Solucionario - Cálculo trascendentes tempranas - 4ta Edición - Dennis g. zill • warren s. wright

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Transcript of Solucionario - Cálculo trascendentes tempranas - 4ta Edición - Dennis g. zill • warren s. wright

  • Single Variable Calculus

    Early Transcendentals

    Complete Solutions Manual

    John David N. Dionisio Brian Fulton Melanie Fulton

    Fourth Edition

  • Contents

    1 Functions 21.1 Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Combining Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3 Polynomial and Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4 Transcendental Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331.5 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431.6 Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 481.7 From Words to Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Chapter 1 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

    2 Limit of a Function 732.1 Limits An Informal Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 732.2 Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 772.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812.4 Trigonometric Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 862.5 Limits that Involve Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 922.6 Limits A Formal Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 982.7 The Tangent Line Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Chapter 2 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

    3 The Derivative 1173.1 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1173.2 Power and Sum Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.3 Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1333.4 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1393.5 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1463.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1533.7 Derivatives of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1633.8 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

    ii

  • CONTENTS iii

    3.9 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1743.10 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180Chapter 3 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

    4 Applications of the Derivative 1944.1 Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1944.2 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2014.3 Extrema of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2104.4 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2164.5 Limits Revisited LHopitals Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 2224.6 Graphing and the First Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2314.7 Graphing and the Second Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . 2394.8 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2484.9 Linearization and Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2634.10 Newtons Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271Chapter 4 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

    5 Integrals 2865.1 The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2865.2 Integration by the u-Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2905.3 The Area Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2985.4 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3095.5 Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318Chapter 5 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330

    6 Applications of the Integral 3356.1 Rectilinear Motion Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3356.2 Area Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3406.3 Volumes of Solids: Slicing Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3516.4 Volumes of Solids: Shell Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3596.5 Length of a Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3666.6 Area of a Surface of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3706.7 Average Value of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3746.8 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3786.9 Fluid Pressure and Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3826.10 Centers of Mass and Centroids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385Chapter 6 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394

  • iv CONTENTS

    B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395

    7 Techniques of Integration 4017.1 Integration Three Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4017.2 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4057.3 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4137.4 Powers of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4327.5 Trigonometric Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4437.6 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4607.7 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4807.8 Approximate Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496Chapter 7 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509

    8 First-Order Differential Equations 5288.1 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5288.2 Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5338.3 Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5398.4 Solution Curves without a Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5478.5 Eulers Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555Chapter 8 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561

    9 Sequences and Series 5709.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5709.2 Monotonic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5769.3 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5819.4 Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5929.5 Comparison Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6019.6 Ratio and Root Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6079.7 Alternating Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6149.8 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6219.9 Representing Functions by Power Series . . . . . . . . . . . . . . . . . . . . . . . . . 6299.10 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6419.11 Binomial Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656Chapter 9 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663

  • CONTENTS 1

    10 Conics and Polar Coordinates 66810.1 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66810.2 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68510.3 Calculus and Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69410.4 Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69910.5 Graphs of Polar Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70410.6 Calculus in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71110.7 Conic Sections in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 723Chapter 10 in Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 731

    A. True/False . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 731B. Fill in the Blanks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 732C. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 733

  • Chapter 1

    Functions

    1.1 Functions and Graphs

    1. f(5) = (5)2 1 = 25 1 = 24f(

    3) = (

    3)2 1 = 3 1 = 2f(3) = (3)2 1 = 9 1 = 8f(6) = (6)2 1 = 36 1 = 35

    2. f(5) = 2(5)2 + (5) = 2(25) 5 = 55f( 12 ) = 2( 12 )2 + ( 12 ) = 2( 14 ) 12 = 1f(2) = 2(2)2 + (2) = 2(4) + 2 = 6f(7) = 2(7)2 + (7) = 2(49) + 7 = 91

    3. f(1) =1 + 1 =

    0 = 0

    f(0) =

    0 + 1 =

    1 = 1

    f(3) =

    3 + 1 =

    4 = 2

    f(5) =

    5 + 1 =

    6

    4. f( 12 ) =

    2( 12 ) + 4 =1 + 4 =

    3

    f( 12 ) =

    2( 12 ) + 4 =

    1 + 4 =

    5

    f( 52 ) =

    2( 52 ) + 4 =

    5 + 4 =

    9 = 3

    f(4) =

    2(4) + 4 =

    8 + 4 =

    12 = 2

    3

    2

  • 1.1. FUNCTIONS AND GRAPHS 3

    5. f(1) = 3(1)(1)2 + 1 =

    31 + 1

    = 32

    f(0) =3(0)

    (0)2 + 1= 0

    f(1) =3(1)

    (1)2 + 1=

    32

    f(

    2) =3(

    2)(

    2)2 + 1=

    3

    22 + 1

    =

    2

    6. f(

    2) =(

    2)2

    (

    2)3 2=

    22

    2 2=

    1

    2 1= 1

    2 + 1

    f(1) = (1)2

    (1)3 2 =1

    1 2 =13 =

    13

    f(0) =(0)2

    (0)3 2 =02 = 0

    f( 12 ) =

    (12

    )2(

    12

    )3 2=

    14

    18 2

    =14

    18 2

    (88

    )=

    21 16 =

    215 =

    215

    7. f(x) = 2x2 + 3xf(2a) = 2(2a)2 + 3(2a) = 2(4a2) + 6a = 8a2 + 6af(a2) = 2(a2)2 = 3(a2) = 2a4 + 3a2

    f(5x) = 2(5x)2 + 3(5x) = 2(25x2) 15x = 50x2 15xf(2a+ 1) = 2(2a+ 1)2 + 3(2a+ 1) = 2(4a2 + 4a+ 1) + 6a+ 3 = 8a2 8a 2 + 6a+ 3

    = 8a2 2a+ 1f(x+ h) = 2(x+ h)2 + 3(x+ h) = 2(x2 + 2xh+ h2) + 3x+ 3h

    = 2x2 4xh 2h2 + 3x+ 3h

    8. f(x) = x3 2x2 + 20f(2a) = (2a)3 2(2a)2 + 20 = 8a3 2(4a2) + 20 = 8a3 8a2 + 20f(a2) = (a2)3 2(a2)2 + 20 = a6 2a4 + 20

    f(5x) = (5x)3 2(5x)2 + 20 = 125x3 2(25x2) + 20 = 125x3 50x2 + 20f(2a+ 1) = (2a+ 1)3 2(2a+ 1)2 + 20 = 8a3 + 3(2a)2 + 3(2a) + 1 2(4a2 + 4a+ 1) + 20

    = 8a3 + 12a2 + 6a+ 1 8a2 8a 2 + 20 = 8a3 + 4a2 2a+ 19f(x+ h) = (x+ h)3 2(x+ h)2 + 20 = x3 + 3x2h+ 3xh2 + h3 2(x2 + 2xh+ h2) + 20

    = x3 + 3x2h+ 3xh2 + h3 2x2 4xh 2h2 + 20

  • 4 CHAPTER 1. FUNCTIONS

    9. Setting f(x) = 23 and solving for x, we find

    6x2 1 = 236x2 = 24

    x2 = 4x = 2.

    When we compute f(2) and f(2) we obtain 23 in both cases, so x = 2 is the answer.

    10. We solve f(x) = 4:x 4 = 4x 4 = 42 = 16

    x = 16 + 4 = 20.

    11. We need 4x 2 0:

    4x 2

    x 12.

    The domain is [12,).

    12. The domain of f(x) =

    15 5x is the set of all x for which 15 5x 0. This is equivalentto:

    15 5x3 xx 3.

    The domain of f(x) is (, 3].

    13. We need 1 x > 0. This implies x < 1, so the domain is (, 1).

    14. We need 3x 1 > 0. This implies x > 13 , so the domain is (13 ,).

    15. The domain of f(x) = (2x 5)/x(x 3) is the set of all x for which x(x 3) 6= 0. Sincex(x 3) = 0 when x = 0 or x = 3, the domain of f(x) is {x | x 6= 0, x 6= 3}.

    16. We need x2 1 6= 0 or x2 6= 1. Thus, x 6= 1 and the domain is (,1) (1, 1) (1,).

    17. We need x2 10x+ 25 6= 0 or (x 5)2 6= 0. Thus, x 6= 5 and the domain is (, 5) (5,).

    18. The domain of f(x) = (x + 1)/(x2 4x 12) is the set of all x for which x2 4x 12 6= 0.Since x2 4x 12 = (x + 4)(x 6) = 0 when x = 4 or x = 6, the domain of f(x) is{x|x 6= 4, x 6= 6}.

  • 1.1. FUNCTIONS AND GRAPHS 5

    19. We need x2 x+ 1 6= 0. Applying the quadratic formula, we have

    x =(1)

    (1)2 4(1)(1)2(1)

    =13

    2.

    Neither value of x is real, so x2 x+ 1 6= 0 for all x, and the domain is (,).20. We need x2 2x 1 6= 0. Applying the quadratic formula, we have

    x =(2)

    (2)2 4(1)(1)2(1)

    =2

    82

    = 1

    2.

    Thus, the domain is (, 1

    2) (1

    2, 1 +

    2) (1 +

    2,).21. The domain of f(x) =

    25 x2 is the set of all x for which 25 x2 0. Since 25 x2 =

    (5 + x)(5 x), we have 25 x2 0 when5 + x 0 and 5 x 0 or 5 + x 0 and 5 x 0

    Rewriting these inequalities for x, we get x 5 and x 5 for the first set of conditions orx 5 and x 5 for the second set of conditions. x 5 and x 5 at the same time cannever happen, so the domain is determined by 5 x 5, or [5, 5].

    22. We need x(4 x) 0, which is true whenx 0 and 4 x 0 or x 0 and 4 x 0

    Rewriting these inequalities for x, we get x 0 and x 4 for the first set of conditions orx 0 and x 4 for the second set of conditions. x 0 and x 4 at the same time can neverhappen, so the domain is determined by 0 x 4, or [0, 4].

    23. We need x2 5x = x(x 5) 0, which is true whenx 0 and x 5 0 or x 0 and x 5 0

    Rewriting these inequalities for x, we get x 0 and x 5 for the first set of conditions orx 0 and x 5 for the second set of conditions. The domain therefore requires that x 5or x 0, so it is (, 0] [5,).

    24. The domain of f(x) =x2 3x 10 is the set of all x for which x2 3x 10 0. Since

    x2 3x 10 = (x+ 2)(x 5), we have x2 3x 10 0 whenx+ 2 0 and x 5 0 or x+ 2 0 and x 5 0

    Rewriting these inequalities for x, we get x 2 and x 5 for the first set of conditions orx 2 and x 5 for the second set of conditions. The domain therefore requires that x 5or x 2, so it is (,2] [5,).

    25. We need (3 x)/(x+ 2) 0 with x 6= 2, which is true when3 x 0 and x+ 2 0 or 3 x 0 and x+ 2 0

    Rewriting these inequalities for x, we get x 3 and x 2 for the first set of conditions orx 3 and x 2 for the second set of conditions. x 3 and x 2 at the same time cannever happen, so the domain is determined by 2 x 3 and x 6= 2, or (2, 3].

  • 6 CHAPTER 1. FUNCTIONS

    26. We need (5 x)/x 0 with x 6= 0, which is true when5 x 0 and x > 0 or 5 x 0 and x < 0

    Rewriting these inequalities for x, we get x 5 and x > 0 for the first set of conditions orx 5 and x < 0 for the second set of conditions. x 5 and x < 0 at the same time can neverhappen, so the domain is determined by 0 < x 5, or (0, 5].

    27. Since the y-axis (a vertical line) intersects the graph in more than one point (three points inthis case), the graph is not that of a function.

    28. This is the graph of a function by the vertical line test.

    29. This is the graph of a function by the vertical line test.

    30. Since the y-axis (a vertical line) intersects the graph in more than one point (three points inthis case), the graph is not that of a function.

