ContentsChapter 0 Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 0 Review Functions and Their Graphs Classes of Functions The Limit of a Function The Derivative of a Function Applications: Graphing Functions; Optimization The Integral of a Function and Applications Other Applications and Extensions of the Integral Calculus of Functions of Two or More Variables 1 58 114 171 206 285 385 439 462 510

Appendix A Graphing Utilities

iii

Chapter 0 Review0.1 Real Numbers1. (a) Natural Numbers: 2, 5 (b) Integers: 6, 2, 5 (c) Rational numbers: 6,

1 , 1.333, 2, 5 2

(d) Irrational numbers: 1 (e) Real numbers: 6, , 1.333, , 2, 5 23. (a) Natural Numbers: 1

(b) Integers: 0, 1 (c) Rational numbers: 0, 1, 1 1 1 , , 2 3 4

(d) Irrational numbers: none 1 1 1 (e) Real numbers: 0, 1, , , 2 3 45. (a) Natural Numbers: none (b) Integers: none (c) Rational numbers: none

(d) Irrational numbers: (e) Real numbers:7. Number: 18.9526 Rounded: 18.953 Truncated: 18.952 11. Number: 0.06291 Rounded: 0.063 Truncated: 0.062 15.

2 , ,

2 + 1, +

1 2

2 , ,

2 + 1, +

1 29.

Number: 28.65319 Rounded: 28.653 Truncated: 28.653 Number: 9.9985 Rounded: 9.999 Truncated: 9.998 521 = 34.73333 15 Rounded: 34.733 Truncated: 34.733 Number:x+2=34

13.

Number:

3 = 0.428571 7 Rounded: 0.429 Truncated: 0.428

17.

19. 3 + 2 = 5

21.

2

SECTION 0.1 25. 29. x2=6

23. 3y = 1 + 2 27.

x =6 2

94+2=5+2=7

31. 6 + 4 3 = 6 + 12 = 6 35.

33. 37.

4+58=98=1 6 [3 5 + 2 (3 2)] = 6 [15 + 2 1] = 6 [15 + 2] = 6 17 = 11 10 [6 2 2 + (8 3)] 2 = 10 [6 4 + 5] 2 = 10 [7] 2 = 10 14 =4 4 + 8 12 = =6 53 2 6 10 2 3 2 5 4 = = 25 27 5 5 3 9 45 5 9 5 5 9 6 25 + 54 79 + = + = = 6 5 65 56 30 30 1 7 1 3 7 5 = 30 18 30 3 18 5 3 35 32 16 = = = 90 90 45 5 18 = 5 27 = 5 9 3 = 15 11 18 11 9 2 11 22 27 x(x 4) = x 2 4x (x 2)(x + 1) = (x 2)x + (x 2)1 = x2 2x + x 2 = x2 x 2

4+

1 12 1 13 + = = 3 3 3 3

39. 2 (3 5) + 8 2 1 = 2 (-2) + 16 1 = 4 + 16 1 = 12 1 = 11

41.

43.

(5 3)

1 1 = (2) = 1 2 2

45.

47.

3 10 30 2 = = 5 21 105 7 3 2 3 5 2 4 15 + 8 23 + = + = = 4 5 45 45 20 20 5 1 52 1 3 10 + 3 13 + = + = = 18 12 18 2 12 3 36 36

49.

51.

53.

55.

57.

59.

3 2 33 24 9 8 1 = = = 20 15 20 3 15 4 60 60

61.

63. 6(x + 4) = 6x + 24 67. (x + 2)(x + 4) = (x + 2)x + (x + 2)4 = x2 + 2x + 4x + 8 = x2 + 6x + 8

65. 69.

SECTION 0.2 71. (x 8)(x 2) = (x 8)x + (x 8)(-2) = x2 8x 2x + 16 = x2 10x + 16 75. Answers will vary. 79. Subtraction is not commutative. Examples will vary. 83. Explanations will vary. 85. There are no real numbers that are both rational and irrational. There are no real numbers that are neither rational nor irrational. Explanations will vary. 87. 0.9999 = 1 To show that 0.9999 = 1, we let n = 0.9999 , then 10n = 9.9999 10n = 9.9999 (1) n = 0.9999 (2) 9n = 9.0000 Subtract (2) from (1). Divide both sides by 9. n=1 73.

3

(x + 2)(x 2) = (x + 2)x + (x + 2)(-2) = x2 + 2x 2x 4 = x2 4 Answers will vary. Division is not commutative. Examples will vary.

77. 81.

0.2 Algebra Review1. 3.

1 >0 2

5.

1>2

7.

> 3.14

9.

1 = 0.5 2 x>0 x1 x>1

11.

2 < 0.67 3 x 0, l > 0, w > 0 domain: C > 0, d > 0 domain: A > 0, x > 0

3 2 x ; 4 4 3 r ; 3

131.

V =

domain: V > 0, r > 0 domain: V > 0, x > 0

133. 135.

V = x3 ;

C = 4000 + 2x (a) If x = 1000 watches are produced, it will cost C = 4000 + 2(1000) = $6000.00 (b) If x = 2000 watches are produced, it will cost C = 4000 + 2(2000) = $8000.00

137.

(a) If actual voltage is x = 113 then |113 115| = | - 2 | = 2 Since 2 < 5, an actual voltage of 113 is acceptable. (b) If actual voltage is x = 109 then |109 115| = | - 6 | = 6

SECTION 0.3

7

Since 6 > 5, an actual voltage of 109 is not acceptable.139.

(a) If the radius is x = 2.999, then | x 3 | = | 2.999 3 | = | - 0.001 | = 0.001 Since 0.001 < 0.010 the ball bearing is not acceptable. (b) If the radius is x = 2.89, then | x 3 | = | 2.89 3 | = | - 0.11 | = 0.11 Since 0.11 > 0.01, the ball bearing is not acceptable. 1 1 0.333 ; > 0.333 3 3 1 = 0.333 ; 0.333 0.333 = 0.000333 3 The answer is no. Student answers should justify and explain why not. Answers will vary.

141.

143. 145.

0.3 Polynomials and Rational Expressions1. 3.

(10 x

5

8 x 2 ) + ( 3x 3 2 x 2 + 6 ) = 10 x 5 + 3x 3 10 x 2 + 62

( x + a)

x 2 = x 2 + 2ax + a 2 x 2 2 = 2ax + a

5. 7.

( x + 8)( 2 x + 1) = 2 x 2 + x + 16 x + 8 = 2 x 2 + 17 x + 8

(x

2

+ x 1)( x 2 x + 1) = x 2 ( x 2 x + 1) + x ( x 2 x + 1) 1( x 2 x + 1)

= x 4 x 3 + x 2 + x 3 x 2 + x x 2 + x 1 = x 4 x 2 + 2x 19.

( x + 1)

3

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

= x 3 + 3x 2 + 3x + 1 x 3 + 3x 2 3x + 1 = 6x 2 + 211.

This is the difference of 2 squares. x 2 36 = ( x 6 )( x + 6 ) This is the difference of 2 squares. 1 4 x 2 = (1 2 x )(1 + 2 x )x 2 + 7 x + 10 = ( x + 5 )( x + 2 )The product of 5 and 2 is 10; the sum of 5 and 2 is 7.

13.

15.

8 17. 19. 21. 23.

SECTION 0.3

This polynomial cannot be factored; it is prime. This polynomial cannot be factored; it is prime.15 + 2 x x 2 = ( 5 x )( 3 + x )The product 5 3 =15; the sum 5 1 + 3 ( 1) = 2. Factor out the common factor of 3. The product 6 2 = 12; the sum 6 + 2 = 4. Factor out the common factor of y 2. The product 5 6 = 30; the sum 5 + 6 = 11.

3x 2 12 x 36 = 3 ( x 2 4 x 12 )= 3 ( x 6 )( x + 2 )

25.

y 4 + 11 y 3 + 30 y 2 = y 2 ( y 2 + 11y + 30 )= y 2 ( y + 5 )( y + 6 )

27.

This polynomial is a perfect square. 2 4 x 2 + 12 x + 9 = ( 2 x + 3)( 2 x + 3) = ( 2 x + 3)3 x 2 + 4 x + 1 = ( 3 x + 1)( x + 1)The products 3 1 = 3 and 1 1 = 1; the sum 3 1 + 1 1 = 4. Treat x 81 as the difference of 2 squares, and factor.4

29. 31.

x 4 81 = ( x 2 9 )( x 2 + 9 )

= ( x 3)( x + 3) ( x 2 + 9 )

x 9 is the difference of 2 squares.

2

33.

Let u = x 3. Then 2 x 6 2 x 3 + 1 = u 2 2u + 1 = ( u 1)( u 1) = ( u 1) Now substitute back for u. That is, replace u with x 3.x 6 2 x 3 + 1 = ( x 3 1)2

35.

x 7 x 5 = x 5 ( x 2 1)= x 5 ( x 1)( x + 1)

Factor out the common factor of x 5.x 1 is the difference of 2 squares.2

37.

5 + 16 x 16 x 2 = (1 + 4 x )( 5 4 x ) 4 y 2 16 y + 15 = ( 2 y 5 )( 2 y 3)

The products 4 ( 4) = 16 and 1 5 = 5; the sum 1 ( 4) + 4 5 = 16.

39.

The products 2 2 = 4 and 5 ( 3) = 15; the sum 2 ( 3) + ( 5) 2 = 16.

41.

Let u = x 2, then 1 8 x 2 9 x 4 = 1 8u 9u 2 = (1 9u )(1 + u ) Now substitute back for u. That is, replace u with x 2. 1 8 x 2 9 x 4 = (1 9 x 2 )(1 + x 2 ) 1 9x 2 is the difference of 2 squares.

SECTION 0.3

9

1 8 x 2 9 x 4 = (1 9 x 2 )(1 + x 2 ) = (1 3 x )(1 + 3 x ) (1 + x 2 )43. 45.x ( x + 3) 6 ( x + 3) = ( x + 3)( x 6 )Factor out the common factor x + 3. Factor out the common factor x + 2. Simplify.3

( x + 2)

2

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

47.

6 x ( 2 x ) 9 x 2 ( 2 x ) = 3x ( 2 x ) 2 ( 2 x ) 3x 4

Factor out the common factors. Simplify. Simplify. Multiply (x 2) 3 by ( 1) 3 and (5x 4) by 1. [( 1) 3 ( 1) = 1].

= 3x ( 2 x ) ( 4 2 x 3x )3

= 3x ( 2 x ) ( 4 5 x )3

= 3x ( x 2 ) ( 5 x 4 )3

49.

x 3 + 2 x 2 x 2 = ( x 3 + 2 x 2 ) ( x + 2 ) Group the polynomial into the difference of 2 binomials. = ( x 2 1) ( x + 2 )= x 2 ( x + 2 ) 1( x + 2 ) Factor the common factors from each binomial.Factor the common factor x + 2.

= ( x 1)( x + 1)( x + 2 ) Factor the difference of 2 squares.

51.

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

Group the polynomial into the sum of 2 binomials. Factor the common factors from each binomial. Factor the common factor x 1.

= ( x 1) ( x 3 + 1)

= ( x 1)( x + 1) ( x 2 x + 1) Factor the sum of 2 cubes.53.3 x 6 x 2 x 6 3 ( x 2 ) ( x 3) ( x + 2 ) = 5x x2 4 5x ( x 2) ( x + 2)

Factor.

=55.

3 ( x 3) 5x

Simplify.

4 x 2 1 x 2 4 x ( 2 x 1) ( 2 x + 1) x ( x 4 ) (2 x 1) x = = x+4 x 2 16 2 x + 1 ( x 4) ( x + 4) 2x + 1x x x x 2 = x 7 x + 6 x 2 x 24 ( x 6 )( x 1) ( x 6 )( x + 4 )2

57.

=

x ( x + 4) x ( x 1) ( x 6 )( x 1)( x + 4 ) ( x 6 )( x + 4 )( x 1)

10

SECTION 0.3

= ==

x 2 + 4x x2 x ( x 6 )( x 1)( x + 4 ) ( x 6 )( x + 4 )( x 1) x 2 + 4x x 2 + x ( x 6 )( x 1)( x + 4 )5x ( x 6 )( x 1)( x + 4 )

59.

4 2 4 2 2 = x 4 x + x 6 ( x 2 )( x + 2 ) ( x + 3)( x 2 )2

= == =

4 ( x + 3) 2 ( x + 2 ) ( x 2 )( x + 2 )( x + 3)

4 ( x + 3) 2 ( x + 2) ( x 2 )( x + 2 )( x + 3) ( x + 3)( x 2 )( x + 2 )

4 x + 12 2 x 4 ( x 2 )( x + 2 )( x + 3)

2 ( x + 4) 2x + 8 = ( x 2 )( x + 2 )( x + 3) ( x 2 )( x + 2 )( x + 3)

61.

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

= = =

x ( x + 1)( x 1) 2 x ( x 1) 3 ( x + 1) 2 + 2 2 x ( x + 1)( x 1) x ( x + 1)( x 1) x ( x 1)( x + 1) x ( x + 1)( x 1) 2 x ( x 1) + 3 ( x + 1) x 2 ( x + 1)( x 1)

x 3 x 2 x 2 + 2 x + 3x + 3 x 2 ( x + 1)( x 1)

x 3 2x 2 + 4x + 3 = 2 x ( x + 1)( x 1)63.

