STANDARD FORM is a QUADRATIC FUNCTION and its graph is a PARABOLA. The domain of a quadratic function is the set of all real numbers.

EXERCISE 2-3 Things to remember: 1. QUADRATIC FUNCTION If a, b, and c are real numbers with a 0, then the function f() = a 2 + b + c STANDARD FORM is a QUADRATIC FUNCTION and its graph is a PARABOLA. The domain of a quadratic function is the set of all real numbers. 2. PROPERTIES OF A QUADRATIC FUNCTION AND ITS GRAPH Given a quadratic function f() = a 2 + b + c, a 0, and the VERTEX FORM obtained b completing the square f() = a( - h) 2 + k. The general properties of f are as follows: a. The graph of f is a parabola: f() f() a > 0 Opens upward k Ais = h } h Verte (h, k) Min f() a < 0 Opens downward Ais = h } h Verte (h, k) Ma f() b. Verte: (h, k) [parabola increases on one side of the verte and decreases on the other] c. Ais (of smmetr): = h (parallel to ais) d. f(h) = k is the minimum if a > 0 and the maimum if a < 0 e. Domain: All real numbers Range: (-, k] if a < 0 or [k, ) if a > 0 f. The graph of f is the graph of g() = a 2 translated horizontall h units and verticall k units. 3. POLYNOMIAL FUNCTION A POLYNOMIAL FUNCTION is a function that can be written in the form f() = a n n + a n-1 n-1 + + a 1 + a 0 for n a nonnegative integer, called the DEGREE of the polnomial. The coefficients a 0, a 1,, a n are real numbers with a n 0. The DOMAIN of a polnomial function is the set of all real numbers. 44 CHAPTER 2 FUNCTIONS AND GRAPHS

4. A RATIONAL FUNCTION is an function that can be written in the form f() = n() d() d() 0 where n() and d() are polnomials. The DOMAIN is the set of all real numbers such that d() 0. 1. f() = 2-4 + 3 = ( 2-4) + 3 = ( 2-4 + 4) + 3-4 (completing the square) = ( - 2) 2-1 (standard form) 3. f() = - 2 + 6-4 = -( 2-6) - 4 = -( 2-6 + 9) - 4 + 9 (completing the square) = -( - 3) 2 + 5 (standard form) 5. From Problem 1, the graph of f() is the graph of = 2 shifted right 2 units and down 1 unit. 7. From Problem 3, the graph of m() is the graph of = 2 reflected in the ais, then shifted right 3 units and up 5 units. 9. (A) m (B) g (C) f (D) n 11. (A) intercepts: 1, 3; intercept: -3 (B) Verte: (2, 1) (C) Maimum: 1 (D) Range: 1 or (-, 1] (E) Increasing interval: 2 or (-, 2] (F) Decreasing interval: 2 or [2, ) 13. (A) intercepts: -3, -1; intercept: 3 (B) Verte: (-2, -1) (C) Minimum: -1 (D) Range: -1 or [-1, ) (E) Increasing interval: -2 or [-2, ) (F) Decreasing interval: -2 or (-, -2] 15. f() = -( - 3) 2 + 2 (A) intercept: f(0) = -(0-3) 2 + 2 = -7 intercepts: f() = 0 -( - 3) 2 + 2 = 0 ( - 3) 2 = 2-3 = ± 2 = 3 ± 2 (B) Verte: (3, 2) (C) Maimum: 2 (D) Range: 2 or (-, 2] EXERCISE 2-3 45

17. m() = ( + 1) 2-2 (A) intercept: m(0) = (0 + 1) 2-2 = 1-2 = -1 intercepts: m() = 0 ( + 1) 2-2 = 0 ( + 1) 2 = 2 + 1 = ± 2 = -1 ± 2 (B) Verte: (-1, -2) (C) Minimum: -2 (D) Range: -2 or [-2, ) 19. = -[ - (-2)] 2 + 5 = -( + 2) 2 + 5 21. = ( - 1) 2-3 23. f() = 2-8 + 12 = ( 2-8) + 12 = ( 2-8 + 16) + 12-16 = ( - 4) 2-4 (standard form) (A) intercept: f(0) = 0 2-8(0) + 12 = 12 intercepts: f() = 0 ( - 4) 2-4 = 0 ( - 4) 2 = 4-4 = ±2 = 2, 6 (B) Verte: (4, -4) (C) Minimum: -4 (D) Range: -4 or [-4, ) 25. r() = -4 2 + 16-15 = -4( 2-4) - 15 = -4( 2-4 + 4) - 15 + 16 = -4( - 2) 2 + 1 (standard form) (A) intercept: r(0) = -4(0) 2 + 16(0) - 15 = -15 intercepts: r() = 0-4( - 2) 2 + 1 = 0 ( - 2) 2 = 1 4-2 = ± 1 2 = 3 2, 5 2 (B) Verte: (2, 1) (C) Maimum: 1 (D) Range: 1 or (-, 1] 46 CHAPTER 2 FUNCTIONS AND GRAPHS

27. u() = 0.5 2-2 + 5 = 0.5( 2-4) + 5 = 0.5( 2-4 + 4) + 3 = 0.5( - 2) 2 + 3 (standard form) (A) intercept: u(0) = 0.5(0) 2-2(0) + 5 = 5 intercepts: u() = 0 0.5( - 2) 2 + 3 = 0 ( - 2) 2 = -6; no solutions. There are no intercepts. (B) Verte: (2, 3) (C) Minimum: 3 (D) Range: 3 or [3, ) 29. f() = 0.3 2 - - 8 (A) f() = 4: 0.3 2 - - 8 = 4 (B) f() = -1: 0.3 2 - - 8 = -1 0.3 2 - - 12 = 0 0.3 2 - - 7 = 0 20 20-10 10-10 10-20 = -4.87, 8.21 = -3.44, 6.78 (C) f() = -9: 0.3 2 - - 8 = -9 0.3 2 - + 1 = 0 2-20 -4 4-2 No solutions. 1000 31. 0 22 0 maimum value f(10.41667) = 651.0417 33. The verte of the parabola is on the ais. EXERCISE 2-3 47

