Calculus for the Life Sciences Marvin L. Bittinger Solution Manuals.pdf (19)

Type:

Solution Manual

Resource:

Calculus for the Life Sciences

Edition: Author(s):

1st Edition Marvin L. Bittinger Neal Brand John Quintanilla

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Contents Chapter 1 Functions and Graphs…………………………………………….1 Chapter 2 Differentiation…………………………………………………..37 Chapter 3 Applications of Differentiation………………………………….87 Chapter 4 Exponential and Logarithmic Functions……………………….209 Chapter 5 Integration……………………………………………………...253 Chapter 6 Matrices………………………………………………………..323 Chapter 7 Functions of Several Variables………………………………...349 Chapter 8 First-Order Differential Equations……………………………..375 Chapter 9 Higher-Order and Systems of Differential Equations…………409 Chapter 10 Probability……………………………………………………455

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Chapter 1

Functions and Graphs Exercise Set 1.1 4. A vertical line at x=10 3

1. Graph y = −4.

Note that y is constant and therefore any value of x we choose will yield the same value for y, which is−4. Thus, we will have a horizontal line at y = −4.

2 y 1 0

2

4

x 6

8

10

–1

4 –2

y 2

–4

–2

0

–3

2 x

4

5. Graph. Find the slope and the y-intercept of y = 3x. −

–2

First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph.

–4

Whenx=0,y=

−3(0) 0, ordered pair (0, −3(1)==3,orderedpair(1, 3) 0) − − Whenx=1,y= When x = −1, y = 3( 1) = 3, ordered pair ( 1,3) − −

2. Horizontal line at y = −3.5

4

4

y 2

y 2

–4

–2

0

2 x

4

–4

–2

2 x

4

–4

–4

3. Graph x = −4.5.

Note that x is constant and therefore any value of y we choose will yield the same value for x, which is 4.5. Thus, we will have a vertical line at x = −4.5.

Compare the equation y = −3x to the general linear equation form of y = mx + b to conclude the equation has a slope of m = −3 and a y-intercept of (0, 0).

Slope of m = 6. −0.5 and y-intercept of (0,0) 4

2

y 2

1

–4 –4x

0 –2

–2

–6

−

–2

–2

0

2

–2 –1

–4 –2

2 x

4

Chapter 1:

Functions and Graphs −2x + 3 to the general linear equation form of y = mx + b to conclude the equation has a slope of Compare the equation y =m = 2 and a y-intercept of (0,3).

2 7. Graph. Find the slope and the y-intercept of y = 0.5x.

First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph. When x = 0, y = 0.5(0) = 0, ordered pair (0, 0) When x = 6, y = 0.5(6) = 3, ordered pair (6, 3)

− 10. Slope of m = 1 − and y-intercept of (0,4) 6

When x = −2, y = 0.5(−2) = −1, ordered pair ( 2, 1) − −

4 y

4

2

y 2

–4

–2

0

–2

6

–4

11.

Compare the equation y = 0.5x to the general linear equa- tion form of y = mx + b to conclude the equation has a slope of m = 0.5 and a y-intercept of (0, 0).

8. Slope of m = 3 and y-intercept of (0, 0)

Graph. Find the slope and the y-intercept of y = x−2. − First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph.

When x = 0, y = (0)−2 = − 2,−ordered pair (0, 2)

−

When x = −2,y= ( 2)−2=0,orderedpair( 2,0) − −

−

When x = 3, y = (3)−2 = − 5,−ordered pair (3, 5)

4

4

y 2

y 2

0

4

4

–4

–2

x

–2

2 x

–2

–4

2

2 x

4

–4

–2

0

–2

–2

–4

–4

9. Graph. Find the slope and the y-intercept of y =−2x+3.

First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph. When x = 0, y = 2(0) + 3 = 3, ordered pair (0,3)

− − − 2(2)+3= −1, ordered pair (2, 1) When x = −2,y= 2( 2)+3=7,orderedpair( 2,7) − − −

Compare the equation y =

4

−x − 2 to the general linear

equation form of y = mx + b to conclude the −2).equation has − a slope of m = 1 and a y-intercept of (0, 12. Slope of m = 3 − and y-intercept of (0,2)

Whenx=2,y=

4 y 2

4 –4

y 2

2 x

–2

0

2 x

4

–2 –4

–2

0

2 x

4

–4

–2 –4

−

13. Find the slope and y-intercept of 2x + y − 2=0.

Exercise Set 1.1

3

Solve the equation for y.

22.

− 2 = y =

2x+y

0 −2x + 2

Compare to y = mx + b to conclude the equation has a slope of m = −2 and a y-intercept of (0, 2).

14. y = 2x + 3, slope of m = 2 and y-intercept of (0, 3)

y=

Plug the given information into the equation y m(x − x1) and solve for y

−2x − 5 5 −x−2

24. y

y

Compare to y = mx + b to conclude the equation has a slope of m = 12 and a y-intercept of (0, −4). −4andy-interceptof(0,4)

3

26. y = 43 x + 7 27. Find the equation of line: with m = 0, containing (2,3) Plug the given information into the equation y m(x − x1) and solve for y

19. Find the equation of the line: with m = −5, −y1containing = m(x−x1) (1, −5) Plug the given information into equation y y −y1 = m(x−x1)

20.

y

=

y

=

−5(x−1) −5x +5 −5x+5−5 −5x

28. y −8 = 0(x−4) −8 = 0 y y = 8 29. Find the slope given ( −4,−2) and (−2,1)

y −7 −7 = = 7(x−1) 7x−7 y

y =

x Use the slope equation m = y2−y1 −x.1 NOTE: Itdoesnot 2 matter which point is chosen as (x1, y1) and which is chosen as (x2,y2) as long as the order the point coordinates are subtracted in the same order as illustrated below

7x

21. Find the equation of line: with m = −2, containing (2, 3) Plug the given information into the equation y − x1) and solve for y m(x y −3== − 2(x−2) −3 2x+4 y = − y = − 2x+4+3 y − 2x+7

−y1=

y −3 = 0(x−2) −3 = 0 y y = 3

and solve for y −(−5) = y+5 =

−6) and

y = mx+b 1 2 x + ( −6) y = 1 2 x −6 y =

x = 2y+8

y

−5(x−5) −5x + 25

m=1 Plug the given information into the equation y = mx + b 2

17. Find the slope and y-intercept of x = 2y + 8.

Solve the equation for y.

1

−0 = y =

25. Find the equation of line: with y-intercept (0,

16. y = x + 2, slope of m = 1 and y-intercept of (0, 2).

4 18. y = −1x+3,slopeofm= 4

−y1=

y −0 = 2(x−3) y = 2x − 6

Compare to y = mx + b to conclude the equation has a slope of m = −1 and a y-intercept of (0,−52).

x x −8 = 2y 1 −4 = 2

−3(x−5) −3x + 15

23. Find the equation of line: with m = 2, containing (3,0)

15. Find the slope and y-intercept of 2x + 2y + 5 = 0.

Solve the equation for y.

2x+2y+5 = 0 2y = y =

−(−2) = y+2 = −3x + 13

y

−y1=

m = = =

1 −(−2) −2 − (−4) 1+2 −2+4 3 2

4

Chapter 1:

Functions and Graphs

39. Find the slope given (x, 2x + 3) and (x + h, 2(x + h) + 3) m = = =

30. m=

6

3−1

2

−2−1) −4 − (−2) −3 −2 3 2

1

−(−2) = 8 = 4 1 52)and(35) 31. Find the slope given (2, −,4 m = = = = = =

4 − 12 5

−3 − 25 8 10 − 510 −15 5 −10 5

= 3x+3h−1−3x+1 =h=3 40 m= [3(x+h)−1]−(3x−1) x+h h 3h

−x

.

Find equation of line containing (−4,−2) and (−2,1)

41.

From 29, that the slope of the line is 3. − we know Using Exercise the point( −2,1)andthevalueoftheslopeinthe y1 = m(x − x1) and solving for y point-slope formula y

2

we get:

3 10 17 5

y y y

3 1 0 5 · 1 7 15 170 3 34 −3−5

3 1 = (x−(−2)) −1 = 3 (x+2) −1 = 2 x+3 yy == 3 x x+ +3 4+1 2 3 2 3 2

−

−3

32. m= − 11 6−6 ( − 3 ) =11 6 = − 316· 4 1= − 34 2

[2(x + h) + 3] −(2x+3) x+h − x 2x+2h+3 − 2 x − 3 h 2h h 2

m = = = =

4

33. Find the slope given (3, −7) and (3,−9) m = =

−9−(−7) 3− 3 − 2 undefinedquantity 0

NOTE: You could use either of the given points and you would reach the final equation. =1 42. Using m

4 and the point (6,3) 1

(x−6) 4 1 x− 6 4 4 3 1 4 x − 2 +3 3 1 4x + 2

y = y −3 −3 =

This line has no slope = y y =

10−2 8 34. m= −4−(−4) = 0 This line has no slope

35. Find the slope given (2,3) and ( −1, 3) −3 3 m = = −1−2 = 0 −3 0 1

1

0 2 − 2 = −1 = 0 36. m = −7−(−6) Find the slope given (x, 3x) and (x + h, 3(x + h)) 37.

m = = = =

38. m=4(x+h)−4x x+h −x =

3(x + h) − 3 x x+h − x 3x + 3h − 3 x h 3h h 3

4x+4h−4x h

4h =h=4

43. Find equation of line containing (2 − 4, 5 ) 5 ,1) 2 and ( 3 From Exercise 31, we know that the slope of the line is 4 − 334andusi ngthepoint (3,5) − y y y

−4 5 4 − 5 4 − 5 y y y

= = = = = =

3

(x−(−3)) 34(x+3) 3 − 34 3 9 − x− 34 34 9 4 3 − 34x − 34+ 5 − 3 45 136 + − x 34 170 170 3 91 − x + 170 34 −

5

Exercise Set 1.1 44. Using m = −13 4andthepoint y−5 8 5 y− 8

= = =

y = y = y

, (−35) 4(8

−13 ( − 3) ) x− 4 −13x−39 4 −13x−39 5+ 4 −13x−39 + 10 4 −13x−29 4

45. Find equation of line containing (3, −7) and (3,−9) From Exercise 33, we found that the line containing (3,−7) −9) has no slope. We notice that the x-coordinate and (3,

52. Using m = 3 and the point (x, 3x −1) y −(3x−1) −(3x−1) = = 3(x−x) 0 y

Change in Life expectancy Change in Time −73.7 − 1990 76.9 2000 3.2 10 = 0.32 peryear

Rate = = =

47. Find equation of line containing (2,3) and (

58.

48. Since the line has a slope of m = 0, it is horizontal. The equation of the line is y = 1 2 49.

Find equation of line containing (x, 3x) and (x+h, 3(x+h))

y −3x −3x = = 3(x−x) 3(0) −3x = =0 y y =

3x

50. Using m = 4 and the point (x, 4x) y −4x −4x = = 4(x−x) 0 y

y =

4x

51. Find equation of line containing (x, 2x+3) and (x+h, 2(x+ h) + 3) From Exercise 37, we found that the line containing (x, 2x + 3) and (x + h, 2(x + h) + 3) had a slope of m = 2.

Using the point (x,3x) and the value of the slope in the point-slope formula y −(2x+3) = 2(x−x) −(2x+3) −(2x+3) = = 2(0) 0 y y

y = 2x+3

o a)F( −10)=9 5·−10)+32=−18+32=14F ( F(0)=9 ·(0)+ 32=0+32=32F o 5 F (10) = 9 ·(10)+32=18+32=50o F 5 F(40) = 9 (40) + 32 = 72 + 32 = 104o F 5· · o b)F(30)=9 (30)+ 32=54+32=86F c) Same temperature in both means F(x) = x. So 5

F(x) = x x+32 = x 9 5 9 −32 5x − x = = = −32 4 = 5x x x 5 −32 · 4 −40o

From Exercise 37, we found that the line containing (x, 3x) and (x+h, 3(x+h)) had a slope of m = 3. Using the point (x, 3x) and the value of the slope in the point-slope formula

y

3x − 1

5 = 0.08. This means the treadmill has a grade 53. Slope = 0.4 of 8%.

54.

The roof has a slope of 2.6≈ 0.3171, or 31.71% 55. 6.2 The slope (or head) of the river is 43.33 1238 = 0.035 = 3.5% 56.

The stairs have a maximum grade of 8.259 =0.916 ≈ 0.9167 = 91.67% The average rate of change of life 57. expectancy at birth is computed by finding the slope of the line containing the two points (1990, 73.7) and (2000, 76.9), which is given by

does not change regardless of the y-value. Therefore, the line in vertical and has the equation x = 3.

46.

Since the the line is line x = has −4. no slope, it is vertical. The equation of −1, 3) From Exercise 35, we found that the line containing (2, 3) and( has a slope of m = 0. We notice that the −1,3) y-coordinate does not change regardless of the x-value.

Therefore, the line in horizontal and has the equation y=3.

y =

59.

a) Since R and T are directly proportional we can write that R = kT , where k is a constant of proportionality.

Using R = 12.51 when T = 3 we can find k.

R = kT 12.51 = k(3) 12.51 = k 3 4.17 = k Thus, we can write the equation of variation as R = 4.17T b) This is the same as asking: find R when T = 6. So, we use the variation equation R = 4.17T = 4.17(6) = 25.02

Chapter 1:

6 60.

We need to find t when D = 6.

64.

a)

D = 293t 6 = 293t

6 293 = ≈ 0.0205 seconds 61.

D(5) D(10) D(20) D(50) D(65)

tt

a) Since B s directly proportional to W we can write B = kW. b) When W = 200 B = 5 means that B = kW 5 = k(200) 5 200 = k 0.025 = k 2.5% = k This means that the weight of the brain is 2.5% the weight of the person. c) Find B when W = 120

65.

1 ·0+5 5 1 = 10 =0.5ft 1 10 115 10 =11.5ft 1 ·10+5 = = 225 =22.5ft 10 = 1 = 10 =55.5ft =72ft 555 1 ·20+5 10 720 1 10 10 1 ·50+5 1 10 1 ·65+5 10

a) M(x) = 2.89x+70.64 M(26) = 2.89(26)+70.64 = 75.14+70.64 = 145.78

= 0.025(120lbs) = 3lbs

The male was 145.78 cm tall.

a)

b) F(x) = 2.75x+71.48 F(26) = 2.75(26)+71.48

M = kW 80 = k(200) 0.4 = k

= 71.5+71.48 = 142.98

Thus, the equation of variation is M = 0.4W b) k = 0.4 = 40% means that 40% of the body weight is the weight of muscles. c)

The female was 142.98 cm tall.

66.

M = 0.4(120) = 48lb 63.

= = = = =

b) c) Since cars cannot have negative speed, and since the car will not need to stop if it has speed of 0 then the domain is any positive real number. NOTE: The domain will have an upper bound since cars have a top speed limit, depending on the make and model of the car.

B = 0.025W

62.

Functions and Graphs

a) The equation of variation is given by N = P + 0.02P = 1.02P. b) N = 1.02(200000) = 204000 c) 367200 = 1.02P 367200 1.02 = P

a) D(0) = 2(0)+115=0+115ft D( −20) = 2(−20)+115=−40+115=75ft D(10) = 2(10)+115=20+115=135ft D(32) = 2(32)+115=64+115=179ft b) The stopping distance has to be a non-negative value.

Therefore we need to solve the inequality 0 ≤ 2F+115 ≤ 2F −115 ≤F −57.5 The 32o limit comes from the fact that for any temperature above that there would be no ice. Thus, the domain of the function is restricted in the interval [−57.5, 32].

360000 = P 67.

a) A(0) = 0.08(0)+19.7=0+19.7=19.7 A(1) = 0.08(1)+19.7=0.08+19.7=19.78 A(10) = 0.08(10)+19.7=0.8+19.7=20.5 A(30) = 0.08(30)+19.7=2.4+19.7=22.1 A(50) = 0.08(50)+19.7=4+19.7=23.7 b) First we find the value of t4, which is 2003 − 1950 = 53. So, we have to find A(53).

A(53) = 0.08(53) + 19.7 = 4.24 + 19.8 = 23.94 The median age of women at first marriage in the year 2003 is 23.94 years.

7

Exercise Set 1.2 3. y = x 2 and y=(x

c)At ()=008t+19.7 .

− 1)2 120 100

23

80 22

60 21

40 20

20 0

10

20

t

30

40

50

68. The use of the slope-intercept equation or the point-slope equation depends on the problem. If the problem gives the slope and the y-intercept then one should use the slopeintercept equation. If the problem gives the slope and a point that falls on the line, or two points that fall on the line then the point-slope equation should be used.

–10 –8

–6

–4

4. y = x 2 and y=(x

–2

and y=

4 x 6

8

10

2

4 x 6

8

10

2

4 x 6

8

10

2

4 x 6

8

10

− 3)2 160 140 120 100 80

Exercise Set 1.2 1 1. y= x2 2

2

60 1 2 − 2x

40 20

40

–10 –8

–6

–4

–2 0

5. y = x 2 and y=(x+1)2

20

120

–10 –8

–6

–4

–2 0

2

4 x 6

8

10

100

–20

80 60

–40

40

2. y = 14 x 2 and y = − 14 x 2

20

20

–10 –8

–6

–4

–2

6. y = x 2 and y=(x+3)2

10

160

–10 –8

–6

–4

–2 0

2

4 x 6

8

10

140 120

–10

100

–20

80 60 40 20 –10 –8

–6

–4

–2 0

Chapter 1:

8 7. y = x 3 and y = x 3 + 1

13. y = x 2 − 4x+3

10

14

8

12

6 y 4

10 y

2 –4

–2

0 –2 –4

2 x

8 6 4

4

–6

–4

–8

–2 2 0

–10

8. y = x 3 and y = x 3 − 1

14. y = x 2 −

4 x 6

8

10

2

4 x 6

8

10

2 x

4

6x+5 14 12

8

10

6 y 4

y8 6 4

2 –2

2

–2 –4

10

–4

Functions and Graphs

0 –2 –4

2 x

4

2 0

–6

–4

–8

–2 –4

–2

–10

9. Since the equation has the form ax2 + bx + c, with a = 0, the graph of the function is a parabola. The x-value of the vertex is given by

15. y = − x 2 +2x − 1 –2 2 –4

4 x = − b =− =−2 2a 2(1)

0 –2 –4

The y-value of the vertex is given by y = =

–6 –8 y –10

( −2)2 + 4(−2) − 7 4− 8 − 7 −11

= Therefore, the vertex is ( −2, 11).

10. Since the equation is not in the form of ax2 + bx + c, the graph of the function is not a parabola.

11. Since the equation is not in the form of ax2 + bx + c, the graph of the function is not a parabola.

12. Since the equation has the form ax2 + bx + c, with a = 0,

–12 –14

16. y = − x 2 − x + 6 8 6 y4

the graph of the function is a parabola. The x-value of the vertex is given by 6 x = − b = − − =1 2a 2(3) The y-value of the vertex is given by y = 3(1)2 6(1) − = =

3− 6

−3

Therefore, the vertex is (1, − 3).

