

5 700 000

P(8) 5 20 1 61(8) 1 3(8)2

P(t) 5 20 1 61t 1 3t2

y 5 2216x 1 486

y 2 (2162) 5 2216(x 2 3)

x 5 3, y 5 2162.

5 2216

5 limhS0

( 2 216 2 108h 2 24h2 2 2h3)

5 limhS0

2216h 2 108h2 2 24h3 2 2h4

h

5 limhS0

22(81 1 108h 1 54h2 1 12h3 1 h4) 1 162

h

m 5 limhS0

22(3 1 h)4 2 (2162)

h

y 5 18x 1 9

y 2 (29) 5 18(x 2 (21))

x 5 21, y 5 29.

5 18

5 limhS0

(18 2 18h 1 6h2)

5 limhS0

18h 2 18h2 1 6h3

h

5 limhS0

6(21 1 3h 2 3h2 1 h3) 2 3 1 9

h

m 5 limhS0

6(21 1 h)3 2 3 2 (26 2 3)

h

y 5 25x 2 5

y 2 5 5 25(x 1 2)

x 5 22, y 5 5.

5 25

5 limhS0

(25 1 h)

5 limhS0

25h 1 h2

h

5 limhS0

4 2 4h 1 h2 1 2 2 h 2 1 2 5

h

m 5 limhS0

(22 1 h)2 2 (22 1 h) 2 1 2 (4 1 2 2 1)

h

y 5 7

y 2 7 5 0(x 2 1)

x 5 1, y 5 7.

5 0

5 limhS0

2h

5 limhS0

2h2

h

5 limhS0

23 2 6h 2 h2 1 6 1 6h 1 4 2 7

h

m 5 limhS0

23(1 1 h)2 1 6(1 1 h) 1 4 2 (23 1 6 1 4)

h

limxS21

f(x)

limxS211

f(x) 2 limxS212

f(x)

limxS212

f(x) 5 5

limxS211

f(x) 5 21

f(x) 5 e 5x2, if x , 21

2x 1 1, if x $ 21

limxS01

0 x 0x

2 limxS02

0 x 0x

limxS01

0 x 0x

5 1

limxS02

0 x 0x

5 21

x S 02 0 x 0 5 2x

limxS0

0 x 0x

x 5 2.



e.

f.


19. a.


b.


c.


d.


20.a.

b.




2.

Slope of secant is

5 213

36 1 3

22 2 15 2

39

3

f(1) 5 5 2 8 5 23

f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36

f(x) 5 5x2 2 8x

limxS11

1

x 2 15 1` 2 lim

xS12

1

x 2 15 2`.

limxS1

1

x 2 1

109 000>h5 109

5 limhS0

(109 1 3h)

5 limhS0

109h 1 3h2

h

5 limhS0

20 1 488 1 61h 1 192 1 48h 1 3h2 2 700

h

5 limhS0

20 1 488 1 61h 1 3(64 1 16h 1 h2) 2 700

h

limhS0

20 1 61(8 1 h) 1 3(8 1 h)2 2 (20 1 488 1 192)

h

5 700 000

P(8) 5 20 1 61(8) 1 3(8)2

P(t) 5 20 1 61t 1 3t2

y 5 2216x 1 486

y 2 (2162) 5 2216(x 2 3)

x 5 3, y 5 2162.

5 2216

5 limhS0

( 2 216 2 108h 2 24h2 2 2h3)

5 limhS0

2216h 2 108h2 2 24h3 2 2h4

h

5 limhS0

22(81 1 108h 1 54h2 1 12h3 1 h4) 1 162

h

m 5 limhS0

22(3 1 h)4 2 (2162)

h

y 5 18x 1 9

y 2 (29) 5 18(x 2 (21))

x 5 21, y 5 29.

5 18

5 limhS0

(18 2 18h 1 6h2)

5 limhS0

18h 2 18h2 1 6h3

h

5 limhS0

6(21 1 3h 2 3h2 1 h3) 2 3 1 9

h

m 5 limhS0

6(21 1 h)3 2 3 2 (26 2 3)

h

y 5 25x 2 5

y 2 5 5 25(x 1 2)

x 5 22, y 5 5.