35. g() = 0.25 2-1.5-7 = 0.25( 2-6 + 9) - 2.25-7 = 0.25( - 3) 2-9.25 (A) intercepts:0.25( - 3) 2-9.25 = 0 ( - 3) 2 = 37-3 = ± 37 = 3 + 37 9.0828, 3-37 -3.0828 intercept: -7 (B) Verte: (3, -9.25) (C) Minimum: -9.25 (D) Range: -9.25 or [-9.25, ) 37. f() = -0.12 2 + 0.96 + 1.2 = -0.12( 2-8 + 16) + 1.92 + 1.2 = -0.12( - 4) 2 + 3.12 (A) intercepts: -0.12( - 4) 2 + 3.12 = 0 ( - 4) 2 = 26-4 = ± 26 = 4 + 26 9.0990, 4-26 -1.0990 intercept: 1.2 (B) Verte: (4, 3.12) (C) Maimum: 3.12 (D) Range: 3.12 or (-, 3.12] 10 10 10 39. -10 10 41. -10 10 43. -10 10-10 = -5.37, 0.37-10 -1.37 < < 2.16 45. f is a quadratic function and min f() = f(2) = 4 Ais: = 2 Verte: (2, 4) Range: 4 or [4, ) intercepts: None -10-0.74 or 4.19 47. (A) g f 48 CHAPTER 2 FUNCTIONS AND GRAPHS

(B) f() = g() -0.4( - 10) = 0.3 + 5-0.4 2 + 4 = 0.3 + 5-0.4 2 + 3.7 = 5-0.4 2 + 3.7-5 = 0 = 3.7 ± 3.72 4(0.4)(5) 2(0.4) 3.7 ± 5.69 = 1.64, 7.61 0.8 (C) f() > g() for 1.64 < < 7.61 (D) f() < g() for 0 < 1.64 or 7.61 < 10 49. (A) g f (B) f() = g() -0.9 2 + 7.2 = 1.2 + 5.5-0.9 2 + 6 = 5.5-0.9 2 + 6-5.5 = 0 6 ± 36 4(0.9)(5.5) = 2(0.9) 6 ± 16.2 = 1.1, 5.57 1.8 (C) f() > g() for 1.10 < < 5.57 (D) f() < g() for 0 < 1.10 or 5.57 < 8 51. f() = 2 + 1 and g() = -( - 4) 2-1 are two eamples. Their graphs are: Their graphs do not intersect the ais. EXERCISE 2-3 49

53. Mathematical model: f() = -0.518 2 + 33.3-481 (A) 28 30 32 34 36 50 Mileage 45 52 55 51 47 (B) 40 f() 45.3 51.8 54.2 52.4 46.5 30 60 20 10 28 32 36 (C) = 31: f(31) = -0.518(31) 2 + 33.3(31) - 481 = 53.502 f(31) 53.50 thousand miles = 35: f(35) = -0.518(35) 2 + 33(35) - 481 49.95 thousand miles (D) The maimum mileage is achieved at 32 lb/in 2 pressure. Increasing the pressure or decreasing the pressure reduces the mileage. 55. Quadratic regression using the data in Problem 53: f() = -0.518 2 + 33.3-481 57. p() = 75-3; R() = p(); 1 20 R() = (75-3) = 75-3 2 (A) 500 400 300 200 100 R 10 20 = -3( 2-25) # = -3 2 25 + 625 & % ( + 1875 $ 4 ' 4 # = -3 25 & 2 % ( + 1875 $ 2 ' 4 = -3( - 12.5) 2 + 468.75 (B) Output for maimum revenue: = 12,500,000 chips; Maimum revenue: $468,750,000 (C) Wholesale price per chip at maimum revenue: p(12.5) = 75-3(12.5) = 37.5 or $37.50 59. Revenue function: R() = (75-3) = 75-3 2 Cost function: C() = 125 + 16 (A) (B) Break-even points: R() = C() 500 75-3 2 = 125 + 16 C 400-3 2 + 59-125 = 0 300 R or 3 2-59 + 125 = 0 200 100 10 20 = 59 ± (59)2 4(3)(125) 2(3) 59 ± 1981 59 ± 44.508 = = 6 6 = 2.415 or 17.251 The compan breaks even at = 2,451,000 chips and 17,251,000 chips. 50 CHAPTER 2 FUNCTIONS AND GRAPHS

(C) Using the results from (A) and (B): Loss: 1 < 2.415 or 17.251 < 20 Profit: 2.415 < < 17.251 61. Revenue function: R() = (75-3) Cost function: C() = 125 + 16 (A) Profit function: P() = (75-3) - (125 + 16) = 75-3 2-125 - 16; P() = 59-3 2-125 400 200 200 C R 10 20 P (B) The coordinates of the intersection points of R and C are the same as the intercepts of P. (C) intercepts of P: -3 2 + 59-125 = 59 ± (59)2 4(3)(125) 6 = 2.415 or 17.251 (See Problem 57) The break-even points are: 2,415,000 chips and 17,251,000 chips. (D) The maimum profit and the maimum revenue do not occur at the same output level and the are not equal. The profit function involves both revenue and cost; the revenue function does not involve the production costs. (E) P() = -3 2 + 59-125 # = -3% 2 59 $ 3 & ( 125 ' = -3 59 2 ) # &, # + % (. 125 + 3 59 & 2 % ( * $ 6 '- $ 6 ' = -3( - 9.833) 2 + 165.083 The maimum profit is $165,083,000; it occurs at an output level of 9,833,000 chips. From Problem 55(B), the maimum revenue is $468,750,000. The maimum profit is much smaller than the maimum revenue. 63. Solve: f() = 1,000(0.04-2 ) = 20 40-1000 2 = 20 1000 2 = 20 2 = 0.02 = 0.14 or -0.14 Since we are measuring distance, we take the positive solution: = 0.14 cm EXERCISE 2-3 51