    31. Projecting the graph onto the x-axis, we see that the domain is [4, 4]. Projecting the graphonto the y-axis, we see that the range is [0, 5].

    32. Projecting the graph onto the x-axis, we see that the domain is [1, 1]. Projecting the graphonto the y-axis, we see that the range is [/2, /2].

    33. Horizontally, the graph extends between x = 1 and x = 9 and terminates at both ends, asindicated by the solid dots. Thus, the domain is [1, 9]. Vertically, the graph extends betweeny = 1 and y = 6, so the range is [1, 6].

    34. Projecting the graph onto the x-axis, we see that the domain is (,1) (1, 1) (1,).Projecting the graph onto the y-axis, we see that the range is (, 0) (1,).

    35. x-intercepts: Solving 12x 4 = 0 we get x = 8. The x-intercept is (8, 0).y-intercept: Since f(0) = 12 (0) 4 = 4, the y-intercept is (0,4).

    36. x-intercepts: We solve f(x) = x2 6x+ 5 = 0:x2 6x+ 5 = 0

    (x 1)(x 5) = 0x = 1, 5.

    The x-intercepts are (1, 0) and (5, 0).y-intercept: Since f(0) = 02 6(0) + 5 = 5, the y-intercept is (0, 5).

    37. x-intercepts: We solve f(x) = 4(x 2)2 1 = 0:4(x 2)2 1 = 0

    4(x 2)2 = 1

    (x 2)2 = 14

    x 2 =

    14

    = 12

    x = 2 12.

  • 1.1. FUNCTIONS AND GRAPHS 7

    Both of these check with the original equation, so the x-intercepts are ( 32 , 0) and (52 , 0).

    y-intercept: Since f(0) = 4(0 2)2 1 = 4(4) 1 = 15, the y-intercept is (0, 15).

    38. x-intercepts: We solve f(x) = (2x 3)(x2 + 8x+ 16) = 0:

    (2x 3)(x2 + 8x+ 16) = 0(2x 3)(x+ 4)2 = 0

    x =32,4.

    The x-intercepts are ( 32 , 0) and (4, 0).y-intercept: Since f(0) = [2(0) 3][(0)2 + 8(0) + 16] = 3(16) = 48, the y-intercept is

    (0,48).

    39. x-intercepts: We solve f(x) = x3 x2 2x = 0:

    x3 x2 2x = 0x(x2 x 2) = 0

    x(x+ 1)(x 2) = 0x = 0,1, 2.

    The x-intercepts are (0, 0), (1, 0), and (2, 0).y-intercept: Since f(0) = 0, the y-intercept is (0, 0).

    40. x-intercepts: We solve f(x) = x4 1 = 0:

    x4 1 = 0(x2 1)(x2 + 1) = 0

    (x 1)(x+ 1)(x2 + 1) = 0x = 1.

    The x-intercepts are (1, 0) and (1, 0).y-intercept: Since f(0) = (0)4 1 = 1, the y-intercept is (0,1).

    41. x-intercepts: We solve x2 + 4 = 0. Since x2 + 4 is never 0, there are no x-intercepts.

    y-intercept: Since f(0) = [(0)2 + 4]/[(0)2 16] = 4/(16) = 14 , the y-intercept is (0, 14 ).

    42. x-intercepts: We solve f(x) = x(x+ 1)(x 6)/(x+ 8) = 0:

    x(x+ 1)(x 6)x+ 8

    = 0

    x(x+ 1)(x 6) = 0, x 6= 8.

    Thus, x = 0, x = 1, and x = 6. The x-intercepts are (0, 0), (1, 0), and (6, 0).y-intercept: Since f(0) = 0, the y-intercept is (0, 0).

  • 8 CHAPTER 1. FUNCTIONS

    43. x-intercepts: We solve f(x) = 32

    4 x2 = 0:

    32

    4 x2 = 0

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

    x = 2.

    The x-intercepts are (2, 0) and (2, 0).y-intercept: Since f(0) = 32

    4 (0)2 = 32 (2) = 3, the y-intercept is (0, 3).

    44. x-intercepts: We solve f(x) = 12x2 2x 3 = 0:

    12

    x2 2x 3 = 0

    x2 2x 3 = 0(x 3)(x+ 1) = 0

    x = 1, 3.

    The x-intercepts are (1, 0) and (3, 0).y-intercept: Since f(0) = 12

    (0)2 2(0) 3 = 12

    3, there is no y-intercept.

    45. To find f(a) for any number a, first locate a on the x-axis and then approximate the signedvertical distance to the graph from (a, 0): f(3) 0.5 (because the graph is so steep atx = 3, f(3) could be reasonably approximated by any number from 0 to 1); f(2) 3.4;f(1) 0.3; f(1) 2; f(2) 3.8; f(3) 2.9. The y-intercept is (0, 2).

    46. The function values are the directed distances from the x-axis at the given value of x.

    f(3) 0; f(2) 3.5; f(1) 0.3; f(1) 2; f(2) 3.8; f(3) 2.8.

    47. To find f(a) for any number a, first locate a on the x-axis and then approximate the signedvertical distance to the graph from (a, 0):

    f(2) 3.6; f(1.5) 2; f(0.5) 3.3; f(1) 4.1; f(2) 2; f(3.2) 4.1.

    The x-intercepts are approximately (3.2, 0), (2.3, 0), and (3.8, 0).

    48. To find f(a) for any number a, first locate a on the x-axis and then approximate the signedvertical distance to the graph from (a, 0):

    f(2) 0; f(1.5) 1.6; f(0.5) 2.3; f(1) 3.8; f(2) 2.2; f(3.2) 0.

    The x-intercepts are approximately (3, 0), (2, 0), (0, 0), and (3.2, 0).

  • 1.1. FUNCTIONS AND GRAPHS 9

    49. Solving x = y2 5 for y, we obtain

    x = y2 5x+ 5 = y2

    y2 = x+ 5

    y = x+ 5

    The two functions are f1(x) = x+ 5 and f2(x) =

    x+ 5. The domains are both [5,).

    50. Solving for y2, we have

    x2 4y2 = 16x2 16 = 4y2

    y2 =14

    (x2 16).

    The two functions are f1(x) = 12x2 16 and f2(x) = 12

    x2 16. The domains are both

    (,4] [4,).51. (a) f(2) = 2! = 2 1 = 2

    f(3) = 3! = 3 2 1 = 6f(5) = 5! = 5 4 3 2 1 = 120f(7) = 7! = 7 6 5 4 3 2 1 = 5040Note that we could have simplified the computation of 7! in this case by writing

    7! = 7 6 5! = 7 6 120 = 5040.

    (b) f(n+ 1) = (n+ 1)! = (n+ 1)n! = n!(n+ 1) = f(n)(n+ 1)

    (c) Using the result from (b), we can simplify as follows:

    f(5)f(4)

    =f(4) 5f(4)

    = 5

    f(7)f(5)

    =f(6) 7f(5)

    =f(5) 6 7f(5)

    = 42

    (d)f(n+ 3)f(n)

    =(n+ 3)!n!

    =(n+ 3)(n+ 2)(n+ 1)n!

    n!= (n+ 1)(n+ 2)(n+ 3)

    52. (a) Identifying n = 100, we have

    S(100) =16

    (100)(100 + 1)(200 + 1) =13

    (50)(101)(201) = 338, 350.

    (b) We can simply try various values of n:

    S(8) = 204, S(9) = 285, S(10) = 385, S(11) = 506.

    We see that n = 10 results in 300 < S(10) < 400.

  • 10 CHAPTER 1. FUNCTIONS

    53. Generally, when the domain is a semi-infinite interval, the function can be one ofx a, or

    a x. One of these should work when the number a is included in the interval. To excludea number a, simply use the reciprocal of

    x a, or a x.

    (a) We try f(x) =

    3 x. Since x = 0 is in the domain of this function, but not in theinterval [3,), this is not the correct choice for f(x). We then try f(x) =

    x 3 and

    see that it does work.(b) Since 3 is not part of the interval, we let f(x) = 1/

    x 3. This function has domain

    (3,).54. (a) A function whose range is a semi-infinite closed interval will be f(x) = x2 + a when the

    range has the form [a,), or f(x) = a x2 when the range has the form (, a]. Thus,a function with range [3,) is f(x) = x2 + 3.

    -3 3

    -3

    3(b) To find a function whose range is a semi-infinite open interval, we

    note the graph of f(x) = 2x shown to the right which as range (0,).The graph of f(x) = 2x + a is the graph of f(x) = 2x shifted upwardby a units. Thus, the range of f(x) = 2x + 3 is (3,).

    55. The graph indicates that f(x) < 0 when x < 1 or x > 3, so the domain of g(x) =f(x)

    must be [1, 3]. Within this domain, 0 f(x) 4, so the range of g(x) must therefore be[0, 2].

    56. We can express M1 as (x, f(x)/2) and M2 as (x/2, f(x)). To find M3(x), note that the pointsM3 are the midpoints of line segments ST . We are told that S = (0, f(x)) and T = (x, 0), soM3(x) = ( 0+x2 ,

    f(x)+02 ) = (x/2, f(x)/2).

    57. g(x) = dxe =

    ...2, 3 < x 21, 2 < x 1

    0, 1 < x 01, 0 < x 12, 1 < x 23, 2 < x 3...

    58.

    -3 3

    -3

    3 59.

    -3 3

    -3

    3

  • 1.2. COMBINING FUNCTIONS 11

    -3 3 6

    3

    660. We discuss two methods for graphing f(x) = |x| + |x 3|. For the first

    method, graph both y = |x| and y = |x3| on the same set of axes. Then,as illustrated on the right (|x| and |x3| are rendered as dotted lines, while|x|+ |x 3| is the solid line), add the vertical distances from the x-axis upto each graph. This will give the point on the graph of f(x) = |x|+ |x3|.For the second method, since

    |x| ={x, x < 0x, x 0 and |x 3| =

    {(x 3), x 3 < 0

    x 3, x 3 0, ={

    3 x, x < 3x 3, x 3,

    then for:

    x < 0, f(x) = (x) + (3 x) = 2x+ 30 x < 3, f(x) = (x) + (3 x) = 3

    x 3, f(x) = (x) + (x 3) = 2x 3and thus, f(x) =

    2x+ 3, x < 03, 0 x < 3

    2x 3, x 3.

    -3 3 6

    3

    6To graph f(x), we note that y = 2x+ 3, where x < 0, is a line with slope2 that approaches the point (0, 3) on the y-axis, while y = 2x 3, wherex 3, is a line with slope 2 that touches the point (3, 3) on the vertical linex = 3. These two partial lines are joined by the horizontal line segmentfrom (0, 3) to (3, 3), shown on the right.

    61. Since x2 9x 3 =

    (x+ 3)(x 3)x 3 = x+ 3, x 6= 3,

    the graph of f(x) is a line with a hole at x = 3. The graph of g(x) is the same line with ahole at x = 3 and a dot at (3, 4). Since x+ 3 = 6 when x = 3, the graph of h(x) is the sameline with no holes.

    62. Since x4 1x2 1 =

    (x2 + 1)(x2 1)x2 1 = x

    2 + 1, x 6= 1,

    the graph of f(x) is a parabola with holes at x = 1. The graph of g(x) is the same parabolawith holes at x = 1 and a dot at (1, 0). Since x2 + 1 = 2 when x = 1, the graph of h(x) isthe same parabola with a hole only at x = 1.

