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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Checkpoint

2.1

2.25

2.2

12.006001

2.3

16 unit2

2.4

limx11x1x1=−1limx11x1x1=−1

2.5

limx2h(x)=−1.limx2h(x)=−1.

2.6

limx2|x24|x2limx2|x24|x2 does not exist.

2.7

a. limx2|x24|x2=−4;limx2|x24|x2=−4; b. limx2+|x24|x2=4limx2+|x24|x2=4

2.8

a. limx01x2=+;limx01x2=+; b. limx0+1x2=+;limx0+1x2=+; c. limx01x2=+limx01x2=+

2.9

a. limx21(x2)3=;limx21(x2)3=; b. limx2+1(x2)3=+;limx2+1(x2)3=+; c. limx21(x2)3limx21(x2)3 DNE. The line x=2x=2 is the vertical asymptote of f(x)=1/(x2)3.f(x)=1/(x2)3.

2.10

Does not exist.

2.11

11101110

2.12

−13;

2.13

1313

2.14

1414

2.15

−1;

2.16

1414

2.17


The graph of a piecewise function with three segments. The first is a linear function, -x-2, for x<-1. The x intercept is at (-2,0), and there is an open circle at (-1,-1). The next segment is simply the point (-1, 2). The third segment is the function x^3 for x > -1, which crossed the x axis and y axis at the origin.


limx−1f(x)=−1limx−1f(x)=−1

2.18

+∞

2.19

0

2.20

0

2.21

f is not continuous at 1 because f(1)=23=limx1f(x).f(1)=23=limx1f(x).

2.22

f(x)f(x) is continuous at every real number.

2.23

Discontinuous at 1; removable

2.24

[−3,+)[−3,+)

2.25

0

2.26

f(0)=1>0,f(1)=−2<0;f(x)f(0)=1>0,f(1)=−2<0;f(x) is continuous over [0,1].[0,1]. It must have a zero on this interval.

2.27

Let ε>0;ε>0; choose δ=ε3;δ=ε3; assume 0<|x2|<δ.0<|x2|<δ.

Thus, |(3x2)4|=|3x6|=|3|·|x2|<3·δ=3·(ε/3)=ε.|(3x2)4|=|3x6|=|3|·|x2|<3·δ=3·(ε/3)=ε.

Therefore, limx23x2=4.limx23x2=4.

2.28

Choose δ=min{9(3ε)2,(3+ε)29}.δ=min{9(3ε)2,(3+ε)29}.

2.29

|x21|=|x1|·|x+1|<ε/3·3=ε|x21|=|x1|·|x+1|<ε/3·3=ε

2.30

δ=ε2δ=ε2

Section 2.1 Exercises

1.

a. 2.2100000; b. 2.0201000; c. 2.0020010; d. 2.0002000; e. (1.1000000, 2.2100000); f. (1.0100000, 2.0201000); g. (1.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.1000000; j. 2.0100000; k. 2.0010000; l. 2.0001000

3.

y=2xy=2x

5.

3

7.

a. 2.0248457; b. 2.0024984; c. 2.0002500; d. 2.0000250; e. (4.1000000,2.0248457); f. (4.0100000,2.0024984); g. (4.0010000,2.0002500); h. (4.00010000,2.0000250); i. 0.24845673; j. 0.24984395; k. 0.24998438; l. 0.24999844

9.

y=x4+1y=x4+1

11.

π

13.

a. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; e. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (1.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; k. −0.99502488; l. −0.99900100

15.

y=x2y=x2

17.

−49 m/sec (velocity of the ball is 49 m/sec downward)

19.

5.2 m/sec

21.

−9.8 m/sec

23.

6 m/sec

25.

Under, 1 unit2; over: 4 unit2. The exact area of the two triangles is 12(1)(1)+12(2)(2)=2.5units2.12(1)(1)+12(2)(2)=2.5units2.

27.

Under, 0.96 unit2; over, 1.92 unit2. The exact area of the semicircle with radius 1 is π(1)22=π2π(1)22=π2 unit2.