2 –4

–2

0 –2 –4 –6 –8

2 x

4

9 − 2x − 2 = 0, then use the quadratic formula, with

Exercise Set 1.2 17. y = 2x2 + 4x

x2 a=1,b=

− 7 10 8 6 y 4

x= =

2 –4

0 –2 –4

–2

−2, and c = −2, to solve for x. −b±√b2−4ac x = 2a

2 x

===

4

=

–6 –8 –10

18. y=3x

2

− 9x+2 8 6 y 4 2 0 –2 –4

–2

2 x

–8

19. y = 12 x 2 +3x − 5 10 6 y 4 20 –2

–2

–4

− (2) −

(−2)2−4(1)(−4) 2(1)

y = 2

4

–6

y = =

–8 –10

= 20. y = 13 x 2 +4x − 2

=

10

=

5 –4

–2

= 2

4

6

x8

10

4=0

23. Use the quadratic formula, with a = 3, b = 8, and c = 2, to solve for y.

8

–6 x –4

2 ±√ittenas x2−x−

± √ 4+16 = 2 ±√2 ±√ = 2 20 2(1 5) = 2 25 = 1 √ ± Thesolutionsare1+ −√ √ 5 Solve 3y2 + 8y + 2 = 0 5and1

–10

–8

2 ± √ 4+8 ±√2 2 12 2 2 2 3 ±√ 2 2(1 ± √ 3)

22. x2 −2x+1=5canberewr x=

4

–6

–10

± √ (−2)2−4(1)(−2) 2(1)

√2 1 ± 3 Thesolutionsare1+ −√ √ 3and1 3

10

–4

− (2) −

12

14

0 –5 y

Thesolutionsare

p = 21. Solve x2 −2x=2 Write the equation so that one side equals zero, that is

−±√2 a 8 (8)2−4(3)(2) −±√2(3) 8 64−24 −±√6 8 40 6 −±√ 8 2 10 6 −±√ 2(4 10) −±√6 4 10 3 √4+10 3

and

− 4 − √10

24. 2p2−5p=1canberewritt ±√enas2

–10 –15

−

−b±√b2−4ac

= =

− (5) − 5 ± √ 25+8 ±√4 5 33 4

3

p2−p− 5

(−5)2−4(2)(−1) 2(2)

1

10

Chapter 1: Thesolutionsa re

√ 5+33and5−√ 4

4

25. Solve x2 −2x+10=0 Using the quadratic formula with a = 1, b = c=10 x = x= = === =

28. x2+4x=3canberewrit−±√tenas

33

−2, and

− ± √ b2 4ac b − 2a (2) − − ± √ (2)2 − − 4(1)(10) 2(1) 2 ± √ 4− 40 2 2 ± √ − 36 2 2 ± 6i 2 2(1 3i)

± 1 ± 3i 2 The solutions are 1 + 3i and 1 3i − 26. x = x= = === =

− ± √ b2 4ac b − √2a (6)2 − 4(1)(10) − 6± 2(1) − 6± √ 36 40 − 2 −± 6√ − 4 62 2i − 2± 2( 3 i)

− ± − 3± i 2 Thesolutionsare 3+iand 3i −

x2 +4 x − 3=0

(4)2−4(1)(−3) 2(1)

−4 ±√

16+12 −±√2 −±√ = 4 28 2(2 = 2 2 √ = −2± 7 =

The solutions are − 2+

√

7and

−−√ 2

7)

7

29. Solvex2+4x+8=0 Using the quadratic formula with a = 1, b = 4, and c = 8 −b±√b2−4ac x = −±√2 a 4 (4)2−4(1)(8) x= −±√2(1) 16−32 4 = −±√2 = 4 −16 2 = −4±4i = 2 = 4(1 ± i ) 2 ±i)=2±2i 2(1 The solutions are 2 + 2i and 2 − 2 i 30. − 10 ± √ (10)2−4(1)(27) ±√2(1) 100−108 − 10 ±√2 −8 − 10 2 √ − 10 ± 2i2

x= = =

− −

27. Solve x2 + 6x = 1 Write the equation so that one side equals zero, that is x2 + 6x − 1 = 0, then use the quadratic formula, with a = 1, b = 6, and c = 1,−to solve for x. − ± √ b2 4ac b − x = √ 2 a x= (6)2 − 4(1)(1) − − 6± = 2(1) === − 6± √ 36+4 = 240 6 − ± √ 2 − 6 ± 2√ 10 2 2( −3 ± √ 10) 2 √ 10 3 − ± − − − √ Thesolutionsare 3+ 10and 3 10 √

4

x=

Functions and Graphs

= = =

2 − ± √ i2) 2(5 2 √ − 5 ±√i 2

The solutions are − 5 +i2and

− 5+i2

√

31. Solve 4x2 = 4x − 1 Write the equation so that one side equals zero, that is 4x2 − 4x − 1 = 0, then use the quadratic formula, with a=4,b= −4, and c = −1, to solve for x. −b±√b2−4ac x = 2a x= = =

− ( − 4) ± √ (−4)2−4(4)(−1) 2(4) 4 ± √ 16+16 4

±√8 32 8

Exercise Set 1.2

11 ± √ 4 4 2 4(1 8 ± √ 2) 8 ± √2 1 2

= = =

Therefore,theaveragepayrollwillbe$100million isthe,1985+23.3111=2007.3111,2007-08season.

NOTE:Wecouldnotchoosethenegativeoptionof thequadraticformulasinceitwouldresultintheresult that is negative which corresponds to a year be-

√

Thesolutionsare1+2and1−√2 2

2

fore1985andthatdoesnotmakesense.

32. −4x2 = 4x − 1 can be rewritten as 0 = 4 2x+4x−1 x

4 (4)2−4(4)(−1) = − ± √ 2(4) = = = − 4± √ 16+16 8 −1±√ − ± √ 4) 4 32 4( = 8 2 8 − ± √ 1 2

35.Solve50=9.41−0.19x+0.09x2.First,letusrewritethe equationas0= −40.59 − 0.19x + 0.09x2 then we can use thequadraticformula √tosolveforx x

= =

33. Findf(7),f(10),andf(12)

= = = = = = = = = =

f(12)

happenduringthe,1990+22.3183=2012.31832012-13 season.NOTE:Wecouldnotchoosethenegativeoption ofthequadraticformulasinceitwouldresultintheresult thatisnegativewhichcorrespondstoayearbefore1990 andthatdoesnotmakephysicalsense.

36.

4.8565+6.8184+102.7584 1144333 .

Theaveragepayrollfor2009-10is$114.4333million b) Solve100=4.8565+0.2841x+0.1784x2. First,let

= =

= =

0.0728(72)2−6.986(72)+289 163.4032pounds

b)Solve170=0.0728h2−6.986h+289,whichcanbe writtenas0.0728h2−±√6.986h+119=0 h = = =

− (6.986) − 6986 .

±√

(−6.986)2−4(0.0728)(119) 2(00728) .

14.1514 0.1456 6.986±3.7618) 0.1456

two 6.986 −3.7618 The possible answers are = 0.1456 .1440 in, which is out side of the domain of the function,and6.986+3.7618 0.1456 =73.8173in,whichisin thedomainintervalofthefunctionw.Therefore,the manisabout73.8inchestall.

37. f(x) = x3 − x 2

us rewrite the equation as 0 = −95.1435+0.2841x+ 0.1784x2thenwecanusethequadraticformulato solveforx x

a) w(72)

a)x=2009 −1985=24 f(24) = 4.8565+0.2841(24)+0.1784(24)2 = =

2 (019)−4(009)(4059) . −. −.

2(0.09) 019 ± √ 0.19 ± √ .00361+146124 14.6485 . . = 0.19 0.18 0.18 ±3.8273 0.18 0.19 0.18 + 3.8273 =22.3183

. −(−019)±

Therefore,theaveragepriceofaticketwillbe$50will

1 1 1 6 (10)3+ 2 (10)2+ 2 (10) 1000 6 +50+5 500 150 15 3 + 3 + 3 665 3 ≈221.6≈222oranges 1 (12) 1 1 6 (12)3+ 2 (12)2+ 2 288+72+6 366oranges

f(10)

34.

1 1 1 (7)3+ (7)2+ (7) 2 2 6 7 343 + 49 + 2 6 + 2 343 147 21 6 6 + 6 511 6 ≈85.16≈85oranges

= =

Thesolutionsare− √1+2and−1−√2 2 2

f(7) =

−0.2841±8.2447 0.3568 −0.2841+8.2447 =22.3111 0.3568

= =

−0.2841± √ 0.28412−4(0.1784)(−95.1435) 2(0.1784) −0.2841±√ 679751 −02841±√0.0807+67.8944 . . = 0.3568 0.3568

3 2 a) For large values of x, x would be larger than x .

3 x=x·x·xandx2=x·xsoforverylargevaluesof 3 3 xthereisanextrafactorofxinxwhichcausesx tobe largerthanx. 2 b)Asxgetsverylargethevaluesofx3becomemuch larger thanthoseof x2 andthereforewecan“ignore” theeffectofx2intheexpressionx3 − x2.Thus,we 3 proximatethefunctiontolooklikexforvery canap largevaluesofx.

12

Chapter 1: c) Below is a graph of x3 − x2 and x3 for 100 ≤ x ≤ 200. It is hard to distinguish between the two graphs confirming the conclusion reached in part b).

Functions and Graphs

0.01

0.005

8e+06 7e+06

–0.01

6e+06

–0.006

5e+06

–0.002

0.002

0.006 x

0.01

–0.005

4e+06 3e+06

–0.01

2e+06

40. f (x)=x3+2x

1e+06 100

38. f (x)=x4

120

140

160

x

180

a) For x values very close to 0, 2x is larger than x3 since for x values less than 1 the higher the degree the smaller the values of the term.

200

−10x3+3x2−2x+7

b) For x values very close to 0, the function will looks like 2x since the x3 term may be “ignored”.

a) For large values of x, x4 will be larger than | −10x3 + − 2x + 7 | since the second term is a third degree 3x2 polynomial (compared to a fourth degree polynomial) and has terms being subtracted.

c) Below − 0 . 0is1 a≤graph x ≤ of x3 + 2x and 2x for 0.01. It is very difficult to distinguish between the two graphs confirming our conclusion in part b).

b) Since the values of x4 “dominate” the function for very large values of x the function will look like x4 for very large values of x.

0.02

c) Below is a graph of x4 −10x3+3x2−2x+7 and x4 for ≤ x ≤ 200. The graphs are close to each other 100 confirming our conclusion from part

0.01

b).

–0.01

–0.006

–0.002

0.002

1.6e+09 1.4e+09 1.2e+09

0.006 x

0.01

–0.01 1e+09

–0.02

8e+08 6e+08

41. f (x) = x

3

4e+08

− x

2e+08 100

120

140

x

160

180

200

39. f (x)=x2+x

x(x

a) For values very close to 0, x is larger than x2 since for values of x less than 1 x2 < x.

f(x) = 0 x3 − x = 0 x(x2 −1) = −1)(x+1) = 0 0 x x x

b) For values of x very close to 0 f(x) looks like x since the x2 can be “ignored”. c) Belowisagraphofx2+xandxfor −0.01 ≤ x ≤ 0.01.

It is very hard to distinguish between the two graphs confirming our conclusion from part b).

0 = 1 − 1 = =

42. x = 2.359 43. x = −1.831, x = 0.856, and x = 3.188 − 44 x = 2.039, and x = 3.594 .

x = −10.153, x = 1.871, − x = −0.821, x = x = 0.098, x = 0.535, x = 1.219, and x = 3.297 45. y = 8.254x − 5.457 46 y = −0.279x + 4.036 . y =1.004x2+1.904x 0.601 − 47 .

48 .

−0.303,

13

Exercise Set 1.3 49. y = 0.942x2 50. y = 0.218x3 51. y = 0.237x4

4. y = √ x and y √ = x − 2

− 2.651x − 27.943 + 0.188x2 − 29.643x + 57.067 − 0.885x3 − 29.224x2 + 165.166x− 210.135

Exercise Set 1.3 1. y = |x |and

y=|

5 4 3 y

x+3 |

2

10 1 8 0 6

2

4

x

6

8

10

2

x4

6

2

x4

6

2

x4

6

5. y = 2x

y 4

6 2

y4 2

–10 –8

–6

–4

–2 0

2

4 x 6

8

10

2. y = | x |and y = | x + 1 |

–6

–4

–2

0 –2

10

–4 8

–6 6

6. y = 3x

y 4

6 2

y4 2

–10 –8

–6

–4

–2 0

2

4 x 6

8

10

3. y = √ x and y = √ x + 1

–6

–4

–2

0 –2

5

–4 4

–6 3

7. y = −x2

y 2

6 1

y4 2

0

2

4

x

6

8

10

–6

–4

–2

0 –2 –4 –6

Chapter 1:

14 8. y = −x3

Functions and Graphs

12. y = x1 | | 6

10

y4 y 5

2

–6

–4

0

–2

2

x4

6

–10 –8

–6

–4

–2

2

–2

4 x 6

8

10

–5

–4 –6

–10

9. y = x12

x2

− 9 . It is important to note here that 13. y= x+3 x = −3 is not in the domain of the plotted function.

10 10 8

8

6 y 4

6 y

2

4 –10 –8

–6

–4

2

–2 0 –2 –4

2

4 x 6

8

10

–6

–4

–2

2 x

–8

4

–10

10. y = x −1 1

x2

−4 14. y= x −2 . Note: x = 2 is not in the domain of the plotted function.

10 8

10

6 y 4

8 6 y 4

2 –4

0 –2 –4

–2

2 x

4

2

–8

–2 0 –2 –4

–10

–6

–10 –8

–6

–4

–6

11.

2

4 x 6

8

10

–8

√3 y= x

–10 x1

–10 –8

–6

–4

4

−1 15. y= x −1 . It is important to note here that in the domain of the plotted function.

y 2

10

–2 0

8

2

4 x 6

8

6 y 4

10

2

–2 –10 –8

–4

–6

–4

–2 0 –2 –4 –6 –8 –10

2

4 x 6

8

10

x=1isnot

15

Exercise Set 1.3 16.y

x2

−25 . Note: = x+5 x = −5 is not in the domain of the plotted function.

10 8

–6

–2 0 –2 –4

–4

2

4 x 6

–6

18.

19.

5 √ x5 = x(2) 3 √5 a3 = a(5) 2

1 7 t = t(7) 21. √ 1 8 c = c(8) 22. √

26.

27.

28.

√13 = t4

8

10

√ 43. 165/2=(16)5=(4)5=1024 √ 44 642/3=(364)3=(4)2=16 √ .

82/3=(38)2=(2)2=4

x2 (x

√ 1 = 1()=t−(12) t

t

1 2

√1 = m

1)=m−(12) m(12

√1x2

=

√1

1

+7

x

()=( 1

(2+7)2

x2 +7) − ( 1 2 )

m6

−1

3)2=

11=

(x3)2 −

3and 2

Which means that the domain is the set of all x -values such that x =3orx =2

x

(x

−5x+6 = =0 0 −3)(x−2) So x = x =

1 3= (1)=( x3 +4) − ( 1 2 ) (x3+4) 2 30. 1 √ x 5=5x 31.

1 √7 t 32. 7 = √ t 2 y 33. y52 23 = 3√ t 5 = t2 34. −2 t 5 = 21 = √ 1t25 t5 35. −2 y3 = 1 2 = √1 3 y2 y3 36.

1 b−13 = 1 = √3 b2 b3 37. b−15 = 1 = 5 1 √ b2 b5 38. e −17 = 11 7 = 6√ 1e 1 7 6 e6 39. − 1 9 m6 1 1 − = 1 9 = 6√ m 1 9

29.

√ w4

.

1 =t−(43) t(43) 24. √ 1 5 = 1 =t−(65) t(65) t6

25.

4

= w5 =

46 √ 255/2=(25)5=(5)5=3125 .

The domain consists of all x-values such that the denom47 inator does not equal 0, that is x − 5 = 0, which leads to x = 5. Therefore, the domain is {x|x = 5} .

50. x + 2 = 0 leads to x = −2. Therefore, the domain is 48 ( −∞, −2) ∪ (−2, ∞). .51.

Solving for the values of the x in the denominator that 49 make it 0.

√ x 3=x(2)3

20. √4 b2 = b(4) = b(12)

23.

1

−4

w5

45. 163/4=(416)3=(2)3=8 √

–8 –10

17.

t12 3 = 3√ t12

√ 42. 93/2=(9)3=(3)3=27

2 –10 –8

.

41.

6 y 4

1 1 1 = √ 4 y+7 (y+7)4

−1

40 (y+7)4=

√1x −3

52.

Solving x2 + 6x + 5 = 0 leads to (x + 3)(x + 2) = 0 which means the domain consists of all real numbers such that x = −3 and x = −2

53. The domain of a square root function is restricted by the value wh√eretheradicantispositive.Thus,thedomainof f(x)= 5x+4canbefoundbyfindingthesolutionto the inequality 5x + 4 ≥0.

5x+4 ≥ 5x x ≥ ≥

0 −4 −4 5

54. Thedomainisthesolutionto2x 6 0.− 2x − 6 ≥ 2x ≥ x ≥

≥

0 6 3

55. To complete the table we will plug the given W values into the equation T(20) = (20)1.31=50.623 51 ≈ T(30) = (30)1.31=86.105 86 ≈ T(40) = (40)1.31=125.516 126 ≈

16

Chapter 1: T(50) = (50)1.31=168.132

b) y(2.7) = 0.73(7)3.63 ≈ 853.156 kg. c)

≈ 168 T(100) = (100)1.31=416.869 41 ≈ T(150) = (150)1.31=709.054 ≈ 7 70 9 Therefore the table is given by

5000 = 0.73(x).

5000 = x3.63 )0.731 5000 3 . 6 3 = x (

15 0 70 9

W 0 10 20 30 40 50 100 T 0 20 51 86 126 168 417 Now the graph

11.393 m ≈

363

x

59. Let V be the velocity of the blood, and let A be the cross sectional area of the blood vessel. Then

700 600

V = k A Using V = 30 when A = 3 we can find k.

500 y400 300 200

30 =

100

(30)(3) = k 90 =

0

20

40

60

80 x

100

120

140

= N 48 =

k

we need to find A when V = 0.026

S

90 A 0.026 = 0.026A = 90

= 3461.538 m2 60. Let V be the velocity of the blood, and let A be the cross sectional area of the blood vessel. Then

16800000 200000 84

a) f(180) = 0.144(180)1/2 = 0.144(13.41640786) 1.932 m2. b) f(170) = 0.144(170)1/2 = 0.144(13.03840481) 1.878 m2. c) The graph 5

90 0.026

=

A

So the variation equation is N = 16800000 S .Now,wehave to find N when S = 200000. = =

k

90 V = A

k 350000 (48)(350000) = k 16800000 = k

N

k 3

Now we can write the proportial equation

56. First find the constant of the variation. Let N represent the number of cities with a population greater than S.

57.

Functions and Graphs

≈ ≈

V = k A Using V = 28 when A = 2.8 we can find k. k 2.8 (28)(2.8) = k 78.4 = k 28 =

Now we can write the proportial equation 4

V = 3

we need to find A when V = 0.025

y 2

0.025 = 1

0

58.