1 1 1 = h x+h x =

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

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

SECTION 0.3 65.2 ( 3 x + 4 ) + ( 2 x + 3) 2 ( 3 x + 4 ) 3 = ( 3 x + 4 ) 2 ( 3 x + 4 ) + 6 ( 2 x + 3) 2

11

= ( 3 x + 4 )( 6 x + 8 + 12 x + 18 ) = ( 3 x + 4 )(18 x + 26 )

= 2 ( 3 x + 4 )( 9 x + 13)

67.

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

= 2 x ( 3x + 5)

69.

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

= ( x + 3)( x 2 )

( 2 x 4 + 3x + 9 ) = ( x + 3)( x 2 ) ( 5 x + 5 ) 2 = 5 ( x + 3)( x 2 ) ( x + 1)2 2

71.

( 4 x 3)

2

+ x 2 ( 4 x 3 ) 4 = ( 4 x 3 ) ( 4 x 3 ) + 8 x = ( 4 x 3)(12 x 3) = 3 ( 4 x 3)( 4 x 1)

73.

2 ( 3 x 5 ) 3 ( 2 x + 1) + ( 3 x 5 ) 3 ( 2 x + 1) 2 = 6 ( 3 x 5 )( 2 x + 1) ( 2 x + 1) + ( 3 x 5 ) 3 2 2 2

= 6 ( 3x 5 )( 2 x + 1) ( 5 x 4 )2

75.

( 2 x + 3) 3 ( 3 x 5 ) 2 = 2 ( 3x 5)x 2 x ( x 2 + 1) 1

6 x + 9 6 x + 10

( 3x 5)+ 1)2

2

=

19

( 3x 5 )=

2

77.

(x

2

+ 1)

2

=

2x 2 x 2 1

(x

2

=

(x

x 2 12

+ 1)

( x 1)( x + 1)

2

(x

2

+ 1)

2

79.

( 3x + 1) 2 x x 2 3 = 6 x 2 + 2 x 3x 2 2 2 ( 3x + 1) ( 3x + 1)

=

3x 2 + 2 x

( 3x + 1)2

2

81.

(x

2

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

(x

2

+ 1)

2

=

3x 2 + 3 6 x 2 8 x

(x

2

+ 1)

=

3x 2 8 x + 3

(x

2

+ 1)

2

12

SECTION 0.4

0.4 Solving Equations1.

3x = 21 x=7 2x 3 = 5 2x = 8 x=4

3.

5x + 15 = 0 5x = 15 x=3 1 5 x= 3 12 5 3 x= 12 4 1 5 x= 4 2(3 + 2x) = 3(x 4) 6 + 4x = 3x 12 x = 18 1 3 x4= x 2 4 1 3 4 x 4 = 4 x 2 4 2x 16 = 3x x = 162 4 + =3 y y 6 =3 y 3y = 6 y=2 z ( z 2 + 1) = 3 + z 3

5.

7.

9.

6 x = 2x + 9 3x = 3 x=18 x ( 2 x + 1) = 3 x 106 x 1 = 3 x 10 3x = 9 x=3

11.

13.

15.

17.

0.9t = 0.4 + 0.1t 0.8t = 0.4 0.4 t= = 0.5 0.8

19.

21.

( x + 7 )( x 1) = ( x + 1)

2

23.

x 2 x + 7x 7 = x 2 + 2x + 1 x 2 + 6x 7 = x 2 + 2x + 1 6x 7 = 2x + 1 4x = 8 x=225.

z3 + z = 3+ z3 z=3

x 2 = 9x x 2 9x = 0 x ( x 9) = 0x = 0 or x 9 = 0 x=9

27.

t 3 9t 2 = 0 t 2 (t 9) = 0 t 2 = 0 or t 9 = 0 t = 0 or t = 9The solution set is {0, 9}.

The solution set is {0, 9}.

SECTION 0.4 29.

13

3 2 = 2x 3 x + 5 3(x + 5) = 2(2x 3) 3x + 15 = 4x 6 x = 21

31.

(x + 2)(3x) = (x + 2)(6) 3x 2 + 6 x = 6 x + 12 3x 2 12 = 0 x2 4 = 0 x2 = 4 x = 2 or x = 2 The solution set is { 2, 2}.

33.

2 ( x + 5) 3( x 2) 10 Write with the common denominator. = + ( x 2 )( x + 5) ( x + 5)( x 2 ) ( x + 5)( x 2 ) 2( x + 5) = 3( x 2) + 10 Solve the equation formed by the numerators. 2x + 10 = 3x 6 + 10 x=6 Check the answer for extraneous solutions. The solution set is {6}.35.2x = 6

2 3 10 = + x 2 x + 5 ( x + 5 )( x 2 )

L.C.D.: (x + 5)(x 2)

37.

2x + 3 = 5

Either 2x = 6 or 2x = 6 x = 3 or x = 3 The solution set is { 3, 3}.39.1 4t = 5

Either 2x + 3 = 5 or 2x + 3 = 5 2x = 2 or 2x = 8 x = 1 or x=4 The solution set is { 4, 1}.41. 2x = 8

Either 1 4t = 5 or 1 4t = 5 4t = 4 or 4t = 6 3 t = 1 or t= 2 3 The solution set is 1, . 2 43.2 x=4

Either 2x = 8 or 2x = 8 x = 4 or x=4 The solution set is { 4, 4}.

45.

2x = 4 x=2

1 2 This equation has no solution. Absolute values are always nonnegative. x2 =

14 47.

SECTION 0.4

x2 4 = 0

x2 4 = 0 x2 = 4 x=2 The solution set is { 2, 2}.49.

x 2 2x = 3 Either x 2 2x = 3 x 2 2x 3 = 0 ( x 3)( x + 1) = 0 x 3 = 0 or x + 1 = 0 x = 3 or x=1 The solution set is { 1, 3}. or x 2 2x = 3 x 2 2x + 3 = 0 a = 1, b = 2, c = 3 The discriminant, b 2 4ac = 4 12 = 8 is negative; the equation has no real solutions.

51.

x 2 + x 1 = 1 Either x 2 + x 1 = 1 x2 + x2 = 0 ( x 1)( x + 2 ) = 0 x 1 = 0 or x = 1 or x+2=0 x=2 or x 2 + x 1 = 1 x2 + x = 0 x ( x + 1) = 0 x = 0 or x + 1 = 0 x = 0 or x = 1

or or

The solution set is { 2, 1, 0, 1}.53.

x 2 = 4x x 2 4x = 0 x(x 4) = 0 x = 0 or x 4 = 0 x=4 The solution set is {0, 4}.

55.

z 2 + 4 z 12 = 0 (z 2)(z + 6) = 0 z 2 = 0 or z + 6 = 0 z = 2 or z=6 The solution set is { 6, 2}.x ( x 7 ) + 12 = 0

57.

2 x 2 5x 3 = 0 (2x + 1)(x 3) = 0 2x + 1 = 0 or x 3 = 0 1 x= or x=3 2 1 The solution set is , 3 . 2

59.

x 2 7 x + 12 = 0 (x 3)(x 4) = 0 x 3 = 0 or x 4 = 0 x = 3 or x=4 The solution set is {3, 4}.

SECTION 0.4 61.

15

4 x 2 + 9 = 12 x 4 x 12 x + 9 = 0 (2x 3)(2x 3) = 0 2x 3 = 0 3 x= 2 3 The solution set is . 2

63.

6 x 2 6 x 5x = 6 2 6x 5x 6 = 0 (3x + 2)(2x 3) = 0 3x + 2 = 0 or 2x 3 = 0 2 3 x= or x = 3 2 2 3 The solution set is , . 3 2 6x 5 =

65.

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

4( x 2) 3 3 + = x3 x x ( x 3)

The lowest common denominator is x(x 3).

Write the equation with the common denominator. Consider the equation formed by the numerator. Simplify. Put the quadratic equation in standard form. Factor. Use the Zero-Product Property. Solve; be sure to check for extraneous solutions.

4 x 8 x + 3x 9 = 3 4 x 2 5x 6 = 0 (4x + 3)(x 2) = 0 4x + 3 = 0 or x 2 = 0 3 x= or x = 2 4 3 The solution set is , 2 . 4 2

67.

x 2 = 25 x = 25 x=5 The solution set is { 5, 5}.

69.

( x 1)

2

=4

x 1 = 4 x 1 = 2 x = 2 + 1 or x = 2 + 1 x=3 or x = 1 The solution set is { 1, 3}.73.

71.

( 2 x + 3)

2

=9

2x + 3 = 9 2x + 3 = 3 2 x = 3 + 3 or 2x = 3 3 2x = 0 or 2x = 6 x=0 or x=3 The solution set is {3, 0}.

x 2 + 8x 2 8 Add = 16. 2 Result x 2 + 8 x + 16

16 75.

SECTION 0.4

1 x 2 2 1 1 Add = 16 4 1 1 Result x 2 + x + 2 16 x2 + x 2 + 4 x = 21 x 2 + 4 x + 4 = 21 + 4 2 ( x + 2 ) = 25 x+2= 5 x = 25 The solution set is { 7, 3}. x2 1 3 x =0 2 16 1 3 x2 x = 2 16 1 1 3 1 x2 x+ = + 2 16 16 16 2 1 4 x = 4 16 x

77.

2 x 3 2 1 1 Add = 9 3 2 1 Result x 2 x + 3 9 x2

79.

Add 4 to both sides. Factor. Use the Square Root Method.

81.

Add

1 16

to both sides.

1 4 = 4 16 1 2 x= 4 4 1 3 The solution set is , . 4 4

83.

1 =0 2 1 1 x2 + x = 0 3 6 1 1 x2 + x = 3 6 1 1 1 1 x2 + x+ = + 3 36 6 36 2 1 7 x+ = 6 36 3x 2 + x x+1 = 6 7 36

SECTION 0.4

17

x=

1 7 6 6

1 7 1+ 7 The solution set is , . 6 6 85.

x 2 4x + 2 = 0 a = 1, b = 4, and c = 2 2 2 The discriminant b 4ac = ( 4 ) 4 (1)( 2 ) = 16 8 = 8 is positive, so there are 2 real solutions to the equation.b b 2 4ac 4 8 42 2 = =2 2 = 2 2 2a The solution set is 2 2, 2 + 2 .

x=

{

}

87.

x 2 5x 1 = 0 a = 1, b = 5, and c = 1 2 2 The discriminant b 4ac = ( 5 ) 4 (1)( 1) = 25 + 4 = 29 is positive, so there are 2 real solutions to the equation. 5 29 b b 2 4ac = 2 2a 5 29 5 + 29 , The solution set is . 2 2 x=

89.

2 x 2 5x + 3 = 0 a = 2, b = 5, and c = 3 2 2 The discriminant b 4ac = ( 5 ) 4 ( 2 )( 3) = 25 24 = 1 is positive, so there are 2 real solutions to the equation.b b 2 4ac 5 1 5 1 = = 4 4 2a 3 The solution set is 1, . 2

x=

91.

4y 2 y + 2 = 02

a = 4, b = 1, and c = 22

The discriminant b 4ac = ( 1) 4 ( 4 )( 2 ) = 1 32 = 31 is negative, so the equation has no real solution.93.

4 x 2 = 1 2 x First we rewrite the equation in standard form. 4x 2 + 2x 1 = 0 a = 4, b = 2, and c = 1 2 2 The discriminant b 4ac = ( 2 ) 4 ( 4 )( 1) = 4 + 16 = 20 is positive, so there are 2 real solutions to the equation.x=

2 20 2 2 5 1 5 b b 2 4ac = = = 8 8 4 2a

18

SECTION 0.4

1 5 1+ 5 , The solution set is . 4 4 95.

x 2 + 3x 3 = 0 solutions to the equation.

a = 1, b = 3 , and c = 3

The discriminant b2 4ac =

( 3)

2

4 (1)( 3) = 3 + 12 = 15 is positive, so there are 2 real

b b 2 4ac 3 15 = 2 2a 3 15 3 + 15 , The solution set is . 2 2

x=

97.

a = 1, b = 5, and c = 7 x 2 5x + 7 = 0 2 2 The discriminant b 4ac = ( 5 ) 4 (1)( 7 ) = 25 28 = 3 is negative, so the equation has no real solution. 9 x 2 30 x + 25 = 0 a = 9, b = 30, and c = 25 2 2 The discriminant b 4ac = ( 30 ) 4 ( 9 )( 25 ) = 900 900 = 0 , so the equation has a repeated solution, a root of multiplicity 2. 3x 2 + 5 x 8 = 0 a = 3, b = 5, and c = 8 2 2 The discriminant b 4ac = ( 5 ) 4 ( 3)( 8 ) = 25 + 96 = 121 is positive, so there are 2 real solutions to the equation. ax b = c ax = b + c b+c x= a x x + =c a b bx ax abc + = ab ab ab bx + ax = abc (b + a)x = abc abc x= a+b 1 1 2 + = x a x + a x 1

99.

101.

103.

105.

107.

SECTION 0.4

19

( x + a )( x 1) + ( x a )( x 1) = 2 ( x a )( x + a ) ( x a )( x + a )( x 1) ( x a )( x + a )( x 1) ( x 1)( x a )( x + a ) ( x + a )( x 1) + ( x a )( x 1) = 2 ( x a )( x + a )x 2 + ax x a + x 2 x ax + a = 2 x 2 + 2ax 2ax 2a 2 2 x 2 2 x = 2 x 2 2a 2 2 x = 2a 2 x = a2109.1 1 1 = + R R1 R 2

R1R 2 RR1R 2

=

RR 2 RR1R 2

+

RR1 RR1R 2

R 1R 2 = RR 2 + RR1 R 1R 2 = R R 2 + R1 R=111.

