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  • 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= − +

    University Calculus Early Transcendentals 3rd Edition Hass Solutions Manual Full Download: http://testbanklive.com/download/university-calculus-early-transcendentals-3rd-edition-hass-solutions-manual/

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  • 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