5 25

5 limhS0

(25 1 h)

5 limhS0

25h 1 h2

h

5 limhS0

4 2 4h 1 h2 1 2 2 h 2 1 2 5

h

m 5 limhS0

(22 1 h)2 2 (22 1 h) 2 1 2 (4 1 2 2 1)

h

y 5 7

y 2 7 5 0(x 2 1)

x 5 1, y 5 7.

5 0

5 limhS0

2h

5 limhS0

2h2

h

5 limhS0

23 2 6h 2 h2 1 6 1 6h 1 4 2 7

h

m 5 limhS0

23(1 1 h)2 1 6(1 1 h) 1 4 2 (23 1 6 1 4)

h

limxS21

f(x)

limxS211

f(x) 2 limxS212

f(x)

limxS212

f(x) 5 5

limxS211

f(x) 5 21

f(x) 5 e 5x2, if x , 21

2x 1 1, if x $ 21

limxS01

0 x 0x

2 limxS02

0 x 0x

limxS01

0 x 0x

5 1

limxS02

0 x 0x

5 21

x S 02 0 x 0 5 2x

limxS0

0 x 0x

x 5 2.



b.

c.

d. f is discontinuous at and 4. a. Average velocity from to

Average velocity from to is 1 km h.b.

Velocity at is 2 km h.5.Average rate of change from to

6.

Slope of the tangent at

Slope of the tangent at is

7. a.

b.

c.

d.

e.

f.

8.

is continuous.Therefore,

b 5 218

5

5b 5 218 25 1 5b 1 a 5 8

a 5 15a 1 3 5 8

f(x)

f(x) 5 • ax 1 3, if x . 5

8, if x 5 5

x2 1 bx 1 a, if x , 5

51

12

51

4 1 4 1 4

5 limxS0

(x 1 8)13 2 2

((x 1 8)13 2 2)((x 1 8)

23 1 2(x 1 8)

13 1 4)

limxS0

(x 1 8)13

2 2

x5 lim

xS0

(x 1 8)13

2 2

(x 1 8) 2 8

51

6

5 limxS3

1

x 1 3

limxS3

a 1

x 2 32

6

x2 2 9b 5 lim

xS3

(x 1 3) 2 6

(x 2 3)(x 1 3)

5 23

4

53

22(2)

limxS21

x3 1 1

x4 2 15 lim

xS21

(x 1 1)(x2 2 x 1 1)

(x 2 1)(x 1 1)(x2 1 1)

5 4

5 limxS5

A!x 2 1 2 2B A!x 2 1 1 2B!x 2 1 2 2

limxS5

x 2 5

!x 2 1 2 25 lim

xS5

(x 2 1) 2 4

!x 2 1 2 2

57

5

limxS2

2x2 2 x 2 6

3x2 2 7x 1 25 lim

xS2

(2x 1 3)(x 2 2)

(x 2 2)(3x 2 1)

5 12

limxS3

4x2 2 36

2x 2 65 lim

xS3

2(x 2 3)(x 1 3)

(x 2 3)

231.x 5 4

5 231

limhS0

f(4 1 h) 2 f(4)

h5 lim

hS0 (231 2 4h)

1 1 2h 1 h2

5 231h 2 4h2

(1 1 2h 1 h2)

54 1 h 2 4 2 32h 2 4h2

1 1 2h 1 h2

f(4 1 h) 2 f(4) 54 1 h

1 1 8h 1 h2 2 4

f(4) 54

1

54 1 h

1 1 8h 1 h2

f(4 1 h) 54 1 h

(4 1 h)2 2 15

x 5 4:

f(x) 5x

x2 2 15

5"16 1 h 2 "16

h

f(5 1 h) 2 f(5)

h

x 5 5 1 h:x 5 5

f(x) 5 "x 1 11

>t 5 3

v(3) 5 limhS0

2h 2 h2

h5 2

5 2h 2 h2

5 24 1 8h 2 9 2 6h 2 h2 2 15

5 8(3 1 h) 2 (3 1 h)2 2 (24 2 9)

s(3 1 h) 2 s(3)

>t 5 5t 5 2

5 1

515 2 12

3

s(5) 2 s(2)

35

(40 2 25) 2 (16 2 4)

3

t 5 5:t 5 2

x 5 2.x 5 1

limxS42

f(x) 5 1

limxS2

f(x) 5 1

limxS1

f(x)


Date post:	05-Nov-2019
Category:	Documents
Upload:	others
View:	0 times
Download:	0 times

C 1 Introduction to Calculus - Ms. Ma's...

Documents