29.

Approximately 1.3333333 unit2

Section 2.2 Exercises

31.

limx1f(x)limx1f(x) does not exist because limx1f(x)=−2limx1+f(x)=2.limx1f(x)=−2limx1+f(x)=2.

33.

limx0(1+x)1/x=2.7183limx0(1+x)1/x=2.7183

35.

a. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999; limx0sin2xx=2limx0sin2xx=2

37.

limx0sinaxx=alimx0sinaxx=a

39.

a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000; limx1(12x)=−1limx1(12x)=−1

41.

a. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889 limx0z1z2(z+3)=limx0z1z2(z+3)=

43.

a. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063; limx212xx24=0.1250=18limx212xx24=0.1250=18

45.

a. −10.00000; b. −100.00000; c. −1000.0000; d. −10,000.000; Guess: limα0+1αcos(πα)=,limα0+1αcos(πα)=, actual: DNE

A graph of the function (1/alpha) * cos (pi / alpha), which oscillates gently until the interval [-.2, .2], where it oscillates rapidly, going to infinity and negative infinity as it approaches the y axis.
47.

False; limx−2+f(x)=+limx−2+f(x)=+

49.

False; limx6f(x)limx6f(x) DNE since limx6f(x)=2limx6f(x)=2 and limx6+f(x)=5.limx6+f(x)=5.

51.

2

53.

1

55.

1

57.

DNE

59.

0

61.

DNE

63.

2

65.

3

67.

DNE

69.

0

71.

−2

73.

DNE

75.

0

77.

Answers may vary.

A graph of a piecewise function with two segments. The first segment is in quadrant three and asymptotically goes to negative infinity along the y axis and 0 along the x axis. The second segment consists of two curves. The first appears to be the left half of an upward opening parabola with vertex at (0,1). The second appears to be the right half of a downward opening parabola with vertex at (0,1) as well.
79.

Answers may vary.

A graph containing two curves. The first goes to 2 asymptotically along y=2 and to negative infinity along x = -2. The second goes to negative infinity along x=-2 and to 2 along y=2.
81.

a. ρ2ρ2 b. ρ1ρ1 c. DNE unless ρ1=ρ2.ρ1=ρ2. As you approach xSFxSF from the right, you are in the high-density area of the shock. When you approach from the left, you have not experienced the “shock” yet and are at a lower density.

Section 2.3 Exercises

83.

Use constant multiple law and difference law: limx0(4x22x+3)=4limx0x22limx0x+limx03=3limx0(4x22x+3)=4limx0x22limx0x+limx03=3

85.

Use root law: limx−2x26x+3=limx−2(x26x+3)=19limx−2x26x+3=limx−2(x26x+3)=19

87.

49

89.

1

91.

5757

93.

limx4x216x4=161644=00;limx4x216x4=161644=00; then, limx4x216x4=limx4(x+4)(x4)x4=8limx4x216x4=limx4(x+4)(x4)x4=8

95.

limx63x182x12=18181212=00;limx63x182x12=18181212=00; then, limx63x182x12=limx63(x6)2(x6)=32limx63x182x12=limx63(x6)2(x6)=32

97.

limx9t9t3=9933=00;limx9t9t3=9933=00; then, limt9t9t3=limt9t9t3t+3t+3=limt9(t+3)=6limt9t9t3=limt9t9t3t+3t+3=limt9(t+3)=6

99.

limθπsinθtanθ=sinπtanπ=00;limθπsinθtanθ=sinπtanπ=00; then, limθπsinθtanθ=limθπsinθsinθcosθ=limθπcosθ=−1limθπsinθtanθ=limθπsinθsinθcosθ=limθπcosθ=−1

101.

limx1/22x2+3x22x1=12+32211=00;limx1/22x2+3x22x1=12+32211=00; then, limx1/22x2+3x22x1=limx1/2(2x1)(x+2)2x1=52limx1/22x2+3x22x1=limx1/2(2x1)(x+2)2x1=52

103.

