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### Transcript of Solutions Manual - CHAPTER 2 LIMITS AND ... Copyright 2016 Pearson Education, Inc. (c) Does not...

• Copyright  2016 Pearson Education, Inc. 43

CHAPTER 2 LIMITS AND CONTINUITY

2.1 RATES OF CHANGE AND TANGENTS TO CURVES

1. (a) (3) (2) 28 9 3 2 1

19f f f x

Δ − − Δ −= = = (b)

(1) ( 1) 2 0 1 ( 1) 2

1f f f x

Δ − − − Δ − −= = =

2. (a) (3) (1) ( )3 1 3 1 2

2g gg x

−Δ − − Δ −= = = (b)

(4) ( 2) 8 8 4 ( 2) 6

0g gg x

− −Δ − Δ − −= = =

3. (a) ( ) ( )34 4

3 4 4 2

1 1 4h hh t

π π

π π π π −Δ − −

Δ − = = = − (b)

( ) ( )2 6 2 6 3

0 3 3 3h hh t

π π

π π π π − − −Δ

Δ − = = =

4. (a) ( ) (0) (2 1) (2 1) 2 0 0

g g g t

π π π π

Δ − − − + Δ − −= = = − (b)

( ) ( ) (2 1) (2 1) ( ) 2

0g g g t

π π π π π

Δ − − − − − Δ − −= = =

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

1R RRθ − + − −Δ

Δ −= = = =

6. (2) (1) (8 16 10) (1 4 5) 2 1 1

2 2 0P PPθ − − + − − +Δ

Δ −= = = − =

7. (a) 2 2 2((2 ) 5) (2 5) 4 4 5 1y h h h

x h h Δ + − − − + + − + Δ = = =

24 4 .h h h

h+ = + As 0, 4 4h h→ + →  at (2, 1)P − the slope is 4. (b) ( 1) 4( 2) 1 4 8y x y x− − = −  + = − y = 4 9x −

8. (a) 2 2 2(7 (2 ) ) (7 2 ) 7 4 4 3y h h h

x h h Δ − + − − − − − − Δ = = =

24 4 .h h h

h− − = − − As 0, 4h h→ − − → 4−  at (2, 3)P the slope is 4.−

(b) 3 ( 4)( 2) 3y x y− = − −  − = 4 8x− +  4 11y x= − +

9. (a) 2 2((2 ) 2(2 ) 3) (2 2(2) 3)y h h

x h Δ + − + − − − − Δ = =

2 24 4 4 2 3 ( 3) 2 2 .h h h h h h h

h+ + − − − − − += = + As 0, 2 2h h→ + →  at (2, 3)P − the slope is 2.

(b) ( 3) 2( 2) 3y x y− − = −  + = 2 4x −  2 7.y x= −

10. (a) 2 2((1 ) 4(1 )) (1 4(1))y h h

x h Δ + − + − − Δ = =

2 21 2 4 4 ( 3) 2 2.h h h h h h h

h+ + − − − − −= = − As 0, 2 2h h→ − → −  at (1, 3)P − the slope is 2.−

(b) ( 3) ( 2)( 1) 3 2 2y x y x− − = − −  + = − +  2 1.y x= − −

11. (a) 3 3 2 3(2 ) 2 8 12 4 8y h h h h

x h h Δ + − + + + − Δ = = =

2 3 212 4 12 4 .h h h h

h h+ + = + + As 0,h → 212 4h h+ + → 12,  at (2, 8)P the slope is 12.

(b) 8 12( 2) 8 12 24y x y x− = −  − = −  12 16.y x= −

12. (a) 3 3 2 32 (1 ) (2 1 ) 2 1 3 3 1y h h h h

x h h Δ − + − − − − − − − Δ = = =

2 3 23 3 3 3 .h h h h

h h− − − = − − − As 0,h → 3− 23 3,h h− − → −  at (1, 1)P the slope is 3.−

(b) 1 ( 3)( 1) 1 3 3y x y x− = − −  − = − +  3 4.y x= − +

• 44 Chapter 2 Limits and Continuity

Copyright  2016 Pearson Education, Inc.

13. (a) 3 3(1 ) 12(1 ) (1 12(1))y h h

x h Δ + − + − − Δ = =

2 3 2 31 3 3 12 12 ( 11) 9 3h h h h h h h h h

+ + + − − − − − + += = 29 3 .h h− + + As 0,h → 29 3h h− + + → 9−  at (1, 11)P − the slope is 9.−

(b) ( 11) ( 9)( 1) 11 9y x y x− − = − −  + = − + 9 9 2.y x = − −

14. (a) 3 2 3 2(2 ) 3(2 ) 4 (2 3(2) 4)y h h

x h Δ + − + + − − + Δ = =

2 3 2 2 38 12 6 12 12 3 4 0 3h h h h h h h h h

+ + + − − − + − += = 23 .h h+ As 20, 3 0h h h→ + →  at (2, 0)P the slope is 0.

(b) 0 0( 2) 0.y x y− = −  =

15. (a) Q Slope of p t

PQ ΔΔ=

1(10,225)Q 650 22520 10 42.5 m/sec − − =

2 (14,375)Q 650 37520 14 45.83 m/sec − − =

3(16.5, 475)Q 650 47520 16.5 50.00 m/sec −

− = 4 (18,550)Q 650 55020 18 50.00 m/sec

− − =

(b) At 20,t = the sportscar was traveling approximately 50 m/sec or 180 km/h.