78.4 A

78.4 A

0.025A = 78.4 50

100 x

150

a) y(2.7) = 0.73(2.7) .363 ≈ 26.864 kg.

200

A

=

78.4 0.025

= 3136m2

17

Exercise Set 1.4 61.

67. x = 2.6458 and x = −2.6458 9 = 0 x 9 x(x+7+ ) = x(0) x x2+7x+9 = 0

68 x = −2andx=3

x+7+

=

and x

= =

The function has no zeros

69. x=1andx=2 70

−7±√49−4(1)(9) 2 −7±√13 2 −7−√13 2

x=

x

.

− 7+

√ 2

Exercise Set 1.4 .

2π 3 rad

1. (120o )(πrad 180o ) = 120

1

13

0.5

62.

1 1 = w = w2 w2− w 1 w2− w − 1 = 0 1 ±√1+4 w = = 2 1±√5

–1

1−

0

–0.5

1

–0.5

–1 5π 6 rad

2. (150o )(πrad 180o ) =

2

1

63. P = 1000t5/4 + 14000

150

0.5

a) t = 37, P = 1000(37)5/4 + 14000 = 105254.0514. t = 40, P = 1000(40)5/4 + 14000 = 114594.6744 t = 50, P = 1000(50)5/4 + 14000 = 146957.3974 –1

≤ t 50. ≤

b)BelowisthegraphofPfor0

0.5

0

–0.5

0.5

1

–0.5

140000 120000

–1

100000 80000

3. (240o )(πrad 180o ) =

60000

1

4π 3 rad

40000

0.5 20000 0

10

20

t

30

40

50

64. At most a function of degree n can have n y-intercepts.

A polynomial of degree n can be factored into at most n linear terms and each of those linear terms leads to a yintercept. This is sometimes called the Fundamental Theorem of Algebra A rational function is a function given by 65. the quotient of two polynomial functions while a polynomial function is +···+ a function that has the form a xn −xn−1 n1 n +a a1x + a0. Since every polynomial function can be written as a quotient of two other polynomial function then every polynomial function is a rational function. x = 1.5 and x =

66. 9.5

–1

–0.5

0

–0.5

240

–1

0.5

1

Chapter 1:

18 4. (300o )(πrad 180o ) =

–1

5π 3 rad

180 8. (7π 6 )( πrad ) = 210o o

1

1

0.5

0.5

0

–0.5

0.5

1

–1

300

180 9. (3π 2 )( πrad ) = 270o o

1

1

0.5

0.5 –0.5

–1 –0.5

1

–1

5. (540o )(πrad 180o ) = 3 πrad

–1

0.5

–0.5

210

540

0

–0.5

–0.5

–1

Functions and Graphs

0.5

0.5

1

1 –0.5

–0.5 –1 270

–1

6. ( − 450o )(πrad 180o ) =

–1

−5π 2

180 ) = 540o 10. (3π )( πrad o

rad

1

1

0.5

0.5

–0.5

0.5

1 540

–0.5

–1

0.5

1

–0.5 –0.5 –1 –450

–1

180 7. (3π 4 )( πrad ) = 135o o

11.

180 ( −3π )( πrad ) = −60o o

1

1

0.5

0.5

135

–1

–0.5

0

0.5

1

–1

–0.5

0

–0.5

–0.5

–1

–1

0.5

–60

1

19 We need to solve θ1 = θ2 + 2π(k) for k. If the solution is an integer then the angles are coterminal otherwise they 17. are not coterminal.

Exercise Set 1.4 ) = −132o 15 )(π180o rad 12. (−11π 1

3π π = +2π(k) 2 2 −π = 2π(k) −π = k 2π −1 = k 2 Since k is not an integer, we conclude that π2 and 3π2 are not coterminal.

0.5

–1

0

–0.5

0.5

1

–0.5 –132

–1

13. We need to solve θ1 = θ2 + 360(k) for k. If the solution is an integer then the angles are coterminal otherwise they are not coterminal.

395 = 15+360(k) 380 = 360(k) 380 = k 360 1.05 = k Since k is not an integer, we conclude that 15o and 395o are not coterminal.

14.

18. π = − 3 π +2π(k) 2 2 2π = 2π(k) 2 = k π 2 = k π1 Since k is an integer, we conclude that π and −3π 2 2 coterminal.

19. We need to solve θ1 = θ2 + 2π(k) for k. If the solution is an integer then the angles are coterminal otherwise they are not coterminal.

7π = −65 π+2π(k) 6 2π = 2π(k) 2π 2π = k = k 1

225 =

−135 + 360(k) 360 = 360(k) 36 0 = k 36 0 = k 1 Since k is an integer, we conclude that 225o and −135o are 15.

coterminal.

We need to solve θ1 = θ2 + 360(k) for k. If the solution is an integer then the angles are coterminal otherwise they are not coterminal.

107 = −107 + 360(k) 214 = 360(k) 21 4 = k 36 0 0.594 = k Since k is not an integer, we conclude that 15o and 395o are not coterminal.

16.

140 = 440+360(k) −300 = 360(k) −300 = k 360 1.61 = k Since k is not an integer, we conclude that 140o and 440o are not coterminal.

are

Since k is an integer, we conclude that 7π6and −5π6 coterminal.

are

20. We need to solve θ1 = θ2 + 2π(k) for k. If the solution is an integer then the angles are coterminal otherwise they are not coterminal.

3π π = − 4 +2π(k) 4 π = 2π(k) π 2π = = k k 1 2 Since k is not an integer, we conclude that 3π and −π 4 4 are not coterminal.

21. sin 34o = 0.5592 22. sin 82o = 0.9903 23. cos 12o = 0.9781 24. cos 41o = 0.7547 25. tan 5o = 0.0875 26. tan 68o = 2.4751

20 27.

28.

29.

30.

31.

32.

o cot34=

tan1 34o = 1.4826

cot56=

tan 56o = 0.6745

o sec23=

cos1 23o = 1.0864

o csc72=

sin172o = 1.0515

sinπ ( )=0.5878 5

33.

cos( 2π )=0.3090 5

34.

tanπ ( )=0.4816 7

35.

cot( 3π ) = 11

36.

sec( 8 ) =

37.

csc( 13 ) =

3π 4π

Functions and Graphs

sin25o

1.4

54.

1

o

Chapter 1:

= =

x 1.4 x sin 25o x = 3.3127 55.

40

cos t =

60 t = cos−1( ) 40 60 t = 48.1897o

1 tan( 11) 3π = 0. 8665 1 . cos( 8)=26131 3π 1 13) = 1. 2151 sin( 4π

56.

20

tant =

25 t = tan−1( ) 20 25 t = 38.6598o

38. sin(2 .3) = 0.7457 39. cos(0 .81) = 0.6895 40 t = sin−1(0.45)=26.7437o

57.

.

t = sin−1(0.87)=60.4586o

41.

t = cos−1(0.34)=70.1231o

42.

t = cos−1(0.72)=43.9455o

9.3 t = tan−1( )

43.

t = tan−1(2.34)=66.8605o

t = 62.6761o

44 .

45.

46 .

t = tan−1(0.84)=40.0302o t = sin−1(0.59)=0.6311

50 t = sin−1 )

t = sin−1(0.26)=0.2630

30 50 t = 36.8699o

t = cos−1(0.60)=0.9273

59. We can rewrite 75o = 30o + 45o then use a sum identity cos(A+B) = cosAcosB − sin Asin B cos75o = cos(30o+45o)

49

51.

30

sin t =

t = cos−1(0.78)=0.6761 t= tan−1(0.11)=0.1096 48 = t . tan−1(1.26)=0.8999

50.

18 9.3

58.

47.

.

18

tant =

= = =

x 40 x = 40sin57o o

sin 57

=

=

√3− √1 2 2 √2 2 − 1+ 3 √ 22

x = 33.5468

52.

15 = x 15 = x tan 20o x = 41.2122

tan20o

o o cs √o30 ocos45−sin30sin45o 3 ·√1−1·√1 2 2 2 √2

60. The x coordinate can be found as follows x o cos20 = 200 x = 200c0s20o = 187.939

53.

15 = x 15 = x cos 50o x = 23.3359

cos50o

The y coordinate sin20o

y 200 y = 200sin20o =

= 68.404

21

Exercise Set 1.4 61. Five miles is the same as 5 · 5280 ft = 26400 ft. The difference in elevation, y, is

b) t

o

y sin 4 = 26400 y = 26400sin4o = 1841.57ft

h

x2+y2 = 316802 (20y)2+y2 = 1003622400 √1003622400 1003622400

=

y

67.

= 1582.02ft 63.

a)

o (1012)tan(24)+

h tan(24o) tan(67o )

(1012)tan(24o) (1012)tan(24o ) )

1−tan(24o = 555.567 ft

a) When we consider the two triangles we have a new triangle that has three equal angles which is the definition of an equilateral triangle. b) The short leg of each triangle is given by 2sin(30) = 2(1 2)=1 c) The long leg (L) is given by

cos40o

x = 150 x = 150cos40o

22 = L2+12 = L2

4 −√ 1

3

= 114.907

d) By considering all possible ratios between the long, short and hypotenuse of small triangles we obtain the o π o trigonometricfunctionsofπ6=30and3=60

b) sin40o

y 150 y = 150sin40o =

68.

= 96.4181 c)

a) Since the triangle has two angles equal in magnitude it should have two sides that are equal as well. (It is an isosceles triangle) b)

z2 = (x+180)2+y2

h2 = 11+11 h2 = √ 2

= (114.907+180)2+(96.4181)2 z2 = √96266.58866 z = 96266.58866

h=

64. v = =

77000·200·sec60o 5000000 15400000 5000000cos 60o

= 6.16cm/sec 65.

77000·100·sec65o v = = 4000000 7700000 4000000cos 65o = 4.55494cm/sec h a)tan(67o)=h x so, =x tan(67o)

2

c) Since the hypotenuse is known then we can use the figure to find the trigonometric functions of4 π = 45o using the ratios of the sides of the triangle.

= 310.268

66.

h 1012 + x h = tan(24o)(1012+x) h = (1012)tan(24o)+xtan(24o) =

h = = = tan(24o ) h(1 − tan(67o)

62. First a grade of 5% means that the ratio of the y coordinate to the x coordinate is 0.05 since tan t = y x. This means y that x = 0.05 = 20y The distance from the base to the top is6 · 5280 ft = 31680 ft. Using the pythagorean theorem

401y2 =

o

69.

a) The tangent of an angle is equal to the ratio of the opposite side to the adjacent side (of a right triangle), and for the small triangle that ratio is 57 . b) For the large right triangle, the opposite side is 10 and the adjacent side is 7+7 = 14. Thus the tangent is 10 14 c) Because the trigonometric functions depend on the ratios of the sides and not the size of triangle. Note that the answer in part b) is equivalent to that in part a) even though the triangle in part b) was larger that that used in part a)

70. Let (x,y) be a non-origin point √thatdefinestheterminal x2+y2bethedistance sideofanangle,t,andletr= fromtheorigintothepoint(x,y).Thenthetrigonometric function are defined as follows:

22

Chapter 1: r sint=y rand csct =y(y =0) = 0) and r cost=x rand sect =x(x x x( x =0)and cott =y(y =0) tant=y From the above definitions and recalling that the reciprox we show that cal of a non zero number x is given by 1 1 sint= csct

cost=

1 =w+x.Usingtheresultswehave e) sin(s+t) = (w+x) obtained from previous parts we can conclude sin(s + t) = w + x = sin(s)cos(t) + cos(s)sin(t) 73.

1

sect and

tant=

()=1=v/v cost b) Consider the triangle made by the sides v, w, and y. The angle vw has a value of 90 − r (completes a straight angle). The sum of angles in any triangle is 180. Therefore a) sin(t)=u 1=u,and

s+90+(90

1 cott

−r) = 180 180 −r =

s + 180

71. sint cost

x/r y ÷x r r y ·r r x y x

=

Thus sint cost

s

y/r

= = =

= tant = tant

x/r y/r x ÷y r x ·r r y x y

= = = =

= y − z. Replacing ur results for y e) cos(s + t) = (y−z) 1 − sin(s)sin(t) and z we get cos(s + t) = cos(s)cos(t) a) cos( π− t) = u1 = u = sin(t) b) v 2 sin(π t 2 −)=1=v=cos(t)

75. Use cos2t + sin2t = 1 as follows cos2t+sin2t = 1 cos2t sin2t 1 cos2 = cos2+t cos2t t 1+tant = sec2t 76.

= cott Thus = cott cost sint

cos2t+sin2t = 1 cos2t sin2t 1 + = sin2t sin2t sin2 cot2t+1 = csc2t

a) sin(t) = u1 = u b) Consider the triangle made by the sides v, w, and y. The angle vw has a value of 90 − r (completes a straight angle). The sum of angles in any triangle is 180. Therefore s+90+(90 s + 180 s

0 r

d) sin(r) = zu, which means z = sin(r)u.

Using results from part a) and part b) we get sin(r) = sin(s) and u = sin(t), therefore z = sin(s)sin(t)

74. cost sint

− r = s =

v, which means y = cos(s)v. c) cos(s) = y But from part a) v = cos(t), therefore y = cos(s)cos(t)

and

72.

Functions and Graphs

77. Let2=t

sina (b += sin(a)cos(b)+cos(a)sin(b) ) sin t =(2)sin(t+t) = sin(t)cos(t)+cos(t)sin(t)

−r) = 180 180 −r = − r = s =

0 r

= 2sin(t)cos(t) 78.

a) cos(2t) = cos(t+t)

c) sin(s) = wv which means that w = sin(s)v. cos(t) = v 1=v Thus, w = sin(s)v = sin(s) cos(t) x 1 d)sin(t)=u= and () =uwhichmeansthat u cosr x = u cos(r).

In part b) we showed that r = s therefore cos(r) = cos(s).

So, x = u cos(r) = sin(t)cos(s)

t+ t

= cos(t)cos(t) sin(t)sin(t) − = cos2(t) sin2(t) − b) cos(2t) = cos2(t) sin2(t) − = cos2(t) (1−cos2(t)) − = 2cos2(t) 1 −

23

Exercise Set 1.5 c) cos(2t) = cos2(t) − sin2(t) = (1 − sin2(t)) − sin2(t)

Exercise Set 1.5 1. 5π/4

1 − 2sin2(t) = 79. Using the result from Exercise 78 part (c)

1

cos(2t) = 1 − 2sin2(t) cos(2t)−1 = −2sin2(t) cos(2t)− 1 = sin2(t) −2 1 −cos(2t) = sin2(t) 2

0.5

–1

cos(2t)+1 = 2cos2(t) cos(2t) 2 +1

0.5

1

–0.5

80. cos(2t) = 2cos2

0

–0.5

5/4*Pi

− 1

–1

2. − 5π/6

= cos2(t)

1

81.

82.

a) V (0) = sinp(0)sinq(0)sinr(0)sins(0) = 0 V(1) = sinp(π π)sins( 2 π) = 1 2 )sinq( 2π) sinr( 2 b) When h = 0 the volume of the tree is zero since there is no height and therefore the proportion of volume under that height is zero. While at the top of the tree, h = 1, the proportion of volume under the tree is 1 since the entire tree volume falls below its height. a)

0.5

–1

–0.5

( )sin48.646 ( √ π) sin−3.728π 4 2 2 9( −123.208 √ π ( )si n86.62 √ π ) × sin 232 242 = 0.8208

1

0.5

1

–0.5

–5/6*Pi

V(0.5)

0.5

=

–1

3. − π 1

b) V(h) 0.5

1 0.8

-Pi

–1

–0.5

0.6

–0.5 0.4

–1

0.2

4. 2π 0

0.2

0.4

h

0.6

0.8

1

1

c) The result from part b) agrees with the definition of V (h) since the values of V (h) are limited between 0 and 1.

0.5

83. π π V( )1= sin−5.621( )sin74.831( √) 2 4 2 2 √ π ( ) sin138.959 ( √ π ) ×n−195.644 si 232 242 = 0.8219

–1

–0.5

0.5 –0.5

–1

1

–2*Pi

Chapter 1:

24

26. t = sin− 1 (−1)=3π2+2nπ

5. 13π/6

−1

27. 2t= sin

1 13/6*Pi

0.5

–1

Functions and Graphs

–0.5

0.5

(0)=nπsot=nπ

2

28. √ = = =− 3 = = −2√ 3

π 2sin(t+ sin(t+ )) 3 π 3 t+ π 3

1

–0.5

= =

t –1

and

6. − 7π/4

t

−1−√3 ( ) sin π 2 − π 3 +2 nπ 3 − − 2π +2 nπ 3 − π + 4π +2 nπ 3 3 π + 2 nπ

1 –7/4*Pi 0.5

29.

–1

–0.5

0.5

−1 cos(3t+ ) π = 4 2 π −1 1 3t+ = cos ( − ) 4 2 = π 2+ π 3t +2 nπ == − 4 3 3t = 5π t and = = 12 2nπ 5π 2 3t nπ 363 t3t

1

–0.5

–1

7. cos(9π/2)=0 8. sin(5π/4)=√−1

2

π 4+ π +2 nπ −4 3 13π +2 nπ 12 13 2 36π + 3 nπ

9. sin( −5 π/ 6)=−12 10. cos( −5 π/ 4) = √−12 11.

cos(5)= π

12.

sin(6)=0 π

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

−1

tan(4− π/3)=

−√

3

30.

−√

3 tan(7− π/3)= −0.5736 cos125o =

cos(2t) = 0 2t = cos−1(0) π 2t t == 2 +2nπ +nπ π 4 3π 2 and 3π 2t = 4 +2nπ = +nπ

sin 164o = 0. 2756 tan( −220o) = −0.8391 cos( −253o) = −0.2924 1 sec286o= cos 286o = 3.62796

t

csc312o= sin1312o =−1.34563 sin(12 . π )=

−0.587785

31.

( 3 ) = −1.37638 23. tan−2 .π 24. cos( −1 .91) = −0.332736 25. sin( −2 .04) = −0.891929 6+2nπ and / t = sin−1(12)=π

5π 6 + 2 nπ

cos(3t) = 1 3t = cos−1(1) 3t = 2nπ 2 t = 3 nπ

25

Exercise Set 1.5 32.

36. √ t 2cos( ) = − 3 2 −√3 t cos( ) = 2 2 −√3 t = cos−1( ) 2 2 +2nπ t 5π +4nπ +2nπ 2 = = 6 = 5π +4nπ = 3 t 7π and 6 7π t 3 2

sin2 t

The only solution comes from sint+1 = 0 t = sin−1( −1) 3π = +2nπ 2 37. y=2sin2t+4 2 =π,mid-liney=4 amplitude = 2, period = 2π maximum = 4 + 2 = 6, minimum = 4 − 2 = 2 38. y = 3cos 2t − 3 amplitude=3,period=2π 2= π,mid-line y = − 3 maximum = − 3+3=0,minimum=−3−3=−6

t 33.