65. Quadratic regression model 0.0000014 2 0.00266 + 5.4 10.6 mph EXERCISE 2-4 Things to remember: 1. EXPONENTIAL FUNCTION The equation f() = b, b > 0, b 1 defines an EXPONENTIAL FUNCTION for each different constant b, called the BASE. The DOMAIN of f is all real numbers, and the RANGE of f is the set of positive real numbers. 2. BASIC PROPERTIES OF THE GRAPH OF f() = b, b > 0, b 1 a. All graphs pass through (0,1); b 0 = 1 for an base b. b. All graphs are continuous curves; there are no holes or jumps. c. The -ais is a horizontal asmptote. d. If b > 1, then b increases as increases. Graph of f() = b, b > 1 e. If 0 < b < 1, then b decreases as increases. Graph of f() = b, 0 < b < 1 52 CHAPTER 2 FUNCTIONS AND GRAPHS

3. PROPERTIES OF EXPONENTIAL FUNCTIONS For a, b > 0, a 1, b 1, and, real numbers: a. EXPONENT LAWS (i) a a = a + (ii) a a = a - (iii) (a ) = a (iv) (ab) = a b! (v) a # a b $ = b b. a = a if and onl if =. c. For 0, a = b if and onl if a = b. 4. EXPONENTIAL FUNCTION WITH BASE e = 2.71828 Eponential functions with base e and base 1/e are respectivel defined b = e and = e -. Domain: (-, ) Range: (0, ) = e - = e 5. Functions of the form = ce kt, where c and k are constants and the independent variable t represents time, are often used to model population growth and radioactive deca. Since (0) = c, c represents the initial population or initial amount. The constant k represents the growth or deca rate; k > 0 in the case of population growth, k < 0 in the case of radioactive deca. 6. COMPOUND INTEREST If a principal P (present value) is invested at an annual rate r (epressed as a decimal) compounded m times per ear, then the amount A (future value) in the account at the end of t ears is given b: A = P 1 + r % $ ' mt # m & 1. (A) k (B) g (C) h (D) f EXERCISE 2-4 53

3. = 5, -2 2 5. = $ 1% ' = 5 -, -2 2 # 5& 2 1 2 25 25 1 5 1 1 5 0 1 0 1 1 1 5 1 5 2 1 2 25 25 7. f() = -5, -2 2 9. = -e -, -3 3 2 25 1 3 # 20 1 1 2 # 7.4 5 1 # 2.7 0 1 0 1 1 5 1 # 0.4 2 25 2 # 0.1 3 # 0.05 11. = 100e 0.1, -5 5 13. g(t) = 10e -0.2t, -5 t 5 g g(t) 5 # 60 5 # 27.2 3 # 74 3 # 18.2 1 # 90 1 # 12.2 0 100 0 10 1 3 # 111 # 135 1 3 # 8.2 # 5.5 5 # 165 5 # 3.7 15. (4 3 ) 2 = 4 6 17. e! 3 19. (2e 1.2t ) 3 = 2 3 e 3(1.2t) = 8e 3.6t e!4 = e(-3)-(-4) = e -3-+4 = e 21. g() = -f(); the graph of g is the graph of f reflected in the ais. 23.g() = f( + 1); the graph of g is the graph of f shifted one unit to the left. f() = 2 g() = -2 f() = 3 g() = 3 +1 54 CHAPTER 2 FUNCTIONS AND GRAPHS

25. g() = f() + 1; the graph of g is the graph of f shifted one unit up. 27. g() = 2f( + 2); the graph of g is the graph of f verticall epanded b a factor of 2 and shifted to the left 2 units. f() = e g() = e + 1 f() = e - g() = 2e -(+2) 29. (A) = f() - 1 (B) = f( + 2) (C) = 3f() - 2 (D) = 2 - f( - 3) 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 31. f(t) = 2 t/10, -30 t 30 33. = -3 + e 1+, -4 2 30 1 4 # 3 8 2 # 2.6 20 1 4 1 2 10 1 0 # 0.3 2 1 # 4.4 0 1 2 # 17.1 10 2 20 4 30 8 35. = e, -3 3 37. C() = e + e!, -5 5 2-3 20.1-1 2.7 0 1 1 2.7 3 20.1 C() -5 74-3 10 0 1 3 10 5 74 EXERCISE 2-4 55

39. = e -2, -3 3-3 0.0001-2 0.0183-1 0.3679 0 1 1 0.3679 2 0.0183 3 0.0001 41. Solve a 2 = a -2 a 2 = 1 a 2 a 4 = 1 a 4-1 = 0 (a 2-1)(a 2 + 1) = 0 a 2-1 = 0 implies a = 1, -1 a 2 + 1 = 0 has no real solutions The eponential function propert: a = a if and onl if = assumes a > 0 and a 1. Our solutions are a = 1, -1; 1 = 1 for all real numbers,, (-1) = (-1) for all even integers. 43. 10 2-3 = 10 5-6 implies (see 3b) 2-3 = 5-6 -8 = -8 = 1 45. 4 5-2 = 4-6 implies 5-2 = -6 or - 2 + 5 + 6 = 0 2-5 - 6 = 0 ( - 6)( + 1) = 0 = 6, -1 47. 5 3 = ( + 2) 3 implies (b propert 3c) 5 = + 2 Thus, = 3. 51. 3e - + 2 e - = 0 e - (3 + 2 ) = 0 3 + 2 = 0 (since e - 0) (3 + ) = 0 = 0, -3 h() 53. h() = 2, -5 0 5 32 5 4 1 4 49. ( - 3)e = 0-3 = 0 (since e 0) = 3 3 3 8 2 1 2 1 1 2 0 0 56 CHAPTER 2 FUNCTIONS AND GRAPHS