16. (a) Q Slope of p t

PQ ΔΔ=

1(5, 20)Q 80 2010 5 12 m/sec − − =

2 (7,39)Q 80 3910 7 13.7 m/sec − − =

3(8.5,58)Q 80 5810 8.5 14.7 m/sec − − =

4 (9.5,72)Q 80 7210 9.5 16 m/sec − − =

(b) Approximately 16 m/sec

17. (a)

(b) 174 62 1122014 2012 2 56 p t

Δ − Δ −= = = thousand dollars per year

(c) The average rate of change from 2011 to 2012 is 62 272012 2011 35 p t

Δ − Δ −= = thousand dollars per year.

The average rate of change from 2012 to 2013 is 111 622013 2012 49 p t

Δ − Δ −= = thousand dollars per year.

So, the rate at which profits were changing in 2012 is approximately 1 2

(35 49) 42+ = thousand dollars per year.

18. (a) ( ) ( 2)/( 2)F x x x= + − x 1.2 1.1 1.01 1.001 1.0001 1

( )F x 4.0− 3.4− 3.04− 3.004− 3.0004− 3−

4.0 ( 3) 1.2 1

5.0;F x

− − −Δ Δ −= = −

3.4 ( 3) 1.1 1

4.4;F x

− − −Δ Δ −= = −

3.04 ( 3) 1.01 1

4.04;F x

− − −Δ Δ −= = −

3.004 ( 3) 1.001 1

4.004;F x

− − −Δ Δ −= = −

3.0004 ( 3) 1.0001 1

4.0004;F x

− − −Δ Δ −= = −

0 2010 2011 2012 2013 2014

40

80

120

160

200

Year

Pr of

it (1

00 0s

)

p

t

• Section 2.1 Rates of Change and Tangents to Curves 45

Copyright  2016 Pearson Education, Inc.

(b) The rate of change of ( )F x at 1x = is 4.−

19. (a) (2) (1) 2 1 2 1 2 1

0.414213g g g x

Δ − − Δ − −= = ≈

(1.5) (1) 1.5 1 1.5 1 0.5

0.449489g g g x

Δ − − Δ −= = ≈

(1 ) (1) 1 1 (1 ) 1

g g h g h x h h

Δ + − + − Δ + −= =

(b) ( )g x x=

1 h+ 1.1 1.01 1.001 1.0001 1.00001 1.000001

1 h+ 1.04880 1.004987 1.0004998 1.0000499 1.000005 1.0000005

( )1 1 /h h+ − 0.4880 0.4987 0.4998 0.499 0.5 0.5

(c) The rate of change of ( )g x at 1x = is 0.5. (d) The calculator gives 1 1 1

20 lim .h

hh

+ − →

=

20. (a) i) 1 1 1 3 2 6(3) (2) 1

3 2 1 1 6 f f

−−− − = = = −

ii) 1 1 2

2 2 2( ) (2) 2 2 2 2 2 ( 2)

T T T Tf T f T

T T T T T

− −− − − − − −= = = =

2 1 2 (2 ) 2

, 2T T T T

T−− − = − =/ (b) T 2.1 2.01 2.001 2.0001 2.00001 2.000001 ( )f T 0.476190 0.497512 0.499750 0.4999750 0.499997 0.499999 ( ( ) (2))/( 2)f T f T− − 0.2381− 0.2488− 0.2500− 0.2500− 0.2500− 0.2500−

(c) The table indicates the rate of change is 0.25− at 2.t = (d) ( )1 12 42lim TT −→ = −

NOTE: Answers will vary in Exercises 21 and 22.

21. (a) 15 0 1 0

[0, 1]: 15 mph;s t

Δ − Δ −= = [1, 2.5]:

s t

Δ Δ =

20 15 10 2.5 1 3

mph;−− = 30 20 3.5 2.5

[2.5, 3.5]: s t

Δ − Δ −= = 10 mph

(b) At ( )12 , 7.5 :P Since the portion of the graph from 0t = to 1t = is nearly linear, the instantaneous rate of change will be almost the same as the average rate of change, thus the instantaneous speed at 1

2 t = is

15 7.5 1 0.5

− − = 15 mi/hr. At (2, 20):P Since the portion of the graph from 2t = to 2.5t = is nearly linear, the

instantaneous rate of change will be nearly the same as the average rate of change, thus 20 20 2.5 2

0 mi/hr.v −−= = For values of t less than 2, we have

Q Slope of s t

PQ ΔΔ=

1(1, 15)Q 15 201 2 5 mi/hr − − =

2 (1.5,19)Q 19 201.5 2 2 mi/hr − − =

3(1.9,19.9)Q 19.9 201.9 2 1 mi/hr − − =

Thus, it appears that the instantaneous speed at 2t = is 0 mi/hr. At (3, 22):P

Q Slope of s t

PQ ΔΔ=

1(4, 35)Q 35 22

4 3 13 mi/hr−− =

2 (3.5, 30)Q 30 22 3.5 3

16 mi/hr−− =

3(3.1, 23)Q 23 22 3.1 3

10 mi/hr−− =

Thus, it appears that the instantaneous speed at 3t = is about 7 mi/hr.

Q Slope of s t

PQ ΔΔ=

1(2, 20)Q 20 222 3 2 mi/hr − − =

2 (2.5, 20)Q 20 222.5 3 4 mi/hr − − =

3(2.9, 21.6)Q 21.6 222.9 3 4 mi/hr − − =

• 46 Chapter 2 Limits and Continuity

Copyright  2016 Pearson Education, Inc.

(c) It appears that the curve is increasing the fastest at 3.5.t = Thus for (3.5, 30)P Q Slope of

s t

PQ ΔΔ=

1(4, 35)Q 35 304 3.5 10 mi/hr −

− =

2 (3.75, 34)Q 34 303.75 3.5 16 mi/hr − − =

3(3.6, 32)Q 32 303.6 3.5 20 mi/hr − − =

Thus, it appears that the instantaneous speed at 3