    1.2 Combining Functions

    1. (f + g)(x) = 2x+ 13 (f g)(x) = 6x 3

    (fg)(x) = 8x2 4x+ 40 (f/g)(x) = 2x+ 54x+ 8 , x 6= 2

    2. (f + g)(x) = 5x2 + 7x 9 (f g)(x) = 5x2 7x+ 9

    (fg)(x) = 35x3 45x2 (f/g)(x) = 5x2

    7x 9 , x 6=97

    3. (f + g)(x) =x2 + x+ 1x2 + x

    (f g)(x) = x2 x 1x2 + x

    (fg)(x) =1

    x+ 1, x 6= 0 (f/g)(x) = x

    2

    x+ 1, x 6= 1, 0

  • 12 CHAPTER 1. FUNCTIONS

    4. (f + g)(x) =9x2 11

    4x2 + 14x+ 6(f g)(x) = 7x

    2 + 74x2 + 14x+ 6

    (fg)(x) =2x2 7x+ 34x2 + 14x+ 6

    (f/g)(x) =8x2 2x2 9 , x 6=

    12,3, 3

    5. (f + g)(x) = 2x2 + 5x 7 (f g)(x) = x+ 1

    (fg)(x) = x4 + 5x3 x2 17x+ 12 (f/g)(x) = x+ 3x+ 4

    , x 6= 4, 1

    6. (f + g)(x) = x2 +x (f g)(x) = x2 x

    (fg)(x) = x5/2 (f/g)(x) = x3/2

    7. (f + g)(x) =x 1 +

    2 x; the domain is [1, 2].

    8. (fg)(x) =

    (x 1)(2 x); the domain is [1, 2].

    9. (f/g)(x) =x 1/

    2 x; the domain is [1, 2).

    10. (g/f)(x) =

    2 x/x 1; the domain is (1, 2].

    11. (f g)(x) = 3x+ 16; (g f)(x) = 3x+ 4

    12. (f g)(x) = 4x2 + 1; (g f)(x) = 16x2 + 8x+ 1

    13. (f g)(x) = x6 + 2x5 + x4; (g f)(x) = x6 + x4

    14. (f g)(x) = 4x+ 9x+ 2

    ; (g f)(x) = 14x+ 12

    15. (f g)(x) = 3x+ 3x

    ; (g f)(x) = 33 + x

    16. (f g)(x) = (x2)2 +x2 = x4 + |x|

    (g f)(x) = (x2 +x)2 = x4 + 2x2x+ xNote that the domain of f g is

  • 1.2. COMBINING FUNCTIONS 13

    20. (f g)(x) = f(2x) = 5 (2x)2 = 5 (4 4x+ x) = 1 + 4x xThe domain of f is all real numbers and the domain of g, determined by x 0, is [0,).Thus, the domain of f g is [0,).

    21. (f (2f))(x) = 2(4x3)3 = 128x9; (f (1/f))(x) = 2(

    12x3

    )3=

    14x9

    22. (f (2f))(x) = 1(2

    x 1

    ) 1

    =x 13 x ; (f (1/f))(x) =

    1(x 1) 1 =

    1x 2

    23. (f g h)(x) = 36x2 36x+ 15

    24. (f g h)(x) =

    2x 2

    25. 2g(x) 5 = 4x+ 13; g(x) = 2x+ 9

    26.

    2g(x) + 6 = 4x2; g(x) = 8x4 3

    27. f(x) = 2x2 x; g(x) = x2

    28. f(x) = 1/x; g(x) = x2 + 9

    29. The point (x, y) on the graph of f corresponds to the point (x, y + 2) on the shifted graph.Thus, (2, 1) corresponds to (2, 3) and (3,4) corresponds to (3,2).

    30. The point (x, y) on the graph of f corresponds to the point (x, y 5) on the shifted graph.Thus, (2, 1) corresponds to (2,4) and (3,4) corresponds to (3,9).

    31. The point (x, y) on the graph of f corresponds to the point (x 6, y) on the shifted graph.Thus, (2, 1) corresponds to (8, 1) and (3,4) corresponds to (3,4).

    32. The point (x, y) on the graph of f corresponds to the point (x + 1, y) on the shifted graph.Thus, (2, 1) corresponds to (1, 1) and (3,4) corresponds to (4,4).

    33. The point (x, y) on the graph of f corresponds to the point (x4, y+1) on the shifted graph.Thus, (2, 1) corresponds to (6, 2) and (3,4) corresponds to (1,3).

    34. The point (x, y) on the graph of f corresponds to the point (x+5, y3) on the shifted graph.Thus, (2, 1) corresponds to (3,2) and (3,4) corresponds to (8,7).

    35. The point (x, y) on the graph of f corresponds to the point (x, y) on the shifted graph.Thus, (2, 1) corresponds to (2, 1) and (3,4) corresponds to (3,4).

    36. The point (x, y) on the graph of f corresponds to the point (x,y) on the shifted graph.Thus, (2,1) corresponds to (1, 1) and (3,4) corresponds to (3, 4).

    37. In (a) the graph is shifted up 2 units; in (b) it is shifted down 2 units; in (c) it is shifted left2 units; in (d) it is shifted right 5 units; in (e) it is reflected in the x-axis; and in (f) it isreflected in the y-axis.

  • 14 CHAPTER 1. FUNCTIONS

    (a) (b) (c)

    (d) (e) (f)

    38. In (a) the graph is shifted up 2 units; in (b) it is shifted down 2 units; in (c) it is shifted left2 units; in (d) it is shifted right 5 units; in (e) it is reflected in the x-axis; and in (f) it isreflected in the y-axis.

    (a) (b) (c)

    (d) (e) (f)

    39. In (a) the graph is shifted up 2 units; in (b) it is shifted down 2 units; in (c) it is shifted left2 units; in (d) it is shifted right 5 units; in (e) it is reflected in the x-axis; and in (f) it isreflected in the y-axis.

  • 1.2. COMBINING FUNCTIONS 15

    (a) (b) (c)

    (d) (e) (f)

    40. In (a) the graph is shifted up 2 units; in (b) it is shifted down 2 units; in (c) it is shifted left2 units; in (d) it is shifted right 5 units; in (e) it is reflected in the x-axis; and in (f) it isreflected in the y-axis.

    (a) (b) (c)

    (d) (e) (f)

    41. In (a) the graph is shifted up 1 unit; in (b) it is shifted down 1 unit; in (c) it is shifted left units; in (d) it is shifted right /2 units; in (e) it is reflected in the x-axis; in (f) it is reflectedin the y-axis; in (g) it is stretched vertically by a factor of 3; and in (h) it is compressedvertically by a factor of 1/2 and then reflected in the x-axis.

  • 16 CHAPTER 1. FUNCTIONS

    (a) (b) (c) (d)

    (e) (f) (g) (h)

    42. In (a) the graph is shifted up 1 unit; in (b) it is shifted down 1 unit; in (c) it is shifted left units; in (d) it is shifted right /2 units; in (e) it is reflected in the x-axis; in (f) it is reflectedin the y-axis; in (g) it is stretched vertically by a factor of 3; and in (h) it is compressedvertically by a factor of 1/2 and then reflected in the x-axis.

    (a) (b) (c) (d)

    (e) (f) (g) (h)

    43. If f(x) is shifted up 5 units and right 1 unit, the new function is y = f(x 1) + 5. Sincef(x) = x3, this becomes y = (x 1)3 + 5.

    44. The function is first multiplied by 3 and then x is replaced by x 2. The equation of the newgraph is thus y = 3(x 2)2/3.

    45. The function is first multiplied by 1 and then x is replaced by x + 7. The equation of thenew graph is thus y = (x+ 7)4.

    46. If f(x) is reflected in the y-axis, then shifted left 5 units and down 10 units, the new functionis y = f(x+ 5) 10. Since f(x) = 1/x, this becomes y = 1/(x+ 5) 10.

  • 1.2. COMBINING FUNCTIONS 17

    47. (a) An even function is symmetricwith respect to the y-axis.

    (b) An odd function is symmetricwith respect to the origin.

    48. (a) An even function is symmetricwith respect to the y-axis.

    (b) An odd function is symmetricwith respect to the origin.

    49. To fill in the bottom row, we use the fact that since f is an even function then f(x) = f(x),and so (f g)(x) = (f g)(x).

    x 0 1 2 3 4f(x) 1 2 10 8 0g(x) 2 3 0 1 4

    (f g)(x) 10 8 1 2 0

    50. To fill in the bottom row, we use the fact that since g is an odd function then g(x) = g(x),and so (g f)(x) = (g f)(x).

    x 0 1 2 3 4f(x) 2 3 0 1 4g(x) 9 7 6 5 13

    (g f)(x) 6 5 9 7 13

    51.

    -4 -2 2 4

    -4

    -2

    2

    4 52.

    -2 2

    -2

    2

    53. 2U(x 0) 2U(x 2) + (1)U(x 2) (1)U(x 3) = 2 3U(x 2) + U(x 3)

  • 18 CHAPTER 1. FUNCTIONS

    54. When compared to the graph of y = f(x3), the graph of y = f(x3)U(x3) is a horizontalline at zero for x < 3, then matches the graph of y = f(x 3) after that.

    y = f(x 3) = (x 3)2

    -3 3

    -3

    3

    y = f(x 3)U(x 3)

    -3 3

    -3

    3

    55.

    -4 -2 2 4

    -4

    -2

    2

    4 56.

    -3 3

    -3

    3

    57. False; let f(x) = x2 and g(x) = h(x) = 1.

    58. [3,1]

    59. If a function is symmetric with respect to the x-axis, then, for any x, both (x, y) and (x,y)are on the graph. If y 6= 0 (as is the case for a function that is nonzero for at least one valueof x), then the vertical line test fails and the graph is not the graph of a function.

    60. A vertical stretch or compression by a factor of c units results in y = cf(x). For a given x,this will result in the same y-value when f(x) = 0. Thus, the points on the graph of y = f(x)are those that also lie on the x-axis.

    When the graph is reflected in the x-axis, the point (x, y) becomes the point (x,y). Thus,the points that remain the same are those for which y = 0; that is, the x-intercepts.

    When the graph is reflected in the y-axis, the point (x, y) becomes the point (x, y). Thus,the only point that remains the same is the y-intercept.

    61. Since |x| = x when x 0 and |x| = x when x < 0, the graph of f(|x|) is the same as thegraph of f(x) when x 0. When x < 0, f(|x|) = f(x), so the graph of f(|x|) in this case isthe reflection of the graph of f(x), x > 0, in the y-axis. To summarize: to obtain the graphof f(|x|) from the graph of f(x), simply ignore the portion of the graph of f(x) to the left ofthe y-axis, and then reflect the portion of the graph of f(x) to the right of the y-axis throughthe y-axis.

  • 1.3. POLYNOMIAL AND RATIONAL FUNCTIONS 19

    62. We see from Figure 1.2.1 in the text that when x is positive:

    (a) if f(x) is near 0, then 1/x is very large;

    (b) if f(x) is large, then 1/x is near 0;

    (c) if f(x) = 1, then 1/x = 1.

    That is, for a fixed x, points (x, y) where y > 1 are reflected through the line y = 1 andcompressed into the interval 0 < y < 1. Also, for a fixed x, points (x, y) where 0 < y < 1 arereflected through the line y = 1 and stretched into the interval y > 1. Similar statementshold for x < 0. The graph of 1/f(x) is shown in blue, and the graph of f(x) is shown in red.

    -3 3

    -3

    363. frac(x) is so named because its value is the non-integer part of x, i.e., thevalue that follows the decimal point. Its graph is shown on the right.

    64. Based on the graph to the right, the graph of the ceiling function g(x) = dxe(red) can be thought of as the reflection of the floor function f(x) = bxc(blue) across the horizontal lines y = f(x)+1/2 = bxc+1/2 (gray, dotted)if x is not an integer. When x is an integer, g(x) = f(x). Thus,

    g(x) =

    {f(x) + 1, x is not an integer

    f(x), x is an integer.