2sin2 t (2sin t + 1)(sin t

39. y = 5cos(t/2) + 1 amplitude = 5, period = 2π1 = 4π, mid-line y = 1 2 maximum = 1 + 5 = 6, minimum = 1 −5=−4 y = 3sin(t/3) + 2 40. amplitude = 3, period = 2π1 = 6, mid-line y = 2

−5sint−3 =0 0 −3) =

The only solution comes from (2sint+1) = 0 sin t =

3

1 −2

t = sin−1( t = and = t

−2sint−3 = =0 0 −3)(sint+1)

(sin t

1 −)2

7π 6 +2nπ 11π 6 +2nπ

34. cos2x+5cosx = 6 cos2x + 5cos − x 6 = 0 −1) = 0 (cos x + 6)(cos x The only solution comes from cosx − 10 = x = cos−1(1) x = 2nπ 35.

maximum = 2 + 3 = 5, minimum = 2 −3=−1 y=1 2 sin(3t)− 3 41. amplitude = 1, period = 2π, mid-line y = − 3 3 3+1=−5 maximum− 2 2,minimum= −3 − 1 2=2 − 7 2 42. y = 12 cos(4t) + 2 2π π 2 amplitude=1,per iod=4=2,mid-liney=2 maximum=2+1=52 2,minimum=2 − 21 = 23 43. y = 4sin(πt) + 2 π = 2, mid-line y = 2 amplitude = 4, period = 2π maximum = 2 + 4 = 6, minimum = 2 −4=−2 44. y = 3cos(3πt) − 2

2 3π=3,mid-line y = −2 amplitude=3,period=2π maximum = − 2+3=1,minimum=−2−3=−5

45. The maximum is 10 and the minimum is -4 so the ampli2 =7.Themid-lineisy=10 tude is 10−(−4) −7=3,and the period is 2π (the distance from one peak to the next one) which means that b = 2π = 1. From the information above, and the graph, we conclude that the function is y=7sint+3

cos2x+5cosx =

−6

cos2x+5cosx+6 = 0 cosx= = = and

=

−5 ±√

25−4(1)(6) 2

−5±1 2 −5−1 = − 3 − 2 2 = −5+1 2

Since both values are larger than one, then the equation has no solutions.

46. The maximum is 4 and the minimum is -1 so the amplitude 5 3 2 is4−(−1)=5.The mid-lineisy=4 − 2 = 2, and the 2 period is 4π which means b = π2. From the information above, and the graph, we conclude that the function is 5 y= cos(t/2)+ 2

3 2

47. The maximum is 1 and the minimum is -3 so the amplitude 2 =2.Themid-lineisy=1 2= is 1−(−3) −1, and the − period is 4π which means b = π2. From the information above, and the graph, we conclude that the function is y=2cos(t/2) 1 −

Chapter 1: Functions and Graphs where f is the · 2π = 1.08071, k· =2π 143.

Therefore,

26 48.

The maximum is -0.5 and the minimum is -1.5 so the amplitudeis−0.5−(−1.5) = 1 Themid-lineis y = 2 12 2. =−1,andtheperiodis1whichmeansthat −0.5−− b=2π1 = 2π. From the information above, and the graph, we conclude that the function is 1 y= sin(2πt) 2

−1

49.

R = 0.339+0.808cos40ocos30o −0.196sin 40o sin 30o − 0.482cos 0o cos 30o = 0.571045 megajoules/m2

59. The amplitude is given as 5.3. b = f frequency, b = 0.172 the function is p(t) = 5.3cos(1.08071t) + 143 60 p(t) = 6.7cos(0.496372t) + 137 .

x = cos(140o), y = sin(140o), (

−0.76604, 0.64279)

61. ( −0.17365, −0.98481) 62.

5 x = cos(9π), y = sin(5),(0.80902, 9π

−0.58779)

63. ( −0.22252, −0.97493) 64 Rewrite 105o = 45o + 60o and use a sum identity.

50.

R = 0.339+0.808cos30ocos20o −0.196sin 30o sin 20o − 0.482cos 180o cos 20o = 1.12788 megajoules/m2

.

sin105o = sin(45o+60o)

65.

= = = =

51.

R = 0.339+0.808cos50ocos55o −0.196sin 50o sin 55o − 0.482cos 45o cos 55o = 0.234721 megajoules/m2

66.

52.

cos165o = cos(120o+45o) o o = cos120ocos √45−sin120sin45 3·√1 = −1 · √ 1 − 2 −2 √ 2 2 = −√1 √3 2 2 √ 2 =

R = 0.339+0.808cos50ocos0o −0.196sin 50o sin 0o − 0.482cos 0o cos 0o = 0.858372 megajoules/m2 53. Period is 5 so b = 2π5 , k = 2500, a = 250. Therefore, the function is 2πt V(t)=250cos +2500 5 3400 54. Periodis2spb=2π 2 =π,a= 2 =1700,k=1700+ 1100 = 2800. Therefore, the function is

V (t) = 1700cos πt + 2800 55. Since our lungs increase and decrease as we breathe then there is a maximum and minimum volume for the air capacity in our lungs. We have a regular period of time at which we breathe (inhale and exhale). These facotrs are reasons why the cosine model is appropriate for describing lung capacity.

The minimum is 35.33 and the maximum is 36.87 so the 56. amplitude is 36.87−35.33 2 =0.77.Theperiodis24sob= 2π 12,k=36.87 − 0.77 = 36.1. Thus, the function is 24 = π T(t)=0.77cos

o

−1+√3 2 2 67.

a) From the graph we can see that the point with angle t has an opposite x and y coordinate than the point with angle t + π. Since the x coordinate corresponds to the cos of the angle which the point makes and the y coordinate corresponds to the sin of the angle which the point makes it follows that sin(t + π) = −sin(t) and cos(t + π) = −cos(t). b) sin(t+π) = sintcosπ+costsinπ = sint

·−1+cost·0

= −sin t and cos(t+π) = costcosπ

− sin tsin π = cost ·−1−sint·0

π +36.1 12

57. The frequency is the reciprocal of the period. Therefore, 2π=440Hz f=b=880π 2π 58. f = 440π 2π =220Hz

o o sin45ocos60o√+cos45sin60 1 3 √1·1 + · 2 2 √ 2 3 √1 √ 2 2 2√ 2 1+√ 3 2 2

=

−cos t

c) tan(t+π) =

sin(t + π) cos(t + π)

Exercise Set 1.5

27 2n/12−1

− sint − cost sint cost = tant

= =

2n/12−1 = 2

a) The amplitude could be thought of as half the difference between the maximum and minimum, = −min. kais max−min 2 ,whichimpliesthat2a=max the average mean of the maximum and the minimum, 2 k = max+min ,whichimpliesthat2k=max+min.

Solving the system of equations above for max and min gives the desired results.

68.

Comparing exponents we can conclude that n −1=1 12 n 12 = = n

2 24

76.

880(2n/12 )π 2π

n/12 − 1

2

c) Half the difference between the maximum and minimum, using the results from part a), implies that the 2a 2 amplitudeis(k+a)−(k−a) =2=a.

= 1320 1320 880

=

2n/12−1 = 1.5 n ( 12 −1)ln(2)=ln(1.5) n − 1 = = 12 n = 12

a) Since the radius of a unit circle is 1, the circumference of the unit circle is 2π. Therefore any point t + 2π will have exactly the same terminal side as the point t, that is to say that the points t and t+2π are coterminal on the unit circle. Therefore, sin t = sin(t + 2π) for all numbers t.

n

ln(1.5) ln(1.5) +1 ln((2) ln(1.5) +1 ) 12

n = 19.01955

b)

There are 19 notes above A above middle C.

g(t + 2π/b) = asin[b(t + 2π/b)] + k = asin(bt+2π)+k from part a) = asin(bt)+k

77.

880(2n/12 )π

by

2π

n/12 − 1

definition g(t+2π/b) = g(t)

2

2π =440 72. f = 880π

n

78.

a) Left to the student b)y=1 2 cos(2t)−12 c) We use the double angle identity obtained in Exercise 79 of Section 1.4 and solve for −sin2(t) to obtain the model in part b).

79.

a) Left to the student b) y = 21cos(2t) + 1 2

2π

= 1760

+1 ln((2) ln(2.5) +1 ) 12

There are 28 notes above A above middle C

880(2n/12

880(2n/12 )π

ln(2.5) ln(2.5)

n = 27.86314

2π 73. f = 880·2−9/12π =261.626 )π 74 Fromtheequation,nh astobe12inorderfor 2π to equal 880. There are 12 notes above A above middle C. .

2200 880

=

n − 1 = = 12 n = 12

70. Since at the apex, L is large, T is small, and d is small, then the basilar membrane is affected mostly by low frequency sounds.

71. Since at the base, L is small, T is large, and d is large, then the basilar membrane is affected mostly by high frequency sounds.

= 2200

2n/12−1 = 2.5 n ( 12 −1)ln(2)=ln(2.5)

c) Since the function evaluated at t+2π/b has the same value as the function evaluated at t and 2π/b =0 then t + 2π/b is evaluated after t. Since we have a periodic function in g(t) it follows that the period of the function is implied to be 2π/b.

75.

1760 880

There are 24 notes above A above middle C.

b) The average mean of the maximum and minimum, using the results from part a), implies that the mid2k 2 line equation is y = (k+a)+(k−a) =2=k.

69.

=

c) We use the double angle identity obtained in Exercise 79 of Section 1.4 and solve for cos2(t) to obtain the model in part b).

Chapter 1:

28 80.

9. y=3x

a) Left to the student

3

b) y = sin(2t)

− 6x+1 300

c) We use the double angle identity for sin(2t) obtained in Exercise 77 of Section 1.4 to obtain the model in part b).

81.

82.

200 100

a) Left to the student b) Left to the student

–4

0

–2

c) The horizontal shift moves every point of the original 4 units to the right. graph π

–100

a) Left to the student b) Left to the student

–300

c) The horizontal shift moves every point of the original 3 units to the left. graph π

2 x

4

–200

10. y = | x + 1 |

83. Left to the student

6

84 Left to the student

5

Left to the student

4

.

Functions and Graphs

85.

3

Chapter Review Exercises

2

a) 100 live births per 1000 women b) 20 years old and 30 years old −2) = 2(−2)2 − (−2) + 3 = 13 2. f ( 1.

1

3.

11.

–2

f (x)=(x

− 2)2

f(1+h) = 2(1+h)2 =

−(1+h)+3 2(1+2h+h2) = − 1 − h + 3

=

(h

=

h

− (2 − h))2 −1)2 −2h+1

0

–1

1

2

–4

2 x

–2 0

25

–2

20

–4

15

–6

10 –8 5

–4

–3

–2

–1

0

x

5

−16 . It is important to note that x = 12. f(x )=x2 x+4 −4 does not belong to the domain of the plotted function.

2

−

4

2

6.

8. f(x)=2x2+3x 1

3

4

5. f ( −5)=(1 ( 5))2=(1+5)2=62=36 − −

7. f(4) = (1 −4)2=( 3)2=9 −

4

6

2h2+3h+4 4. f(0)=2(0)2 (0)+3=3 −

f(2 h) − = (1

2 x

8

−1−h+3 =

2+4h+2h2

0

–4

13.

1

x

2

3

a) f (2) = 1.2 −3 b) x =

4

29

Chapter Review Exercises 20.

14. x = − 2

x2

2

−7x+12 = −3)(x−4) =0 0 0

(x

3 x − 3 = 0 x = 4 = −Or 4 x x =

1

–4

–2

2 x

4

–1

21. x2+2x=8 x2 + 2x − 8 = 0 −2) = 0 (x + 4)(x x + 4 = 0

–2

15. y = 4 − 2x 10

x = = Or = − 2 x

8

x

6

22. x2+6x=20

4

x2 + 6x −20 = 0

2 –3

–2

0

–1

1

2

x

x

3

16. m =

y

= −117

=

y −y1 = m(x−x1) = − 7(x−4) y − −(2) −7x + 28 = −2

−6 ±√

36+80 −±√2 6 2 29 2 √ −3± 29

23. x3 + 3x2 x2(x + 3)

= 0 0 − (−xx+−33) =

(x + 3)(x2 −1) = −1)(x+1) = 0 0 (x + 3)(x x+3 = 0

−7x + 6

y= 17.

= =

–2 − 2− 5 4− ( − 7)

−4 0 2

Use the slope-point equation y −y1 = m(x−x1) y −11 = 8(x− )2

x

1

y = 8x −4+11 y = 8x+7

x = = Or = − 1 x Or

x+1 = 0 x

18. Slope = −16,y-intercept(0,3)

−3 0 1

=

−1

24.

19. x2+5x+4 = 0 (x+1)(x+4) = 0 x+1 = 0 x = −1 Or x+4 = 0 x

=

−4

x4 + 2x3 x3(x + 2)

= 0 0 − (−xx+−22) =

(x + 2)(x3 −1) = 0 x

=

− 2 − 1

x = 25. Using the points (one could use any two points on the line) (0,50, and (4,350) the rate of change is 350 −50 300 = =75pagesperday 4− 0 4

Chapter 1:

30 26.

/ 36. tan(7π 4)= −1

The rate of change is 20 −100 −80 −20 = = meterspersecond 12 − 0 12 3

37. s

27. The variation equation is M = kW, with k constant.

When W = 150, M = 60 means 60 = k(150) 60 = k 150 = k 2 5

x 127

=

x = 127sin(70o) = 119.341

39. t=tan−1( √ 3)=π 3+nπ 40 2t = cos−1(2), No solution .

M

41.

2 5W (210) 2 5 = 84lbs

= =

28. 5x2 − x − 7 = 0 = =

12cos2(2t − π )) 4) cos2(2t − π ) 4 cos2(2t − π π cos(2t− 4

= = =

9 9 1 2 3 4

=±3/4√

Two solutions π 2t− 4 = = 2t − π 4

1 ±√1+140 10 1 ±√141

cos−1(3/4) √ π π +2nπ 6 π 2t = 2t = = = = 5π+ +2nπ 12 4 = 5pi+2nπ 24

10 √y 29. y1/ 6 = 6 √ 2 03 3 20 x = x / √ 31. 27/2 3 = ( 3 27)2=32=9 30.

t and

32.

+nπ

π 2t− 4

cos−√1(−3/4) +2nπ −ππ 2t − π 4 −π4 2t 6 + +2nπ π 12+2nπ 2t = t = pi +nπ 24

4

2

–2

2

x 4

6

–2

33.

o

38. t = sin−1(1) = π 2 + 2nπ

Find M when W = 210

x

Functions and Graphs

42.

(2 sin(t)

18 − 74 = 97 a) m = 92 23 9 = 14 −

G

− 74 = G G

= =

7 =85.6 b)G(18) = 97 (18) + 437 G(25) = 97 (25) + 437 7 =94.6 34. sin(2 π/3)= 35. cos( − π )=−1

√

3 2

9 (x−9) 9 7 9 x 81 7 +74 − 7 437 x+ 7

−1)(sin(t)+4) = 0 sin(t)+4 = 0 sin(t) = = −4 Nosolution sin(t) = 0 2 sin(t) − 1 1 2 t = sin−1(1/2) π t = 6 +2nπ t = 5π 6 +2nπ

43. y = 2 sin(t/3) 4 − 2π amplitude = 2, period = (1/3) = 6π mid-line y = 4, max = −4+2=2,min= 4 2= − − −

−6

31

Chapter 1 Test 44. y

1 = cos(2πt)+3 2 amplitude = 1, period = 2π 2π=1 2 mid-line y = 3, max = 3 + 1 2= 7, 2min = 3

6. m = 10−(−5) −3−2 =3 2−1=5

2 =2,period=π,mid-linevalue=3 45. Amplitude = 5−1 y = 2sin(2t) + 3 46 Amplitude = 1−(−5) 2 =3,period=2,mid-linevalue= −2 y = 3cos(πt) − 2 .

Amplitude = 135−1 2 =67,period=1/2meansthat 47. a) 2π b= (1/2) = 4π, mid-line value = 1 + 67 = 68 Since the heel begins on the top of the eye we will use a cosine model h(t) = 67 cos(4πt) + 68

7. Use the points (0,30) and (3,9) −21 3 =−7 Average rate of change = 9−30 3−0 = The computer loses $700 of its value each year.

8. Rate of change = 3−0 6−0=2

9. Variation equation F = kW. Use F = 120 when to find k

The equation of variation is F = 2 3 W

√1 =1 32 48 (645/3)−1/2 = 64−5/6 = (664)5

10.

49

x=0,x= −2, and x = 2 x = ±√10 and x = ± √2 2

.

( −1.8981,0.7541), (−0.2737,1.0743), and (2.0793,0.6723)

50.

a) G(x) = 0.6255x + 75.4766 b) G(18) = 0.6255(18) + 75.4766 = 86.7356 G(25) = 0.6255(25) + 75.4766 = 91.1141

51.

11.

52.

c) In Exercise 33, G(18) = 85.6 and G(25) = 94.6. The results obtained with the regression line are close to those obtained in Exercise 33.

53.

a) w(h) = 0.003968x2 + 3.269048x − 76.428571 b)

w(67) = 0.003968(67)2+3.269048(67) = 160.415 lbs

Chapter 1 Test 1.

2.

W =180

120 = k(180) 120 180 = k 2 = k 3

b) h(10) = 67 cos(40π) + 68 = 101.5 m

.

1

a) f(1) = − 4 b)x= −3andx=3 x2 + 4x − 2 = x = = = =

12.

0 − 4± √ 16+8 2 − 4± √ 24 2 − 4 ± 2√ 6 2 √ 2 − ± 6

20

− 76.428571

y10

–10 –8

–6

–4

a) Approximately 1150 minutes per month b) About 62 years old

–2 0

2

4 x 6

8

10

–10

f(x) = x2 + 2 –20

a)f( −3) = (−3)2 + 2 = 11 b)f(x+h)=(x+h)2+2=x2+2xh+h2+2 3. f(x) = 2x2 + 3 a)f( −2)=2(−2)2+3=8+3=11 b)f(x+h)=2(x+h)2+3=2(x2+2xh+h2)+3= 2x2 + 4xh + 2h2 + 3

13. √ 1 /t=1/t1/2=t−1/2 t− 3/5=1/t3/5=15/ √ t3 14.

15. f(x) = x2−1 x+1 . It is important to note that in the domain of the plotted function 4

4. Slope = −3, y-intercept (0,2) 5.

2 –4

y −y1 = m(x x1) − 1 (x 8) y −(−5) = 4 1 − = 2 5 y 4x = x − 7− y 1 4 −

2 x

–2

–2 –4 –6

4

x = −1 is not

Chapter 1:

32 16.

17.

6)= sin(11π/

4)= cos( 3−π/ 18. ()=0 tanπ 19.

1

27.

−2

Functions and Graphs

25

sqrt2 2

20 15

tan(40o) = = x

3.28 x

10

3.28 tan(40o ) = 3.909

5

–4

0

–2

28.

20.

2 x

4

2 x

4

4

tan(t) = ± √ 3 t = tan− √1( 3)

2

π t = 3 +2nπ and t = tan− 1(− √ 3) t = − π +2nπ 3

–4

–2 –2

21.

–4

29.

cos2(t) = 2

cos(t) = ± √ 2 cos(t) = 1.414

1 a) Find the slope m = 176−170 80−50 =5 Use slope-point equation

M −M1 = m(r−r1) − 1 170 = = 1 (r−50) M 5 1 r −10+170 M 5 r+160 M=

No solution, cos(t) cannot have values larger than 1.

22.