55. N = 100 1 + e!t, 0 t 5 t N 0 50 1 73.1 2 88.1 3 95.3 5 99.3 57. f() = 4-7 59. f() = 2 + 3 + 10 Solve 4-7 = 0 Solve 2 + 3 + 10 = 0 1.40-0.73 61. A = P 1 + r % $ ' # m & mt, we have: (A) P = 2,500, r = 0.07, m = 4, t = 3 4 A = 2,500 1 + 0.07 % $ ' 4 3/4 = 2,500(1 + 0.0175) 3 = 2,633.56 # 4 & Thus, A = $2,633.56. (B) A = 2,500 1 + 0.07 % $ ' 4 15 = 2,500(1 + 0.0175) 60 = 7079.54 # 4 & Thus, A = $7,079.54. 63. With P = 7,500 and r = 0.0835, we have: A = 7,500e 0.0835t (A) A = 7,500e (0.0835)5.5 = 7,500e 0.45925 11,871.65 Thus, there will be $11,871.65 in the account after 5.5 ears. (B) A = 7,500e (0.0835)12 = 7,500e 1.002 20,427.93 Thus, there will be $20,427.93 in the account after 12 ears. mt 65. Use A = P 1 + r % $ ' ; P = 10,000, t = 1. # m & (A) Stonebridge Bank: r = 0.0540, m = 12 A = 10,000 1 + 0.0540 % $ ' 12(1) = $10,553.57 # 12 & (B) Deep Green Bank: r = 0.0495, m = 365 A = 10,000 1 + 0.0495 % $ ' 365(1) = $10,507.42 # 365 & (C) Provident Bank: r = 0.0515, m = 4 A = 10,000 1 + 0.0515 % 4(1) $ ' = $10,525.03 # 4 & EXERCISE 2-4 57

67. Given N = 2(1 - e -0.037t ), 0 t 50 t N 0 10 0 0.62 30 1.34 N approaches 2 as t increases without bound. 69.(A) Eponential regression model = ae b ; 2115 is ear 25; (25) $7,674,000 (B) According to the model, (10) $1,773,000. Inclusion of the ear 2000 data would increase the estimated average salar in 2010 since the actual ear 2000 average is higher than that estimated b the model. Inclusion of the ear 2000 data gives: 71. Given I = I 0 e -0.23d (A) I = I 0 e -0.23(10) = I 0 e -2.3 I 0 (0.10) (25) $7,997,000 Thus, about 10% of the surface light will reach a depth of 10 feet. (B) I = I 0 e -0.23(20) = I 0 e -4.6 I 0 (0.010) Thus, about 1% of the surface light will reach a depth of 20 feet. 73. (A) Model: N(t) = 65e 0.02t (B) Infected prior to 2002: t = -4 N(-4) = 65e 0.02(-4) 65e -0.08 60,000,000 Infected prior to 2014: t = 8 N(8) = 65e 0.02(8) = 65e 0.16 76,000,000 (C) 90 60 30 0 N 2 4 6 8 Years since 2006 t 58 CHAPTER 2 FUNCTIONS AND GRAPHS

75. (A) Model: P(t) = 6.5e 0.0114t (B) In the ear 2015, t = 9; P(9) = 6.5e 0.0114(9) = 6.5e 0.1026 7.2 billion In the ear 2030, t = 24; P(24) = 6.5e 0.0114(24) = 6.5e 0.2736 8.5 billion (C) P 10 8 6 4 2 77.(A) Eponential regression model (21) 24,238,000,000 24.238 billion 0 6 12 18 24 Years since 2006 (B) The model implies that the number of internet hosts will be more than 3 times larger than the estimated world population. t EXERCISE 2-5 Things to remember: 1. ONE-TO-ONE FUNCTIONS A function f is said to be ONE-TO-ONE if each range value corresponds to eactl one domain value. 2. INVERSE OF A FUNCTION If f is a one-to-one function, then the INVERSE of f is the function formed b interchanging the independent and dependent variables for f. Thus, if (a, b) is a point on the graph of f, then (b, a) is a point on the graph of the inverse of f. Note: If f is not one-to-one, then f DOES NOT HAVE AN INVERSE. 3. LOGARITHMIC FUNCTIONS The inverse of an eponential function is called a LOGARITHMIC FUNCTION. For b > 0 and b 1, Logarithmic form Eponential form = log b is equivalent to = b The LOG TO THE BASE b OF is the eponent to which b must be raised to obtain. [Remember: A logarithm is an eponent.] The DOMAIN of the logarithmic function is the range of the corresponding eponential function, and the RANGE of the logarithmic function is the domain of the corresponding eponential function. Tpical graphs of an eponential function and its inverse, a logarithmic function, for b > 1, are shown in EXERCISE 2-5 59

the figure below: = b = = log b Base b > 1 4. PROPERTIES OF LOGARITHMIC FUNCTIONS If b, M, and N are positive real numbers, b 1, and p and are real numbers, then: a. log b 1 = 0 e. log b MN = log b M + log b N M b. log b b = 1 f. log b N = log b M - log b N c. log b b = g. log b M p = p log b M d. b log b =, > 0 h. log b M = log b N if and onl if M = N 5. LOGARITHMIC NOTATION; LOGARITHMIC-EXPONENTIAL RELATIONSHIPS Common logarithm log means log 10 Natural logarithm ln means log e log = is equivalent to = 10 ln = is equivalent to = e 1. 27 = 3 3 (using 3) 3. 1 = 10 0 5. 8 = 4 3/2 7. log 7 49 = 2 9. log 4 8 = 3 2 11. log b A = u 13. log 10 1 = is equivalent to 10 = 1; = 0. 15. log e e = is equivalent to e = e; = 1. 17. log 0.2 0.2 = is equivalent to (0.2) = 0.2; = 1. 19. log 10 10 3 = 3 21. log 2 2-3 = -3 23. log 10 1,000 = log 10 10 3 = 3 (using 2a) 25. log b P Q = log b P - log b Q 27. log b L5 = 5 log b L 29. 3 p log 3 q = 3 log 3 qp = q p. (using 4g and 4d) 60 CHAPTER 2 FUNCTIONS AND GRAPHS