    1.3 Polynomial and Rational Functions

    1. The form of the equation of the line is y = 23x+b. Letting x = 1 and y = 2, we have 2 =23 +b,

    so b = 2 23 = 43 . The equation of the line is y = 23x+ 43 .

    2. Letting m = 110 , x1 = 1, and y1 = 2, we obtain from the point-slope form of the equation ofa line that

    y 2 = 110

    (x 1) = 110x 1

    10

    y =110x+

    1910.

    3. A line with slope 0 is horizontal and has the form y = b. In this case, b = 2, so the equationif y = 2.

    4. The form of the equation of the line is y = 2x + b. Letting x = 1 and y = 2, we have2 = 2 + b, so b = 2 + 2 = 4. The equation of the line is y = 2x+ 4.

  • 20 CHAPTER 1. FUNCTIONS

    5. Letting m = 1, x1 = 1, and y1 = 2, we obtain from the point-slope form of the equation ofa line that

    y 2 = 1(x 1) = x+ 1y = x+ 3.

    6. When the slope is undefined, the line is vertical and has the form x = a. In this case a = 1,so the equation is x = 1.

    7. To find the x-intercept we set y = 0. This gives 3x + 12 = 0 or x = 4. The x-intercept is(4, 0). Now, write the equation in slope-intercept form by solving for y:

    -4 4

    -4

    43x 4y + 12 = 04y = 3x 12

    y =34x+ 3.

    The slope of the line is m = 34 and the y-intercept is (0, 3).

    -6 -3 3 6

    -6

    -3

    3

    68. Solving for y, we have 3y = 12x 3 or y = 16x 1, so the slope is 16 andthe y-intercept is (0,1). Setting y = 0, we have 0 = 16x 1 or x = 6, sothe x-intercept is (6, 0).

    -4 4

    -4

    49. Solving for y, we have 3y = 2x 9 or y = 23x 3, so the slope is 23 and they-intercept is (0,3). Setting y = 0, we have 0 = 23x 3 or x = 92 , so thex-intercept is ( 92 , 0).

    10. To find the x-intercept we set y = 0. This gives 4x + 6 = 0 or x = 32 . The x-intercept is( 32 , 0). Now, write the equation in slope-intercept form by solving for y:

    -4 4

    -4

    44x 2y + 6 = 0

    2y = 4x 6y = 2x+ 3.

    The slope of the line is m = 2 and the y-intercept is (0, 3).

    11. The slope of the line is m =5 36 2 =

    84

    = 2. The form of the equation of the line isy = 2x+ b. Letting x = 2 and y = 3, we have 3 = 2(2) + b, so b = 3 + 4 = 7. The equationof the line is y = 2x+ 7.

  • 1.3. POLYNOMIAL AND RATIONAL FUNCTIONS 21

    12. The slope of the line is m =0 (6)

    4 5 =61 = 6, and we identify x1 = 5 and y1 = 6. Then,

    using the point-slope form of the equation of a line, we have y(6) = 6(x5) = 6x+30,or y = 6x+ 24.

    13. Solving 3x + y 5 = 0 for y, we obtain y = 3x + 5. The slope of this line is m = 3, sothe form of the line through (2, 4) is y = 3x + b. Letting x = 2 and y = 4, we have4 = 3(2) + b, so b = 4 6 = 2 and the equation of the line is y = 3x 2.

    14. Lines parallel to the y-axis are vertical, so the equation is x = 5.

    15. Solving for x 4y+ 1 = 0 for y, we obtain 4y = x 1 and y = 14x+ 14 . Thus, the slope ofa line perpendicular to this one is m = 1/(1/4) = 4. We identify x1 = 2 and y1 = 3 anduse the point-slope form of the equation of a line:

    y 3 = 4(x 2) = 4x+ 8y = 4x+ 11.

    16. The line through (1, 1) and (3, 11) has the slope m =11 13 1 =

    102

    = 5, so the desired line

    has slope 15 . It then has the form y = 15x+ b. Since it passes through (5,4), we have4 = 15 (5) + b or b = 4 1 = 5. The equation of the line is then y = 15x 5.

    17. Letting x = 1, y = 5, and x = 1, y = 6 in f(x) = ax+ b, we have5 = a(1) + b6 = a(1) + b

    ora+ b = 5a+ b = 6.

    Adding these equations, we find 2b = 11, so b = 112 . Then a = 6 b = 6 112 = 12 and thefunction is f(x) = 12x+

    112 .

    18. We are looking for a function having the form f(x) = ax+ b. The first condition implies

    a(1) + b = 1 + a(2) + ba+ b = 2a+ b+ 1

    3a = 1; a = 13.

    The second condition implies

    a(3) + b = 4[a(1) + b]3a+ b = 4a+ 4b; 0 = a+ 3b.

    Since we have already determined that a = 13

    ,

    0 = 13

    + 3b

    13

    = 3b; b =19.

    The linear function is f(x) = 13x+

    19

    .

  • 22 CHAPTER 1. FUNCTIONS

    19. When x = 1, the corresponding point on the blue curve has y-coordinate y = (1)2 +1 = 2.When x = 2, the corresponding point on the blue curve has y-coordinate y = 22 + 1 = 5.The slope of the line through (1, 2) and (2, 5) is m = (5 2)/[2 (1)] = 3/3 = 1. Theform of the equation of the line is then y = 1x+ b = x+ b. Using x = 1 and y = 2 we have2 = 1 + b or b = 3. Thus, the equation of the line is y = x+ 3.

    20. The center of the circle is at (2, 3), and since the circle is tangent to the y-axis, its radius is2. Thus, the equation of the circle is (x 2)2 + (y 3)2 = 4. The point on the circle hasx-coordinate 3, so (3 2)2 + (y 3)2 = 4 or (y 3)2 = 4 1 = 3 and y =

    3 + 3. (We use

    the positive square root of 3 because the point on the circle if above the center of the circle.)The point on the circle is then (3,

    3 + 3). The slope of the line through the center of the

    circle and this point is m = [(

    3 + 3) 3]/(3 2) =

    3/1 =

    3, so the slope of the tangentline, which is perpendicular to this line, is 1/

    3. The equation of the tangent line then has

    the form y = (1/

    3)x+ b. Using x = 3 and y =

    3 + 3 we have

    3 + 3 = (1/

    3)(3) + bor b =

    3 + 3 +

    3 = 2

    3 + 3. Thus, the equation of the tangent line is:

    y = 13x+ 2

    3 + 3.

    21. (a) x-intercepts: Solving x(x+ 5) = 0 we get x = 0,5, so the x-intercepts are (0, 0) and(5, 0).

    y-intercept: Since f(0) = 0, the y-intercept is (0, 0).

    (b) f(x) = x(x+ 5) = x2 + 5x =

    [x2 + 5x+

    (52

    )2](

    52

    )2= (x+ 5/2)2 (25/4)

    -6 -3 3 6

    -6

    6(c) Identifying h = 52

    and k = 254

    in part (b), we see that the vertex

    is(5

    2,25

    4

    )and the axis of symmetry is x = 5

    2.

    (d) Since a = 1 > 0 in part (b), the parabola opens up.

    (e) The range of f(x) is [25/4,).(f) f(x) is increasing on [5/2,) and decreasing on (,5/2].

    22. (a) x-intercepts: Solving x2 + 4x = x(x + 4) = 0 we get x = 0, 4, so the x-interceptsare (0, 0) and (4, 0).

    y-intercept: Since f(0) = 0, the y-intercept is (0, 0).

    (b) f(x) = x2 + 4x = [x2 4x+ (2)2] + (2)2 = (x 2)2 + 4

    -6 -3 3 6

    -6

    6(c) Identifying h = 2 and k = 4 in part (b), we see that the vertex is(2, 4) and the axis of symmetry is x = 2.

    (d) Since a = 1 < 0 in part (b), the parabola opens down.(e) The range of f(x) is (, 4].(f) f(x) is increasing on (, 2] and decreasing on [2,).

    23. (a) x-intercepts: Solving (3x)(x+1) = 0 we get x = 3,1, so the x-intercepts are (1, 0)and (3, 0).

  • 1.3. POLYNOMIAL AND RATIONAL FUNCTIONS 23

    y-intercept: Since f(0) = (3 0)(0 + 1) = 3, the y-intercept is (0, 3).(b) f(x) = (3 x)(x+ 1) = x2 + 2x+ 3 = (x2 2x) + 3

    = (x2 2x+ 1) + 3 + 1 = (x 1)2 + 4

    -4 4

    -4

    4(c) Identifying h = 1 and k = 4 in part (b), we see that the vertex is

    (1, 4) and the axis of symmetry is x = 1.

    (d) Since a = 1 < 0 in part (b), the parabola opens down.(e) The range of f(x) is (, 4].(f) f(x) is increasing on (, 1] and decreasing on [1,).

    24. (a) x-intercepts: Solving (x 2)(x 6) = 0 we get x = 2, 6, so the x-intercepts are (2, 0)and (6, 0).

    y-intercept: Since f(0) = 12, the y-intercept is (0, 12).

    (b) f(x) = (x 2)(x 6) = x2 8x+ 12 = [x2 8x+ (4)2] (4)2 + 12 = (x 4)2 4

    -4 4

    -4

    4(c) Identifying h = 4 and k = 4 in part (b), we see that the vertex is

    (4,4) and the axis of symmetry is x = 4.(d) Since a = 1 > 0 in part (b), the parabola opens up.

    (e) The range of f(x) is [4,).(f) f(x) is increasing on [4,) and decreasing on (, 4].

    25. (a) x-intercepts: Solving x23x+2 = (x1)(x2) = 0 we get x = 1, 2, so the x-interceptsare (1, 0) and (2, 0).

    y-intercept: Since f(0) = 2, the y-intercept is (0, 2).

    (b) f(x) = x2 3x+ 2 = [x2 3x+ (3/2)2] (3/2)2 + 2 = (x 3/2)2 (1/4)

    -4 4

    -4

    4(c) Identifying h = 3/2 and k = 1/4 in part (b), we see that the vertex

    is (3/2,1/4) and the axis of symmetry is x = 3/2.(d) Since a = 1 > 0 in part (b), the parabola opens up.

    (e) The range of f(x) is [1/4,).(f) f(x) is increasing on [3/2,) and decreasing on (, 3/2].

    26. (a) x-intercepts: Factoring, we obtain

    f(x) = x2 + 6x 5 = (x2 6x+ 5) = (x 1)(x 5).

    Solving (x 1)(x 5) = 0 we get x = 1, 5, so the x-intercepts are (1, 0) and (5, 0).y-intercept: Since f(0) = 5, the y-intercept is (0,5).

    (b) To complete the square and obtain the standard form, we start by factoring 1 from thetwo x-terms:

    f(x) = x2 + 6x 5 = (x2 6x) 5 = (x2 6x+ 9) 5 + 9 = (x 3)2 + 4

  • 24 CHAPTER 1. FUNCTIONS

    -4 4

    -4

    4(c) Identifying h = 3 and k = 4 in part (b), we see that the vertex is

    (3, 4) and the axis of symmetry is x = 3.

    (d) Since a = 1 < 0 in part (b), the parabola opens down.(e) The range of f(x) is (, 4].(f) f(x) is increasing on (, 3] and decreasing on [3,).

    27. The vertex of the graph is (10, 0), so the graph of f(x) is the graph of y = x2 shifted to theright by 10 units.

    28. The vertex of the graph is (6, 0), so the graph of f(x) is the graph of y = x2 shifted to theleft by 6 units.

    29. The vertex of the graph is (4, 9), so the graph of f(x) is the graph of y = x2 reflected in thex-axis, compressed by a factor of 1/3, shifted to the left by 4 units, and shifted up by 9 units.