2sin3 (2t) sin(2t)(2sin(2t)

−3sin2(2t)−2sin(2t) −1)(sin(2t)+2) = =0 0 t = Or = t= Or t

nπ 2 −π 12 +nπ

b) M(62) = 1 5 (62) + 160 = 172.4 M(75) = 15(75) + 160 = 175 30.

7π 12 +nπ

=π,mid-liney=4 23. Amplitude = 4, period = 2π2 −4=0 max=4+4=8,min=4 24.

2π Amplitude = 6, period = (1/3) = 6π, mid-line y = −10 −10+6=−4,min=−10−6=−16 25. max= 3, 2 2 2π Amplitude −0.5−(−1.5) = 1, period b= (2π/3) ==3, mid-line value is -1 = 2π Thus, equation of the line is y = 12 cos(3t)− 1 = 3 2,period=1, 26. Amplitude = 4−1 2 b=2π1 = 2π, mid-line value is 2.5 Thus, equation of the line is y = 3sin(2πt) +52 2

8

−1 = 0 x 3x2 − x + 8 = 0 1 ± √ 1−96 x = ±√6 1 i95 = 6 = 1 √ 95 ± i 6 6 3x+

31. x = − − 1.2543 32. There are no real zeros for this function.

33. ( −1.21034, 2.36346) 34.

a) M(r) = 0.2r + 160 b) M(62) = 172.4 M(75) = 174 c) The results from the regression model are exactly the same as the result obtained in Exercise 29.

33

Technology Connection Linear Model: y = 37.57614x + 294.47744 Quadratic Model: y = −0.59246x2 + 74.60681x − − − 2.60421x2 0.32399x3 + + 117.72472 y = 0.02203x3 Cubic Model: 125.71434x−439.64751 Quartic Model: y = 0.00284x4 11.45714x2−88.51211x+507.83874

35. a)

b)

-3 -2 -1 0 1 2 -29 -15 -5 1 3 1 -5 -15 -29

2.

1.x

= 4 .4149

2.x = 0.618034andx=1.618034 − = 0.37393x+1.02464 −

2.y = 0 .46786x2 − 3.36786x + 5.26429 3.y = 0 .975x3 − 6.031x2 + 8.625x − 3.055 • Page 28: = = =−2andx=1 −1.414214, x 1.x =2andx= −5 = 0,−4andx=6 and x == 1.414214

500

2.x 0

10

20

30

x

40

50

c) By consider the graph in part b) and the scatter plot of the data points, it seems like quartic model best fits the data. The reason for this conclusion is because the scatter plot and the quartic model have the least amount of deviation (sometimes called residue) between them compared to the other models. d) Left to the student (answers vary).

3.x 4.x 5.x

=0andx=700 = =

095574 6.x = 3−2.079356, x = 3.1164004 −3.095574, x x= = 0.46295543, −0.6460838, and x = 0.6460838, and 7.x x .

8.x = −1andx=1 x = 1.414214, x = 1, and x = 1.414214 −3,x=−1,x=2,andx=3 9.x −2, = −0.3874259 = = = 6 and x = 1.7207592 10.x 11.x 1329332

Technology Connection

12.x

• Page 5:

1. [0, ∞ )

Page 7: line will look like a vertical line.

1. The 2. The line will look like a horizontal line.

3. The line will look like a vertical line.

4. The line will look like a horizontal line.

• Page 10: 1. Graphs are parallel 2. The function values differ by the constant value added.

3.

Graphs are parallel • Page 19: (5)=6,f( −4.7) = 3.99, f(11) f(11) = 150, f(2/3) = − −21.3, f(−4.7) = −12.3, = −117.3, − 1556 .

2.f (5)= − f (2 /3) = 3.2556

1.f

−75, f(−4.7) = −45.6, f(11) = 420.6, − 3.f (5)= (2 3) / = 1.6889 f P a g e 2 1 : 0 4

1 -1

.

• Page 37:

• Left to the student

-1 11

5

• Page 23:

1.y

1000

1.

4

• Page 27:

2000 1500

•

3

2 -4

3 -5

4 -4

5 -1

6 4

7 11

2. [− 2,∞ ) 3. − ∞, ∞ ) ( 4. ( − ∞, ∞ ) 5. [1, ∞ ) 6. ( − ∞, ∞ )

7. − 3,∞ ) 8. [ − ∞, ∞ )

9. ( − ∞ ∞ ) , 10. ( − ∞, ∞ ) ( 11. Not correct 12. Correct • Page 46: = 6 .89210o 2.t = 46 .88639o 1.t

3. No solution 4.t = 1 .01599 5.t = 0 .66874 6. 0. 46677

Chapter 1:

34

Functions and Graphs

b)

• Page 56: Number 2 equation: shifts the cos(πx) graph up by 1 unit Number 3 equation: shifts the cos(πx) graph up by 1 unit and shrinks the period by a factor of 2 Number 4 equation: shifts the cos(πx) graph up by 1 unit, shrinks the period by a factor of 2, and increases the am- plitude by a factor of 3 Number 5 equation: shifts the cos(πx) graph up by 1 unit, shrinks the period by a factor of 2, increases the amplitude by a factor of 3, and shifts the graph to the right by 0.5 units

370 360 350 340 330 320 10

0

Extended Life Science Connection 1.

20x

30

40

c) January 1990 corresponds to t = 31 y = 0.0122(31)2 + 0.8300(31) + 314.8104 = 352.29 January 2000 corresponds to t = 41 y = 0.0122(41)2 + 0.8300(41) + 314.8104 = 369.39 The estimates seem to be reasonable when compared to the data.

a) y = 1.343450619x + 311.3019556 b) 360

d) January 2010 corresponds to t = 51 y = 0.0122(51)2 + 0.8300(51) + 314.8104 = 388.94 January 2050 corresponds to t = 91 y = 0.0122(91)2 + 0.8300(91) + 314.8104 = 491.57 Find x when y = 500

350 340 330

e)

320

500 = 0.0122x2+0.8300x+ 0

10

20 x

30

40

c) January 1990 corresponds to t = 31 y = 1.343450619(31) + 311.3019556 = 352.95 January 2000 corresponds to t = 41 y = 1.343450619(41) + 311.3019556 = 366.38 The estimates seem to be reasonable when compared to the data. d) January 2010 corresponds to t = 51 y = 1.343450619(51) + 311.3019556 = 379.82 January 2050 corresponds to t = 91 y = 1.343450619(91) + 311.3019556 = 433.56 The estimates seem to be reasonable when compared to the data.

Find x when y = 500 e) y = 1.34345x+311.30196

314.8104 0 = 0.0122x2+0.8300x

− 185.1896 0.8300 2(0.0122) + √(0.8300)2−4(0.0122)(−185.1896)

x

=

x

≈ 161.63

2(0.0122) The carbon dioxide concentration will reach 500 parts per million sometime in the year 2120.

3.

a) y = −0.000307x3 + 0.031536x2 + 0.509387x + 315.8660781 b) 370

500 = 1.34345x+311.30196 500 −311.30196 = 1.34345x 500 −311.30196 = x 1.34345 140.5 ≈ x

2.

The carbon dioxide concentration will reach 500 parts per million sometime in the year 2099. a) y = 0.0122244281x2 + 0.8300246407x + 314.8103665

360 350 340 330 320 0

10

20x

30

40

c) January 1990 corresponds to t = 31 y = −0.000307x3 + 0.031536x2 + 0.509387 x + 315.8660781 = 352.82 January 2000 corresponds to t = 41 y = −0.000307x3 + 0.031536x2 + 0.509387 x + 315.8660781 = 368.60

Extended Life Science Connection The estimates seem to be reasonable when compared to the data d) January 2010 corresponds to t = 51 y = −0.000307x3 + 0.031536x2 + 0.509387x + 315.8660781 = 383.15 January 2050 corresponds to t = 91 y = −0.000307x3 + 0.031536x2 + 0.509387x + 315.8660781 = 392.02 e)

Find x when y = 500. The maximum of the cubic function does not intersect the line y = 500 therefore under this model the carbon dioxide concentration will never reach 500 parts per million.

380

4.

a) 370 360

• QUADRATIC MODEL: This model also resembles the original data set’s scatter plot. At relatively small values of t it allows for a longer time for the increase in the concentration of carbon dioxide since it is a parabola. As time increases though the level at which the concentration of carbon dioxide will increase will be quicker than the linear model.

• CUBIC MODEL: This modelalso resembled the original data set’s scattor plot indicates that there is a level after which the concentration of carbon diox- ide will not increase. It is the only model that did not allow the concentration level of carbon dioxide to reach 500 parts per million. This model suggests that as time increased the concentration of carbon dioxide will begin to decrease indefinitely.

• PERIODIC MODEL: This model, as the other, modeled the data set to a very good degree of accuracy. It was the only model that allowed for oscillating behavior in the future, which is more likely to happen than what the other models suggested.

350 340 330 320 0

10

20

30

t

40

50

b) The graph represents a steady increase in the concentration of carbon dioxide. c) 0.4 0.2 0

46

47

t

48

49

50

–0.2 –0.4

d) The graph shows an oscillating behavior for the concentration of carbon dioxide. e) 378 376 374 372 370 45

46

47

t

48

49

50

f ) The graph behavior shows that there is a periodic fluctuation in the concentration of carbon dioxide.

5.

35 the scatter plot of the original data sets. Under this model, the concentration levels of carbon dioxide will increase with time indefinitely.

This model is the easiest math• LINEAR MODEL: ematically to compute and explain. It does resemble

Chapter 2

Differentiation Exercise Set 2.1 f ) Since lim g(x)

=g(−2),thenthefunctionisnot x→ − 2 continuous at x = − 2

1. The function is not continuous at x = 1 since the limit from the left of x = 1 is not equal to the limit from the right of x = 1 and therefore the limit of the function at x = 1 does not exist.

7.

a) As we approach the x-value of 1 from the right we notice that the y-value is approaching a value of 2.

Thus, lim h(x)=2. Asweapproachthex-valueof1 x→ 1 +

from the left we notice that the y-value is approaching a value of 2. Thus, lim h(x) = 2. Since x→1+ x→ 1 − lim limh(x) h(x)== 2 then lim h(x) = 2.

2. The function is not continuous at x = −2 since the limit −2 is not equal to the limit from the −2 and therefore the limit of the function at from the left of x = right of x =

b) Reading the value from the graph h(1) = 2.

x = −2 does not exist.

c) Since the lim h(x) = 2 and = h(1) = 2 then h(x) is x →1 continuous at x = 1. d) As we approach the x-value of -2 from the right we notice that the y-value is approaching a value of 3.

Thus, lim h(x)=0. Asweapproachthex-value

3. The function is continuous at every point in the given plot.

Note that the graph can be traced without a jump from one point to another.

4. The function is not continout at x = −2 since the value of −2 is undefined.

5. the function at x = a) As we approach the x-value of 1 from the right we notice that the y-value is approaching a value of 1. Thus, lim f(x)= −1. Asweapproachthex-

x→ − 2 +

of -2 from the left we notice that the y-value is approaching a value of 0. Thus, lim h(x) = 0. Since x→ − 2 −

lim h(x)= lim h(x)=0then lim h(x)=0.

x→ 1 +

f)

x→ 1 −

lim f(x) = lim f(x) then lim f(x) does not exist. x→1

x→1−

b) Reading the value from the graph f(1) =

−1.

8.

c)

x → 1t(x) = t(1) Since lim d) atx=1 x→ − 2 +

6.

a) lim g(x)= x→ 1 +

x→ − 2

x→−2

limg(x)=−2,

−2,

x→1−

lim f(x) = − 2

x→ 1

b) g(1) =

f ) Since lim t(x) does not exist, then the function is x→ − 2

−1=g(1)=−2,theng(x)isconc) Since lim = x →g(x) 1 tinuous at x = 1 d) lim g(x)=4, lim g(x)= x→ − 2 +

not continuous at x = − 2 9.

a) As we approach the x value of 1 from the right we find that the y value is approching 3. Thus lim f(x) = 3 x→ 1 + b) As we approach the x value of 1 from the left, we find c) thattheyvalueisapproching3. Thus lim f(x)=3 Since lim f(x) = 3 and lim f(x) = 3 then

−2

does not exist −2) = −3 e)g(

lim t(x)=undefined,

x→−2−

e) t( −2) = undefined

e) Reading the value from the graph f( −2) = 3. −2)=3,thenf(x)iscontherefore

≈0.25,thent(x)iscontinuous

thus lim t(x)doesnotexists

x→ − 2 −

x → −f(x) 2 = 3 and f( f) Since lim tinuous at x = −2.

limt(x)≈0.25,

x→1−

lim t(x)=undefined,

x→ − 2 +

of -2 from the left we notice that the y-value is approaching a value of 3. Thus, lim f(x) = 3. Since x→−2−

a) lim t(x) ≈0.25, x →1+ lim t(x) ≈0.25 b) t(1) = ≈ 0.25

d) As we approach the x-value of -2 from the right we notice that the y-value is approaching a value of 3.

Thus, lim f(x)=3. Asweapproachthex-value

lim f(x)= lim f(x)=3then lim f(x)=3.

−2)=0,thenh(x)iscontherefore

→ − h(x)=0andh( 2 Since xlim tinuous at x = −2.

x→ 1

c) Since the lim x →f(x) 1 − does not exist, then f(x) is not continuous at x = 1.

x→ − 2 +

x→−2

x→−2−

x→ − 2 +

e) Reading the value from the graph h(−2) = 0.

value of 1 from the left we notice that the y-value is approachingavalueof2. Thus, lim f(x)=2. Since x→ 1 +

x→1

x→ 1 −

x→−2−

−3,thus

limg(x) x→−2

x→ 1 +

lim f(x) = 3

x→ 1 −

x→1−

x→ 1

d) From the given conditions f(1) = 2 e) f(x) is not continuous at x = 1 since lim f(x) x→ 3

f ) f(x) is continuous at x = 2 since lim f(x) = lim f(x)=2=f(2)

x→ 2 −

x→ 2 +

=f(1)

Chapter 2: Differentiation

38 10.

a) b) c)

lim f(x)=0

f) False. Since lim f(x) = 3 and f(4) =

lim f(x)=0

g) False. Since lim f(x)

lim f(x)=0

h) True. Since lim f(x) = 4 and f(0) = 4

x→ 4

x→ − 2 +

x

x→ 0

→−2

i) True. Since lim f(x) = 3 = lim f(x)

−2 f(x) is not continuous at x = f(x) is continuous at x =

j) False. Since lim f(x) does not exist

d) f ( −2) = 3 e) f) 11.

x→ 2

15.

a) 1

d) True. Since lim f(x) = 4 = lim f(x)

x→2−

x→ 2 +

e) False. Since lim f(x) x→ 0 −

=limf(x) x→0+

x→ 2 +

x→2−

e) True. Since lim f(x)= lim f(x)=0

f) True. Since lim f(x) = lim f(x) = 4

f ) False. Since lim f(x)

g) False. Since lim f(x) does not exist

x→ 0 +

x→0−

x→ 3 +

=limf(x) x→3−

x→ 0

h) True. Since lim f(x) = 4 = f(2)

g) True. Since lim f(x) = 0 = f(0)

x→ 2

x→ 0

h) False. Since lim f(x) does not exist a) b) c) d) e)

16.

x→ 3

a) True. lim g(x) = 0 = g(0) x→ 0 +

True. Since lim g(x) = 1 = g(2)

b) True. lim g(x) = 0 = g(0)

False. Since lim g(x) =

c) True.

x→ 2 +

x→ 2 −

−1 =1=g(2)

False. Since lim g(x)=1and lim g(x)= x→2−

x→ 2 +

False. Since lim g(x) x→ 2 +

x→ 0

e) True. lim g(x) = 0 = g(0)

=limg(x) x→2−

x→ 0

x→ 2

False. As we approach the x value of -2 from the right b) we find that the y value is approching 2.

True. As we approach the x value of -2 from the left, c) we find that the y value is approching 0. lim f(x)=0 d) False.Since lim f(x)=1and 2+ e) False.Since lim f(x) 2+

x→ 0 −

d) True. lim g(x) = 0

−1

False. Since lim g(x) does not exist

a)

x→ −

x→ 1 −

= lim f(x)

17. The function p is not continuous at x = 1 since the p is continuous at x = 1.5 limp(x) does not exist. x→ 1

since lim p(x) = 0.6 = p(1.5). p is not continuous at x →1.5 x = 1 since the lim p(x) does not exist. p is continuous at x→ 2

x = 2.01 since lim p(x) = 0.8 = p(2.01). x →2.01

18. p is continuous at x = 2.99 since lim p(x) = 0.8 = x →2.99 p(2.99). p is not continuous at x = 3 since the lim p(x) x→ 3

f ) False. Since lim f(x) does not exist

p is continuous at x = 3.04 since does not exist. lim p(x) = 1 = p(3.04). p is not continuous at x = 4

g) True. Since lim f(x)= lim f(x)=0

since the lim p(x) does not exist.

x→ −

x→ − 2 −

x→ − 2

x → 0+

h)

14.

x→0−

x→ 0 +

x→3−

x→ 3 +

a) True. As we approach the x value of 0 from the right we find that the y value is approching 0, which is the value of the function at x = 0. b) False. As we approach the x value of 0 from the left, we find that the y value is approching 2 instead of 0. c) False. Since lim f(x) = 0 and lim f(x) = 2

True. The values of y as we approch x = 0 from the c) left is the same as the value of the function at x = 0, x→0− which is 0 x → 0 + d) True.

False. Since Since lim lim f(x)=0and f(x)=3 andlim limf(x)=0 f(x)=1

13.

x→5

x→ 3

True. The values of y as we approch x = 0 from the right is the same as the value of the function at x = b) 0, which is 0

12.

=f(4)

x→ 4

x→ − 2 +

−1

x→ 4

x→ 0 −

True. The graph indicate a point (solid dot) at (0, 2) i) False.Since lim f(x) = lim f(x) x → − 2+ x→ − 2 − j) False. Since lim f(x) =f(0) x→ 0 a) True.Since lim f(x)= lim f(x)=f( −1) x → − 1+ x→ − 1 − b) False. Since lim f(x) = 0 c) x → 2− d) False. Since lim f(x) = 3 e) False.

x → 3.04

x → 2+

19.

lim p(x) = 0.4, lim p(x) = 0.6, therefore lim p(x) does x→1

x → 1−

not exist

x→ 1 +

does 20. lim p(x) = 0.6, lim p(x) = 0.8, therefore lim p(x) x→2 x → 1−

not exist

21.

lim p(x) = 0.8,

x → 2.6−

lim p(x)=0.8

x→ 2 +

lim p(x) = 0.8, therefore

x →2.6+

x → 2.6

22. limp(x) does not exist since lim p(x) = 0.8 and x→ 3

False. Since lim f(x)= lim f(x) x → 2+

x → 2−

x → 4+

x → 4−

True. Since lim f(x)= lim f(x)=3

x → 3−

lim p(x)=1

x → 3+

23.

lim p(x)=1since lim p(x)=1, lim p(x)=1

x → 3.4

x →3.4−

x3.4+ →

39

Exercise Set 2.1 24. C is continuous at x = 0.1 since lim C(x) = 2.3 = x →0.1 C(01). . Cisnotcontinuousatx=0.2sincethe lim C(x)

x →0.2

C is continuous at x = 0.25 since does not exist. lim C(x) = 3 = C(0.25). C is continuous at x = 0.267 x →0.25

since lim C(x)=3=C(0.267). x →0.267

25. If we continue the pattern used for the taxi fare function, we see that for x = 2.3, which falls in the range of 2.2 and 2.4 miles, the fare will be $5.60, for x = 5, which falls between the range 2.4 and 2.6 miles, the fare is $5.90.