31. log 3 = 2 33. log 7 49 = 35. log b 10-4 = -4 = 3 2 log 7 7 2 = 10-4 = b -4 = 9 2 = This equalit implies Thus, = 2. b = 10 (since the eponents are the same). 37. log 4 = 1 39. log 2 1/3 9 = 41. log b 1,000 = 3 2 = 4 1/2 9 = 1! # 3$ log b 10 3 = 3 2 = 2 3 2 = (3-1 ) 3 log b 10 = 3 2 43. False; countereample: = 2. 3 2 = 3 - log b 10 = 1 2 This inequalit 10 = b 1/2 implies that Square both sides: 2 = - or = -2. 100 = b, i.e., b = 100. 45. True; if g is the inverse of f, then f is the inverse of g so g must be one-to-one. 47. True; if = 2, then = 2 implies = 2. 49. False; if 0 < b < 1, then log b is decreasing. 51. False; f() = ln is one-to-one; domain of f = (0, ) range of f = (-, ). 53. log b = 2 3 log b 8 + 1 2 log b 9 - log b 6 = log b 82/3 + log b 9 1/2 - log b 6 = log b 4 + log b 3 - log b 6 = log b 4! 3 6 log b = log b 2 = 2 55. log b = 3 2 log b 4-2 3 log b 8 + 2 log b 2 = log b 43/2 - log b 8 2/3 + log b 2 2 = log b 8 - log b 4 + log b 4 = log b 8 log b = log b 8 = 8 EXERCISE 2-5 61

57. log b + log b ( - 4) = log b 21 59. log 10 ( - 1) - log 10 ( + 1) = 1 # 1& log b ( - 4) = log b 21 log 10 % ( = 1 $ + 1' Therefore, ( - 4) = 21 Therefore,! 1 + 1 = 101 = 10 2-4 - 21 = 0-1 = 10( + 1) ( - 7)( + 3) = 0-1 = 10 + 10 Thus, = 7. -9 = 11 [Note: = -3 is not a solution = - 11 9 since log b (-3) is not defined.] There is no solution, since # log 10 % 11 $ 9 1 & ' ( = log # 20 & % 10 $ 9 ( ' is not defined. Similarl, # log 10 % 11 $ 9 + 1 & ' ( = log # 2 & % 10 $ 9 ( ' is not defined. 61. = log 2 ( - 2) = 2-2 = 2 9-2 4 = 2 5 + 2-1 2 3 0 4 1 6 2 18 4 63. The graph of = log 2 ( - 2) is the graph of = log 2 shifted to the right 2 units. 65. Since logarithmic functions are defined onl for positive inputs, we must have + 1 > 0 or > -1; domain: (-1, ). The range of = 1 + ln( + 1) is the set of all real numbers. 67. (A) 3.54743 69. (A) log = 1.1285 (B) -2.16032 = 13.4431 (C) 5.62629 (B) log = -2.0497 (D) -3.19704 = 0.0089 (C) ln = 2.7763 = 16.0595 (D) ln = -1.8879 = 0.1514 71. 10 = 12 (Take common logarithms of both sides) log 10 = log 12 1.0792 1.0792 (log 10 = log 10 = ; log 10 = 1) 62 CHAPTER 2 FUNCTIONS AND GRAPHS

73. e = 4.304 (Take natural logarithms of both sides) ln e = ln 4.304 1.4595 1.4595 (ln e = ln e = ; ln e = 1) 75. 1.03 = 2.475 (Take either common or natural logarithms of both sides; we use common logarithms) log(1.03) = log 2.475 log 2.475 = log 1.03 30.6589 77. 1.005 12t = 3 (Take either common or natural logarithms of both sides; here we'll use natural logarithms.) ln 1.005 12t = ln 3 ln 3 12t = ln 1.005 220.2713 t = 18.3559 79. = ln, > 0 81. = ln, > 0 0.5-0.69 1 0 2 0.69 4 1.39 5 1.61 increasing (0, ) 0.5 0.69 1 0 2 0.69 4 1.39 5 1.6 decreasing (0, 1] increasing [1, ) 83. = 2 ln( + 2), > -2 85. = 4 ln - 3, > 0-1.5-1.39-1 0 0 1.39 1 2.2 5 3.89 10 4.61-3 5 3 0.5-5.77 1-3 5 3.44 10 6.21 10-5 5 10-5 -10 increasing (-2, ) increasing (0, ) 87. For an number b, b > 0, b 1, log b 1 = is equivalent to b = 1 which implies = 0. Thus, log b 1 = 0 for an permissible base b. EXERCISE 2-5 63

89. log 10 - log 10 c = 0.8 log 10 c = 0.8 Therefore, c = 100.8 (using 1) and = c 10 0.8. 91. 1 16 A function f is larger than a function g on an interval [a, b] if f() > g() for a b. r() > q() > p() for 1 16, that is > > ln for 1 < 16 93. From the compound interest formula A = P(1 + r) t, we have: 2P = P(1 + 0.2136) t or (1.2136) t = 2 Take the natural log of both sides of this equation: ln(1.2136) t = ln 2 [Note: the common log could have been used instead of the natural log.] t ln(1.2136) = ln 2 ln 2 t = ln 1.2136 0.69135 = 3.58 4 ears 0.19359 95. A = A = P 1 + r % mt $ ' # m &, r = 0.06, P = 1000, A = 1800. Quarterl compounding: m = 4 1800 = 1000 1 + 0.06 % 4t $ ' # 4 & = 1000(1.015) 4t (1.015) 4t 1800 = 1000 = 1.8 4t ln(1.015) = ln(1.8) t = ln(1.8) 4 ln(1.015) 9.87 $1000 at 6% compounded quarterl will grow to $1800 in 9.87 ears. Dail compounding: m = 365 1800 = 1000 1 + 0.06 % 365t $ ' # 365 & = 1000(1.0001644)365t (1.0001644) 365t 1800 = 1000 = 1.8 365t ln(1.0001644) = ln 1.8 ln 1.8 t = 365 ln(1.0001644) 9.80 $1000 at 6% compounded dail will grow to $1800 in 9.80 ears. 64 CHAPTER 2 FUNCTIONS AND GRAPHS