(

)

R1R 2 R1 + R 2

mv 2 R FR = mv 2 mv 2 R= F F= a 1 r S (1 r ) = a S Sr = a Sr = S a S a r= S S=

113.

115. The roots of the quadratic function ax 2 + bx + c = 0 are

x1 =

b b 2 4ac 2a

and

x2 =

b + b 2 4ac 2a

The sum x 1 + x 2 =

b b 2 4ac b + b 2 4ac + 2a 2a

b b 2 4ac + ( b ) + b 2 4ac = 2a 2b b = = 2a a

20

SECTION 0.5

117. If kx 2 + x + k = 0 has a repeated real solution, then its discriminant is zero. a = k, b = 1, and c = k discriminant: b 2 4ac = 1 2 4(k)(k) = 0 1 4k 2 = 0 4k 2 = 1 2k = 1 1 1 So the equation has one repeated root if k = or k = . 2 2 119. The real solutions of the equation ax 2 + bx + c = 0 are

x1 =

b b 2 4ac 2a

and x 2 =

b + b 2 4ac 2a

=

b + b 2 4ac 2a

=

b b 2 4ac 2a

The real solutions of the equation ax 2 bx + c = 0 arex3 =b b 2 4ac 2a

and

x4 =

b + b 2 4ac 2a

So x 1 = x 4 and x 2 = x 3.121. (a) x 2 = 9 and x = 3 are not equivalent. The solution set of x 2 = 9 is { 3, 3}, but the solution set of x = 3 is {3}.

(b) x =

9 and x = 3 are equivalent since they both have the same solution set, {3}.

(c) (x 1)(x 2) = (x 1) 2 and x 2 = x 1 are not equivalent. The solution of the first equation is {1}, but the second equation has no solution.123. 127. Answers will vary.

0.5 Intervals; Solving Inequalities1. 5.

The graph represents [0, 2] or 0 x 2. The graph represents [0, 3) or 0 x < 3. [4, 6)

3. 7.

The graph represents ( 1, 2) or 1 < x < 2. [0, 4]

9.

11.

[4, )

SECTION 0.5 13. ( , 4) 15.

21

2x5

17. 3 < x < 2

19.

4x 3 5 1>8 (d) ( 2)(4) < ( 2)( 3) 8 ( 2)(2) 4x 2 > 431.

(c) (3)(3) < (3)(5) 9 < 1525. 4 > 3 (a) 4 + 3 > 3 + 3 7>0

(c) (3)(4) > (3)( 3) 12 > 927. 2x + 1 < 2 (a) (2x + 1) + 3 < 2 + 3 2x + 4 < 5

(c) (3)(2x + 1) < (3)(2) 6x + 3 < 629. 0 x>6 2x < ( 2)(6) 2x < 12

33.

35.

2x > 6 2x 6 > 2 2 x>3

22 41.

SECTION 0.5

1 x3 2 1 ( 2 ) x ( 2 )( 3) 2 x6

43.

x+1 9 3x 9 > 3 3 x>3 The solution set is {x | x > 3} or the interval (3, ).

49.

3x 1 3 + x 3x 1 + 1 3 + x + 1 3x 4 + x 3x x 4 + x x 2x 4 x2 The solution set is {x | x 2} or the interval [2, ).

51.

2(x + 3) < 8 2 ( x + 3) 8 > 2 2 x+3>4 x+33>43 x>7 The solution set is {x | x > 7} or the interval ( 7, ).

53.

4 3 (1 x ) 3 4 3 + 3x 3 1 + 3x 3 1 + 3x 1 3 1 3x 2 2 x 3 The solution set is x x 2 interval , . 3

55.

2 or the 3

1 ( x 4) > x + 8 2 1 2 ( x 4 ) > 2 ( x + 8) 2 x 4 > 2x + 16 x 4 x > 2x + 16 x 4 > x + 16 4 16 > x + 16 16 20 > x or x < 20 The solution set is {x | x < 20} or the interval ( , 20).

SECTION 0.5

23

57.

x x 1 2 4 x x 4 4 1 2 4 2x 4 x 2x + x 4 x + x 3x 4 4 x 3 4 The solution set is x x or the 3 4 interval , . 3

59.

0 2x 6 4 is equal to the two inequalities 0 2x 6 0 + 6 2x 6 + 6 6 2x 3x and 2x 6 4 2x 6 + 6 4 + 6 2x 10 x5

The solution set consists of all x for which x 3 and x 5 which is written either as {x | 3 x 5} or as the interval [3, 5].

61.

5 4 3x 2 is equal to the two inequalities 5 4 3x 9 3x 3x

and

4 3x 2 3x 2 2 x 3

The solution set consists of all x for which x x 2 x 3 or as the interval 3 2 3 , 3 .

2 and x 3 which is written either as 3

24 63.

SECTION 0.5

3 0 for x < 4 or x > 3. We write the solution set either as {x | x < 4 or x > 3} or as all x in the interval ( , 4) or (3, ).

77. First we solve the equation x ( x 7 ) = 12 and use the solutions to separate the real number line. x ( x 7 ) = 12

x 2 7 x = 12 x 2 7 x + 12 = 0 ( x 3) ( x 4 ) = 0 x3=0 x=3

or or

x4=0 x=4

We separate the number line into the intervals (3, 4) (4, ) ( , 3) We select a number in each interval and evaluate the expression x 2 7 x + 12 at that value. We choose 0, 3.5, and 5. For x = 0: 0 2 7 ( 0 ) + 12 = 12 , a positive number. For x = 3.5: For x = 5:

( 3.5 )2

2

7 ( 3.5 ) + 12 = 0.25 , a negative number.

5 7 ( 5 ) + 12 = 2 , a positive number.

The expression x 2 7 x + 12 > 0 for x < 3 or x > 4. We write the solution set either as {x | x < 3 or x > 4} or as all x in the interval ( , 3) or (4, ).

SECTION 0.5 79. First we solve the equation 4 x 2 + 9 = 6 x and use the solutions to separate the real number line. 4x 2 + 9 = 6x 4x 2 6x + 9 = 0

27

This equation has no real solutions. Its discriminant, b 2 4ac = 36 144 = 108, is negative. The value of 4 x 2 6 x + 9 either is always positive or always negative. To see 2 which is true, we test x = 0. Since 4 ( 0 ) 6 ( 0 ) + 9 = 9 is positive, we conclude that expression is always positive, and the inequality 4 x 2 + 9 < 6 x has no solution.81. First we solve the equation ( x 1) ( x 2 + x + 1) = 0 and use the solutions to separate the

real number line.

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

x 1 = 0 or x 2 + x + 1 = 0 x=1 x = 1 is the only solution, since the equation x 2 + x + 1 = 0 has a negative discriminant. We use x = 1 to separate the number line into two parts: and 1 0, a 1 are equivalent. If a2 = 1.6, then log a 1.6 = 2. 5. We use the fact that y = log a x and x = a y, a > 0, a 1 are equivalent. If 1.12 = M, then log 1.1 M = 2. 7. We use the fact that y = log a x and x = a y, a > 0, a 1 are equivalent. If 2x = 7.2, then log 2 7.2 = x. 9. We use the fact that y = log a x and x = a y, a > 0, a 1 are equivalent. If x 2 = , then log x = 2 . 11. We use the fact that y = log a x and x = a y, a > 0, a 1 are equivalent. If e x = 8, then log e 8 = x or ln 8 = x.

140 SECTION 2.4 13. We use the fact that x = a y, a > 0, a 1 and y = log a x are equivalent. If log 2 8 = 3, then 23 = 8. 15. We use the fact that x = a y, a > 0, a 1 and y = log a x are equivalent. If log a 3 = 6, then a 6 = 3. 17. We use the fact that x = a y, a > 0, a 1 and y = log a x are equivalent. If log 3 2 = x, then 3 x = 2. 19. We use the fact that x = a y, a > 0, a 1 and y = log a x are equivalent. If log 2 M = 1.3, then 21.3 = M. 21. We use the fact that x = a y, a > 0, a 1 and y = log a x are equivalent.

If log 2 = x , then

( 2)

x

= .

23. We use the fact that x = a y, a > 0, a 1 and y = log a x are equivalent. If ln 4 = x, then e x = 4. 25. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 2 1

2 y =1 2 y = 20 y=0 Therefore, log 2 1 = 0

Write in exponential form. 1 = 20 Equate exponents.

27. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 5 25

5 y = 25 5 y = 52 y=2 Therefore, log 5 25 = 2 .

Write in exponential form. 25 = 5 2 Equate exponents.

29. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 1/ 2 16

1 = 16 2 y 4 1 1 = 2 2 y=4 Therefore, log 1/ 2 16 = 4 .

y

Write in exponential form. 16 =

1 2

4

Equate exponents.

SECTION 2.4 141 31. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 10 10

10 y = 10

Write in exponential form.10 = 101/ 2

10 = 10 1 y= 2

y

1/ 2

Equate exponents.

Therefore, log 10 10 =

1 . 2

33. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = log 2 4

( 2) = 4 ( 2) = ( 2)y y

Write in exponential form.4

4=2 =2

( )2

2

=

2

( 2)

4

Therefore, log

2

y=4 4 = 4.

Equate exponents.

35. To find the exact value of the logarithm, we change the expression to its equivalent exponential expression and simplify. y = ln e

ey = e

Write in exponential form.e =e1/ 2

e =e 1 y= 2 1 Therefore, ln e = . 2

y

1/ 2

Equate exponents.

37. The domain of a logarithmic function is limited to all positive real numbers, so for f ( x ) = ln ( x 3) , x 3 > 0. The domain is all x > 3, or using interval notation (3, ). 39. The domain of a logarithmic function is limited to all positive real numbers, so for F ( x ) = log 2 x 2 , x 2 > 0. x 2 is positive except when x = 0, meaning the domain is the

set {x | x 0}.41.

The domain of f (x) = 3 2log 4

x x is restricted by log 4 which is defined only when 2 2

x is positive. So the domain of f is all x > 0, or using interval notation (0, ). 243.

There are two restrictions on the domain of f ( x ) = ln x .

142 SECTION 2.4

First, y = x is defined only when x is nonnegative, that is when x 0. Second, y = ln x is defined only when x is positive, that is when x > 0. However, on the interval (0, 1), ln x < 0, so f ( x ) = ln x is not defined. So the domain of f is all x 1, or using interval notation [1, ).45.

ln

5 = 0.511 3

47.

ln

10 3 = 30.099 0.04

49. If the graph of f contains the point (2, 2), then x = 2 and y = 2 must satisfy the equation y = f ( x ) = log a x

2 = log a 2 a2 = 2 a= 251. The graph of y = log 3 x has the properties:

The domain is (0, ). The range is (, ). The x-intercept is (1, 0). The y-axis is a vertical asymptote. The graph is increasing since a = 3 > 1. 1 The graph contains the points (3, 1) and , 1 . 3 The graph is continuous.

53. The graph of y = log 1/ 5 x has the properties:

The domain is (0, ). The range is ( , ). The x-intercept is (1, 0). The y-axis is a vertical asymptote. 1 The graph is decreasing since a = < 1. 5 1 The graph contains the points (5, 1) and , 5 The graph is continuous.55. B. This graph is reflected over the y-axis.

1 .

57. D. This graph was reflected about both the x- and y-axes. 59. A. This graph has not been shifted or reflected. 61. E. This graph was shifted vertically down 1 unit.

SECTION 2.4 143 63.

y = ln x

Replace x by x + 4; horizontal shift 4 units to the left.

f (x) = ln(x + 4)

The domain of f is all x > 4, or in interval notation, ( 4, ). The range is all real numbers or ( , ). The vertical asymptote is the line x = 4.65.

y = ln x

Add 2; vertical shift up 2 units.

f (x) = 2 + ln x

The domain of f is all x > 0, or in interval notation, (0, ). The range is all real numbers or ( , ). The vertical asymptote is the y-axis, that is the line x = 0.67.

y = log x

Replace x with x 4; horizontal shift right 4 units.

f (x) = log (x 4)

The domain of f is all x > 4, or in interval notation, (4, ). The range is all real numbers or ( , ). The vertical asymptote is the line x = 4.

144 SECTION 2.4 69.

y = log x

Add 2; vertical shift up 2 units.

f (x) = log x + 2

The domain of f is all x > 0, or in interval notation, (0, ). The range is all real numbers or ( , ). The vertical asymptote is the y-axis, that is the line x = 0.71.

log 3 x = 2

x = 32 = 973.log 2 ( 2 x + 1) = 3

Change to exponential form and simplify.

2x + 1 = 2 3 2x + 1 = 8 2x = 7 7 x= 275.

Change to exponential form. Solve the linear equation.

log x 4 = 2 4 = x2 x=2Change to exponential form. Solve using the Square Root Method. (x 2, the base is positive.)

77.

ln e x = 5 ex= e5 x=5 log 4 64 = x 64 = 4 x 43 = 4x 3=x

Change to exponential form. Since the bases are equal, the exponents are equal.

79.

Change to exponential form. Since the bases are equal, the exponents are equal.

81.

log 3 243 = 2 x + 1 3 2x + 1 = 243 3 2x + 1 = 3 5 2x + 1 = 5 2x = 4 x=2Change to exponential form. Since the bases are equal, the exponents are equal. Solve the linear equation.