For x = 2.6 and x = 3 we need to be careful since they act as a boundary of two possible fares. Therefore C is continuous at x = 2.3 since lim C(x) = 5.60 = C(2.3). C x →2.3

is continuous at x = 2.5 since lim C(x) = 5.90 = C(2.5). x →2.5 Cisnotcontinuousatx=2.6since lim C(x)=5.90 x →2.6−

and lim C(x)=6.20thus, lim C(x)doesnotexist. x→2.6

x →2.6+

C is not continuous at x = 3 since lim C(x)6.50 and x→ 3 −

lim C(x) = 6.80 thus, lim C(x) does not exist.

x→ 3 +

26.

x→3

lim C(x) = $2.60, lim C(x) = $2.60, therefore

x →1/4−

x→1/4+

lim C(x)=$2.60

x →1/4

27.

lim C(x) = $2.30, lim C(x) = $2.60, therefore

x →0.2−

lim C(x) does not exist

x→0.2+

x →0.2

28.

lim C(x) = $2.90, lim C(x) = $3.20, therefore

x →0.6−

lim C(x) does not exist

x→0.6+

x →0.6

29.

lim C(x)=$2.90

38.

lim p(t) = 35

t→0.6+

lim p(t) = 33

t→0.6−

39. From the graph, the “I’ve got it” experience seems to occur after spending 20 hours on the task.

40 Once the task mastered one should get 100 correct trials out of 100 trials. . lim N(t) = 100, lim N(t) = 30, therefore limN(t) t→20− t→ 41. t→20+ does not exist 42.

lim N(t) = 100, lim N(t) = 100, therefore

t→30−

lim N(t) = 100

t→30+

t→30

43. N(t) is discontinuous at t = 20 since lim N(t) does not t→20 exist. N(t) is continuous at t = 30 since lim N(t) = 100 = t→30 N (30) 44. N(t) is discontinuous at t = 10 since lim N(t) does not t→10 exist. N(t) is continuous at t = 26 since lim N(t) = 100 = t→26 N (26) 45. A function may not be continuous if the function is not defined at one of the points in the domain, it also may not be continuous if the limit at a point does not exist, it also may not be continuous if the limit at a point is different than the value of the function at that point.

NOTE: See the graphs on page 77.

46. f x( ) is continuous by C1 and C2, lim f(x) = 19

lim C(x) = $2.90, lim C(x) = $2.90, therefore

x →0.5−

37.

x→0.5+

x→ 3

47. f x( ) is continuous by C1 and C2, lim f(x) = 0 x→ 1

x →0.5

30.

lim C(x) = $2.60, lim C(x) = $2.90, therefore

x →0.4−

lim C(x) does not exist

x→0.4+

x →0.4

31. The population function, p(t), is discontinuous at t∗ = 0.5, t∗ = 0.75, t∗ = 1.25, t∗ = 1.5, and at t∗ = 1.75 since at these points the population function has a “jump” which means that the l→imt∗p(t)doesnotexist t 32. There was a jump in the population at t = 0.5, t = 0.75, t = 1.5, and t = 1.75 due to births. While there was a decline in the populaion at t = 1.25 due to deaths.

33.

34.

lim p(t) = 12

48. g(x) is continuous by C4, lim g(x) = x→ − 1

49. g(x) is continuous by C4, lim g(x) = 1 x→ 1

50. tan x is continuous by C5, lim tan x = 1 x→ π 4

51. cot x

is continuous by C5, lim cot x = x→ π 3

52. sec x 53. csc x

is continuous by C5, lim sec x = x→ π 6

iscontinuousbyC5,limcscx= x→ π 4

t→1.5+

lim p(t) = 11

t→1.5−

35. The population function, p(t), is discontinuous at t∗ = 0.1, t∗ = 0.3, t∗ = 0.4, t∗ = 0.5, t∗ = 0.6, and at t∗ = 0.8 since at these points the population function has a “jump” which means that the l→imt∗p(t)doesnotexist t 36. There was a jump in the population at t = 0.1, t = 0.4, t = 0.5, and t = 0.6 due to births. While there was a decline in the populaion at t = 0.3 and t = 0.8 due to deaths.

1 2

54. f x()iscontinuousbyC5,limf(x)=

x→ 3

55. f x( ) is continuous by C5, lim f(x) = x→ π 3

56. g (x) is continuous by C3, lim f(x) = x→ π 4

1 √ 3 2 √ 3 √ 2 √ 19 √43 √2 1 2

57. g (x) is continuous by C5, dslimx →πg(x)=−1 6 58. Limit approaches 0.92857 59. Limit approaches 0

2

Chapter 2: Differentiation

40 60.

61.

62.

63.

x2 −25 Therfore,lim =lim(x −5)=−5−5=−10 x→ − 5 x + 5 x → − 5

Limit approaches 0 Limit approaches 1

x2 −16 (x−4)(x+4) +4 = 8. lim x→ − 4 x x +4 = − 4−4=−8

Limit approaches 0 Limit does not exist

Exercise Set 2.2 1. x2−3 is a continuous function (it is a polynomial). Therefore, we can use direct substitution lim (x2 −3)

x→ 1

= = =

(1)2−3 1 −− 23

9. Since 5x is continuous at x = −2 we can use direct substitution.

5 5 5 = − =2 − 2 lim x→ − 2 x 2 − 2 − 2 = =5 5 10. lim x − →−5x 11. The function x2+x−6 x −2 isdiscontinuousatx=2,butitcan be simplified algebraically. The limit is then computed as follows:

2. lim(x2+4)=(1)2+4=1+4=5

lim

x→ 1

3. The function f(x) = 3 x is not continuous at x −0 since the denominator equals zero. There are no algebraic simplifications that can be done to the function. To find the limit, we can either plug points that are approaching 0 from the right and the left and detrmine the limit from each side, or we can use the graph of the function to determine the limit (if it exists). Looking at the graph, we see that as x approches 0 from the left the y values are becoming more and more negative, and as x approches 0 from the right, the y values are becoming more and more positive. There3 3 3 thenlim doesnotexist. fore,since lim =lim x → 0 + x x→0−x x→0x

x→ 2

x2+x − 6 x− 2

= lim(x+3)

6 y 4

–2 0 –2 –4

2+3 5

12. x

lim →−4

x2 − x − 2 0 x+4

=

= lim(x −5) x→ − 4 − 4 −5 − 9 =

lim3x2−17

2

4 x 6

8

√352−17 √3 √25−17 38

= = = =

x→5 √

10

2

–6 –8

14. lim √ x2+5=

–10

− 4 doesnotexistsincethelimitfromtheleftof 4. lim x→ 0 + x x = 0 does not equal the limit from the right of x = 0

15.

5. 2x+5 is a continuous function (it is a polynomial). There-

16.

fore, we can use direct substitution

17.

lim(2x+5) = 2(3)+5

x→ 3

= =

18.

6+5 11

19.

6. lim (5 −3x)=5−3(4)=5−12=−7 x→ 4

7. The function x2−25 x+5 isdiscontinuousatx= be simplified algebraically. x2 −25 = x+5

(x−5)(x+4) x+4

lim

x→−4

=

2 –4

x→ 2

= =

√ 13. Since3x2 − 17 is continuous at x = 5 we can use direct substitution

8

–6

(x−2)(x+3) x−2

= limx→2

10

–10 –8

=lim(x −4) x→−4

(x−5)(x+5) x+5

−5, but it can

=x − 5

√ 22+5=

π lim(x+sinx)= +sin = + 4 √3

lim(cosx+tanx)= →π6

2

x

lim

1+sinx

lim

x→ 0

1+cosx cosx

= =

π

4+5= π 4

4

x→ π 4

x →01−sinx

√

+

√

9=3

1 √ 2

1 √3

1+0 =1 1−0 =2 1+1 1 1

Using the graph of x −2 we find that the limit as x approaches 2 does not exist since the limit from the left of x = 2 does not equal the limit from the right of x = 2

Exercise Set 2.2

41 Since we have a limit in terms of h, we can treat x as a constant. To evaluate the limit we can use direct substi25. tution (we have a polynomial in h, which is continuous for all values of h).

8 6 y4

lim (6x2 + 6xh + 2h2)

2 –4

h→ 0

0

–2

2 x

= 6x2 + 6x(0) + 2(0)2 =6x2+0+0 =6x2

4

–2 –4

26. lim (10x + 5h) = 10x + 5(0) = 10x

–6

h→ 0

–8

20. The limit to the left of x = 1 is ∞ and the limit to the 1 right of x = 1 is ∞ .Thus,lim doesnotexist x →1 (x − 1)2

27. Since we have a limit in terms of h, we can treat x as a − 2 x − h is continuous at h = 0 we can use constant. Since x2(x+h)2 direct substitution −2x−h lim →0x2 (x+h)2

21. Since 3x2−4x+2 7x2 −5x+3 is continuous at x = 2 we can use direct substitution 3x2 −4x+2 lim →27x2− x 5x+3

= = = =

4x2 + 5x − 7 lim →−13x2− 2x+1

= = = =

12 −8+2 28 −10+3 6 21 2 7

4(−1)2+5(−1)−7 3(−1)2−2(−1)+1 45−7 − 3+2+1 −8 6 4 −3

x2

+x−6 23. Thefunction x2 −4 isdiscontinuousatx=2.Butwecan simplify it algebraically first then find the limit as follows

lim

x→ 2

x2+x − 6 x2− 4

= = = =

(x−2)(x+3) lim x→2(x−2)(x+2) (x + 3) lim x →2 (x + 2) 2+3 2+2 5 4

24. lim

x2 −16

x →4x2−x−

= 12 = = =

=

3(2)2−4(2)+2 7(2)2−5(2)+3

22. x

= =

h

(x−4)(x+4) lim x→4(x−4)(x+3) (x + 4) lim x →2 (x + 3) 4+4 4+3 8 7

=

−5 28. lim h→0x(x+h) 29.

=

−5 x(x+0)

−2x−0 x2(x+0)2 −2x x2 (x)2 −2x x4 −2 x3

= −5 x2

tanx sinx cosx=1 =lim x→0 x x sinx lim =1. x→ 0 x 1 x = =1 30. limxcscx=lim x→ 0 1 x→0sinx lim → 0 x

·1=1Recallthat

sinh 31. limh →sinxsinh 0 =sinxlim h→0 h h =sinx1=sinx · cosh−1 sin x(cos h−1) h =sinxlim =sinx ·0=0 32. lim h→0 h→ 0 h 33. x2+3x lim →0x−2x4

x

= = = =

x(x+3) limx→0x(1−2x3) lim

(x + 3)

x →0 (1 − 2x)

(0 + 3) (1 −0) 3 =3 1

x(1−2x2) x2 −2x =lim 34. lim x→0 (x+3) x →0x2+3x

0 = 3 =0

35. √ xx lim →0x+x2

x

= = =

√ xx limx→0x( √1+x) x lim x →√0(1+x) 0 =0 (1 + 0)

Chapter 2: Differentiation

42 x+ 2 x(1√+x) 36. lim √x =lim x→0 x x x →0x x =lim 1√ + x √ 1 √ = + x Thelimitdoesnotexist x→ 0 x x −2

37.

√3

x−2 = lim x→2(x−2)(x+1) 1 = = xlim →2 (x + 1)

x lim x →2x2−x− 2

1 (2+1)

1 = 3

38. x2 − 1 lim x →−1x+1

43.

=

lim

= lim(x =

− √3

√ √ 3 · √3− x + √ 3 x+ 3 3− √3 − x − 3 x( 3 − x+ 3)√ −1 √ √ 3− x + 3 √

= = =

√4+

x

− √ 4−x

√ 4+x − √ 4−

= = 2=

x

lim

x2 −9 6

= = = =

( −1−1)=−2

lim lim

Thus, lim √ √ x →0 4+x+ 4 − x

(x−3)(x+3) 2(x−3)

(x + 3) 2 (3 + 3) 2 6 2=3

x→ 3

3x2 + 5x − 2 (x+2)(3x−1) lim → − 2 x 2 − 3 x − = limx→−2(x+2)(x−5) 10 (3x −1) = = x lim →−2 (x − 5) = (3( −2)−1) ( −2 − 5)

4

46 .

√7+ 2x − √ 7

= = =

x

2 Thus,lim √ x →0 7+2x+ 7 47.

√a2 − 4 a2+5 − 3

√a2 +5+3 √a2−4 a2 a2+5 −√ 3 · √ +5+3 (a − 4)( a2+5+3) (a2+√5−9) 2 − 4)( a2+5+3) (a2−4) √(a

= = = =

2

Thus,lim(a2+5+3)=6 a→−2√ 42.

√

x→ 1

x+1

48.

7−√49−x2 x

= = =

(√

Thus,lim √ x →07+ 49 2

√ 2 x 2+ x 4− x · 2 + √ x 4 −x (4 − x)(2+ x)√ 1 2+ √x

= = =

1 = 4

1 Thus,lim √ x →42+ x

√x − √1 1)( x+1) x− √1x+1

x− 1 = x− 1 =1 1 =√

Thus, lim √

a2+5+3

√ 7+2x − √ 7 √ 7+2x+ 7 √ √ · √ x 7+2x+ 7 7+2x 7− √ x( √ 7=2x+ 7) 2 √ √ x 3− + 3 = 1 √ √ 7 − √

2− √ x 4− x

− 7 =1 −7 41.

1= 2

45. Limit approaches 3

40. x

√ √ x · √4+x+ √ 4− x 4+x+ x) 4− x

x 4+x (4− √ − x( √ 4+x+ 4 − x) 2 √4+x+ √ 4 − x

39. x→ 3 2 x −

− √

44.

−1)

x→ − 1

3− x x

1 Thus,limx √ − −1 √ = − √ →0 3 x+ 3 2 3

(x−1)(x+1) x+1

x→1

− x x

x

− x2

7

− √

√ 49 − x2 7+ 49 − x2 x 7+ √ 49 − x2 ·

49 −(49 √ − x2) x(7+ 49 − x2) x √ 7+ 49 − x2 =0

Exercise Set 2.3

43 First we obtain the expression for f(x + h) with 3.

a)

Exercise Set 2.3 1.

f (x) = − 7 x 2 f(x+h) = −7(x + h)2 −7(x2 2xh +−h2) −7x2 −+14xh 7h2 =

a) First we obtain the expression for f(x + h) with f(x) = 7x 2 f(x+h) = 7(x+h)2

f(x+h) −f(x) h

Then f(x+h) −f(x) h

= = =

(7x2+14xh+7h2)−7x2 h

14xh + 7h2 h h(14x + 7h) h = 14x+7h

−14xh−7h2 h h( −14x−7h) h −14x − 7h

Forx=4andh=1, −14x − 7h = −14(4) − 7(1) = −56 − 7 = −63 For x = 4 and h = 0.1,

Forx=4andh=1,

−14x − 7h = −14(4) − 7(0.1) = −56 − 0.7 = −56.7

14x + 7h = 14(4) + 7(1) = 56 + 7 = 63

For x = 4 and h = 0.01,

For x = 4 and h = 0.1,

−14x−7h = −14(4)−7(0.01) = −56−0.07 = −56.07

14x + 7h = 14(4) + 7(0.1) = 56 + 0.7 = 56.7 4.

For x = 4 and h = 0.01,

a) f(x+h) −f(x)

14x + 7h = 14(4) + 7(0.01) = 56 + 0.07 = 56.07

h

a) h

(−7x2−14xh−7h2)−(−7x2) h

−14x − 7h = −14(4) − 7(2) = −56 − 14 = −70

14x + 7h = 14(4) + 7(2) = 56 + 14 = 70

f(x+h) −f(x)

= = = =

b) Forx=4andh=2,

b) Forx=4andh=2,

2.

=

Then

= 7(x2+2xh+h2) = 7x2+14xh+7h2

= = =

(5x2+10xh+5h2)−5x2 h

10xh + 5h2 h h(10x + 5h) h = 10x+5h

Forx=4andh=1, −10x − 5h = −10(4) − 5(1) = −40 − 5 = −45 For x = 4 and h = 0.1,

Forx=4andh=1,

−10x − 5h = −10(4) − 5(0.1) = −40 − 0.5 = −40.5

10x + 5h = 10(4) + 5(1) = 40 + 5 = 45

For x = 4 and h = 0.01,

For x = 4 and h = 0.1,

10x + 5h = 10(4) + 5(0.01) = 40 + 0.05 = 40.05

−10xh−5h2 h h( −10x−5h) h −10x − 5h

−10x − 5h = −10(4) − 5(2) = −40 − 10 = −50

10x + 5h = 10(4) + 5(2) = 40 + 10 = 50

For x = 4 and h = 0.01,

(−5x2−10xh−5h2)−(−5x2) h

b) Forx=4andh=2,

b) Forx=4andh=2,

10x + 5h = 10(4) + 5(0.1) = 40 + 0.5 = 40.5

= = = =

−10x−5h = −10(4)−5(0.01) = −40−0.05 = −40.05 5.

a) First we obtain the expression for f(x + h) with f(x) = 7x3 f(x+h) = 7(x+h)3 = 7(x3+3x2h+3xh2+h3) = 7x3+21x2h+21xh2+7h3

44

Chapter 2: Differentiation For x = 4 and h = 0.01,

Then f(x+ ) h −f(x) h

(7x3+21x2h+21xh2+7h3)−7x

= = =

3

h 21x2h + 21xh2 + 7h3 h h(21x2 + 21xh + 7h2 h

15x2+15xh+5h2 = 15(4)2+15(4)(0.01)+5(0.01)2 = 240+.6+0.0005 = 240.6005 7.

a) First we obtain the expression for f(x + h) with f(x) = 5 x

= 21x2+21xh+7h2

5

f(x+h) =

b) Forx=4andh=2,

(x + h)

21x2+21xh+7h2 = 21(4)2+21(4)(2)+7(2)2 = 336+168+28 =

Then

532

5

f(x+ ) h − f(x)

Forx=4andh=1,

= = = = =

h

21x2+21xh+7h2 = 21(4)2+21(4)(1)+7(1)2

5 (x+h)

= 336+84+7 =

21x2+21xh+7h2 = 21(4)2+21(4)(0.1)2 = 336+8.4+0.07 = 344.47

−5 x(x+h)

21x2+21xh+7h2 = 21(4)2+21(4)(0.01)+7(0.01)2 = 336+0.84+0.0007

−5 x(x+h)

a) h

(5x3+15x2h+15xh2+5h3)−5x

3

−5 x(x+h)

−5 x(x+h)

b) Forx=4andh=2,

=

380

Forx=4andh=1, 15x2+15xh+5h2 = 15(4)2+15(4)(1)+5(1)2 = 240+60+5 =

305

For x = 4 and h = 0.1, 15x2+15xh+5h2 = 15(4)2+15(4)(0.1)+5(0.1)2 = 240+6+0.05 = 246.05

=

−5 4(4+2)

5 = − ≈ − 0.208 26

=

−5 4(4+1)

5 = − ≈ − 0.25 20

=

−5 4(4+0.1)

5 = − ≈ − 0.305 16.4

For x = 4 and h = 0.01,

= 15x2+15xh+5h2

= 240+120+20

5

− x · x(x+h) h 1 ·x(x+h)

For x=4andh=0.1,

h 15x2h + 15xh2 + 5h3 h h(15x2 + 15xh + 5h2 h

15x2+15xh+5h2 = 15(4)2+15(4)(2)+5(2)2

x(x+h)

For x=4andh=1,

= 336.8407

= = =

·

b) For x=4andh=2,

For x = 4 and h = 0.01,

f(x+h) −f(x)

h

5

5x − 5(x+h) hx(x + h) − 5h hx(x + h) −5 x(x + h)

427

For x = 4 and h = 0.1,

6.