97. (A) Logarithmic regression model, Table 1: (B) Logarithmic regression model, Table 2: To estimate the demand at a price level of $50, we solve the equation a + b ln = 50 for. The result is 5,373 screwdrivers per month. To estimate the suppl at a price level of $50, we solve a + b ln = 50 for. The result is 7,220 screwdrivers per month. (C) The condition is not stable, the price is likel to decrease since the demand at a price level of $50 is much lower than the suppl at this level. 99. I = I 0 10 N/10 Take the common log of both sides of this equation. Then: log I = log(i 0 10 N/10 ) = log I 0 + log 10 N/10 = log I 0 + N 10 log 10 = log I 0 + N 10 (since log 10 = 1) So, N 10 = log I - log I 0 = log I # I 0 % & and N = 10 log I # I 0 101. Logarithmic regression model! $! $ &. % The ield in 2015: (115) 153.9 bushels/acre. 103. Assuming that the current population is 6.5 billion and that the growth rate is 1.14% compounded continuousl, the population after t ears will be P(t) = 6.5e 0.0144t Given that there are 1.68 10 14 = 168,000 billion square ards of land, solve 6.5e 0.0114t = 168,000 for t: e 0.0114t = 168,000 25,846 6.5 0.114t = ln(25,846) t = ln(25,846) 0.0114 891 It will take approimatel 891 ears. EXERCISE 2-5 65

CHAPTER 2 REVIEW 1. 2. 2 = 2 : 3 2 1 0 1 2 3 ±3 ±2 ±1 0 ±1 ±2 ±3 5 (2-1) 5 5 5 (2-1) 3. 2 = 4 2 : 3 2 1 0 1 2 3 ±6 ±4 ±2 0 ±2 ±4 ±6 6 6 6 4. (A) Not a function; fails vertical line test (B) A function (C) A function (D) Not a function; fails vertical line test (2-1) 5. f() = 2-1, g() = 2-2 (A) f(-2) + g(-1) = 2(-2) - 1 + (-1) 2-2(-1) = -2 (B) f(0) g(4) = (2 0-1)(4 2-2 4) = -8 (C) g(2) f(3) = 22! 2 2 2 3! 1 = 0 (D) f(3) g(2) 6 (2-1) not defined because g(2) = 0 (2-1) 6. u = e v 7. = 10 8. ln M = N v = ln u (2-5) = log (2-5) M = e N (2-5) 9. log u = v u = 10 v (2-5) 10. 5 +4 5 4 = 5+4-(4-) = 5 2 (2-4) 66 CHAPTER 2 FUNCTIONS AND GRAPHS

u # 11. eu & % $ e u ( ' = (e u+u ) u = (e 2u ) u = e 2u2 (2-4) 12. log 3 = 2 13. log 36 = 2 14. log 2 16 = 15. 10 = 143.7 = 3 2 = 9 (2-5) 2 = 36 2 = 16 = log 143.7 = 6 (2-5) = 4 (2-5) 2.157 (2-5) 16. e = 503,000 17. log = 3.105 = ln 503,000 13.128 (2-5) = 10 3.105 1273.503 (2-5) 18. ln = -1.147 = e -1.147 0.318 (2-5) 19. (A) = 4 (B) = 0 (C) = 1 (D) = -1 or 1 (E) = -2 (F) = -5 or 5 (2-1) 20. (A) (B) (C) (D) (2-2) 21. f() = - 2 + 4 = -( 2-4) = -( 2-4 + 4) + 4 = -( - 2) 2 + 4 (standard form) The graph of f() is the graph of = 2 reflected in the ais, then shifted right 2 units and up 4 units. (2-3) 22. (A) g (B) m (C) n (D) f (2-2, 2-3) CHAPTER 2 REVIEW 67

23. = f() = ( + 2) 2-4 (A) intercepts: ( + 2) 2-4 = 0 ( + 2) 2 = 4 + 2 = -2 or 2 = -4, 0 intercept: 0 (B) Verte: (-2, -4) (C) Minimum: -4 (D) Range: -4 or [-4, ) (E) Increasing interval [-2, ) (F) Decreasing interval (-, -2] (2-3) 24. (A) This correspondence specifies a function; each person has eactl one mother. 25. (B) This correspondence does not specif a function; a person ma have no children, or two or more children. (2-1) 10 26. = 4 + 3 2 = 3 2 + 4; quadratic function (2-3) 5 10 5 (2-1) 27. = 1 + 5 = 5 6 6 + 1 6 ; linear function (2-3) 28. = 7 4 2 = 7 2-2; none of these (2-1), (2-3) 29. = 8 + 2(10 4) = 8 + 20 8 = 20; constant function (2-1) 30. log( + 5) = log(2-3) 31. 2 ln( - 1) = ln( 2-5) + 5 = 2-3 ln( - 1) 2 = ln( 2-5) - = -8 ( - 1) 2 = 2-5 = 8 (2-5) 2-2 + 1 = 2-5 -2 = -6 = 3 (2-5) 32. 9-1 = 3 1+ 33. e 2 = e 2-3 (3 2 ) -1 = 3 1+ 2 = 2-3 3 2-2 = 3 1+ 2-2 - 3 = 0 2-2 = 1 + ( - 3)( + 1) = 0 = 3 (2-4) = 3, -1 (2-4) 68 CHAPTER 2 FUNCTIONS AND GRAPHS

34. 2 2 e = 3e 2 2 = 3 (divide both sides b e ) 2 2-3 = 0 (2-3) = 0 = 0, 3 2 (2-4) 37. log 9 = 3 2 9 3/2 = = 27 (2-5) 35. log 1/3 9 = ' 1 $ % # 3& = 9 1 3 = 9 3 = = -2 (2-5) 36. log 8 = -3-3 = 8 1 3 = 8 3 = 1 8 = 1 2 (2-5) 38. = 3(e 1.49 ) 13.3113 (2-5) 39. = 230(10-0.161 ) 158.7552 (2-5) 40. log = -2.0144 41. ln = 0.3618 0.0097 (2-5) 1.4359 (2-5) 42. 35 = 7(3 ) 43. 0.01 = e -0.05 3 = 5 ln(0.01) = ln(e -0.05 ) = -0.05 ln 3 = ln 5 ln 3 = ln 5 = ln 5 ln 3 1.4650 (2-5) Thus, = ln(0.01) -0.05 44. 8,000 = 4,000(1.08) 45. 5 2-3 = 7.08 (1.08) = 2 ln(5 2-3 ) = ln 7.08 ln(1.08) = ln 2 (2-3)ln 5 = ln 7.08 ln 1.08 = ln 2 2 ln 5-3 ln 5 = ln 7.08 = ln 2 ln 1.08 9.0065 = ln 7.08 + 3 ln 5 2 ln 5 92.1034 (2-5) ln 7.08 + ln 53 (2-5) = 2 ln 5 = ln[7.08(125)] 2 ln 5 2.1081 (2-5) 46. = log 2 7 = log 7 log 2 2.8074 or = log 2 7 = ln 7 ln 2 2.8074 (2-5) log 5.321 47. = log 0.2 5.321 = log 0.2-1.0387 ln 5.321 or = log 0.2 5.321 = ln 0.2-1.0387 (2-5) CHAPTER 2 REVIEW 69