    30. The vertex of the graph is (2,1), so the graph of f(x) is the graph of y = x2 stretched by afactor of 10, shifted to the right by 2 units and shifted down by 1 unit.

    31. Since f(x) = (x 6)2 4 = (x+ 6)2 4, the vertex of the graph is (6,4), so the graphof f(x) is the graph of y = x2 shifted to the left by 6 units and shifted down by 4 units.

    32. Since f(x) = (1x)2 + 1 = (x1)2 + 1, the vertex of the graph is (1,1), so the graph off(x) is the graph of y = x2 reflected in the x-axis, shifted to the right by 1 unit, and shiftedup by 1 unit.

    33. End behavior: For large x, the graph is like that of y = x3.

    -4 4

    -4

    4Symmetry: Since the powers of x are all odd, the graph is symmetric

    with respect to the origin.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Solving f(x) =x3 4x = x(x 2)(x + 2) = 0, we see that the x-intercepts are(2, 0), (0, 0), and (2, 0).

    Graph: From f(x) = x1(x 2)1(x+ 2)1 we see that 0, 2, and 2 are all simple zeros.

    34. End behavior: For large x, the graph is like that of y = x3.

    -10 -5 5 10

    -10

    -5

    5

    10Symmetry: Since the powers of x are all odd, the graph is symmetricwith respect to the origin.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Solving f(x) =9x x3 = x(3 + x)(3 x) = 0, we see that the x-intercepts are(3, 0), (0, 0), and (3, 0).

    Graph: From f(x) = x1(x+ 3)1(x 3)1 we see that 0, 3, and 3 are all simple zeros.

    35. End behavior: For large x, the graph is like that of y = x3.

  • 1.3. POLYNOMIAL AND RATIONAL FUNCTIONS 25

    -5 5

    -5

    5Symmetry: Since the powers of x are both even and odd, the graph has

    no symmetry with respect to the origin or y-axis.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Solving f(x) =x3 + x2 + 6x = x(x+ 2)(x 3) = 0, we see that the x-interceptsare (0, 0), (2, 0), and (3, 0).

    Graph: From f(x) = x1(x+ 2)1(x 3)1 we see that 0, 2, and 3 are all simple zeros.

    36. End behavior: For large x, the graph is like that of y = x3.

    -5 5

    -5

    5Symmetry: Since the powers of x are both even and odd, the graph hasno symmetry with respect to the origin or y-axis.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Solving f(x) =x3 + 7x2 + 12x = x(x + 3)(x + 4) = 0, we see that the x-interceptsare (0, 0), (3, 0), and (4, 0).

    Graph: From f(x) = x1(x+ 3)1(x+ 4)1 we see that 0, 3, and 4 are all simple zeros.

    37. End behavior: For large x, the graph is like that of y = x3.

    -5 5

    -5

    5Symmetry: Since the powers of x in f(x) = (x + 1)(x 2)(x 4) =

    x35x2 +2x+8 are both even and odd, the graph has no symmetrywith respect to the origin or y-axis.

    Intercepts: Since f(0) = 8, the y-intercept is (0, 8). Solving f(x) =(x + 1)(x 2)(x 4) = 0, we see that the x-intercepts are (1, 0),(2, 0), and (4, 0).

    Graph: From f(x) = (x+ 1)1(x 2)1(x 4)1 we see that 1, 2, and 4 are all simple zeros.

    38. End behavior: For large x, the graph is like that of y = x3.

    -5 5

    -5

    5Symmetry: Since the powers of x in f(x) = (2 x)(x + 2)(x + 1) =x3x2 +4x+4 are both even and odd, the graph has no symmetrywith respect to the origin or y-axis.

    Intercepts: Since f(0) = 4, the y-intercept is (0, 4). Solving f(x) =(2 x)(x + 2)(x + 1) = 0, we see that the x-intercepts are (2, 0),(2, 0), and (1, 0).

    Graph: From f(x) = (x 2)1(x + 2)1(x + 1)1 we see that 2, 2, and 1 are all simplezeros.

    39. End behavior: For large x, the graph is like that of y = x4.

    -5 5

    -5

    5Symmetry: Since the powers of x are both even and odd, the graph hasno symmetry with respect to the origin or y-axis.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Solving f(x) =x4 4x3 + 3x2 = x2(x 1)(x 3) = 0, we see that the x-interceptsare (0, 0), (1, 0), and (3, 0).

  • 26 CHAPTER 1. FUNCTIONS

    Graph: From f(x) = x2(x 1)1(x 3)1 we see that 1 and 3 are simple zeros and the graphis tangent to the x-axis at x = 0.

    40. End behavior: For large x, the graph is like that of y = x4.

    -3 3

    -3

    3Symmetry: Since the powers of x in f(x) = x2(x 2)2 = x4 4x3 + 4x2

    are both even and odd, the graph has no symmetry with respect tothe origin or y-axis.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Solving f(x) =x2(x 2)2 = 0, we see that the x-intercepts are (0, 0) and (2, 0).

    Graph: From f(x) = x2(x 2)2 we see that the graph is tangent to the x-axis at x = 0 andx = 2.

    41. End behavior: For large x, the graph is like that of y = x4.

    -3 3

    -3

    3Symmetry: Since the powers of x are all even, the graph is symmetric

    with respect to the y-axis.

    Intercepts: Since f(0) = 1, the y-intercept is (0,1). Solving f(x) =(x42x2 +1) = (x+1)2(x1)2 = 0, we see that the x-interceptsare (1, 0) and (1, 0).

    Graph: From f(x) = (x + 1)2(x 1)2 we see that the graph is tangent to the x-axis atx = 1 and x = 1.

    42. End behavior: For large x, the graph is like that of y = x5.

    -5 5

    -5

    5Symmetry: Since the powers of x are all odd, the graph is symmetricwith respect to the origin.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Solving f(x) =x5 4x3 = x3(x + 2)(x 2) = 0, we see that the x-intercepts are(0, 0), (2, 0), and (2, 0).

    Graph: From f(x) = x3(x+2)1(x2)1 we see that 2 and 2 are simple zeros and the graphis tangent to but passes through, x-axis.

    To solve Problems 4348, first note that all of the functions have zeros at x = 0and x = 1. Next, note whether the exponents of x and x 1 are 1, even, or oddand greater than 1. Finally, use the facts that when the exponent of x a is1: the graph passes directly through the x-axis at x = a;

    a or a

    even: the graph is tangent to, but does not pass through the x-axis at x = a;

    a or a

  • 1.3. POLYNOMIAL AND RATIONAL FUNCTIONS 27

    odd and greater than 1: the graph is tangent to, and passes through the x-axis.

    a or a

    The forms of the functions in (a)-(f) are

    (a) xeven(x 1)even (b) xodd(x 1)1(c) xodd(x 1)odd (d) x1(x 1)odd(e) xeven(x 1)1 (f) xodd(x 1)even

    Because each of these is distinct, it is not necessary for this set of problems toconsider whether the lead coefficient is positive or negative.

    43. The form of the function must be xodd(x 1)even, so this graph corresponds to (f).

    44. The form of the function must be xodd(x 1)odd, so this graph corresponds to (c).

    45. The form of the function must be xeven(x 1)1, so this graph corresponds to (e).

    46. The form of the function must be xeven(x 1)even, so this graph corresponds to (a).

    47. The form of the function must be xodd(x 1)1, so this graph corresponds to (b).

    48. The form of the function must be x1(x 1)odd, so this graph corresponds to (d).

    -10 -5 5 10

    -10

    -5

    5

    1049. Vertical asymptotes: Setting 2x + 3 = 0 we see that x = 3/2 is a

    vertical asymptote.

    Horizontal asymptote: The degree of the numerator equals the degreeof the denominator, so y = 4/2 = 2 is the horizontal asymptote.

    Intercepts: Since f(0) = 3, the y-intercept is (0,3). Setting 4x9 = 0we see that x = 9/4, so (9/4, 0) is the x-intercept.

    -10 -5 5 10

    -10

    -5

    5

    1050. Vertical asymptotes: Setting x 2 = 0 we see that x = 2 is a vertical

    asymptote.

    Horizontal asymptote: The degree of the numerator equals the degreeof the denominator, so y = 2/1 = 2 is the horizontal asymptote.

    Intercepts: Since f(0) = 2, the y-intercept is (0,2). Setting 2x+4 = 0we see that x = 2, so (2, 0) is the x-intercept.

    -3 3

    -3

    351. Vertical asymptotes: Setting (x1)2 = 0 we see that x = 1 is a vertical

    asymptote.

    Horizontal asymptote: The degree of the numerator is less than thedegree of the denominator, so y = 0 is the horizontal asymptote.

    Intercepts: Since f(0) = 1, the y-intercept is (0, 1). The numerator isnever zero, so there are no x-intercepts.

  • 28 CHAPTER 1. FUNCTIONS

    -9 -6 -3 3 6

    -9-6-3

    369

    52. Vertical asymptotes: Setting (x + 2)3 = 0 we see that x = 2 is avertical asymptote.

    Horizontal asymptote: The degree of the numerator is less than thedegree of the denominator, so y = 0 is the horizontal asymptote.

    Intercepts: Since f(0) = 4/(23) = 1/2, the y-intercept is (0, 1/2). Thenumerator is never 0, so there are no x-intercepts.

    Graph: The left branch must lie entirely in the third quadrant since there are no x-intercepts,and f(x) < 0 for x < 2. The right branch must lie above the x-axis because it passesthrough (0, 1/2) and there are no x-intercepts.

    -4 4

    -4

    453. Vertical asymptotes: Setting x2 1 = (x + 1)(x 1) = 0 we see thatx = 1 and x = 1 are vertical asymptotes.

    Horizontal asymptote: The degree of the numerator is less than thedegree of the denominator, so y = 0 is the horizontal asymptote.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Since the numeratoris simply x, (0, 0) is also the only x-intercept.

    Graph: We use the facts that the only x-intercept is (0, 0) and the x-axis is a horizontalasymptote. For x < 1, f(x) < 0, so the left branch is below the x-axis. For 1 < x < 0,f(x) > 0, and for 0 < x < 1, f(x) < 0, so the middle branch passes through the origin(as opposed to being tangent to the origin and lying strictly above or below the x-axis).For x > 1, f(x) > 0, so the right branch is above the x-axis.

    -4 4

    -4

    454. Vertical asymptotes: Setting x2 4 = (x + 2)(x 1) = 2 we see thatx = 2 and x = 2 are vertical asymptotes.

    Horizontal asymptote: The degree of the numerator equals the degreeof the denominator, so y = 1/1 = 1 is the horizontal asymptote.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Setting x2 = 0 wesee that x = 0, so (0, 0) is also the only x-intercept.

    -4 4

    -4

    455. Vertical asymptotes: Setting x2 = 0 we see that x = 0, or the y-axis, isa vertical asymptote.

    Horizontal asymptote: The degree of the numerator equals the degreeof the denominator, so y = 1/1 = 1 is the horizontal asymptote.

    Intercepts: Since the y-axis is a vertical asymptote, there is no y-intercept. Setting 1 x2 = (1 + x)(1 x) = 0 we see that x = 1and x = 1, so (1, 0) and (1, 0) are x-intercepts.

  • 1.3. POLYNOMIAL AND RATIONAL FUNCTIONS 29

    -5 5

    -5

    556. Vertical asymptotes: Setting x2 9 = (x + 3)(x 3) = 0 we see that

    x = 3 and x = 3 are vertical asymptotes.

    Horizontal asymptote: The degree of the numerator equals the degreeof the denominator, so y = 1/1 = 1 is the horizontal asymptote.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Setting x(x 5) = 0we see that x = 0 and x = 5, so (0, 0) and (5, 0) are x-intercepts.

    57. Vertical asymptotes: Setting the denominator equal to zero, we see that x = 0, or they-axis, is a vertical asymptote.