SECTION 2.4 145 83.

e 3 x = 10 ln 10 = 3x ln 10 x= 3 0.768 e 2 x +5 = 8 ln 8 = 2x + 5 2x = 5 + ln 8 5 + ln 8 x= 2 1.460log 3 ( x 2 + 1) = 2

Change to a logarithmic expression. Exact solution. Approximate solution.

85.

Change to a logarithmic expression. Exact solution. Approximate solution.

87.

x2 + 1 = 32 x2 + 1 = 9 x2 = 8 x= 8=2 289.log 2 8 x = 3

Change to an exponential expression.

Use the Square Root Method. Simplify.

8x= 23 (2 3) x = 2 3 2 3x = 2 3 3x = 3 x=191.

Change to an exponential expression.

Since the bases are equal, the exponents are equal.

(a) When H + = 0.1, pH = log (0.1) = 1. (b) When H + = 0.01, pH = log (0.01) = 2. (c) When H + = 0.001, pH = log (0.001) = 3. (d) As the hydrogen ion concentration decreases, pH increases. (e) If pH = 3.5, then 3.5 = log x log x = 3.5 10 3.5 = x x = 0.000316 (f) If pH = 7.4, then 7.4 = log x log x = 7.4 10 7.4 = x x = 3.981 10 8

93.

(a) If p = 320 mm, 320 = 760 e 0.145 h

146 SECTION 2.4

320 8 = 760 19 8 0.145h = ln 19 1000 8 h= ln 145 19 5.965

e 0.145 h =

Divide both sides by 760. Change to a logarithmic expression. 0.145 =145 1000

The aircraft is at an altitude of approximately 5.965 kilometers above sea level. (b) If p = 667 mm, 667 = 760 e 0.145 h 667 e 0.145 h = 760 667 0.145h = ln 760 1000 667 h= ln 145 760 0.900

Divide both sides by 760. Change to a logarithmic expression. 0.145 =145 1000

The mountain is approximately 0.9 kilometer above sea level.95. (a) We want F(t) = 0.50. 0.50 = 1 e 0.1t e 0.1t = 0.50 0.1t = ln 0.5 t = 10 ln 0.5 6.93 After approximately 6.9 minutes the probability that a car arrives at the drive-thru reaches 50%.

(b) We want F(t) = 0.80. 0.80 = 1 e 0.1t e 0.1t = 0.20 0.1t = ln 0.2 t = 10 ln 0.2 16.09 After approximately 16.1 minutes the probability that a car arrives at the drive-thru reaches 80%.97. We want D = 2. 2 = 5 e 0.4 h 2 e 0.4 h = = 0.4 5 0.4h = ln 0.4

SECTION 2.4 147

5 ln 0.4 2 2.29 The drug should be administered every 2.3 hours (about 2 hours 17 minutes).

h=

99. We want I = 0.5 12 10 / 5 t 0.5 = 1 e ( ) = 1.2 (1 e 2t ) 10 0.5 5 = 1 e 2t = 1.2 12 7 e 2t = 12 7 2t = ln 12 1 7 t = ln 2 12 0.269 It takes about 0.27 seconds to obtain a current of 0.5 ampere.

Next we want I = 1.0 1.0 = 1.2 (1 e 2t ) 1.0 5 = 1.2 6 1 e 2t = 6 1 2t = ln 6 1 1 t = ln 2 6 0.896 It takes about 0.9 seconds to obtain a current of 1ampere. 1 e 2t =101. We are interested in the population when t = 11, so we want P(11). P(10) = 298,710,000 + 10,000,000 log 11 = 309,123,926.9 On January 1, 2020 the population (to the nearest thousand) will be 309,124,000. 103. Normal conversation: x = 10 7 watt per square meter, 10 7 L (10 7 ) = 10 log 12 = 50 decibels 10 105. Amplified rock music: x = 10 1 watt per square meter, 10 1 L (10 1 ) = 10 log 12 = 110 decibels 10

148 SECTION 2.5 107. Mexico City, 1985: x = 125,892 mm. 125,892 M(125,892) = log = 8.1 = 8.1 3 10 109. (a) Since R = 10 when x = 0.06, we can find k. 10 = 3 e 0.06k 10 0.06k = ln 3 100 10 50 10 k= ln = ln 20.066 6 3 3 3

(b) If x = 0.17, then the risk of having a car accident is

R = 3e = 90.91% (c) If the risk is 100%, then R = 100.3e 3 50 10 ln x 3

50 10 ln ( 0.17 ) 3 3

= 100

100 3 50 10 100 ln x = ln 3 3 3 x 0.1747

e

50 10 ln x 3 3

=

(d)e

R = 15 = 3 e 50 10 ln x 3 3

50 10 ln x 3 3

=5

50 10 ln x = ln 5 3 3 x 0.080 A driver with a blood alcohol concentration greater than or equal to 0.08 should be arrested and charged with DUI. (e) Answers will vary.111. Answers may vary.

2.5 Properties of Logarithms1. log 3 371 = 71 5.log a a r = ralog a M

3.

ln e 4 = 4

log a a r = r

2 log 2 7 = 7

=M

SECTION 2.5 149 7. 9.

log 8 2 + log 8 4 = log 8 ( 2 4 ) = log 8 8 = 1 18 log 6 18 log 6 3 = log 6 = log 6 6 = 1 3

11. Use a change of base formula, ln 6 ln 4 ln 4 log 2 6 log 6 4 = = ln 2 ln 6 ln 2

Next simplify ln 4,

ln 4 ln 2 2 2 ln 2 = = =2 ln 2 ln 2 ln 2 So, log 2 6 log 6 4 = 213.3log 3 5 log 3 4

=3

5 log 3 4

=

5 4

15. First we write the exponent as y = log e 2 16 , and express it as an exponential.

y = log e 2 16 e 2y = 16Then, eloge2

or

e y = 4. So y = ln 4.

16

= e y = e ln 4 = 4.19.

17. ln 6 = ln (2 3) = ln 2 + ln 3 = a + b 21. ln 8 = ln 23 = 3 ln 2 = 3a 23.

ln 1.5 = ln

3 = ln 3 ln 2 = b a 2

ln

5

6=

1 1 1 1 ln 6 = ln (2 3) = (ln 2 + ln 3) = (a + b) 5 5 5 5

25. 27.

log 5 ( 25 x ) = log 5 ( 5 2 x ) = log 5 52 + log 5 x = 2log 5 5 + log 5 xlog 2 z 3 = 3log 2 z

29. ln (ex) = ln e + ln x = 1 + ln x 31. ln (xe x) = ln x + ln e x = ln x + x ln e = ln x + x 33. 35.

log a ( u 2v 3 ) = log a u 2 + log a v 3 = 2log a u + 3log a v ln x 2 1 x = ln x 2 + ln (1 x )

(

)

1/ 2

1 = 2ln x + ln (1 x ) 2

150 SECTION 2.5 37. x3 3 log 2 = log 2 x log 2 ( x 3) = 3log 2 x log 2 ( x 3) x 3 x ( x + 2) 2 log = log x ( x + 2 ) log ( x + 3) = log x + log ( x + 2 ) 2log ( x + 3) 2 ( x + 3)

39.

41.

x 2 x 2 ln 2 ( x + 4)

1/ 3

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

43.

ln

5 x 1 + 3x

( x 4)

3

1 3 = ln 5 x 1 + 3x ln ( x 4 ) = ln 5 + ln x + ln (1 + 3x ) 3ln ( x 4 ) 2

45. 47.

3 log 5 u + 4 log 5 v = log 5 u 3 + log 5 v 4 = log 5 ( u 3 v 4 ) log 3 x 5 x log 3 x 3 = log 3 3 = log 3 x 5 / 2 = log 3 x 5 / 2 or log 3 x 2 x

49.

x 2 1 5 log 4 ( x 2 1) 5 log 4 ( x + 1) = log 4 ( x 2 1) log 4 ( x + 1) = log 4 5 ( x + 1) ( x 1) ( x + 1) x 1 = log 4 = log 4 4 ( x + 1) 5 4 ( x + 1)

51.

x x +1 1 x x +1 ln + ln ln ( x 2 1) = ln x x 1 x 1 x ( x 1) ( x + 1) 1 2 = ln = ln ( x 1) = 2 ln ( x 1) 2 ( x 1) 4 8log 2 3x 2 log 2 + log 2 4 = log 2 x

53.

(

3x 24

)

= log 2 ( 3 x 2 )

4 log 2 + log 2 4 x 4 log 2 + log 2 4 x8

SECTION 2.5 151

( 3x 2 ) 4 4 4 = log 2 = log 2 x ( 3x 2 ) 4 x 55.2 1 2log a ( 5 x 3 ) log a ( 2 x + 3) = log a ( 5 x 3 ) log a 2 x + 3 2

= log a57.

(5x )3

2

2x + 3

= log a2

25 x 6 2x + 3

2log 2 ( x + 1) log 2 ( x + 3) log 2 ( x 1) = log 2 ( x + 1) log 2 ( x + 3) log 2 ( x 1)

( x + 1) = log ( x + 1) = log 2 2 x 2 + 2x 3 ( x + 3)( x 1)2 2

59.

log 3 21 =

ln 21 = 2.771 ln 3

61.

log 1/ 3 71 =

ln 71 = 3.880 1 ln 3

63.

log

2

7=

ln 7 = 5.615 ln 2 ln x ln 4

65.

log e =

ln e = 0.874 ln ln ( x + 2 ) ln 2

67.

y = log 4 x =

69.

y = log 2 ( x + 2 ) =

71.

y = log x 1 ( x + 1) =

ln ( x + 1) ln ( x 1)

73.

ln y = ln x + ln C ln y = ln ( Cx ) y = Cx

152 SECTION 2.5

75.

ln y = ln x + ln ( x + 1) + ln C

77.

ln y = ln x ( x + 1) C y = x ( x + 1) C79.ln ( y 3) = 4 x + ln C ln ( y 3) = ln e 4 x + ln C

ln y = 3x + ln C ln y = ln e 3 x + ln C ln y = ln Ce 3 x y = Ce 3 x

ln ( y 3) = ln ( Ce 4 x ) y 3 = Ce 4 x y = Ce 4 x + 381.

3 ln y =

1 1 ln ( 2 x + 1) ln ( x + 4 ) + ln C 2 3 1/ 2 1/ 3 ln y 3 = ln ( 2 x + 1) ln ( x + 4 ) + ln C1/ 2 1/ 3 ln y 3 = ln C ( 2 x + 1) ln ( x + 4 ) C ( 2 x + 1) 1/ 2 3 ln y = ln 1/ 3 ( x + 4)

y =3

C ( 2 x + 1)

1/ 2

( x + 4)

1/ 3

C ( 2 x + 1) 1/ 2 y = 1/ 3 ( x + 4)

1/ 3

=

C ( 2 x + 1)

1/ 6

( x + 4)

1/ 9

C 1/ 3 is still a positive constant. We write C.

83. We use the change of base formula and simplify. log 2 3 log 3 4 log 4 5 log 5 6 log 6 7 log 7 8

= =

log 3 log 2

log 4 log 3

log 5 log 4

log 6 log 7 log 8 log 5 log 6 log 7

log 8 log 23 = log 2 log 2

SECTION 2.5 153

=

3 log 2 log 2

=3

85. We use the change of base formula and simplify. log 2 3 log 3 4 log 4 5 log 5 6 log 6 7 log 7 8 log n ( n + 1) log n +1 2

Noticing that the first n 1 factors follow the pattern of problem 83, we get log 2 3 log 3 4 log 4 5 log 5 6 log 6 7 log 7 8 log n ( n + 1) log n +1 2 log 3 log 4 log 5 log ( n + 1) log 2 ... = log n log ( n + 1) log 2 log 3 log 4 =187.

log a x + x 2 1 + log a x x 2 1 = log a x + x 2 1 = log a x 2 x 2 1

(

)

(

)

(

(

= log a x 2 x 2 1 = log a 1 = 089.

= log a ( x 2 x 2 + 1)

(

)

)( x ) 2

x 2 1

)

ln (1 + e 2 x ) = 2 x + ln (1 + e 2 x ) We work from the complicated side (the right) and simplify to get to the left side of the equation. 2 x + ln (1 + e 2 x ) = ln e 2 x + ln (1 + e 2 x ) = ln e 2 x (1 + e 2 x ) = ln e 2 x + e 2 x e 2 x

= ln e 2 x + 1 = ln (1 + e 2 x )

91. We show that f ( x ) = log 1/ a x by using the Change of Base Formula.f ( x ) = log a x so f ( x ) = log a x =

ln x ln a

=

ln x ln a ln x ln x = = 1 1 ln a ln a

154 SECTION 2.6

Using the Change of base formula in reverse, the last expression becomes ln 1/ a x . So f ( x ) = log 1/ a x

93. If f ( x ) = log a x , then

1 1 f = log a = log a x 1 = log a x = f ( x) x x95.

If A = log a M and B = log a N , then (writing each expression as an exponential) we get a A = M and a B = N . aA M log a = log a B N a = log a a A B

= A B = log a M log a N97.6 5 5

6

6

6

Y1 = log ( x

5

2

)

Y2 = 2log ( x )

5

Explanations may vary.