48. (A) 2 - - 6 = 0 at = -2, 3 Domain: all real numbers ecept = -2, 3 (B) 5 - > 0 for < 5 Domain: < 5 or (-, 5) (2-1) 49. Function g multiplies a domain element b 2 and then subtracts three times the square root of the domain element from the result. (2-1) 50. f() = 4 2 + 4-3 = 4( 2 + ) - 3 = 4 2 + + 1 % $ '- 3-1 # 4& = 4 + 1 % 2 $ ' - 4 (standard form) # 2& (A) Intercepts: intercept: f(0) = 4(0) 2 + 4(0) - 3 = -3 intercepts: f() = 0 4 + 1 % 2 $ ' - 4 = 0 # 2& + 1 % 2 $ ' = 1 # 2& + 1 2 = ±1 = - 1 2 ± 1 = - 3 2, 1 2 # (B) Verte: % 1 $ 2, 4 & ( (C) Minimum: -4 (D) Range: -4 or [-4, ) (2-3) ' 51. f() = e - 1, g() = ln( + 2) 2-3 3-2 52. f() = 50 2 + 1 : f 3 2 1 0 1 2 3 f() 5 10 25 50 25 10 5 g Points of intersection: (-1.54, -0.79), (0.69, 0.99) (2-4, 2-5) 50 6 6 (2-1) 70 CHAPTER 2 FUNCTIONS AND GRAPHS

53. f() = 66 2 + 2 : 3 2 1 0 1 2 3 6 6 f() 6 8.25 22 66 22 8.25 6 10 40 (2-1) For Problems 54 57, f() = 5 + 1. 54. f(f(0)) = f(5(0) + 1) = f(1) = 5(1) + 1 = 6 (2-1) 55. f(f(-1)) = f(5(-1) + 1) + f(-4) = 5(-4) + 1 = -19 (2-1) 56. f(2 1) = 5(2 1) + 1) = 10 4 (2-1) 57. f(4 ) = 5(4 ) + 1 = 20 5 + 1 = 21 5 (2-1) 58. f() = 3-2 (A) f(2) = 3-2(2) = 3-4 = -1 (B) f(2 + h) = 3-2(2 + h) = 3-4 - 2h = -1-2h (C) f(2 + h) - f(2) = -1-2h - (-1) = -2h f(2 + h)! f(2) (D) = - 2h = -2 (2-1) h h 59. f() = 2-3 + 1 (A) f(a) = a 2-3a + 1 (B) f(a + h) = (a + h) 2-3(a + h) + 1 = a 2 + 2ah + h 2-3a - 3h + 1 (C) f(a + h) - f(a) = a 2 + 2ah + h 2-3a - 3h + 1 - (a 2-3a + 1) = 2ah + h 2-3h (D) f(a + h)! f(a) h = 2ah + h2! 3h h = h(2a + h! 3) h = 2a + h - 3 (2-1) 60. The graph of m is the graph of = reflected in the ais and shifted 4 units to the right. (2-2) 61. The graph of g is the graph of = 3 verticall contracted b a factor of 0.3 and shifted up 3 units. (2-2) 62. The graph of = 2 is verticall epanded b a factor of 2, reflected in the ais and shifted to the left 3 units. Equation: = -2( + 3) 2 (2-2) CHAPTER 2 REVIEW 71

63. Equation: f() = 2 + 3-1 (2-2) 64. True; p() = p() 1 is a rational function for ever polnomial p. (2-3) 65. False; f() = 1 = -1 is not a polnomial function. (2-3) 66. False; f() = 1 2 + 1 has no vertical asmptotes. (2-3) 67. True: let f() = b, (b > 0, b 1), then the positive -ais is a horizontal asmptote if 0 < b < 1, and the negative -ais is a horizontal asmptote if b > 1. (2-4) 68. True: let f() = log b (b > 0, b 1). If 0 < b < 1, then the positive -ais is a vertical asmptote; if b > 1, then the negative -ais is a vertical asmptote. (2-5) 69. False; as increases without bound, log b if b > 1 and log b - if 0 < b < 1. (2-5) 70. True; f() = has vertical asmptote = 1 and horizontal! 1 asmptote = 1. (2-3) 71. False; the domain of f() = b (b > 0, b 1) is (-, ). (2-4) 72. = 2-1, -2 4 73. f(t) = 10e -0.08t, t 0-2 1 8-1 1 4 0 1 2 1 1 2 2 4 8 increasing [-2, 4] (2-4) t f(t) 0 10 10 4.5 20 2 30 0.9 40 0.4 decreasing [0, ) (2-4) 72 CHAPTER 2 FUNCTIONS AND GRAPHS