    -5 5

    -5

    5Slant asymptote: Since the degree of the numerator is one greater than

    the degree of the denominator, the graph of f(x) possesses a slantasymptote. From f(x) = (x2 9)/x = x 9/x, we see that y = x isa slant asymptote.

    Intercepts: Since the y-axis is a vertical asymptote, the graph has noy-intercept. Setting x2 9 = (x+ 3)(x 3) = 0 we see that x = 3and x = 3, so (3, 0) and (3, 0) are the x-intercepts.

    Graph: We need to determine if the graph crosses the slant asymptote. To do this, we solve

    x2 9x

    = x; x2 9 = x2; 9 = 0.

    Since there is no solution, the graph does not cross its slant asymptote.

    58. Vertical asymptotes: Setting the denominator equal to 0, we see that x = 0 or the y-axisis a vertical asymptote.

    -5 5

    -5

    5Slant asymptote: Since the degree of the numerator is one greater than

    the degree of the denominator, the graph of f(x) possesses a slantasymptote. From f(x) = (x2 3x 10)/x = x 3 10/x, we seethat y = x 3 is a slant asymptote.

    Intercepts: Since the y-axis is a vertical asymptote, the graph has noy-intercept. Setting (x+ 2)(x5) = 0 we see that x = 2 and x = 5or (2, 0) and (5, 0) are the x-intercepts.

    Graph: We can just about find the graph from the asymptotes and intercepts, but we needto determine if the graph crosses the slant asymptote. To do this, we solve

    x2 3x 10x

    = x 3; x2 3x 10 = x2 3x; 10 = 0.

    Since there is no solution, the graph does not cross its slant asymptote.

    59. Vertical asymptotes: Setting x+ 2 = 0, we see that x = 2 is a vertical asymptote.

  • 30 CHAPTER 1. FUNCTIONS

    -5 5

    -10

    -5

    5Slant asymptote: Since the degree of the numerator is one greater thanthe degree of the denominator, the graph of f(x) possesses a slantasymptote. Using synthetic division we see that f(x) = x2/(x+2) =x 2 + 4/(x+ 2), and the slant asymptote is y = x 2.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Setting x2 = 0 wesee that x = 0, so (0, 0) is also the only x-intercept.

    Graph: We need to determine if the graph crosses the slant asymptote. To do this, we solve

    x2

    x+ 2= x 2; x2 = x2 4; 0 = 4.

    Since there is no solution, the graph does not cross its slant asymptote.

    60. Vertical asymptotes: Setting x+ 2 = 0, we see that x = 2 is a vertical asymptote.

    -10 -5 5 10

    -15

    -10

    -5

    5Slant asymptote: Since the degree of the numerator is one greater

    than the degree of the denominator, the graph of f(x) possessesa slant asymptote. Using synthetic division we see that f(x) =(x2 2x)/(x + 2) = x 4 + 8/(x + 2), and the slant asymptoteis y = x 4.

    Intercepts: Since f(0) = 0, the y-intercept is (0, 0). Setting x2 2x =x(x 2) = 0 we see that x = 0 and x = 2, so (0, 0) and (2, 0) are thex-intercept.

    Graph: We need to determine if the graph crosses the slant asymptote. To do this, we solve

    x2 2xx+ 2

    = x 4; x2 2x = x2 2x 8; 0 = 8.

    Since there is no solution, the graph does not cross its slant asymptote.

    61. Vertical asymptotes: Setting x 1 = 0, we see that x = 1 is a vertical asymptote.

    -5 5

    -5

    5Slant asymptote: Since the degree of the numerator is one greater than

    the degree of the denominator, the graph of f(x) possesses a slantasymptote. Using synthetic division we see that f(x) = (x2 2x 3)/(x 1) = x 1 + 4/(x 1), and the slant asymptote is y = x 1.

    Intercepts: Since f(0) = 3/(1) = 3, the y-intercept is (0, 3). Setting(x+ 1)(x3) = 0 we see that x = 1 and x = 3 or (1, 0) and (3, 0)are the x-intercepts.

    Graph: We can just about find the graph from the asymptotes and intercepts, but we needto determine if the graph crosses the slant asymptote. To do this, we solve

    x2 2x 3x 1 = x 1; x

    2 2x 3 = x2 2x+ 1; 3 = 1.

    Since there is no solution, the graph does not cross its slant asymptote.

    62. Vertical asymptotes: Setting x+ 2 = 0, we see that x = 2 is a vertical asymptote.

  • 1.3. POLYNOMIAL AND RATIONAL FUNCTIONS 31

    -10 10

    -10

    10

    20Slant asymptote: Since the degree of the numerator is one greaterthan the degree of the denominator, the graph of f(x) possessesa slant asymptote. Using synthetic division we see that f(x) =(x 1)2/(x + 2) = (x2 + 2x 1)/(x + 2) = x + 4 9/(x + 2)and the slant asymptote is y = x+ 4.

    Intercepts: Since f(0) = 1/2, the y-intercept is (0,1/2). Setting(x1)2 = 0 we see that x = 1, so (1, 0) is also the only x-intercept.

    Graph: We need to determine if the graph crosses the slant asymptote. To do this, we solve

    (x 1)2x+ 2

    = x+ 4; x2 + 2x 1 = x2 + 2x+ 8; 1 = 8.

    Since there is no solution, the graph does not cross its slant asymptote.

    63. Set f(x) = 1: Set f(x) = 2:2x 1x+ 4

    = 1 2x 1x+ 4

    = 2

    2x 1 = x 4 2x 1 = 2x+ 8x = 1 1 = 8

    Thus, 1 is in the range and 2 is not.64. The degree of the numerator equals the degree of the denominator, so y = 1 is the horizontal

    asymptote. To determine the points where the graph of f(x) crosses the horizontal asymptotey = 1 we solve

    (x 3)2x2 5x = 1

    x2 6x+ 9 = x2 5x6x+ 9 = 5x; 9 = x.

    Thus, f(x) crosses its horizontal asymptote at (9, 1).

    65. Begin by calculatingTFTC

    =140 3260 0 =

    95

    . Then:

    TF 32 =95

    (TC 0)

    TC =95TC + 32

    Try it out: when TC = 100, TF =95

    (100) + 32 = 212.

    66. Begin by calculatingTKTC

    =300 273

    27 0 = 1. Then:

    TK 273 = TC 0TK = TC + 273

    Try it out: when TC = 100, TK(100) = 100 + 273 = 373 K.

  • 32 CHAPTER 1. FUNCTIONS

    67. Identifying t = 20, P = 1000, and r = 0.034, we have

    A = P + Prt = 1000 + 1000(0.034)(20) = 1000 + 680 = $1680.

    Assuming that P and r remain the same, we solve 220 = 1000+ 1000(0.034)t for t. This givest = 12/0.34 35.29 years.

    68. If x is the number of years, then A(x) = ax + b for appropriate choices of a and b. SinceA = 20, 000 when x = 0, we have 20, 000 = a(0) + b = b. Since A = 0 when x = 25, we have0 = a(25)+b = 25a+20, 000, so 25a = 20, 000 and a = 400. Thus, A(x) = 400x+20, 000.For x = 10 years, the item will have a value of A(10) = 400(10)+20, 000 = 4000+20, 000 =16, 000.

    69. The ball is on the ground when s(t) = 0. Solving 16t2 + 96t = 0, we find t = 0 seconds and6 seconds.

    70. The ball is 80 feet above the ground when s(t) = 80. Solving 16t2 + 96t = 80, we find t = 1second and 5 seconds. The ball reaches its highest point when t = 3 seconds. This is s(3) =144 feet.

    71. The slope of a line is its rate of change, which means the change in output for each (positive)unit change in input. In this case, the slope is 5/2, so when x is changed by one unit, y willchange by 5/2 = 2.5 units. When x is changed by 2 units, y will change by 2(5/2) = 5 units,and when x is changed by n units, y will change by (5/2)n units.

    72. Using f(x) = ax+ b, we have

    f

    (x1 + x2

    2

    )= a

    (x1 + x2

    2

    )+ b =

    12

    (ax1 + ax2) + b

    =12

    (ax1 + b+ ax2 + b) =f(x1) + f(x2)

    2.

    Geometrically, this says that for a > 0, the point on the line connecting (x1, f(x1)) and(x2, f(x2)) that lies directly above the midpoint of the interval [x1, x2] lies level with themidpoint of the interval [f(x1), f(x2)] on the y-axis.

    73. First, find the slope of the line through ( 12 , 10) and (32 , 4); this is m = 6. The slope of a

    perpendicular line is then 1/m; in this case, 16 . To find the point the line passes through,find the midpoint of ( 12 , 10) and (

    32 , 4); this is (1, 7). Now, find the equation of the line through

    the midpoint with slope 1/m; this is y = 16x+ 416 .

    74. First, find the slopes of the lines through each pair of points; these are 2, 12 , and 1. Ifthe product of any two of these slopes is 1, then two of the sides are perpendicular andthe triangle is a right triangle. In this case, 2

    ( 12)

    = 1, so (2, 3), (1,3), and (4, 2) arevertices of a right triangle.

  • 1.4. TRANSCENDENTAL FUNCTIONS 33

    1.4 Transcendental Functions

    1.

    -2

    2

    2

    2.

    -2

    2

    2

    3.

    -2

    2

    2

    4.

    3

    6

    2

    5.

    -6

    -3

    3

    2

    6.

    -2

    2

    2

    1 2

    -4

    -2

    2

    47. The amplitude of y = 4 sinx is A = 4 and the period is

    2

    = 2.

    -5

    5

    2

    34

    8. The amplitude of y = 5 sin x2

    is A = |5| = 5 and the period is 2( 12

    ) = 4.

    0.5 1

    -3

    39. The amplitude of y = 3 cos 2x is A = | 3| = 3 and the period is22

    = 1.

    -2

    2

    2

    4

    10. The amplitude of y =52

    cos 4x is A =52

    and the period is24

    =

    2.

  • 34 CHAPTER 1. FUNCTIONS

    -2

    2

    4

    6

    2

    11. The amplitude of y = 2 4 sinx is A = | 4| = 4 and the period is21

    = 2.

    0.5 1 1.5 2

    2

    412. The amplitude of y = 2 2 sinx is A = | 2| = 2 and the period is2

    = 2.

    2

    4

    2 3

    13. The amplitude of y = 1 + cos2x3

    is A = 1 and the period is2(23

    ) = 3.

    1 2 3 4

    -2

    -1

    14. The amplitude of y = 1 + sin x2

    is A = 1 and the period is2(2

    ) = 4.

    15. We begin with y = A sinx since the graph flattens out at /2 and 3/2. The amplitude is12 [3 (3)] = 3 and the graph has been reflected through the line y = 0, so A = 3. Thus,y = 3 sinx.

    16. Since the graph flattens out at x = 0 we use y = A cosx. The amplitude is 12[

    14

    (14

    )]= 14 .

    Thus y = 14 cosx.

    17. Since the graph flattens out at x = 0, we use y = A cosx+D. The amplitude is 12 [4(2)] = 3and the graph has been reflected through the line y = 1, so A = 3 and D = 1. Thusy = 3 cosx+ 1.

    18. We begin with y = A sinx+D since the graph flattens out at /2 and 3/2. The amplitudeis 12 [0 (1)] = 12 and the graph has been reflected through the line y = 12 , so A = 12 andD = 12 . Thus, y = 12 sinx 12 .

    19. Since the y-intercept is (0, 0), the equation has the form y = A sinBx. The amplitude of thegraph is A = 3 and the period is = 2/B, so B = 2 and y = 3 sin 2x.