2.6 Continuously Compounded Interest1. If $1000 is invested at 4% compounded continuously, the amount A after 3 years is 0.04 3 A = Pe rt = 1000e ( )( ) = 1000e 0.12 = $1127.50 3. If $500 is invested at 5% compounded continuously, the amount A after 3 years is 0.05 3 A = Pe rt = 500e ( )( ) = 500e 0.15 = $580.92 5. The present value of $100 that will be invested at 4% compounded continuously for 6 months is 0.04 0.5 P = Ae rt = 100e ( )( ) = 100e 0.02 = $98.02 7. The present value of $500 that will be invested at 7% compounded continuously for 1 year is 0.07 1 P = Ae rt = 500e ( )( ) = 500e 0.07 = $466.20

SECTION 2.6 155 9. If $1000 is invested at 2% compounded continuously, the amount A after 1 year is 0.02 1 A = Pe rt = 1000e ( )( ) = 1000e 0.02 = $1020.20

A P = $20.20 interest was earned.11. We need to deposit the present value of $5000 that will be invested at 3% compounded continuously for 4 years. 0.03 4 P = Ae rt = 5000e ( )( ) = 5000e 0.12 = $4434.60

We need to deposit the present value of $5000 that will be invested at 3% compounded continuously for 8 years. 0.03 8 P = Ae rt = 5000e ( )( ) = 5000e 0.24 = $3933.1413. If P is invested, it will double when amount A = 2P. A = 2 P = Pe rt 2 = e 3r 3r = ln 2 ln 2 r= 0.2310 3 It will require an interest rate of about 23.1% compounded continuously to double an investment in 3 years. 15. If P is invested, it will triple when amount A = 3P. A = 3P = Pe rt 3 = e 0.10t 0.10t = ln 3 ln 3 t= 10.986 0.10 It will take about 11 years for an investment to triple at 10 % compounded continuously. 17. We need the present value of $1000 that will be invested at 9% compounded continuously for 1 year. 0.09 1 P = Ae rt = 1000e ( )( ) = 1000e 0.09 = $913.93

We need the present value of $1000 that will be invested at 9% compounded continuously for 2 years. 0.09 2 P = Ae rt = 1000e ( )( ) = 1000e 0.18 = $835.2719. Tami and Todd need to invest the present value of $40,000 that will earn 3% compounded continuously for 4 years. 0.03 4 P = Ae rt = 40,000e ( )( ) = 40,000e 0.12 = $35,476.82 21. If P is invested, it will triple when amount A = 3P. A = 3P = Pe rt 3 = e 5r

156 CHAPTER 2 REVIEW

5r = ln 3 ln 3 r= 0.2197 5 It will require an interest rate of about 22% compounded continuously to triple an investment in 5 years.23. The rule of 70: (a) The actual time it takes to double an investment if r = 1% is ln 2 t= = 69.3147 years. 0.01 The estimated time it takes to double an investment using the Rule of 70 if r = 1% is 0.70 t= = 70.0 years. 0.01 The Rule of 70 overestimates the time to double money by 0.6853 year, about 8 months.

(b) The actual time it takes to double an investment if r = 5% is ln 2 t= = 13.8629 years. 0.05 The estimated time it takes to double an investment using the Rule of 70 if r = 5% is 0.70 t= = 14 years. 0.05 The Rule of 70 overestimates the time to double money by 0.1371 year, about 1.6 months. (c) The actual time it takes to double an investment if r = 10% is ln 2 t= = 6.9315 years. 0.10 The estimated time it takes to double an investment using the Rule of 70 if r = 10% is 0.70 t= = 7.0 years. 0.10 The Rule of 70 overestimates the time to double money by 0.0685 year, almost 1 month.

Chapter 2 ReviewTRUE-FALSE ITEMS 1. True 7. False 3. True 9. False 5. False

CHAPTER 2 REVIEW 157 FILL IN THE BLANKS 1. parabola 7. (0, ) REVIEW EXERCISES 1. 3. x = 9. one

b 2a

5. one

First we expand the function. 2 f ( x ) = ( x 2) + 2 = x 2 4x + 4 + 2 = x 2 4 x + 6 and find that a = 1, b = 4, c = 6.

( 4 ) = 2 , and the y-coordinate is b = 2 (1) 2a 2 f ( 2) = ( 2) 4 ( 2) + 6 = 2 . So the vertex is (2, 2) and the axis of symmetry is the line x = 2.The x-coordinate of the vertex is Since f (0) = c = 6, the y-intercept is (0, 6). The x-intercepts are found by solving f (x) = 0. Since the discriminant of f b 2 4ac = ( 4) 2 4(1)(6) = 8 is negative, the equation f (x) = 0 has no real solution, and therefore, the parabola has no x-intercept. To graph the function, we use symmetry. If we choose the y-intercept, we obtain its symmetric point (4, 6).

a > 0; the parabola opens up.

3.

1 , b = 0, c = 16 4 a > 0, the parabola opens up.

a=

The x-coordinate of the vertex is

b = 0, and the y-coordinate is 2a

1 2 f ( 0 ) = ( 0 ) 16 = 16 . 4 So the vertex is (0, 16), and the axis of symmetry is the line x = 0. Since f (0) = c = 16, the y-intercept is also (0, 16) The x-intercepts are found by solving f (x) = 0.

158 CHAPTER 2 REVIEW

1 2 x 16 = 0 4 x 2 64 = 0 x 2 = 64 x=8 The x-intercepts are (8, 0) and ( 8, 0).

5. a = 4, b = 4, c = 0 a < 0, the parabola opens down.

The x-coordinate of the vertex is

b 4 1 = = , and the y-coordinate is 2 ( 4) 2 2a2

1 1 f = 4 + 4x = 1 2 21 1 So the vertex is , 1 , and the axis of symmetry is the line x = . 2 2 Since f (0) = c = 0, the y-intercept is (0, 0). The x-intercepts are found by solving f (x) = 0. 4 x 2 + 4x = 0 4 x ( x 1) = 0

x = 0 or x = 1The x-intercepts are (0, 0) and (1, 0).

7.

a=

9 , b = 3, c = 1 2

a > 0; the parabola opens up.The x-coordinate of the vertex is

b 3 1 = = , and the y-coordinate is 3 9 2a 2 22

1 1 9 1 1 f = + 3 + 1 = 2 3 2 3 3 1 1 1 So the vertex is , , and the axis of symmetry is the line x = . 3 3 2

CHAPTER 2 REVIEW 159

Since f (0) = c = 1, the y-intercept is (0, 1). The x-intercepts are found by solving f (x) = 0. Since the discriminant of f 9 b 2 4ac = 3 2 ( 4 ) (1) = 9 2 is negative, the equation f (x) = 0 has no real solution, and therefore, the parabola has no x-intercept. To graph the function, we use symmetry. If we choose the y-intercept, we obtain its 2 symmetric point , 1 . Because this point is so close to the vertex, we also use the 3 5 point 1, . 2

9. a = 3, b = 4, c = 1

a > 0; the parabola opens up.The x-coordinate of the vertex is

b 4 2 = = , and the y-coordinate is 2a 2 ( 3) 32

7 2 2 2 f = 3 + 4 1 = 3 3 3 3 7 2 2 So the vertex is , and the axis of symmetry is the line x = . 3 3 3

Since f (0) = c = 1, the y-intercept is (0, 1). The x-intercepts are found by solving f (x) = 0. 3x 2 + 4 x 1 = 0x= 4 4 2 4 ( 3)( 1) 2 ( 3) = 4 28 6

x=

42 7 2 7 = 6 3

The x-intercepts are difficult to graph, so we choose another point and use symmetry. 2 We chose ( 2, 3) and by symmetry , 3 . 3

160 CHAPTER 2 REVIEW

11. a = 3, b = 6, c = 4 The quadratic function has a minimum since a = 3 is greater than zero. b 6 x= = =1 2a 2 ( 3)

f (1) = 3 (1) 6 (1) + 4 = 12

The minimum value is 1, and it occurs at x = 1.13. a = 1, b = 8, c = 4 The quadratic function has a maximum since a = 1 is less than zero. b 8 x= = =4 2a 2 ( 1)

f ( 4 ) = ( 4 ) + 8 ( 4 ) 4 = 122

The maximum value is 12, and it occurs at x = 4.15. a = 3, b = 12, c = 4 The quadratic function has a maximum since a = 3 is less than zero. b 12 x= = =2 2a 2 ( 3)

f ( 2 ) = 3 ( 2 ) + 12 ( 2 ) + 4 = 162

The maximum value is 16, and it occurs at x = 2.17. Answers will vary. 19.

y = x4

Add 2; vertical shift 2 units up.

f ( x) = x 4 + 2

CHAPTER 2 REVIEW 161 21.

y = x5

Multiply y by 1; reflect about x-axis.

y = x5

Add 1; vertical shift up 1 unit.

f ( x) = x 5 + 1

23. f is a polynomial function. Its degree is 5. 25. f is not a polynomial function. The exponent on the middle term is not a positive integer. 27. The power function that models the end behavior of the function f is p(x) = 2x 4. 29. The domain of a rational function is all real numbers except those that make the denominator zero. The denominator of R(x), x 2 9 = 0 when x = 3 or when x = 3. The domain of R is {x | x 3 or x 3}. 31. The domain of a rational function is all real numbers except those that make the denominator zero. The denominator of R(x), (x + 2) 2 = 0, when x = 2. The domain of R is {x | x 2}. 33. (a) f (4) = 3 4 = 81

(b) g(9) = log 3 9 = log 3 3 2 = 2 1 1 1 g = log 3 = log 3 3 (d) 27 27 3 3 = log 3 3 = 337.

(c) f ( 2) = 3

2

1 = 9

35.

52 = z log 5 z = 2

log 5 u = 13

5 13 = u

162 CHAPTER 2 REVIEW 39.

2 The domain of f consists of all x for which 3x 2 > 0, that is for all x > , or using 3 2 interval notation, , . 3 The domain of H consists of all x for which 3x + 2 > 0, that is for all x < 2 interval notation, , . 3 2 , or using 3

41.

43.

1 log 2 = log 2 ( 2 3 ) = 3 log 2 2 = 3 8 ln e 2 = 2ln e = 2log 2 0.4

45.

log 3 81 = log 3 ( 3 4 ) = 4 log 3 3 = 4

47. 51. 53.

49.

ln e

2

= 2 ln e = 2

2

= 0.4

uv 2 2 log 3 = log 3 ( uv ) log 3 w w = log 3 u + log 3 v 2 log 3 w

= log 3 u + 2 log 3 v log 3 w55.

log x 2 x 3 + 1 = log x 2 + log = 2 log x +

(

)

x 3 +1

1 log ( x 3 + 1) 2

57.

x 3 x 2 +1 ln = ln x x3

(

3

x 2 + 1 ln ( x 3)3

)

= ln x + ln

x 2 + 1 ln ( x 3)

= ln x +59.

1 ln ( x 2 + 1) ln ( x 3) 3

3 log 4 x 2 +

3 1 log 4 x = log 4 ( x 2 ) + log 4 x 2 = log 4 x 6 + log 4 x 1/ 4

( )

1/ 2

= log 4 ( x 6 x 1/ 4 )= log 4 x 25 / 4

CHAPTER 2 REVIEW 163 61.

x 1 x x 1 x 2 2 ln + ln ln ( x 1) = ln x + 1 ln ( x 1) x x +1 x x 1 x +1 = ln 2 ( x 1) x 1 1 = ln 2 x + 1 x 1

1 x 1 = ln x + 1 ( x 1) ( x + 1) 1 2 = ln ( x + 1) = ln 2 ( x + 1) = 2 ln ( x + 1)63.

2 log 2 + 3 log x

1 1 log ( x + 3) + log ( x 2 ) = 2 log 2 + 3 log x log ( x + 3)( x 2 ) 2 2 = log 2 2 + log x 3 log ( x + 3)( x 2 ) = log ( 4 x 3 ) log ( x + 3)( x 2 ) 1/ 2 1/ 2

4x 3 = log ( x + 3)( x 2 ) 1/ 2 4x 3 = log 2 x + x6 65.

log 4 19 =

ln 19 = 2.124 ln 4

67.

y= 2x

Replace x by x 3; horizontal shift right 3 units.

f ( x ) = 2 x 3

164 CHAPTER 2 REVIEW

The domain of the function f is all real numbers, or in interval notation ( , ); the range is all {y | y > 0}. The x-axis is a horizontal asymptote as f becomes unbounded in the negative direction.69.

y=ex

y= ex

Multiply y by 1; reflect about the x-axis.

Add 1; vertical shift up 1 unit.

f ( x) = 1 e x

The domain of the function f is all real numbers, or in interval notation ( , ); the range is all {y | y < 1}. The line y = 1 is a horizontal asymptote as f becomes unbounded in the negative direction.71.

y = ln xAdd 3; vertical shift up 3 units.

f (x) = 3 + ln x

The domain of the function f is {x | x > 0}, or in interval notation (0, ); the range is all real numbers. The y-axis is a vertical asymptote.73.

(2 )

4 1 2 x = 2=2Set exponents equal.

2 1 2 x 2 (1 2 x )

2 = 21 2 4x = 1 4x = 1 1 x= 475.

3x

2

+x

= 3 = 3 1/ 2 1 x2 + x = 2

CHAPTER 2 REVIEW 165

1 =0 2 2x 2 + 2x 1 = 0 x2 + xx= 2 2 2 4 ( 2 )( 1) 2 ( 2) = 2 4 + 8 2 12 2 2 3 1 3 = = = 4 4 2 42

1 + 3 1 3 x= , 2 2 77.

log x 64 = 3 x 3 = 64 = 4 33 1 1 1 x = = 4 4 1 x= 4 33

79.