−∞

105.

−∞

107.

limx62f(x)g(x)=2limx6f(x)limx6g(x)=72limx62f(x)g(x)=2limx6f(x)limx6g(x)=72

109.

limx6(f(x)+13g(x))=limx6f(x)+13limx6g(x)=7limx6(f(x)+13g(x))=limx6f(x)+13limx6g(x)=7

111.

limx6g(x)f(x)=limx6g(x)limx6f(x)=5limx6g(x)f(x)=limx6g(x)limx6f(x)=5

113.

limx6[(x+1)f(x)]=(limx6(x+1))(limx6f(x))=28limx6[(x+1)f(x)]=(limx6(x+1))(limx6f(x))=28

115.


The graph of a piecewise function with two segments. The first is the parabola x^2, which exists for x<=3. The vertex is at the origin, it opens upward, and there is a closed circle at the endpoint (3,9). The second segment is the line x+4, which is a linear function existing for x > 3. There is an open circle at (3, 7), and the slope is 1.


a. 9; b. 7

117.


The graph of a piecewise function with two segments. The first segment is the parabola x^2 – 2x + 1, for x < 2. It opens upward and has a vertex at (1,0). The second segment is the line 3-x for x>= 2. It has a slope of -1 and an x intercept at (3,0).


a. 1; b. 1

119.

limx−3(f(x)3g(x))=limx−3f(x)3limx−3g(x)=0+6=6limx−3(f(x)3g(x))=limx−3f(x)3limx−3g(x)=0+6=6

121.

limx−52+g(x)f(x)=2+(limx−5g(x))limx−5f(x)=2+02=1limx−52+g(x)f(x)=2+(limx−5g(x))limx−5f(x)=2+02=1

123.

limx1f(x)g(x)3=limx1f(x)limx1g(x)3=2+53=73limx1f(x)g(x)3=limx1f(x)limx1g(x)3=2+53=73

125.

limx−9(xf(x)+2g(x))=(limx−9x)(limx−9f(x))+2limx−9(g(x))=(−9)(6)+2(4)=−46limx−9(xf(x)+2g(x))=(limx−9x)(limx−9f(x))+2limx−9(g(x))=(−9)(6)+2(4)=−46

127.

The limit is zero.

The graph of three functions over the domain [-1,1], colored red, green, and blue as follows: red: theta^2, green: theta^2 * cos (1/theta), and blue: - (theta^2). The red and blue functions open upwards and downwards respectively as parabolas with vertices at the origin. The green function is trapped between the two.
129.

a.

A graph of a function with two curves. The first is in quadrant two and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to negative infinity. The second is in quadrant one and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to infinity.


b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

Section 2.4 Exercises

131.

The function is defined for all x in the interval (0,).(0,).

133.

Removable discontinuity at x=0;x=0; infinite discontinuity at x=1x=1

135.

Infinite discontinuity at x=ln2x=ln2

137.

Infinite discontinuities at x=(2k+1)π4,x=(2k+1)π4, for k=0,±1,±2,±3,…k=0,±1,±2,±3,…

139.

No. It is a removable discontinuity.

141.

Yes. It is continuous.

143.

Yes. It is continuous.

145.

k=−5k=−5

147.

k=−1k=−1

149.

k=163k=163

151.

Since both s and y=ty=t are continuous everywhere, then h(t)=s(t)th(t)=s(t)t is continuous everywhere and, in particular, it is continuous over the closed interval [2,5].[2,5]. Also, h(2)=3>0h(2)=3>0 and h(5)=−3<0.h(5)=−3<0. Therefore, by the IVT, there is a value x=cx=c such that h(c)=0.h(c)=0.

153.

The function f(x)=2xx3f(x)=2xx3 is continuous over the interval [1.25,1.375][1.25,1.375] and has opposite signs at the endpoints.

155.

a.