  • 1.4. TRANSCENDENTAL FUNCTIONS 35

    20. Since the y-intercept is (0,2) and not (0, 0), the equation has the form y = A cosBx. Theamplitude is 2, and the graph has been reflected through the line y = 0, so A = 2. Theperiod is = 2/B, so B = 2, and y = 2 cos 2x.

    21. Since the y-intercept is (0, 12 ) and not (0, 0), the equation has the form y = A cosBx. Theamplitude of the graph is A = 12 and the period is 2 = 2/B, so B = and y =

    12 cosx.

    22. Since the y-intercept is (0, 2) and not (0, 0), the equation has the form y = A cosBx. Theamplitude of the graph is A = 2 and the period is 4 = 2/B, so B = /2 and y = 2 cos(x/2).

    23. Since the y-intercept is (0, 0), the equation has the form y = A sinBx. The amplitude of thegraph is 1, and the graph has been reflected through the line y = 0, so A = 1. The period is2 = 2/B, so B = and y = sinx.

    24. Since the y-intercept is (0, 3) and not (0, 0), the equation has the form y = A cosBx. Theamplitude of the graph is A = 3 and the period is 8 = 2/B, so B = /4 and y = 3 cos(x/4).

    -1

    1

    6

    136

    23

    53

    25. The amplitude of y = sin(x /6) is A = 1 and the period is 2/1 = 2.The phase shift is | /6|/1 = /6. Since C = /6 < 0, the shift is tothe right.

    -1

    1

    12

    34

    512

    26. The amplitude of y = sin(3x /4) is A = 1 and the period is 2/3. Thephase shift is | /4|/3 = /12. Since C = /4 < 0, the shift is to theright.

    -1

    1

    74

    34

    4

    27. The amplitude of y = cos(x+ /4) is A = 1 and the period is 2/1 = 2.The phase shift is |/4|/1 = /4. Since C = /4 > 0, the shift is to theleft.

    -2

    -1

    1

    2

    1312

    712

    12

    28. The amplitude of y = 2 cos(2x /6) is A = | 2| = 2 and the period is2/2 = . The phase shift is | /6|/2 = /12. Since C = /6 < 0, theshift is to the right.

  • 36 CHAPTER 1. FUNCTIONS

    -4-3-2-1

    1234

    74

    54

    34

    29. The amplitude of y = 4 cos(2x3/2) is A = 4 and the period is 2/2 = .The phase shift is | 3/2|/2 = 3/4. Since C = 3/2 < 0, the shift isto the right.

    -3

    3

    78

    8

    38

    30. The amplitude of y = 3 cos(2x+/4) is A = 3 and the period is 2/2 = .The phase shift is |/4|/2 = /8. Since C = /8 > 0, the shift is to theleft.

    -3

    3143

    23

    83

    31. The amplitude of y = 3 sin(x/2 /3) is A = 3 and the period is2/(1/2) = 4. The phase shift is | /3|/(1/2) = 2/3. SinceC = /3 < 0, the shift is to the right.

    -3

    3

    624

    32. The amplitude of y = cos(x/2 ) is A = | 1| = 1 and the period is2/(1/2) = 4. The phase shift is | |/(1/2) = 2. Since C = < 0,the shift is to the right.

    2 4 6

    -3

    333. The amplitude of y = 4 sin(x/3/3) is A = |4| = 4 and the period is2/(/3) = 6. The phase shift is |/3|/(/3) = 1. Since C = /3 < 0,the shift is to the right.

  • 1.4. TRANSCENDENTAL FUNCTIONS 37

    -1 1

    -2

    2

    23

    13

    16

    34. The amplitude of y = 2 cos(2x 4/3) = 2 cos(2x + 4/3) is A = 2and the period is 2/2 = 1. The phase shift is |4/3|/2 = 2/3. SinceC = 4/3 > 0, the shift is to the left.

    35. y = 5 sin[

    (x 1

    2

    )]= 5 sin

    (x

    2

    )

    36. y = 8 + cos[

    12

    (x 2

    3

    )]= 8 + cos

    (x2

    3

    )

    37. Setting 1 + sinx = 0 we have sinx = 1, which is true for x = /2 in [0, 2]. By periodicity,the x-intercepts of 1 + sinx are (/2 + 2n, 0), where n is an integer.

    38. Setting 1 2 cosx = 0 we have 2 cosx = 1 or cosx = 1/2. Since cosx is positive, x is inthe first or fourth quadrant. Since cosx = 1/2 for x = /3, we see that 1 2 cosx = 0 forx = /3 and x = 5/3 in [0, 2]. By periodicity, then, the x-intercepts of 1 + 2 cosx are at

    3+ 2n =

    (13

    + 2n) and

    53

    + 2n =(

    53

    + 2n),

    so the intercepts are(

    3+ 2n, 0

    )and

    (53

    + 2n, 0)

    , where n is an integer.

    39. Setting sinx = 0, we have by (3) in this section of the text that the x-intercepts are deter-mined by x = n, n an integer. Thus, x = n, and the x-intercepts are (n, 0), where n is aninteger.

    40. Setting cos 2x = 0, we have by (4) in this section of the text that the x-intercepts aredetermined by 2x = (2n+ 1)/2, n an integer. Thus, x = (2n+ 1)/4, so the intercepts are(/4 + n/2, 0), where n is an integer.

    41. Setting 10 cos(x/2) = 0, we have by (4) in this section of the text that the x-intercepts aredetermined by x/2 = (2n + 1)/2, n an integer. Thus, x = (2n + 1), so the intercepts are( + 2n, 0), where n is an integer.

    42. Setting 3 sin(5x) = 0, we have by (3) in this section of the text that the x-interceptsare determined by 5x = n, n an integer. Thus, x = n/5, and the x-intercepts are(n/5, 0), where n is an integer.

    43. Setting sin(x /4) = 0, we have by (3) in this section of the text that the x-intercepts aredetermined by x /4 = n, n an integer. Thus, x = n + /4 = (n + 1/4), and thex-intercepts are (/4 + n, 0), where n is an integer.

    44. Setting cos(2x ) = 0, we have by (4) in this section of the text that the x-intercepts aredetermined by 2x = (2n+ 1)/2, n an integer. Thus,

    x =12

    [(2n+ 1)

    2+

    ]=

    34

    +n

    2,

    so the intercepts are (3/4 + n/2, 0), where n is an integer.

  • 38 CHAPTER 1. FUNCTIONS

    -1 1

    -3

    345. The period of y = tanx is / = 1. Since

    tanx =sinxcosx

    ,

    the x-intercepts of tanx occur at the zeros of sinx; namely, at x = nor x = n for n an integer. The vertical asymptotes occur at the zeros ofcosx; namely, at

    x =(2n+ 1)

    2or x =

    12

    (2n+ 1) for n an integer.

    -6

    -3

    3

    6

    46. The period of y = tan(x/2) is /(1/2) = 2. Since

    tanx

    2=

    sin(x/2)cos(x/2)

    ,

    the x-intercepts of tan(x/2) occur at the zeros of sin(x/2); namely, atx/2 = n or x = 2n for n an integer. The vertical asymptotes occur atthe zeros of cos(x/2); namely, at

    x

    2=

    (2n+ 1)2

    or x = (2n+ 1) = + 2n for n an integer.

    Since the graph has vertical asymptotes at and (using n = 1 and n = 0), we graphone cycle on the interval (, ).

    -6

    -3

    3

    6

    2

    47. The period of y = cot 2x is /2. Since

    cot 2x =cos 2xsin 2x

    ,

    the x-intercepts of cot 2x occur at the zeros of cos 2x; namely, at

    2x =(2n+ 1)

    2or x =

    (2n+ 1)4

    for n an integer.

    The vertical asymptotes occur at the zeros of sin 2x; namely, at

    2x = n or x =n

    2for n an integer.

    3

    -6

    -3

    3

    648. The period of y = cot(x/3) is /(/3) = 3. Since

    cot x3

    = cos(x/3)sin(x/3)

    ,

    the x-intercepts of cot(x/3) occur at the zeros of cos(x/3); namely, at

    x

    3=

    (2n+ 1)2

    or x =32

    (2n+ 1) for n an integer.

  • 1.4. TRANSCENDENTAL FUNCTIONS 39

    The vertical asymptotes occur at the zeros of sin(x/3); namely, at

    x

    3= n or x = 3n for n an integer.

    Since A = 1, the graph of y = cot(x/3) is the graph of y = cot(x/3) reflected throughthe x-axis.

    -3

    3

    2

    32

    2

    49. The period of y = tan(x/2 /4) is /(1/2) = 2. Since

    tan(x

    2

    4

    )=

    sin(x/2 /4)cos(x/2 /4) ,

    the x-intercepts of tan(x/2 /4) occur at the zeros of sin(x/2 /4);namely, at

    x

    2

    4= n or x = 2n +

    2for n an integer.

    The vertical asymptotes occur at the zeros of cos(x/2 /4); namely, at

    x

    2

    4=

    (2n+ 1)2

    or x = (2n+ 1) +

    2=

    32

    + 2n for n an integer.

    Since the graph has vertical asymptotes at /2 and 3/2 (using n = 1 and n = 0), wegraph one cycle on the interval (/2, 3/2).

    -2

    2

    2

    2

    50. The period of y = 14 cot(x /2) is /1 = . Since

    14

    cot(x

    2

    )=

    14 cos(x /2)

    sin(x /2) ,

    the x-intercepts of 14 cot(x/2) occur at the zeros of cos(x/2); namely,at

    x 2

    =(2n+ 1)

    2or x = (n+ 1) for n an integer.

    The vertical asymptotes occur at the zeros of sin(x /2); namely, at

    x 2

    = n or x = n +

    2for n an integer.

    -1 1 2

    -4

    -2

    2

    14

    51. The period of y = 1 + cotx is / = 1. To find the x-intercepts, wesolve 1 + cotx = 0, or cotx = cosx/ sinx = 1, which is equivalentto solving for cosx = sinx. This occurs when

    x = (4n+ 1)

    4or x = n+

    14

    for n an integer.

    The vertical asymptotes occur at the zeros of sinx; namely, x = n or x = n, for n aninteger.

  • 40 CHAPTER 1. FUNCTIONS

    -3

    3

    3

    2

    23

    6

    52. The period of y = tan(x+ 5/6) is . Since

    tan(x+ 5/6) =sin(x+ 5/6)cos(x+ 5/6)

    ,

    the x-intercepts of tan(x+5/6) occur at the zeros of sin(x+5/6); namely,at

    x+56

    = n or x = n 56

    for n an integer.

    The vertical asymptotes occur at the zeros of cos(x+ 5/6); namely, at

    x+56

    = (2n+ 1)

    2or x = (2n+ 1)

    2 5

    6=

    3+ n for n an integer.

    Since the graph has vertical asymptotes at /3 and 2/3 (using n = 0 and n = 1), we graphone cycle on the interval (/3, 2/3).

    -2 -1 1 2

    -6

    -3

    3

    6

    53. The period of y = 3 cscx is 2/ = 2. Since 3 cscx = 3/ sinx, thevertical asymptotes occur at the zeros of sinx; namely, at x = n orx = n, for n an integer. We plot one cycle of the graph on (1, 1), sincethe period of the function is 2 = 1 (1) and vertical asymptotes occurat x = 1 and x = 1 (taking n = 1 and 1).

    -6

    -3

    3

    6

    3 6

    54. The period of y = 2 csc(x/3) is 2/(1/3) = 6. Since

    2 csc x3

    = 2sin(x/3)

    ,

    the vertical asymptotes occur at the zeros of sin(x/3); namely, at x/3 = nor x = 3n, for n an integer.

    -3

    3

    6

    23

    3

    55. The period of y = sec(3x /2) is 2/3. Since

    sec(

    3x 2

    )=

    1cos(3x /2) ,

    the vertical asymptotes occur at the zeros of cos(3x /2); namely, at

    3x 2

    = (2n+ 1)

    2or x = (n+ 1)

    3for n an integer.