−

8.

=

−5 4(4+0.01)

=

−5 0.312 16.04 ≈ −

a) f (x +h) − f(x) h

4

= = = = =

4

− (

h 4 (x+h)·xx +

4 )− x ·x(x+h) h h 1 ·x(x+h)

4x − 4(x+h) hx(x + h) − 4h hx(x + h) −4 x(x + h)

45

Exercise Set 2.3 b) For x =4andh=2, −4

b) Forx=4andh=2, −4 4(4+2)

= − 1 ≈−0.167 6

−4 = 4(4+1) x(x+h)

= −1 ≈−0.2 5

=

x(x+h)

2x+h −1=2(4)+2−1=8+2−1=9 Forx=4andh=1,

Forx=4andh=1, −4

2x+h −1=2(4)+2−1=8+1−1=8 For x = 4 and h = 0.1, 2x+h −1=2(4)+0.1−1=8+0.1−1=7.1

For x = 4 and h = 0.1, −4 x(x+h)

=

−4 4(4+0.1)

For x = 4 and h = 0.01,

= − 1 ≈−0.244 4.1

2x+h −1=2(4)+2−1=8+0.01−1=7.01

For x = 4 and h = 0.01, −4 x(x+h) 9.

=

12.

−4 4(4+0.01)

= − 4 ≈−0.249 4.04

a) f(x+h) −f(x) h

2xh+h2+h h h(2x + h + 1) h = 2x+h − 1

a) First we obtain the expression for f(x + h) with f (x) = −2x + 5 f(x+h) = −2(x + h) + 5 −2x−2h+5 =

Then f(x+h) −f(x) h

b) Forx=4andh=2, 2x+h+1=2(4)+2+1=8+2+1=11

(−2x−2h+5)−(−2x+5) h

= =

Forx=4andh=1,

−2h h −2

=

2x+h+1=2(4)+2+1=8+1+1=10 For x = 4 and h = 0.1, 2x+h+1=2(4)+0.1+1=8+0.1+1=9.1

b) Since the difference quotient is a constant, then the value of the difference quotient will be -2 for all the values of x and h.

10.

For x = 4 and h = 0.01, 2x+h+1=2(4)+2+1=8+0.01+1=9.01

a) f(x+h) −f(x)

= = =

h

13.

(2x+2h+3)−(2x+3) h 2h h 2

x2

a) f(x + h) with f(x) = x2 − x

Then = = =

14.5 12 ≈ 1.20834poundspermonth

b) For the average growth rate during the second year we use the points (12,22.4) and (24,27.8)

(x2+2xh+h2−x−h)−(x2 2xh+h2 − h h h(2x+h 1)−

h = 2x+h 1 −

h

= =

x2

−(x+h) = x2=2xh+h2 − x − h − x)

22.4−7.9 12−0

= =

y2 − −y x1 1

f(x+h) = (x+h)2

f(x+h) −f(x) h

a) For the average growth rate during the first year we use the points (0,7.9) and (12,22.4) y2 − −y x1 1

b) Since the difference quotient is a constant, then the value of the difference quotient will be 2 for all the values of x and h. First we obtain the expression for 11.

(x2+2xh+h2+x+h)−(x2+x) h

= = =

27.8−22.4 24−12

5.4 12 = 0.45poundspermonth

c) For the average growth rate during the third year we use the points (24,27.8) and (36,31.5) y2 − −y x1 1 x2

= =

31.5−27.8 36−24 3.7 12

≈ 0.30834poundspermonth

46

Chapter 2: Differentiation 102−102.4 =−0.4degreesperday 8− 7 100−102 =−2degreesperday −8 9 98.6 −100 =−1.4degreesperday 10− 9 98.6−98.6 −10 =0degreesperday 11

d) For the average growth rate during his first three years we use the points (0, 7.9) and (36, 31.5) y2 − y 1 − x1 x2

= =

31.5−7.9 36−0 23.6 36

c) The temperature began to rise on day 1, reached the peak on day 5, began to subside on day 6, and was d) back to normal on day 10

≈ 1.967poundspermonth e) The graph indicates that the highest growth rate out of the first three years of a boy’s life happens at birth (that is were the graph is the steepest).

14.

a) b) c)

15.

20.5 −7.9 9−0 =1.4poundspermonth 17.4 −7.9 ≈1.583poundspermonth 6− 0 ≈1.767poundspermonth 13.2 −7.9 − 0 3

17.

Average rate of change from t = 0 to t = 8 10−0 N2 − 8−0 −N t 11 = = t2 10 8 =1.25wordsperminute

d) Based on the the calculated average growth rates above, we can estimate the average growth rate to be larger than 1.767 for the first few weeks of a typical boy’s life.

For the average growth rate between ages 12 and 18 a) months y2 − y 1 − x1 x2

= =

By examining the graph, we notice an increase in tem- peraure after day 1 which reaches a maximum value of 102.5o on day 5 and continues at the max temp through day 6 at which time the temperature begins to drop until it reaches the normal level of a) 98.6o on day 10.

Average rate of change from t = 8 to t = 16 N2 − −N t 11 t2

25.9−22.4 18−12 3.5 6

N2 − −N t 11 t2

b) For the average growth rate between ages 12 and 14 (we use the point at 15 months) = =

24.5−22.4 15−12

16.

≈

N2 − −N t 11 t2

b)

N2 − −N t 11 t2

≈ 0.831 ≈ 0.83poundspermonth

99.5−98.6 =0.9degreesperday −1 1012−99.5 =1.5degreesperday −2 3−101 102−3 =1degreesperday 4 102.5 −102 =0.5degreesperday 5− 4 102.5−102.5 =0degreesperday −5 6 =−0.1degreesperday 102.4−102.5 −6 7

= =

25−25 32−24 0 8 =0wordsperminute

Average rate of change from t = 32 to t = 36

23.23−22.4

d) The average growth of a typical boy when he is 12 months old is about 0.9 pounds per month a) 98.6 −98.6 10−1 =0degreesperday. Usingthisrateof change we would not be able to conclude the person was sick.

25−20 24−16 5 8 =0.625wordsperminute

= =

Average rate of change from t = 24 to t = 32

2.1 3 = 0.7poundspermonth

c) For the average growth rate between ages 12 and 13 (we can approximate the value of y when x = 13 by reading it from the graph) y2 − y 1 x 2 − x1

10 8 =1.25wordsperminute

Average rate of change from t = 16 to t = 24

≈ 0.583poundspermonth

y2 − y 1 − x1 x2

20−10 16−8

= =

= =

25−25 36−32 0 4 =0wordsperminute

b) The rate of change becomes 0 after 24 minutes because the number of words memorized does not change and remains at 25 words, that means that there is no change in the number of words memorized after 24 minutes. s(2) = 10(2)2 = 10(4) = 40 miles.

18.

a) s(5) = 10(5)2 = 10(25) = 250 miles b) s(5)

19.

− s(2) = 250 − 40 = 210 miles. This represents the distance travels between t = 2 and t = 5 seconds. c) 250 3=70milesperhour −40 = 210 5−2 a) When t = 3, s = 16(3)2 = 16(9) = 144 feet b) When t = 5, s = 16(5)2 = 16(25) = 400 feet

47

Exercise Set 2.3 = 256 2 =128feetper c) Average velocity = 400−144 5−3 =14.5milespergallon second =14.5milespergallon 20. a) 3097020 − 30680 b) 3097020 − 30680 21. a) Population A: The average growth rate = 500−0 4− 0 = = 500 4 =125millionperyear Population B: The average growth rate = 500−0 4− 0 125 million per year b) We would not detect the fact that the population grow at different rates. The calculation shows the populations growing at the same average growth rate, since for either population we used the same points to calculate the average growth rate (0, 0), and (4, 500).

25.

c) Population A: Between t = 0 and t = 1, Average Growth Rate = 290−0 1−0 =290millionpeopleperyear Between t = 1 and t = 2, Average Growth Rate = 250−290 2−1 =−40millionpeopleperyear Between t = 2 and t = 3, Average Growth Rate = 200−250 3−2 =−50millionpeopleperyear Between t = 3 and t = 4, Average Growth Rate = 500−200 4−3 =300millionpeopleperyear

26.

Population B: Between t = 0 and t = 1, Average Growth Rate = 125−0 1−0 =125millionpeopleperyear Between t = 1 and t = 2, Average Growth Rate = 250−125 2−1 =125millionpeopleperyear Between t = 2 and t = 3, Average Growth Rate = 375−250 3−2 =125millionpeopleperyear Between t = 3 and t = 4, Average Growth Rate = 400−375 4−3 =125millionpeopleperyear d) It is clear from part (c) that the first population has different growing rates depending on which interval of time we choose. Therefore, the statement “the population grew by 125 million each year” does not convey how population went through periods were the population increased and periods were the population decreased.

f (x + h) − f(x) h

=

= = =

f (x + h) − f(x) h

24. f(x+h) −f(x) h

= = = =

m(x+h)+b−(mx+b) h mx+mh+b −mx−b h mh h m

( + h)3+( bx + h)2 − ( ax3+ bx2) ax h 3 2 + 3 axh + 3 axh2 + h3 + h ax 2 2 2 bx + 2 bxh+ bh − ax3 − bx h 2 3axh + 3 axh2 + h3 + 2 bxh+ bh2 h 3ax2 + 3axh + h2 + 2bx+ bh (3 ax2+2 bx)+ h(3 ax +) b

= =

= = = 27. f (x + h) − f ()x h

√

−√ x + h −√ x + h √ x h x· √√ +x h √ x +√ + h x+ h x √x+ h − √x h x + h+ x 1 √ √ x + h+ x + )4

= = = =

28. f (x + h) − f(x) h

22. In the period between 1850 and 1860 the deer population is decreasing which corresponds to a negative rate of change.

In the period between 1890 and 1960 the deer population is increasing which corresponds to a positive rate of change.

23. The rate of change in the period between 1800 and 1850 is similar to that of 1930 to 1950 is the sense that they both exhibit steady increase in the population. The drastic drop in the population shortly after 1850 is similar to the drop in population seen near 1975.

a( x + h)2+( b x+ h)+ c − h ( ax2 + bx+ )c h 2 + 2 axh + ah2 + bx+ bh+ c ax h − ax2 + bx+ c h 2axh + ah2 + bh h 2ax + ah + b a(2 x + h)+ b

=

= = = =

4

(x

h − x h 4 3 + 6 2xh 2 + 4 xh3 + h4 − x4 x + 4 xh h 4xh 3 + 6 xh 2 2 + 4 xh3 + h4 h 4x3 + 6 2xh + 4 xh2 + h3

29. f (x + h) − f ()x h

1

= = = = =

−

1

h 2 x − ( x + h)2 2 hx ( x + h)2 2 2 2 x − x − 2xh − h 2 hx ( x + h)2 − 2x − h x 2 ( x + )2 h − 2x + h x 2 ( x + h )2

Chapter 2: Differentiation

48 30. f (x + h) − f(x) h

= = = =

1 1− (x+h)

1

− 1− x

h (1 − x ) −(1 (x+h)) − h(1 − x)(1 (x+h)) − 1 − x −1 + x + h h(1 − x )(1 (x+h)) − 1 (1 − x )(1 (x+h)) −

Exercise Set 2.4 1. a-b) f (x)=5x

2

100 50

–4

31. f (x + h) − f(x) h

(x+h) (1+x+h)

= = = =

–100

c) f′

(x)

2(x+h) − √ 3− 2x h √ −3 2(x+h) (3 2x) − √−− √− − h( 3 2(x+h)+ 3 2x) −2 3 2(x+h)+ 3 2x √ − √ − 3

= = = = =

1 √1 √ + xh − x

) ( ) fx( + h−fx h 5(x + h)2 − 5 x 2 lim h→ 0 h 5x2 + 10xh + 5h2 − 5 x 2 lim h→ 0 h h(10x + 5h) lim h h→ 0 = lim10x+5h = = = =

lim

h→ 0

h→ 0

=

10x

d) f′( −2) = 10(−2) = −20

33. f (x + h) − f(x) h

4

x 1+x

h (x+h)(1+x) − x(1+x+ ) h h(1 + x)(1 + x + h) x2+x+xh+h − x − x2 −xh h(1 + x)(1 + x + h) 1 (1+x)(1+x+h)

= = =

2 x

–50

−

32. f (x + h) − f ()x h

0

–2

2. a-b)

h x+h √ √−√√ x h x x+h √ √ √ √ x√ − √ x + h x + x + h h x x + h √ x + √x + h

f′(0) = 10(0) = 0 f′(1) = 10(1) = 10. These slopes are in agreement with the slopes of the tangent lines drawn in part (b). f(x) = 7x2 50 40 30 y 20

√ √ x (x+h) − √ √ h x x+h( x+ x+h) √√ −1 √ x x+h( x+ x+h)√

10

–4

–3

–2

–1

34.

0

1

2 x

3

4

–10

f (x + h) − f(x) h

= = = =

2(x+h) x+h 1 −2xx1 − −

h 2(x+h)(x 1) − 2x(x+h − 1) − h(x 1)(x+h 1) − − 2x2 −2x+2xh 2h 2x2 − 2xh+2x − − h(x 1)(x+h 1) − − −2 (x 1)(x+h 1) − −

c) f ′ ( x )= = = =

lim

h→ 0

lim

f (x + h) −()f x h 7( x + h2 ) − 7 x2

h 7x2 +14 +7 xh lim h→ 0 h h(14x+7 ) h lim h→ 0 h = lim14 +7 x h h→ 0

=

h→ 0

14x

h2 − 7x 2

49

Exercise Set 2.4 d) f′( 2) − = 14( −2) = −28 f′(0) = 14(0) = 0 f′(1) = 14(1) = 14. These slopes are in agreement with the slopes of the tangent lines drawn in part 3. a-b) (b). f(x)=−5x2

= =

d) f′( −2) = −14(−2) = 28 −14(0) = 0 −14(1) = −14. These slopes are in f ′(0) = fagreement ′(1) = with the slopes of the tangent lines

10 –4

–3

–2

–1

1

x 2

lim −14x−7h −14x h→ 0

drawn in part (b). f(x) = x3 3

4

5. a-b) 10

–10

8

–20 y

6 y 4

–30

20 –2 –4

–40

–8

–4 x

–6

–2

2

4

–6

–50

–8

c)

–10 ′

f(x)

= = = = = =

lim

h→ 0

f(x+h)−f(x) h

−5(x+h)2−(−5x2) h→ 0 h −5x2−10xh−5h2−(−5x2) lim h→ 0 h h( −10x−5h) lim h→ 0 h lim −10x−5h h→ 0 −10x

d) f′( 2)= −2) = 20 − 10(− f′(0)= 10(0)=0 − f′(1)= 10(1)= −10. These slopes are in agreement − with the slopes of the tangent lines drawn in part (b).

4. a-b) f(x)= 7x2 − 10 –4

–3

–2

c)

lim

–1

1

x 2

3

4

′

f(x)

= = = =

f(x+h)−f(x) h

lim

h→ 0

lim

h→ 0

lim

(x + h)3 − x 3 h x3 + 3x2h + 3xh2 + h3 − x 3

h h(3x2 + 3xh + h2) lim h h→ 0 h→ 0

= lim3x2+3xh+h2 h→ 0

=

3x2

d) f′( 2)=3( − − 2)2=12 f′(0) = 3(0)2 = 0 f′(1) = 3(1)2 = 3. These slopes are in agreement with the slopes of the tangent lines drawn in part (b). f(x)= x3 6. a-b)

−

–10

10 8

–20 y

6 y 4

–30 –40 –8

–4 x

–6

–50

–2

20 –2 –4

2

–6

c)

–8

f ′ ( x )= = = =

lim

h→ 0

f(x+h) − f(x) h

−7(x + h)2 − (−7x2 ) h 2 −7x2 − 14xh − 7h2 − −( 7 x ) lim h→ 0 h h( −14x − 7h) lim h→ 0 h lim

h→ 0

–10

c) f ′ ( x )= =

f (x + h) − f(x) h h ( + ) 3 − (−x3 ) lim − x h→ 0 h lim

h→ 0

4

50

Chapter 2: Differentiation

= = = =

−x3−3x2h−3xh2−h3−(−x3) h

lim

h→ 0

lim

=

h( −3x2−3xh−h2)

h lim −3x2−3xh−h2 h→ 0 −3x2 h→ 0

= =

d) f′( −2) = −3(−2)2 = −12 f ′(0) = f′(1) = with the slopes of the tangent lines −3(0)2 −3(1)2 = =0 −3. These slopes are in agreement drawn in part (b). f(x) = 2x + 3 7. a-b)

–4

=

–2

−2(x+h)+5−(−2x+5) h

lim

h→ 0

−2x−2h+5+2x−5 h lim − 2 h h→ 0 h −2 lim

h→ 0

2)= 2 − d) f′( − f′(0)= 2 − f′(1)= 2.Theseslopesareinagreementwiththe − slopes of the tangent lines drawn in part (b). f(x)= 4x 9. a-b) −

4

4

y 2

y 2

0

2 x

4

–4

0

–2

–2

–2

–4

–4

c)

2 x

4

c) ′

f(x)

= = = = =

lim

h→ 0

f(x+h)−f(x) h

′

f(x)

–4

–2

lim

h→ 0

=

2(x+h)+3 −(2x+3) h h→ 0 2x+2h+3 − 2 x − 3 lim h→ 0 h 2h lim h →0h 2 lim

d) f′( −2)=2 f′(0) = 2 f′(1) = 2. These slopes are in agreement with the slopes of the tangent lines drawn in part (b). f(x)= 2x+5 8. a-b) −

=

−4(x+h)−(−4x) h lim −4x−4h+4x h→ 0 h − 4 h lim h→ 0 h −4 lim

h→ 0

= = =

d) f′( −2) = −4 −4 f ′(0) = f′(1) = slopes of the tangent lines −4. These slopes are in agreement with the drawn in part (b). f(x) = 1 10. a-b)

2x

4

4

y 2

y 2

0

2 x

4

–4

0

–2

–2

–2

–4

–4

c)

f(x+h)−f(x) h

2 x

4

c) f ′ ( x )=

lim

h→ 0

f (x + h) − f(x) h

f ′ ( x )=

lim

h→ 0

f (x + h) − f(x) h

Exercise Set 2.4

51 1

lim2 (x + h) − 12x h→ 0 h 1 x − 12h − 12x lim 2 h→ 0 h 1 2h lim h →0h 1 2

= = = =

c) ′

f(x)

= = = = = =

lim

x2 + 2xh + h2 h −x−h−x2+x lim h→ 0 2xh + h2 h − h lim −1) h→ 0 h h(2x + h lim h→ 0 h 1 − 2x

d) f′( −2)=2( 2)−1= 5− − f′(0)=2(0) 1=−1 − f′(1)=2(1) 1=1.Theseslopesareinagreement − with the slopes of the tangent lines drawn in part (b).