(3 )2

9 2 x = 27 3 x 42x

= (3 3 )

3x 4

3 4x = 3 ( ) 4 x = 3 ( 3x 4 )3 3x4

4 x = 9 x 12 5x = 12 12 x= 5

81.

log 3 ( x 2 ) = 2

32 = x 2 9=x2 x = 1183. If $100 is invested at 10% compounded continuously, the amount A after 2.25 years (2 years and 3 months) is 0.1 2.25 A = Pe rt = 100e ( )( ) = 100e 0.225 = $125.23 85. I need to invest the present value of $1000 at 4% compounded continuously for 2 years. 0.04 2 P = Ae rt = 1000e ( )( ) = 1000e 0.08 = $923.12 87. If P is invested, it will double when amount A = 2P. A = 2 P = Pe rt 2 ( 220,000 ) = 220,000e 0.06t

2 = e 0.06t 0.06t = ln 2 ln 2 11.55 t= 0.06 It will take about 11 and one half years for an investment to double at 6 % compounded continuously.

166 CHAPTER 2 REVIEW 89. When T = 0 and P 0 = 760 mm. of mercury, 760 h ( x ) = 8000log x 760 h ( 300 ) = 8000log = 3229.54 300 The Piper Cub is flying at an altitude of approximately 3230 meters above sea level. 91. The perimeter of the pond is P = 200 = 2l + 2w 200 2w l= = 100 w 2

If x denotes the width of the pond, then A = lw = (100 x)x =100x x 2 A = A(x) is a quadratic function with a = 1, so the maximum value of A is found at the vertex of A. b 100 x= = = 50 2a 2 ( 1)A ( 50 ) = 100(50) 502 = 2500 square feet.

The dimensions should be 50 feet by 50 feet.93. The perimeter of the window is 100 feet and can be expressed as P = 2l + d = 2l + x = 100

Solving for l in terms of x, we find 100 x x l= = 50 2 2 The area A of the rectangle is a quadratic function given by x x2 A ( x ) = 50 x = 50 x 2 2 The maximum area is found at the vertex of A. The width of the rectangle with maximum area is 50 50 b = = feet x= 2a 2 2 The length of the rectangle is 50 x l = 50 = 50 = 50 25 = 25 feet 2 295. (a) A 3.5 inch telescope has a limiting magnitude of L = 9 + 5.1 log 3.5 = 11.77

CHAPTER 2 REVIEW 167

(b) If the stars magnitude is 14, the telescope must have a lens with a diameter of 9 + 5.1 log d = 14 5.1 log d = 5 5 log d = 5.1 d = 10 5 / 5.1 = 9.56 inches.97.

(a) If $620.17 grows to $5000 in 20 years when interest is compounded continuously, the interest rate is 10.436% A = Pe rt 5000 = 620.17 e 20 r 5000 e 20 r = 620.17 5000 20r = ln 620.17 1 5000 r= ln = 0.10436 20 620.17 (b) An investment of $4000 will have a value A in 20 years if it is invested at 10.436% compounded continuously. A = Pe rt = 4000 e (0.10436)(20) = $32, 249.24

99. (a) The Calloway Company will minimize marginal cost if it produces ( 617.4 ) = 63 golf clubs. b x= = 2a 2 ( 4.9 )

(b) The marginal cost of making the 64 golf club is 2 C ( 63) = 4.9 ( 63) 617.4 ( 63) + 19,600 = $151.90

168 CHAPTER 2 PROJECT CHAPTER 2 PROJECT 1.

3. (a) t = 150 represents the year 2000. The projected population at t = 150 is P (150 ) = 2.4 1.59 150 /10 = 2.4 1.59 15 = 2518.7 thousand persons.

(b) The projected population overestimates the actual population by 565.1 thousand persons, or by 28.9%. (c) Explanations will vary. (d) t = 160 represents the year 2010. Using the exponential growth function, the predicted population is P (160 ) = 2.4 1.59 160 /10 = 2.4 1.59 16 = 4004.7 thousand persons. (e) t = 200 represents the year 2050. Using the exponential growth function, the predicted population is P ( 200 ) = 2.4 1.59 200 /10 = 2.4 1.59 20 = 25,594 The predicted population is 25.594 million people.5.

The growth rate for Houston is P ( t ) = 2.4 e

ln1.59 ( t /10)

7. Answers will vary. 9. Answers will vary. 11. 3. To do this problem with the result from above, use TBLSET Indpnt: Ask, TABLE. (a) The population of Houston in 2000 (t = 50) is predicted to be 2159.1 thousand.

(b) The prediction overestimates the actual population by 205.5 thousand, or 10.5%.

MATHEMATICAL QUESTIONS 169

(c) Answers will vary. (d) The population of Houston in 2010 (t = 60) is predicted to be 2702.0 thousand. (e) The population of Houston in 2050 (t = 100) is predicted to be 6627.0 thousand. 4. If P = P0 a t /10 = P0 e k t /10 , where P0 = 703.4597 and a = 1.0227 then a t /10 = e k t /10 k = ln a = ln 1.0227 = 0.022446 5. The growth rate for Houston is P ( t ) = 703.4597 e Explanations will vary.MATHEMATICAL QUESTIONS FROM PROFESSIONAL EXAMS 1.ln1.0227 ( t /10 )

(a) 4 27 = 2 2 3 3 = 6 2 3 =

62 3 2 62 6 63 = = 2 2 2

63 log 6 ( 4 27 ) = log 6 = log 6 ( 6 3 ) log 6 ( 2 ) = 3 b 2

3.

(c)

e 2x

e x e x y= 2 x 2y = e e x 2 ye x = e x e x e x e x 2 ye x = e 2 x 1 2 ye x 1 = 0 2y 4y 2 + 4 e = 2x

Multiply both sides by e x. a = 1, b = 2y, c = 1

2y 2 y 2 +1 = 2 = y y 2 +1 Since e x e x > 0 , y > 0, so e x = y + y 2 + 1 . 2 x = ln y + y 2 + 15.

(

)

(b)

log a ( log a b ) ( log b a ) = log b log b = 1 a log

170 MATHEMATICAL QUESTIONS 7. (a)

e e

2 ln ( x 1) ln ( x 1)2

=4

=4 ( x 1) = 42

x1=2 x=3

Section 2.1 Quadratic Functions47. Analyzing the Motion of a Projectile A projectile is fired from a cliff 200 feet above the water at an inclination of 45 to the horizontal, with a muzzle velocity of 50 feet per second. The height h of the projectile above the water is given by 32 x 2 h ( x) = + x + 200 2 ( 50 ) where x is the horizontal distance of the projectile from the base of the cliff. (d) Using a graphing utility, graph the function h, 0 x 200 .

Enter the formula for h into the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try 0 y 300 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.

(e)

When the height of he projectile is 100 feet above the water, how far is it from the cliff?

Return to the function editor, input the equation y = 100 , and graph the two equations.

57

The point where the graph of h and the horizontal line y = 100 intersect represents the point where the projectile is 100 feet above the water. Recall that you can find the point of intersection on your calculator using the intersect function, which can be found in the CALC menu. The x-coordinate of the point of intersection appears to be close to x = 125 .

Thus, the point of intersection is approximately (135.698,100 ) . When the projectile is 100 feet above the water it is approximately 135.698 feet from the cliff.

58

53.

Hunting The function H ( x ) = 1.01x 2 + 114.3 x + 451.0 models the number of individuals who engage in hunting activities whose annual income is x thousand dollars.

(b)

Using a graphing utility, graph H = H ( x ) . Are the number of hunters increasing or decreasing for individuals earning between $20,000 and $40,000?

Enter the formula for H into the function editor. Since x is in thousands, the earnings range of $20,000 to $40,000 corresponds to 20 x 40 . Go to WINDOW and enter limits 20 x 40 for x. We must determine limits for y, so try 0 y 1000 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.

There is no graph in our viewing rectangle, so we will need to find a better window. Lets try 0 y 5000 .

59

This time we can see the graph. The graph is increasing on this interval, so the number of hunters is increasing for individuals earning between $20,000 and $40,000.Male Murder Victims The function M ( x ) = 0.76 x 2 107.00 x + 3854.18 models

55.

the number of male murder victims who are x years of age ( 20 x 90 ). (c) Using a graphing utility, graph M = M ( x ) .

Enter the formula for h into the function editor. Go to WINDOW and enter limits for x. We must determine limits for y, so try 0 y 4000 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.

Section 2.3 Exponential Functions63.Exponential Probability Between 12:00 PM and 1:00 PM, cars arrive at Citibanks drive-thru at the rate of 6 cars per hour (0.1 car per minute). The following formula from probability can be used to determine the probability that a car will arrive within t minutes of 12:00 PM: F ( t ) = 1 e 0.1t

60

(d)

Graph F using your graphing utility.

Enter the formula for F into the function editor using x instead of t. Since x is in minutes, the time range 12:00 PM to 1:00 PM corresponds to 0 x 60 . Go to WINDOW and enter the limits 0 x 60 for x. We must determine limits for y, so try 0 y 1 . If this does not work, then we can go back to WINDOW and adjust the limits on y until we find a good window.

(e)

Using TRACE, determine how many minutes are needed for the probability to reach 50%.

The TRACE feature uses the points on the curve that the calculator generated to draw the curve. As you TRACE along the curve, the distance between successive x-coordinates of points is the same. It is determined byx = Xmax Xmin 94

Because of this, as we trace along the curve, we may not get nice values for x (with the window used for the problem we will not get nice values). We must trace until we find the value of x that corresponds to the value y that is closest to 0.5. Select the TRACE feature.

61

r

Note that your cursor may not be in the same spot as shown above. Use the | and ~ keys to move the cursor along the curve. Move the cursor until the value of y changes from more than 0.5 to less than 0.5.

Since y is less than 0.5 when x 6.383 , and y is more than 0.5 when x 7.021 , then y is equal to 0.5 when 6.383 < x < 7.021 . We can improve our approximation if we TRACE over a smaller viewing rectangle. We can obtain a smaller viewing window by zooming in on the graph using Zoom In.

q

Select Zoom In.

62

Be sure the y-value of the cursor is close to 0.5. Zoom in on the graph.

Use TRACE and move the cursor so that the y-value is close 0.5.

Since y is less than 0.5 when x 6.861 , and y is more than 0.5 when x 7.021 , then y is equal to 0.5 when 6.383 < x < 7.021 . We can continue to improve our solution by zooming in on our graph.Zoom In again, then TRACE.

. Since y is less than 0.5 when x 6.902 , and y is more than 0.5 when x 6.914 , then y is equal to 0.5 when 6.902 < x < 6.914 . We can continue to improve our solution by zooming in on our graph.

63

Zoom In again, then TRACE.

. Since y is less than 0.5 when x 6.921 , and y is more than 0.5 when x 6.932 , then y is equal to 0.5 when 6.921 < x < 6.932 . We can continue to improve our solution by zooming in on our graph.Zoom In again, then TRACE.

. Since y is less than 0.5 when x 6.929 , and y is more than 0.5 when x 6.932 , then y is equal to 0.5 when 6.929 < x < 6.932 . We can continue to improve our solution by zooming in on our graph.Zoom In again, then TRACE.

. Since y is less than 0.5 when x 6.9309 , and y is more than 0.5 when x 6.9315 , then y is 0.5 when x is approximately 6.931. Thus, the probability will reach 50% after 6.931 minutes.

64

79.

The hyperbolic sine function, designated by sinh x, is defined as 1 sinh x = ( e x e x ) 2 (b) Graph f ( x ) = sinh x using a graphing utility.

Enter the formula for f into the function editor. Go to WINDOW and enter limits for x and y. Since we are not given limits for x or y, use the standard window, 10 x 10 and 10 y 10 .

________________________________________________________________________

Section 2.5 Properties of LogarithmsIn Problems 6772, graph each function using a graphing utility and the Change-of-Base formula. 67. y = log 4 x

log b M . Because we will log b a use our calculator to graph the equation, we must use 10 or e for the new base b. Use e as the new base. Thus

Recall that the Change-of-Base formula states that log a M =

65

y = log 4 x =

log e x ln x = log e 4 ln 4

Enter y =

ln x into the function editor. Go to WINDOW and enter limits for x and y. Since ln 4 we are not given limits for x or y, use the viewing rectangle 0 x 10 and 10 y 10 .

69.

y = log 2 ( x + 2 )

Using the Change-of-Base formula we obtainy = log 2 ( x + 2 ) = ln ( x + 2 ) ln 2

Enter y =

ln ( x + 2 ) into the function editor. Go to WINDOW and enter limits for x and y. ln 2 Since we are not given limits for x or y, use the viewing rectangle 3 x 10 and 10 y 10 .

66

71.

y = log x 1 ( x + 1)

Using the Change-of-Base formula we obtainy = log x 1 ( x + 1) = ln ( x + 1) ln ( x 1)

Enter y =

ln ( x + 1) into the function editor. Go to WINDOW and enter limits for x and y. ln ( x 1) Since we are not given limits for x or y, use the viewing rectangle 0 x 10 and 10 y 10 .

67

97.

Graph Y1 = log ( x 2 ) and Y2 = 2 log ( x ) on your graphing utility. Are they equivalent? What might account for any differences in the two functions?

Enter Y1 = log ( x 2 ) into the function editor. Go to WINDOW and enter limits for x and y.

Since we are not given limits for x or y, use the viewing rectangle 10 x 10 and 10 y 10 .