    -3

    3

    4

    4

    56. The period of y = csc(4x+ ) is 2/4 = /2. Since

    csc(4x+ ) =1

    sin(4x+ ),

    the vertical asymptotes occur at the zeros of sin(4x+ ); namely, at

    4x+ = n or x = (n 1)4

    = 4

    +n

    4for n an integer.

  • 1.4. TRANSCENDENTAL FUNCTIONS 41

    We plot one cycle of the graph on (/4, /4) since the period of the function is /2 =/4 (/4) and vertical asymptotes occur at x = /4 and x = /4 (taking n = 0 and 2).

    369121518

    2

    2

    25.570813.5708

    57. The amplitude is A = 12 (18 6) = 6. The tidal period is 2/B = 12, soB = /6, and the average depth D = (18 + 6)/2 = 12. Thus, the functionis

    d(t) = 12 + 6 sin

    6

    (t

    2

    ).

    58. (a) Since 8 A.M. is 8 hours after midnight,

    T (8) = 50 + 10 sin

    12(8 8) = 50

    and the temperature at 8 A.M. is 50 F.

    (b) We solve 50 + 10 sin

    12(t 8) = 60

    sin

    12(t 8) = 1

    12(t 8) =

    2t 8 = 6; t = 14.

    Since 14 hours after midnight is noon plus 2 hours, the temperature will be at 60 F at2 P.M.

    6 12 18 24102030405060

    2

    (c) The amplitude of the graph is A = 10 and the graph is verticallycentered at T = 50. The period is 2/(/12) = 24 and the phaseshift is |C|/B = (8/12)/(/12) = 8. Since C = 8/12 > 0, thegraph of y = 50 + 10 sint/12 is shifted 8 units to the left to obtainthe graph of

    T = 50 + 10 sin(

    12t 8

    12

    ).

    (d) The minimum temperature will be 40 F and this will occur when (t 8)/12 = 3/2,since x = 3/2 where y = sinx first reaches its minimum. Solving for t, we obtaint = 26. Since the period is 24, the temperature is first minimum when t = 2 or at 2 A.M.The maximum temperature will be 60 F and this will occur when (t 8)/12 = /2,since x = /2 where y = sinx first reaches its maximum. Solving for t, we obtaint = 14. Thus, the temperature is first maximum when t = 14 or at 2 P.M. (Note:We could have reached the same conclusion about the first occurrence of the maximumtemperature from the information in parts (b) and (c).)

    59. (a) When = 0, sin = 0 and sin 2 = 0, so g = 978.0309 cm/s2.(b) At the north pole, = 90, so sin = 1 and sin 2 = sin 180 = 0, so g = 978.0309 +

    5.18552 = 983.2164 cm/s2.(c) When = 45, sin =

    2/2 and sin 2 = sin 90 = 1, so

    g = 978.0309 + 5.18552(

    2/2)2 0.00570(1)2

    = 978.0309 + 2.59276 0.00570 = 980.618 cm/s2.

  • 42 CHAPTER 1. FUNCTIONS

    60. (a) The range of a shot put released from a height of 2.0 m above the ground is

    R =13.7 cos 40

    9.81

    [13.7 sin 40 +

    (13.72 sin2 40 + 2(9.81)(2)

    ]= 20.98 m,

    while the range of a shot put released from a height of 2.4 m above the ground is

    R =13.7 cos 40

    9.81

    [13.7 sin 40 +

    (13.72 sin2 40 + 2(9.81)(2.4)

    ]= 21.36 m.

    Thus, an additional 0.4 m above the ground results in an additional 21.36 20.98 =0.38 m in range.

    (b) If h increases while the other parameters remain fixed, the 2gh term increases and, as aresult, the product increases and the range R increases.

    (c) This demonstrates that a taller shot-putter will achieve a longer range.

    61. The period of sin 12x is 2/(1/2) = 4. The period of sin 2x is 2/2 = . The period of asum of periodic functions is equal to the least common multiple of the individual periods, sothe period of f(x) is 4.

    62. The graph of the absolute value of a function is the graph of the function with any portionsof the graph that lie below the x-axis reflected through the x-axis.

    1

    2

    2

    y = | sinx|

    1

    2

    2

    y = | cosx|63. The graph of the absolute value of a function is the graph of the function with any portions

    of the graph that lie below the x-axis reflected through the x-axis.

    1

    2

    3

    4

    5

    2 3

    y = | secx|

    1

    2

    3

    4

    5

    2

    2 3

    y = | cscx|

    64. (a) We have 9 cscx = 9/ sinx = 1, or sinx = 9, which is not possible for any real number xsince 1 sinx 1.

    (b) We have 10 secx = 10/ cosx = 7, or cosx = 10/7, which is not possible for any realnumber x since 1 cosx 1.

    (c) We have secx = 1/ cosx = 10.5 = 11/2, or cosx = 2/11, which is possible for somereal number x since 1 cosx 1.

  • 1.5. INVERSE FUNCTIONS 43

    65. The graph of y = cotx is shown below as a red curve, while the graphs of y = A tan(x+ C),for various choices of A and C, are shown as blue curves.

    -4

    -2

    2

    4

    2

    2

    A = 1, C = /2

    -4

    -2

    2

    4

    2

    2

    A = 1, C = /2-4

    -2

    2

    4

    2

    2

    A = 1, C = /2

    We see from the third graph that cotx = tan(x /2).

    -5

    5

    2

    22

    66. Comparing the graphs of y = secx, shown in blue, and y = cscx, shownin red, we see that the graph of the cosecant is equivalent to the graph ofthe secant shifted /2 units to the right. Thus, cscx = sec(x /2), andwe identify A = 1 and C = /2.

    1.5 Inverse Functions

    1. For f(x) = 1 + x(x 5), the value y = 1 in the range of f occurs at either x = 0 or x = 5 inthe domain of f . Thus, f(x) is not one-to-one.

    2. For f(x) = x4 + 2x2, the value y = 3 in the range of f occurs at either x = 1 or x = 1 inthe domain of f . Thus, f(x) is not one-to-one.

    3.

    5 10

    -5

    5

    2

    Not one-to-one.

    4.

    -2 2

    -2

    2

    2

    One-to-one.

    5.

    -4 -2 2 4

    -4

    -2

    2

    4

    2

    One-to-one.

    6.

    -4 -2 2 4

    -4

    -2

    2

    4

    2

    Not one-to-one.

    7.-8 -4 4 8

    -12

    -8

    -4

    4

    2

    One-to-one.

    8.

    -4 4

    -4

    4

    2

    Not one-to-one.

    9. y = 3x3 + 7, x =(y 7

    3

    )1/3; f1(x) =

    (x 7

    3

    )1/3

    10. y = 3

    2x 4, x = 12y3 + 2; f1(x) =

    12x3 + 2

  • 44 CHAPTER 1. FUNCTIONS

    11. y =2 x1 x, x =

    y 2y 1 ; f

    1(x) =x 2x 1

    12. y = 5 2x, x =

    25 y ; f

    1(x) =2

    5 x

    13. f(f1(x)) = f(

    15x+ 2

    )= 5

    (15x+ 2

    ) 10 = x+ 10 10 = x

    f1(f(x)) = f1(5x 10) = 15

    (5x 10) + 2 = x 2 + 2 = x

    14. f(f1(x)) = f(

    1 xx

    )=

    11 xx

    + 1=

    11 xx

    + 1(xx

    )=

    x

    1 x+ x = x

    f1(f(x)) = f1(

    1x+ 1

    )=

    1 1x+ 11

    x+ 1

    =1 1

    x+ 11

    x+ 1

    (x+ 1x+ 1

    )=x+ 1 1

    1= x

    15. Domain: [0,); Range: [2,)

    16. Domain: [3,); Range: [1/2,)

    17. Domain: (, 0) (0,); Range: (,3) (3,)

    18. Domain: (, 1) (1,); Range: (, 4) (4,)

    19. When x = 2, y = f(2) = 2(2)3 + 2(2) = 2(8) + 4 = 20, so the point on the graph of f is(2, 20). The corresponding point on the graph of f1 is then (20, 2).

    20. When x = 5, y = f(5) = 37, so the point on the graph of f is (5, 37). The correspondingpoint on the graph of f1 is then (37, 5).

    21. Since f(9) = 12, (9, 12) is a point on the graph of f . The corresponding point on the graphof f1 is then (12, 9), and so f1(12) = 9.

    22. Since f(

    12

    )=

    4(1/2)1/2 + 1

    =2

    3/2=

    43

    , (1/2, 4/3) is a point on the graph of f . The corresponding

    point on the graph of f1 is then (4/3, 1/2), and therefore f1(4/3) = 1/2.

    -3 -2 -1 1 2 3

    -3-2-1

    123

    2

    2

    23. The graph of f1, shown in red, is obtained from the graph of f , shownin blue, by reflection through the line y = x.

  • 1.5. INVERSE FUNCTIONS 45

    -5 5

    -5

    5

    2

    24. The graph of f1, shown in red, is obtained from the graph of f , shownin blue, by reflection through the line y = x.

    !

    !

    2

    !

    2

    -11

    -22

    1

    2

    -1

    -2

    25. To graph f , shown in blue, we use the fact that f is the inverse of f1,shown in red, and that the graph of the inverse of a function is the reflectionof the graph of the original function reflected through the line y = x.

    -1 1

    -1

    1

    2

    2

    26. To graph f , shown in blue, we use the fact that f is the inverse of f1,shown in red, and that the graph of the inverse of a function is the reflectionof the graph of the original function reflected through the line y = x.

    27. By restricting f s domain to x 5/2, we get f1(x) = 5x

    2, x 0.

    28. By restricting f s domain to x 0, we get f1(x) =(x 9

    3

    )1/2, x 9.

    29. f(x) can be rewritten as (x+ 1)2 + 3. Thus, by restricting the domain of f to x 1, we getf1(x) =

    x 3 1.

    30. f(x) can be rewritten as (x 4)2 + 16. By restricting the domain of f to x 4, we getf1(x) =

    16 x+ 4.

    31. For f(x) = x3 and g(x) = 4x + 5, we get (f g)(x) = (4x + 5)3 and thus (f g)1(x) =14 (x

    1/3 5). f1(x) = x1/3 and g1(x) = 14 (x 5), so (g1 f1)(x) = 14 (x1/3 5), which isthe same as (f g)1.

    32. Solving y = 3x 3y for x, we get x = (y + 3y)3. Thus, f1(x) = (x+ 3x)3.

    33. 3/4

    34. /3

    35. /4

    36. /3

    37. 3/4

  • 46 CHAPTER 1. FUNCTIONS

    38.

    39. /3

    40. 5/6

    41. sin(

    arctan43

    )=

    45

    5

    3

    4

    42. cos(

    sin125

    )=

    215

    5 2

    21

    43. tan(

    cot112

    )= 2

    1

    25

    44. csc(

    tan123

    )=

    132

    3

    213

    3 1

    8

    45. sin(

    2 sin113

    )= 2 sin

    (sin1

    13

    )cos(

    sin113

    )= 2

    (13

    )(8

    3

    )=

    4

    29

    4

    3

    746. cos(

    2 cos134

    )= cos2

    (cos1

    34

    ) sin2

    (cos1

    34

    )

    =(

    34

    )2(

    74

    )2=

    18

    13

    2

    3

    2

    547. sin

    (arcsin

    3

    3+ arccos

    23

    )

    = sin

    (arcsin

    3

    3

    )cos(

    arccos23

    )+ cos

    (arcsin

    3

    3

    )sin(

    arccos23

    )

    =

    (3

    3

    )(23

    )+

    (6