6 y4

13. a-b) f(x)=2x2+3x 2

2

–6

–4

–2

(x + h)2 −(x+h)−(x2−x)

h→ 0

d) f′( −2) = 1 2 (0)=1 2 f′ f(1)=1 2. These slopes are in agreement with the ′ slopes of the tangent lines drawn in part (b).

11. a-b) f(x) = x2 + x

f(x+h)−f(x) h

lim

h→ 0

0

2

x4

− 6

6

–2

y4

–4

2

–6

–4

–2

c)

0

2 x

4

–2 ′

f(x)

= = = = =

f(x+h)−f(x) h

lim

h→ 0

lim

h→ 0

lim

–4

(x+h)2+(x+h) h x2+2xh+h2+x+h

h→ 0

2xh+h2+h h h(2x + h + 1) lim h h→ 0 = 2x+1

−x2−x

h

lim

2

–4 –6

2

h 2(x + h)2 + 3(x + h)

4xh + 2h2 + 3h h h(4x + 3) lim h h→ 0 = 4x+3

−2−(2x2+3x h

− 2)

d) f′( −2)=4(2)+3=−5 − f′(0) = 4(0) + 3 = 3 f′(1) = 4(1) + 3 = 7. These slopes are in agreement with the slopes of the tangent lines drawn in part (b).

14. a-b) f(x)=5x2 2x+7 −

y4

–2

lim

f(x +) h −fx ()

lim

6

0

lim

h→ 0

h→ 0

12. a-b) (b). f(x)=x2− x

–2

= = = =

h→ 0

d) f′( −2)=2(2)+1=−3 − f′(0) = 2(0) + 1 = 1 f′(1) = 2(1) + 1 = 3. These slopes are in agreement with the slopes of the tangent lines drawn in part

–4

c) f′ (x)

h→ 0

–6

–6

−(x2+x)

x4

6

52

Chapter 2: Differentiation 1(2)2=−1 d) f′( −2)=− − f′(0) = does not exist f′(1) = − 1(1)2 = −1. These slopes are in agreement with the slopes of the tangent lines drawn in part (b).

25 20 15

16. a-b) f(x) = 5

y 10

x

10 8

5

–4

6 y 4

0

–2

2 x

4

2

c)

–4

f′

(x)

= = = = =

lim

h→ 0

lim

h→ 0

lim

h→ 0

lim

( fx+h)−f(x) h

10xh + 5h2 h h(10x + 5h

–8

h −2h −2)

c) ′

f(x)

= = = = =

d) f′( −2) = 10(−2) − 2 = −22 −2=−2 − 2 = 8. These slopes are in agreement f′(0) = 10(0) with the slopes of the tangent lines drawn in part f′(1) = 10(1) (b). f(x) = 1 15. a-b) x

= =

f(x+h)−f(x) h

lim

h→ 0

lim

h→ 0

lim

5 (+)−5 xh 5 ( xh + )

h

h→ 0

x

h−5 x·x(x+h) ·xx(+) h·x(x+h)

lim

5x −5(x+h) hx(x+h)

lim

−5h)

h→ 0

4

h→0 hx(x + h)

y 2

h→0 x(x + h)

−5 lim −5x2

0

–2

4

–10

h

10x − 2

2 x

–6

5(x + h)2 −2(x+h)+7−(5x2−2x+7)

h→ 0

–4

0 –2 –4

–2

2 x

4

–2 –4

5 5 d)f′ (2)= − − f′ (0) = does not exist f′ (1)= − 5(1)2 = −5. These slopes are in agreement with the slopes of the tangent lines drawn in part (b).

17. fx( ) = mx

c) f ′ ( x )= = = = = = =

lim

h→ 0

lim

h→ 0

1 (+h) x

h

1 (x+h)

f(x)

− ( 1

= = = = =

lim

h→ 0

lim

h→ 0

lim

1 h) − x · x( x +h)

·xx+ h→ 0 h · x (x + h) x − ( x + h) lim ( + h) h→ 0 hxx − h) lim h→ 0hxx ( + h) −1 lim ) h→ 0 x( x+ h − 1 x2 lim

′

f(x+h) − f(x) h

h→ 0

f(x+h)−f(x) h m(x + h) −mx

h mx + mh−mx mh

h

lim →0h m h

18. f (x)=ax2+bx+c f ′ x()

= =

lim

h→ 0

lim

h→ 0

f(x+h) −f(x) h a(x+h)2+b(x+h)+c

−(ax2+bx+c) h

Exercise Set 2.4 = = =

53 ax2 + 2axh + ah2 + bx + bh −ax2−bx−c

lim

h→ 0

2axh + ah2 + bh lim h h→ 0 h(2ax + ah + b) lim h h→ 0 = lim2ax+ah+b

h

52 21. FromExercise14,f′(x)= − x.Forthepoint(15)we , f ′ 512=−5=m.Sotheequationofthe have (1)= − tangent line is

y −y1 = m(x − x1) y −5 = − 5(x 1) − y = − 5x+5+5 y = − 5x+10

h→ 0

= 2ax+b 19. f(x) = x2. From Example 3, f′(x) = 2x. For the point (3,9) we have f′(3) = 2(3) = 6 = m. So the equation of the tangent line is y

y

−y1 = m(x y −9 = 6(x 3) − y = 6x 18+9 −

6x − 9

For the point ( −1,1) we have f′( 1)− = 2(−1) = −2. So the equation of the tangent line is y −y1 = m(x − x1) y −1 − 2(x (−1))− −1 = = − 2x − 2 y y = − 2 x −2 + 1 y =

For the point (100,0.05), which can be rewritten as 5 1 20 (100,1),wehavef′ (10)= − 1002= − 2000.Sotheequation of the tangent line is y −y1 = m(x − x1) (x 1 + − 1 = y − x − 100) 20 = = y = 1 1 y y − + 2000 20 20 2 − x + 2000 20 1 − x + 2000 10

− 2x − 1

For the point (10,100) we have f′(10) = 2(10) = 20. So the equation of the tangent line is − x1) y −100 = 20(x 10) − y = 20x 200+100 − y = 20x 100− y −y1 = m(x

20. f(x) = x3. From Example 4, f′(x) = 3x2. For the point ( −2,−8) we have f′(−2) = 12 = m. So the equation of the tangent line is y −y1 = m(x y+8 = 12(x+2)

− x1)

y = 12x+24 8 y = 12x+16

−

For the point (0,0) we have f′(0) = 0. So the equation of the tangent line is y −y1 = m(x y −0 = 0(x 0) y = 0

− x1) −

For the point (4,64) we have f′(4) = 48. So the equation of the tangent line is − x1) −64 = 48(x 4) − y y = 48x 192+64 − y = 48x 128− y −y1 = m(x

− x1) 5(x (−1))− −(−5) = − y+5 = − 5(x+1) = − 5x − 5 − 5 y = − 5x − 10 y

y −y1 = m(x

− x1)

y =

− 5(−1)2 = −5.

For the point ( −1,−5) we have f′( 1) − = So the equation of the tangent line is

22. f(x) = 2 x. Using the difference quotient we find − f(x) h 2 − x

f(x+h)

′

f(x) = lim

h→ 0

= lim(x+h)

2

h→ 0 h = lim(x+h) (2 + ) · x x h→ 0

= = = =

lim

h→ 0

2

h − x ·x(x+h) hx(x+h) ·

2x − 2(x+h) hx(x+h)

lim

− 2h)

lim

−2

h→0 hx(x + h) h→0 x(x + h)

−

2 x2

2 Forthepoint( 1,−2)wehavef′( 1)= − − − ( −1)2 = −2 = m. So the equation of the tangent line is

y − y1 = m(x − x1) = 2(x y − (2) − − − (1)) − y+2 = 2(x+1) − = y − 2x − 2 − 2 = y − 2x − 4

54

Chapter 2: Differentiation For the point (2,1) we have f′(2) = equation of the tangent line is

1 −222=−2.Sothe

24. ′

y −y1 = m(x−x1) 1 − = y y −(x−2) = 2 = y x − 2 +1+1 +2 −x 2 2 1 f′ = Forthepoint(10,1 5) we have (10) −102 = −50. So the equation of the tangent line is y −y1 = m(x−x1) 1 −1 = y − x (x−10) 5 = = − 50 1 1 x + 5+ 5 y y 2 − +50 5 23. First, let us find the expression for f′(x). f(x+h)−f(x) h h→ 0 4 −(x+h)2−(4−x2) lim = h→ 0 h 4 = −x2−2xh−h2−4+x2 lim h→ 0 h −2xh−h2 lim h→ 0 h h( − 2 x − h ) lim h→ 0 h = lim( − 2 x − h ) h→ 0 −2x =

′

f(x)

= = =

lim

For the point ( −1, 3) we have f′(−1) = −2(−1) = 2 = m.

So the equation of the tangent line is y −y1 = m(x−x1) y −3 −3 = = 2(x−(−1)) y

−2x+2 y = 2x+2+3 y = 2x+5

For the point (0,4) we have f′(0) =

−2(0) = 0. So the

4

For the point (5, −21) we have f′(5) = −2(5) = −10. So the equation of the tangent line is y

= = = = =

f(x+h)−f(x) h

lim

h→ 0

lim

(x + h)2 −2(x+h)−(x2−2x)

x2 + 2xh + h2 h −2x−2h−x2+2x lim h→ 0 2xh + h2 h −2h −2) lim h→ 0 h h(2x + h lim h→ 0 h = lim(2x+h −2) h h→ 0

=

→0

2x − 2

For the point ( −2,8) we have f′(−2) = −6 = m. So the equation of the tangent line is y −y1 = m(x−x1) y −8 −8== −6(x−(−2)) −6x−12 −6x−12+8 −6x − 4 = y y = y For the point (1, −1) we have f′(1) = 0. So the equation of the tangent line is y −y1 = m(x−x1) y −(−1) = 0(x−0) y = 0− 1 −1 y = For the point (4,8) we have f′(4) = 6. So the equation of the tangent line is y −y1 = m(x−x1) y −8 = 6(x−4) y = 6x −24+8 −16 y =

equation of the tangent line is y −y1 = m(x−x1) y −4 = 0(x−0) y = 0+4 y =

f(x)

y −y1 = m(x−x1) −(−21) = − 10(x−5)

y+21 = − 10x+50 y = − 10x+50 −21 y = − 10x+29

6x

25. The function is not differentiable at x0 since it is discontinuous, x3 since it has a corner, x4 since it has a corner, x6 since it has a corner, and x12 since it has a vertical tangent.

The function is not differentiable at x2, x4, x5, x7 and x8.

26.

The function is not differentiable at interger values of x 27. since the function is not continuous at integer values of x.

28. The function is not differentiable at values of x of the form ···.

29. x = 0.2n where n = 1, 2, 3, The function is differentiable for all values in the domain.

30.

The function is not differentiable in the year 1850, 1860, 31. 1865, 1880, 1910, 1960, 1975, and 1980.

The function is differentiable for all values in the domain.

55

Exercise Set 2.4 32.L4 and L6 aretheonlylinesthatappeartobetangent since the others intersect the curve and are no “tangent” to it.

= = =

33. Asthepoints getclosertoPthesecantlinesaregetting Q closer to the tangent line at point P. 34. f(x+h) −f(x) = lim = h = h→ 0 (x + h)4 − x 4 lim h→ 0 h 4x3h + 6x2h2 + 4xh3 + h4 lim h h→ 0 = lim4x3+6x2h+4xh2+h3

′ f(x)

=

38. f ′ ( x )= = = = = = =

h→ 0

4x3

35. ′

f(x)

= = = = = = =

lim

h→ 0

lim

h→ 0

lim

f(x+h)−f(x) h

1 1 x2 +2xh+h2 − x 2

h x2 −x2−2xh−h2

h→0 hx2(x2 + 2xh + h2)

lim

−h(2x−h)

lim

−2x+h

f ′ ( x )= = = = = = =

h→0 hx2(x2 + 2xh + h2) h→0 x2(x2 + 2xh + h2)

36. ′

= = = = =

lim

h→ 0

lim

f(x+h)−f(x) h 1 1 1 − x − h −1−x

h 1 −x−1+x+h lim h→0 h(1 − x)(1 − x − h) 1 lim h→0(1−x)(1−x−h) 1 (1 − x ) 2 h→ 0

′

f(x)

= = = =

lim

h→ 0

f(x+h)−f(x) h

x+h

x

lim 1+x+h− h→ 0 h (x + h)(1 + x) −x(1+x+h) lim h(1+x)(1+x+h) h→ 0 x+x2+h+hx −x−x2−xh lim h(1+x)(1+x+h) h→ 0

f( x+ h) − f (x ) h 1 1 lim √ − √ h→ 0 x + h x √ √ x√ − √ x + h lim h →0h x x+ h √ √ √ √−√√ + x x h x+ x+ h lim ·√ √ h →0h x x+ h x+ x+ h x− √x− h lim √ √ h +h ( x + √ x + h) x x →0h ) −√ 1 lim √ √ √ h→ 0 x x + h( x + x+ h 1 √− 2 x3 lim

h→ 0

40. f ′ ( x )= = =

37.

f (x + h) − f (x ) h √ + − √ x h x lim h→ 0 √ x + h − √x √ + + √ x h x lim · √x h √ x h→ 0 h − x + + hh √ x lim ++ + √ ) h → 0 (h x x h+ √ x h lim √ h ( + h x) →0h ) 1 lim √ √ h→ 0 ( x + h + x 1 2√ x lim

h→ 0

39.

−2x x2 (x2 ) −2 x3

f(x)

h lim (1+ )(1+ →0h x +) x h 1 lim (1+ h→ 0 x )(1+ x+ ) h 1 (1+ x ) 2 h

= = = =

lim

h→ 0

lim

f (x + h) − f (x ) h 3(+) x h − 3x ++5 xh x+5

h (3 x+3 )( h +5)x3 ( + +5) − x x h lim x2 h→ 0 h(+5)(+ x x h +5) 32+15 x x +3 +15 xh h − 3 − 3xh − 15x lim h→ 0 h(+5)(+ x x h +5) 15h lim h→0h ( +5)( x + +5) x h 15 lim h→0( x+5)( + +5) x h 15 (x + 5)2 h→ 0

Chapter 2: Differentiation

56 41.

42.

43.

44.

45.

1

The function f(x) will is not differentiable at x = −3.

dy 13. dx= cosx − 1−2= x

f′(3) = 6, f′( 1)− =

f′( 2)=12,f′(0)=0,f′(4)=48 −

d 1/2 1 dy 14. dx=dx( x − x −1/2 )=2 x 2−3/2 √12 = 12 −1/2 + 1 or x + x x )2−1

f′(1)= 5,f′( − 1)= −

dy +1 15. dx=2(2 x

−2, f′(10) = 20

−5, f′(100) = −12000

x2 cosx −1/2−1 1

(2)=4(2 +1) x

46.

f′( 1)= − 2,f′(2)= − −1,f′(10)= 2−2

47.

f′( 1)=2,f′(0)=0,f′(5)= −

48.

f′( 2)= − 6,f′(1)=0,f′(4)=6 −

5. 11.5 18. f′(x) = 0 32(12 . x = 4 11.5 x

a) 30

19. f′(x)=10

dy 16. dx=2(3 x − 2)(3)=6(3 x − 2)

100

2. 3.2−1) = 0 8. 2.2 17. f′(x) = 0 25(3 . x x

−10

21. f′(x)=

1

− 2

23. Rewrite as f( )x= 5 −1x+ x,

15 10 5

0

10

15

t20

25

30

35

b) y = 0.41x + 16.067 c) Rate of change is 0.41 • (12,w(12) and (24,w(24)): y = 0.472416x+15.318, Rate of change = 0.47241 • (12,w(12) and (18,w(18)): y = 0.5685x + 14.165, Rate of change = 0.0.5685 • (12,w(12) and (15,w(15)): y = 0.6326x + 13.395, Rate of change = 0.0.6326 e) The slope of the tangent line at (12,w(12)) appears to be approaching 0.71

dy = 7 x 7−1 =7 x 6 1. dx 7 dy 2. dx=8 x 2 dy 3. dx=3 · 2x − 1 = 6 x dy = 20 x 3 4. dx dy

5. dx=4 ·3 x

3− 1 =12

x

2 2/3 − 1

x

= 2 x − 1/3

1 8. dy − 2 · 2 −1/2 =− dx= x 9. Rewriteas =y x 3/4,

10. Rewriteas =y dy 11. dx=4 cosx dy 12. dx= − 3sinx

x 4/7,

− x

1/2

29. f(x) = 2 + x f′(x)= − x − 2 30. f(x)=3 −1 4 −2 x − x f′(x)= 3 −2+8 x −3 x − 31. −1

f(x)= +2+3−1+4−2 x x x 32. f′(x)=1 3−2−8−3 x − x

′ q(x)=8

dy= 3 3/4 − 1 = 3 −1/4 x 4x dx 4 dy −3/7 dx= 4x 7

1

f ′ ( x) = − 5 x− 2 + 5

x

37. r(x)=12 cosx +8 sinx − 5 r′(x)= 12− sinx +8 cosx 38. s(x)=3 √ − √ 2cosx 2 2sinx − √ − √ s′(x)= 3 2 sinx 2 2 cosx 39. f′(x)=3 2 cosx +58 sinx

2

dy ()cosx 6. dx=10

· 3

5

24. f′(x)= f′ 2−5/3+3−1/4 + 84 x 1/ 5 − 8 x − 4 4x − x 5 3 7 −3/4 1−/ 5 4 25. −1/2 −32/ + x x x x (x)= 2 − − − 2 7/2 26. f(x)= x f′(x)=7 5/2 2x 27. f(x)= x 18/5 − x 13/5 138/5 28. f′(x)=1813/5 − 5x 5 x

33. f(x)=4 +3x2 −1 − x f′(x) = 4 + 2 −2 x 34. g(x)=2 −1 4 −2+6 −3 x − x x g′(x)= 2 −2+8 x −3 18 −4 x − − 35. p(x)=6 −4x 2 −3 − x 36. p′(x)= 24−−5+6 x −4 x

Exercise Set 2.5

dy

cosx −√ 3 9 sinx

22. f′(x)=−2 x2

20

7. dx=3

cosx +12 sinx

20. f′(x)= 5 √

25

d)

+21 x−/1 2− 1

3

( √35) cosx + 8 ( √7)5 sinx

4 W(x)=x 3+3 2 − x−1 cosx +3 4 ′ W(x)=1+−2 x − 3 sinx 3

− U(x)= sinx 21/2+3−1/2 x x U′(x)= cosx − x −1/2 − 3−3/2 2x