A graph of the given piecewise function containing two segments. The first, x^3, exists for x < 1 and ends with an open circle at (1,1). The second, 3x, exists for x > 1. It beings with an open circle at (1,3).


b. It is not possible to redefine f(1)f(1) since the discontinuity is a jump discontinuity.

157.

Answers may vary; see the following example:

A graph of a piecewise function with several segments. The first is an increasing line that exists for x < -8. It ends at an open circle at (-8,-8). The second is an increasing curve that exists from -8 <= x < -6. It begins with a closed circle at (-8, 0 ) and goes to infinity as x goes to -6 from the left. The third is a closed circle at the point (-6, 3). The fourth is a line that exists from -6 < x <= 3. It begins with an open circle at (-6, 2) and ends with a closed circle at (3,2). The fifth is an increasing line starting with an open circle at (3,3). It exists for x > 3.
159.

Answers may vary; see the following example:

The graph of a piecewise function with two parts. The first part is an increasing curve that exists for x < 1. It ends at (1,1). The second part is an increasing line that exists for x > 1. It begins at (1,3).
161.

False. It is continuous over (,0)(0,).(,0)(0,).

163.

False. Consider f(x)={xifx04ifx=0.f(x)={xifx04ifx=0.

165.

False. Consider f(x)=cos(x)f(x)=cos(x) on [π,2π].[π,2π].

167.

False. The IVT does not work in reverse! Consider (x1)2(x1)2 over the interval [−2,2].[−2,2].

169.

R=0.0001519mR=0.0001519m

171.

D=63.78kmD=63.78km

173.

For all values of a,f(a)a,f(a) is defined, limθaf(θ)limθaf(θ) exists, and limθaf(θ)=f(a).limθaf(θ)=f(a). Therefore, f(θ)f(θ) is continuous everywhere.

175.

Nowhere

Section 2.5 Exercises

177.

For every ε>0,ε>0, there exists a δ>0,δ>0, so that if 0<|tb|<δ,0<|tb|<δ, then |g(t)M|<ε|g(t)M|<ε

179.

For every ε>0,ε>0, there exists a δ>0,δ>0, so that if 0<|xa|<δ,0<|xa|<δ, then |φ(x)A|<ε|φ(x)A|<ε

181.

δ0.25δ0.25

183.

δ2δ2

185.

δ1δ1

187.

δ<0.3900δ<0.3900

189.

Let δ=ε.δ=ε. If 0<|x3|<ε,0<|x3|<ε, then |x+36|=|x3|<ε.|x+36|=|x3|<ε.

191.

Let δ=ε4.δ=ε4. If 0<|x|<ε4,0<|x|<ε4, then |x4|=x4<ε.|x4|=x4<ε.

193.

Let δ=ε2.δ=ε2. If 5ε2<x<5,5ε2<x<5, then |5x|=5x<ε.|5x|=5x<ε.

195.

Let δ=ε/5.δ=ε/5. If 1ε/5<x<1,1ε/5<x<1, then |f(x)3|=5x5<ε.|f(x)3|=5x5<ε.

197.

Let δ=3N.δ=3N. If 0<|x+1|<3N,0<|x+1|<3N, then f(x)=3(x+1)2>N.f(x)=3(x+1)2>N.

199.

0.033 cm, ε=8,δ=0.33,a=12,L=144ε=8,δ=0.33,a=12,L=144

201.

Answers may vary.

203.

0

205.

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

207.

Answers may vary.

Chapter Review Exercises

209.

False

211.

False. A removable discontinuity is possible.

213.

5

215.

8/78/7

217.

DNE

219.

2/32/3

221.

−4;

223.

Since −1cos(2πx)1,−1cos(2πx)1, then x2x2cos(2πx)x2.x2x2cos(2πx)x2. Since limx0x2=0=limx0x2,limx0x2=0=limx0x2, it follows that limx0x2cos(2πx)=0.limx0x2cos(2πx)=0.

225.

[2,][2,]

227.

c=−1c=−1

229.

δ=ε3δ=ε3

231.

0m/sec0m/sec

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