Return to the function editor, clear the first function, enter Y2 = 2 log ( x ) and graph on the same view rectangle.

68

just x > 0 . The range of x 2 is x 0 , so the domain of log ( x 2 ) is all real numbers except 0.

No, they are not equivalent, because the graphs are not identical. The difference is due to the range of x and x 2 . The range of x is all real numbers, but the domain of log ( x ) is

Chapter 2 Review99.Minimizing Marginal Cost The marginal cost of a product can be thought of as the cost of producing one additional unit of output. For example, if the marginal cost of producing the 50th product is $6.20, then it cost $6.20 to increase production from 49 to 50 units of output. Callaway Golf Company has determined that the marginal cost C of manufacturing x Big Bertha golf clubs may be expressed by the quadratic function C ( x ) = 4.9 x 2 617.4 x + 19, 600

(a)

How many golf clubs should be manufactured to minimize the marginal cost?

Enter the formula for C into the function editor. Go to WINDOW and enter limits for x and y. Since we are not given limits for x or y, use the viewing rectangle 0 x 150 and 0 y 30000 .

69

Recall that we can use the minimum function to find the local minimum. Note that the xcoordinate of the local minimum is between x = 50 and x = 75 .

The local minimum is approximately ( 63.0, 151.9 ) . Thus, Callaway should produce 63 Big Bertha clubs. (b) At this level of production, what is the marginal cost?

The marginal cost for producing 63 Big Bertha clubs is $151.90.

70

SummaryThe commands introduced in this chapter were:TRACE Zoom In

71

72

Chapter 3 The Limit of a Function3.1 Finding Limits Using Tables and Graphs1. Here f (x) = 2x, and c = 1. We complete the table by evaluating the function f at each value of x. x 0.9 0.99 0.999 f ( x) = 2x 1.8 1.98 1.998 x 1.1 1.01 1.001 f ( x) = 2x 2.2 2.02 2.002 We infer from the table that lim f ( x ) = lim 2 x = 2 .x 1 x 1

3. Here f (x) = x 2 + 2, and c = 0. We complete the table by evaluating the function f at each value of x. x 0.1 0.01 0.001 f ( x ) = x2 + 2 x f ( x) = x + 22

2.01 0.1 2.01x 0

2.0001 0.01 2.0001x 0

2.0000 0.001 2.0000

We infer from the table that lim f ( x ) = lim ( x 2 + 2 ) = 2. 5. Here f (x) = x2 4 , and c = 2. We complete the table by evaluating the function f at x+2 each value of x. x 2.1 2.01 2.001 f ( x) = x f ( x) = x2 4 x+2 x2 4 x+2 4.1 1.9 3.9 4.01 1.99 3.99 4.001 1.999 3.999

x2 4 We infer from the table that lim f ( x) = lim = 4 . x 2 x 2 x+2

7. Here f (x) =

x3 + 1 , and c = 1. We complete the table by evaluating the function f at x +1 each value of x.

172 SECTION 3.1 x f ( x) = x f ( x) = x +1 x +13

1.1 x +1 x +13

1.01 3.0301 0.99 2.9701

1.001 3.0030 0.999 2.9970

3.31 0.9 2.71

x3 + 1 We infer from the table that lim f ( x) = lim = 3. x 1 x 1 x +1

9. Here f (x) = 4x 3 , and c = 2. We choose values of x close to 2, starting at 1.99. Then we select additional numbers that get closer to 2, but remain less than 2. Next we choose values of x greater than 2, starting with 2.01, that get closer to 2. Finally we evaluate the function f at each choice to obtain the table: x 1.99 1.999 1.9999 2.0001 2.001 2.01f ( x) = 4 x3 31.522 31.952 31.995 32.005 32.048 32.482

We infer that as x gets closer to 2, f gets closer to 32. That is, lim f ( x ) = lim ( 4 x3 ) = 32x 2 x2

11. Here f (x) = x + 1 , and c = 0. We choose values of x close to 0, starting at 0.01. Then x2 + 1 we select additional numbers that get closer to 0, but remain less than 0. Next we choose values of x greater than 0, starting with 0.01, that get closer to 0. Finally we evaluate the function f at each choice to obtain the table: x 0.01 0.001 0.0001 0.0001 0.001 0.01 x +1 f ( x) = 2 0.9899 0.9990 0.9999 1.0001 1.001 1.0099 x +1

We infer that as x gets closer to 0, f gets closer to 1. That is, x +1 lim f ( x ) = lim 2 =1 x 0 x 0 x + 1 13.

Here f (x) =

x2 4 x , and c = 4. We choose values of x close to 4, starting at 3.99. Then x4 we select additional numbers that get closer to 4, but remain less than 4. Next we choose values of x greater than 4, starting with 4.01, that get closer to 4. Finally we evaluate the function f at each choice to obtain the table: x 3.99 3.999 3.9999 4.0001 4.001 4.01 f ( x) = x2 4 x x4 3.99 3.999 3.9999 4.0001 4.001 4.01

SECTION 3.1 173

We infer that as x gets closer to 4, f gets closer to 4. That is, x2 4 x lim f ( x ) = lim =4 x 4 x4 x4 15. Here f (x) = ex + 1, and c = 0. We choose values of x close to 0, starting at 0.01. Then we select additional numbers that get closer to 0, but remain less than 0. Next we choose values of x greater than 0, starting with 0.01, that get closer to 0. Finally we evaluate the function f at each choice to obtain the table: x 0.01 0.001 0.0001 0.0001 0.001 0.01 f ( x) = e x + 1 1.9900 1.9990 1.9999 2.0001 2.0010 2.0101

We infer that as x gets closer to 0, f gets closer to 2. That is, lim f ( x ) = lim ( e x + 1) = 2x 0 x 0

17.

To determine the lim f ( x ) we observe that as x gets closer to 2, f (x) gets closer to 3. Sox 2

We conclude that

lim f ( x ) = 3x 2

19.

To determine the lim f ( x ) we observe that as x gets closer to 2, f (x) gets closer to 4.x 2

We conclude that

lim f ( x ) = 4x 2

21.

To determine the lim f ( x ) we observe that as x gets closer to 3, but remains less than 3,x 3

the value of f gets closer to 3. However, we see that as x gets closer to 3, but remains greater than 3, the value of f gets closer to 6. Since there is no single number that the values of f are close to when x is close to 3, we conclude that lim f ( x ) does not exist.x 3

23.

We conclude from the graph that lim f ( x ) = lim ( 3 x + 1) = 13x 4 x 4

174 SECTION 3.1 25.

We conclude from the graph that lim f ( x ) = lim 1 x 2 = 3x 2 x 2

(

)

27. We conclude from the graph thatx3

lim f ( x ) = lim ( x 2 ) = 1x3

29. We conclude from the graph thatx0

lim f ( x ) = lim e x = 1x0

31. We conclude from the graph that 1 lim f ( x ) = lim = 1 x 1 x 1 x

33. We conclude from the graph thatx0

lim f ( x ) = 0

Notice that lim f ( x ) = lim x 2 = 0x0 x0

and

x0+

lim f ( x ) = lim+ ( 2 x ) = 0x0

35. We conclude from the graph that lim f ( x ) does not exist.x 1

Notice that lim f ( x ) = lim 3 x = 3 x 1 x 1

but lim f ( x ) = lim ( x + 1) = 2 + +x 1 x 1

SECTION 3.1 175 37. We conclude from the graph that lim f ( x ) = 0x0

Notice that lim f ( x ) = lim x = 0x0 x0

and lim+ f ( x ) = lim+ ( 3 x ) = 0x0 x0

39. We conclude from the graph that lim f ( x ) = 0x0

Notice that lim f ( x ) = lim ( e x 1) = 1 1 = 0x0 x0

and

x0

lim+ f ( x ) = lim+ x 2 = 0x0

41. To find the limit we create the tables shown below and conclude x 3 x 2 + x 1 2 lim 4 = 3 x 1 x x + 2x 2 3

43. To find the limit we create the tables shown below and conclude x3 2 x 2 + 4 x 8 lim = 1.6 x 2 x2 + x 6

45. To find the limit we create the tables shown below and conclude x3 + 2 x 2 + x =0 lim 4 x 1 x + x 3 + 2 x + 2

176 SECTION 3.2

3.2 Techniques for Finding Limits of Functions1. Using formula (1) (p. 243), we find lim 5 = 5x1

3. Using formula (2) (p. 243), we find lim x = 4x 4

5.

lim ( 3 x + 2 ) = lim ( 3x ) + lim 2 = lim 3 lim x + lim 2 x 2 x 2 x 2 x 2 x 2 x 2 = (3) (2) + 2 = 6 + 2 = 8 lim 3x 2 5 x = 3 ( 1) 5 ( 1) = 3 1 + 5 = 3 + 5 = 82

7.

x 1

(

)

9.

lim 5 x 4 3 x 2 + 6 x 9 = 5 14 3 12 + 6 1 9x 1

(

)

=5131+69 = 5 3 3 = 111.3 3 lim x 2 + 1 = lim x 2 + 1 = ( 2 ) = 8 x1 x 1

(

)

(

)

3

13.

lim 5 x + 4 = lim ( 5 x + 4 ) = 9 = 3x 1 x 1

15. The limit we seek is the limit of a rational function whose domain is {x | x 2, x 2}. Since 0 is in the domain, we use formula (12). x 2 4 02 4 4 lim 2 = 2 = = 1 x 0 x + 4 0 +4 4 17.

lim ( 3x 2 )x 2

5 2

5 = lim ( 3x 2 ) 2 = ( 4 ) 2 = ( 2 ) = 32 x 2

5

5

19. The domain of the rational function R(x) =

x2 4 is {x | x 0, x 2}. Since 2 is not x2 2x

in the domain, we cannot evaluate R(2), but we notice that the function can be factored asx 2 4 ( x + 2 )( x 2 ) = x ( x 2) x2 2x

SECTION 3.2 177

Since x is near 2, but x 2, we can cancel the (x 2)s. Formula (11), can then be used to find the limit of the function as x approaches 2. ( x + 2 ) ( x 2 ) lim ( x + 2 ) 4 x2 4 = x 2 = =2 lim 2 = lim x 2 x 2 x x 2 lim ( x ) 2 x ( x 2) x 221. The domain of the rational function R(x) =

x 2 x 12 is {x | x 3, x 3}. Since 3 is x2 9 not in the domain, we cannot evaluate R(3), but we notice that the function can be factored as x 2 x 12 ( x 4 )( x + 3) = x2 9 ( x 3)( x + 3) Since x is near 3, but x 3, we can cancel the (x + 3)s. Formula (11), can then be used to find the limit of the function as x approaches 3. ( x 4 ) ( x + 3) xlim3 ( x 4 ) ( 3) 4 7 7 x 2 x 12 lim = lim = = = = 2 x 3 x 3 x 3 x 9 ( ) ( x + 3) lim ( x 3) ( 3) 3 6 6x 3

23. The domain of the rational function R(x) =

x3 1 is {x | x 1}. Since 1 is not in the x 1 domain, we cannot evaluate R(1), but we notice that the function can be factored as 2 x 3 1 ( x 1) x + x + 1 = x 1 x 1 Since x is near 1, but x 1, we can cancel the (x 1)s. Formula (11), can then be used to find the limit of the function as x approaches 1. ( x 1) x 2 + x + 1 lim x 2 + x + 1 12 + 1 + 1 x3 1 = x1 = =3 lim = lim x 1 x 1 x 1 lim (1) 1 x 1

(

)

(

)

(

)

x 1

25. The domain of the rational function R(x) =

is {x | x 1, x 1}. Since 1 is not x2 1 in the domain, we cannot evaluate R(1), but we notice that the function can be factored as 2 2 ( x + 1) ( x + 1) = x 2 1 ( x 1)( x + 1) Since x is near 1, but x 1, we can cancel (x + 1)s. Formula (11), can then be used to find the limit of the function as x approaches 1.x 1

( x + 1)

2

lim

( x + 1)

2

x2 1

( x + 1) = lim x 1 x 1 ( ) ( x + 1)2

= lim

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

178 SECTION 3.2 27.

The limit of the denominator of this function as x approaches 1 is zero, so formula (11), cannot be used directly. We first factor the function by grouping. 2 x 2 ( x 1) + 1 ( x 1) ( x 1) x + 1 x3 x 2 + x 1 = = x 4 x3 + 2 x 2 x3 ( x 1) + 2 ( x 1) ( x 1) x3 + 2

( (

) )

Since x is near 1, but x 1, we can cancel (x 1)s. Then using formula (11), we get 2 ( x 1) x 2 + 1 lim x + 1 2 x3 x 2 + x 1 x 1 lim 4 = lim = = x 1 x x 3 + 2 x 2 x 1 x 1 ( ) x3 + 2 lim x3 + 2 3

( (

) )

x 1

( (

) )

29. The limit of the denominator of this function as x approaches 2 is zero, so formula (11), cannot be used directly. We first factor the function, using grouping to factor the numerator.2 x2 + 4 ( x 2) x3 2 x 2 + 4 x 8 x ( x 2 ) + 4 ( x 2 ) = = x2 + x 6 ( x + 3)( x 2 ) ( x + 3)( x 2 ) Since x is near 2, but x 2, we can cancel (x 2)s. Then using formula (11), we get lim 2 x2 + 4 ( x 2) x 2 x + 4 22 + 4 8 x3 2 x 2 + 4 x 8 = = = lim = lim x 2 x 2 ( x + 3) ( x 2 ) lim ( x + 3) 2+3 5 x2 + x 6

(

)

(

)

(

)

x 2

31. The limit of the denominator of this function as x approaches 1 is zero, so formula (11), cannot be used directly. We first factor the function, using grouping to factor the denominator. x x2 + 2 x + 1 x ( x + 1)( x + 1) x3 + 2 x 2 + x = 3 = 3 4 3 x + x + 2 x + 2 x ( x + 1) + 2 ( x + 1) x + 2 ( x + 1)

(

)

(

)

Since x is near 1, but x 1, we can cancel (x + 1)s. Then using formula (11), we get x ( x + 1) ( x + 1) xlim1 x ( x + 1) ( 1) ( 1) + 1 ( 1) ( 0 ) x3 + 2 x 2 + x = = = =0 lim 4 = lim 3 3 3 3 x 1 x + x + 2 x + 2 x 1 x + 2 1 ( x + 1) lim x + 1 ( 1) + 2

(

)

x 1

(

)

33. The average rate of change of f from 2 to x is y f ( x ) f ( 2 ) ( 5 x 3) ( 5 2 3) 5 x 3 7 5 x 10 5 ( x 2 ) = = = = = x x2 x2 x2 x2 x2 The limit of the average rate of change as x approaches 2 is 5 ( x 2) f ( x ) f ( 2) ( 5 x 3) 7 5 x 10 =5 lim = lim = lim = lim x 2 x 2 x 2 x 2 x 2 x2 x2 x2 35. The average rate of change of f from 3 to x is y f ( x ) f ( 3) x 2 32 ( x + 3)( x 3) = = = x x3 x3 x3

SECTION 3.2 179

The limit of the average rate of change as x approaches 3 is ( x + 3) ( x 3) f ( x ) f ( 3) x 2 32 =6 lim = lim = lim x 3 x 3 x 3 x 3 x 3 x337. The average rate of change of f from 1 to x is 2 2 2 2 x 2 + 2 x ( 1) + 2 ( 1) y f ( x ) f ( 1) = x + 2 x ( 1) = x + 2 x + 1 = ( x + 1) = = x x ( 1) x +1 x +1 x +1 x +1 The limit of the average rate of change as x approaches 1 is 2 x 2 + 2 x ( 1) f ( x ) f ( 1) ( x + 1) x2 + 2 x + 1 = ( 1) + 1 = 0 lim = lim = lim = lim x 1 x 1 x 1 x 1 x ( 1) x +1 x +1 x +1

(

)

(

)

(

)

39. The average rate of change of f from 0 to x is 3 2 2 y f ( x ) f ( 0 ) 3 x 2 x + 4 ( 4 ) 3 x 3 2 x 2 x ( 3 x 2 ) = = = = x x0 x x x The limit of the average rate of change as x approaches 0 is f ( x ) f ( 0) x 2 ( 3x 2 ) 3x3 2 x 2 lim = lim = lim =0 x 0 x 0 x 0 x0 x x

(

)

41. The average rate of change of f from 1 to x is 1 1 1 x 1 x ( 1) ( x 1) y f ( x ) f (1) x 1 x x 1 x = = = = x = = x 1 x 1 x 1 x 1 x ( x 1) x ( x 1) x Find a common denominator Simplify Factor out (1)

The limit of the average rate of change as x approaches 1 is ( 1) ( x 1) f ( x ) f (1) 1 x = 1 lim = lim = lim x 1 x 1 x ( x 1) x 1 x 1 x ( x 1)43. The average rate of change of f from 4 to x is y f ( x ) f ( 4 ) x 4 x 2 = = = x x4 x4 x +2 x 2

(

)(

)

Factor the denominator

The limit of the average rate of change as x approaches 4 is f ( x ) f ( 4) x 2 1 1 x 4 = = lim = lim = lim x 4 x 4 x 4 x4 x4 4 +2 4 x +2 x 2

(

)(

)

180 SECTION 3.3 45. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim 2 f ( x ) = lim 2 lim f ( x ) = 2 5 = 10 x c x c x c x c x c 3

3 47. Since lim f ( x ) = 5 and lim g ( x ) = 2 , lim g ( x ) = lim g ( x ) = 23 = 8 x c x c x c x c

49. Since lim f ( x ) = 5 and lim g ( x ) = 2 , limx c x c

x c

lim 4 4 4 x c = = f ( x ) lim f ( x ) 5x c

51. Since lim f ( x ) = 5 and lim g ( x ) = 2 ,x c x c

lim 4 f ( x ) 5 g ( x ) = lim 4 f ( x ) lim 5 g ( x ) x c x c x c

= lim 4 lim f ( x ) lim 5 lim g ( x ) x c x c x c x c = 4 5 5 2 = 20 10 = 10

3.3 One-Sided Limits; Continuous Functions1. The domain of f is {x | 8 x < 3 or 3 < x < 4 or 4 < x 6} or the intervals [8, 3) or ( 3 , 4) or (4, 6]. 3. The x-intercepts of the graph of f are ( 8, 0) and ( 5, 0). At these points the graph of f either crosses or touches the x-axis. 5. f ( 8) = 0 and f ( 4) = 2 7. To find lim f ( x ) , we look at the values of f when x is close to 6, but less than 6. x 6

Since the graph of f is approaching y = 3 for these values, we have lim f ( x ) = 3 .x 6

9. To find lim f ( x ) , we look at the values of f when x is close to 4, but less than 4. x 4

Since the graph of f is approaching y = 2 for these values, we have lim f ( x ) = 2 .x 4

11. To find lim f ( x ) , we look at the values of f when x is close to 2, but less than 2. Since x 2

the graph of f is approaching y = 1 for these values, we have lim f ( x ) = 1 .x 2

SECTION 3.3 181 13. The lim f ( x) exists because both lim f ( x) = 0 and lim+ f ( x) = 0. Since both onex 4 x 4 x 4

sided limits exist and are equal, the limit of f as x approaches 4 exists and is equal to the one-sided limits. That is, lim f ( x) = 0.x 4

15. The function f is not continuous at x = 6, because lim f ( x ) does not exist. (The onex 6

sided limits are not equal. See Problems 7 and 8.)17. The function f is continuous at x = 0. The function is defined at zero, f (0) = 3, and lim f ( x) = f (0) = 3. (See Problem 14.)x 0

19. The function f is not continuous at x = 4. The function is not defined at 4. That is, 4 is not part of the domain of f . 21. To find the one-sided limit we look at values of x close to 1, but greater than 1. Since f (x) = 2x + 3 for such numbers, we conclude that lim+ f ( x ) = lim+ ( 2 x + 3) = 5x 1 x 1

23. To find the one-sided limit we look at values of x close to 1, but less than 1. Since f (x) = 2x 3 + 5x for such numbers, we conclude that lim f ( x ) = lim 2 x3 + 5 x = 7x 1 x 1

(

)

25. To find the one-sided limit we look at values of x close to 0, but less than 0. Since f (x) = ex for such numbers, we conclude that lim f ( x ) = lim e x = 1x 0 x 0

( )

27. To find the one-sided limit we look at values of x close to 2, but greater than 2. Since x2 4 for such numbers, we first factor the function. f (x) = x2 ( x + 2 )( x 2 ) x2 4 = f (x) = x2 x2 We can then conclude that ( x + 2) ( x 2) x2 4 =4 lim+ f ( x ) = lim+ = lim+ x 2 x 2 x 2 x 2 x2 29. To find the one-sided limit we look at values of x close to 1, but less than 1. Since x2 1 for such numbers, we first factor the function. f (x) = 3 x +1 ( x + 1)( x 1) x2 1 f (x) = 3 = x + 1 ( x + 1) x 2 x + 1

(

)

182 SECTION 3.3

We can then conclude thatx 1

lim f ( x ) = lim x 1

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

(

)

31. To find the one-sided limit we look at values of x close to 2, but greater than 2. Since x2 + x 2 for such numbers, we first factor the function. f (x) = 2 x + 2x x 2 + x 2 ( x + 2 )( x 1) = f (x) = 2 x + 2x x ( x + 2) We can then conclude that ( x + 2 ) ( x 1) x2 + x 2 x 1 1 lim+ f ( x) = lim 2 = lim = lim+ = + + x2 x2 x2 x2 x + 2x 2 x x ( x + 2) 33. f (x) = x 3 3x 2 + 2x 6 is continuous at c = 2 because f is a polynomial, and polynomials are continuous at every number. 35. f (x) =

x2 + 5 is a rational function whose domain is {x | x 6}. f is continuous at c = 3 x6 since f is defined at 3.

37. f (x) =

x+3 is a rational function whose domain is {x | x 3}. f is not continuous at c = x3 3 since f is not defined at 3. x3 + 3x f (x) = 2 is a rational function whose domain is {x | x 0, x 3}. f is not x 3x continuous at c = 0 since f is not defined at 0. if x 0 if x = 0

39.

x3 + 3x 41. To determine whether f ( x ) = x 2 3x 1 f when x = 0. f (0) = 1

is continuous at c = 0, we investigate

x x2 + 3 3 x3 + 3x = = 1 lim f ( x ) = lim 2 = lim x 0 x 0 x 3 x x 0 x ( x 3) 3 x x2 + 3 3 x3 + 3x = = 1 lim+ f ( x ) = lim+ 2 = lim+ x 0 x 0 x 3 x x 0 x ( x 3) 3

(

)

(

)

Since lim f ( x ) = 1 f (0) = 1, the function f is not continuous at c = 0.x 0

SECTION 3.3 183

x3 + 3x 43. To determine whether f ( x ) = x 2 3x 1 f when x = 0. f (0) = 1

if x 0 if x = 0

is continuous at c = 0, we investigate

x x2 + 3 3 x3 + 3x = = 1 lim f ( x ) = lim 2 = lim x 0 x 0 x 3 x x 0 x ( x 3) 3 x x2 + 3 3 x3 + 3x = = 1 lim+ f ( x ) = lim+ 2 = lim+ x 0 x 0 x 3 x x 0 x ( x 3) 3

(

)

(

)

Since lim f ( x ) = 1 = f (0), the function f is continuous at c = 0.x 0

x3 1 2 x 1 45. To determine whether f ( x ) = 2 3 x +1 when x = 1. f (1) = 2

if x < 1 if x = 1 is continuous at c = 1, we investigate f if x > 1

( x 1) x 2 + x + 1 1 + 1 + 1 3 x3 1 = = lim f ( x ) = lim 2 = lim x 1 x 1 x 1 x 1 1+1 2 ( x 1) ( x + 1)x 1

(

)

lim+ f ( x ) = lim+x 1

3 3 = x +1 2

Since lim f ( x ) =x 1

3 f (1), the function f is not continuous at c = 1. 2

2e x 47. To determine whether f ( x ) = 2 3 2 x + 2x x2 investigate f when x =0. f (0) = 2 lim f ( x ) = lim 2e x = 2x 0 x 0

if x < 0 if x = 0 is continuous at c = 0, we if x > 0

x2 ( x + 2) x3 + 2 x 2 =2 lim f ( x ) = lim+ = lim+ x 0 + x 0 x 0 x2 x2 The lim f ( x ) exists, and lim f ( x ) = f (0) = 2. So we conclude that the function f isx 0 x 0

continuous at c = 0.

184 SECTION 3.3 49. 51.

f (x) = 2x + 3 is a first degree polynomial function. Polynomial functions are continuous at all real numbers. f (x) = 3x 2 + x is a second degree polynomial function. Polynomial functions are continuous at all real numbers.

53. f (x) = 4 ln x is the product of a constant function h(x) = 4, which is continuous at every number, and the logarithmic function g(x) = ln x, which is continuous for every number in the domain (0, ). So f (x) = 4 ln x is continuous for all values x > 0. 55.

f (x) = 3e x is the product of a constant function h(x) = 3, which is continuous at every number, and the exponential function g(x) = e x, which is continuous for every number in the domain ( , ). So f (x) = 3e x is continuous for all real numbers.2x + 5 is a rational function. Rational functions are continuous at every number in x2 4 the domain. The domain of f is {x | x 2, x 2}, and f is continuous at all those numbers. f is discontinuous at x = 2 and x = 2.

57.

f (x) =

59.

x3 is the quotient of a polynomial function, which is continuous at all real ln x numbers and the logarithmic function, which is continuous at all numbers in the domain (0, ). So f is continuous at all positive numbers or for x > 0. f (x) = The pieces of f , that is, y = 3x + 1, y = x 2, and y = 1 x 5 , are each continuous for 2 every number since they are polynomials. So we only need to investigate x = 0 and x = 2, the two points at which the pieces change. For x = 0: f (0) = 3(0) + 1 = 1 lim f ( x ) = lim ( 3x + 1) = 1x 0 x 0 x 0 + x 0

61.

lim f ( x ) = lim+ ( x 2 ) = 0x 0

Since lim+ f ( x ) f (0), we conclude that the function f is discontinuous at x = 0. For x = 2: f (2) = 2 2 = 4 lim f ( x ) = lim ( x 2 ) = 4x 2

x 2

1 lim+ f ( x ) = lim+ x 5 = 4 x 2 x 2 2 ( x ) = f ( 2 ) = 4 , we conclude that f is continuous at x = 2. Since lim fx 2

SECTION 3.3 185

39.99 63. The cost function is C(x) = 0.25 x 47.51 (a) lim C ( x ) = lim ( 39.99 ) = 39.99x 350 x 350

if 0 &