74. = ln( + 1), -1 < 10-0.5-0.7 0 0 4 1.6 8 2.2 10 2.4 increasing (-1, 10] (2-5) 15 80 75. 10 76. 60 40 5 20 10 20 30 40 (2-2) 10 20 30 40 50 (2-2) 77. = -( - 4) 2 + 3 (2-2, 2-3) 78. f() = -0.4 2 + 3.2-1.2 = -0.4( 2-8 + 16) + 7.6 = -0.4( - 4) 2 + 7.6 (A) intercept: 1.2 intercepts: -0.4( - 4) 2 + 7.6 = 0 ( - 4) 2 = 19 = 4 + 19 8.4, 4-19 -0.4 (B) Verte: (4.0, 7.6) (C) Maimum: 7.6 (D) Range: 7.6 or (-, 7.6] (2-3) 79. -10 10 10 (A) intercept: 1.2 intercepts: -0.4, 8.4 (B) Verte: (4.0, 7.6) (C) Maimum: 7.6 (D) Range: 7.6 or (-, 7.6] (2-3) -10 80. log 10 π = π log 10 = π (see logarithm properties 4.b & g, Section 2-5) 10 log 2 = is equivalent to log = log 2 which implies = 2 Similarl, ln e π = π ln e = π (Section 2-5, 4.b & g) and e ln 2 = implies ln = ln 2 and = 2. (2-5) CHAPTER 2 REVIEW 73

81. log - log 3 = log 4 - log( + 4) log 3 = log 4 + 4 3 = 4 + 4 ( + 4) = 12 2 + 4-12 = 0 ( + 6)( - 2) = 0 = -6, 2 Since log(-6) is not defined, -6 is not a solution. Therefore, the solution is = 2. (2-5) 82. ln(2-2) - ln( - 1) = ln 83. ln( + 3) - ln = 2 ln 2 # ln 2 2 & % ( = ln ln + 3 % $ ' = ln(2 2 ) $ 1 ' # & # 2( 1) & ln $ % 1 ' ( = ln + 3 = 4 ln 2 = ln + 3 = 4 = 2 (2-5) 3 = 3 = 1 (2-5) 84. log 3 2 = 2 + log 9 85. ln = -5t + ln c log 3 2 - log 9 = 2 ln - ln c = -5t! log 32 $ # & = 2 9 % ln c = -5t log % $ ' = 2 # 3 & c = e-5t 3 = 102 = 100 = ce -5t (2-5) = 300 (2-5) 86. Let be an positive real number and suppose log 1 =. Then 1 =. But, 1 = 1, so = 1, i.e., = 1 for all positive real numbers. This is clearl impossible. (2-5) 87. The graph of = 3 is verticall epanded b a factor of 2, reflected in the ais, shifted 1 unit to the left and 1 unit down. Equation: = -2 3 + 1-1 (2-2) 88. G() = 0.3 2 + 1.2-6.9 = 0.3( 2 + 4 + 4) - 8.1 = 0.3( + 2) 2-8.1 (A) intercept: -6.9 intercepts:0.3( + 2) 2-8.1 = 0 ( + 2) 2 = 27 = -2 + 27 3.2, -2-27 -7.2 (B) Verte: (-2, -8.1) (C) Minimum: -8.1 (D) Range: -8.1 or [-8.1, ) (E) Decreasing: (-, -2]; Increasing: [-2, ) (2-3) 74 CHAPTER 2 FUNCTIONS AND GRAPHS

89. -10 10-10 10 90. (A) f() = 0.181 2 4.88 + 271 consumption f() 0 271 271 5 255 251 10 233 240 15 236 239 20 256 246 25 257 262 (C) Year 2012 corresponds to = 32 f(32) 300 eggs (A) intercept: -6.9 intercept: -7.2, 3.2 (B) Verte: (-2, -8.1) (C) Minimum: -8.1 (D) Range: -8.1 or [-8.1, ) (E) Decreasing: (-, -2] Increasing: [-2, ) (2-3) (B) 300 250 200 150 100 50 0 5 10 15 20 25 Years since 1980 (D) Egg consumption fell annuall from 1980 until sometime between 1990 and 1995. From that point on, egg consumption has increased but has not et reached the 1980 consumption. (2-3) 91. Quadratic regression model = 0.181 2 4.88 + 271 (2-3) 92. (A) S() = 3 if 0 20; S() = 3 + 0.057( - 20) = 0.057 + 1.86 if 20 < 200; S(200) = 13.36 S() = 13.36 + 0.0346( - 200) = 0.0346 + 6.34 if 200 < 1000; S(1000) = 40.94 S() = 40.94 + 0.0217( - 1000) = 0.0217 + 19.24 if > 1000 # 3 if 0 20 % Therefore, S() = 0.057 + 1.86 if 20 < 200 $ 0.0346 + 6.34 if 200 < 1000 % & 0.0217 + 19.24 if > 1000 CHAPTER 2 REVIEW 75

(B) 50 40 30 20 10 S 500 1,000 (2-2) 93. A = P 1 + r % mt $ ' ; P = 5,000, r = 0.0535, m = 4, t = 5. # m & A = 5000 1 + 0.0535 % 4(5) $ ' = 5000(1.013375) 20 6,521.89 # 4 & After 5 ears, the CD will be worth $6,521.89 (2-4) 94. A = 1 + r % mt $ ' ; P = 5,000, r = 0.0482, m = 365, t = 5 # m & A = 5000 1 + 0.0482 % 365(5) $ ' 5000(1.0001321) 6,362.50 # 365 & After 5 ears, the CD will be worth $6,362.50. (2-4) 95. A = P 1 + r % mt $ ', r = 0.0659, m = 12 # m & Solve P 1 + 0.0659 % 12t $ ' = 3P or (1.005492) 12t = 3 # 12 & for t: 12t ln(1.005492) = ln 3 ln 3 t = 16.7 ears (2-4) 12 ln(1.005492) 96. A = P 1 + r % mt $ ', r = 0.0739, m = 365 # m & Solve P 1 + 0.0739 % 365t $ ' = 2P or (1.0002025) 365t = 2 # 365 & for t: 365t ln(1.0002025) = ln 2 ln 2 t = 9.37 ears (2-4) 365 ln(1.0002025) 97. p() = 50-1.25 Price-demand function C() = 160 + 10 Cost function R() = p() = (50-1.25) Revenue function 76 CHAPTER 2 FUNCTIONS AND GRAPHS

(A) R C (B) R = C (50-1.25) = 160 + 10-1.25 2 + 50 = 160 + 10-1.25 2 + 40 = 160-1.25( 2-32 + 256) = 160-320 -1.25( - 16) 2 = -160 ( - 16) 2 = 128 = 16 + 128 27.314, 16-128 4.686 R = C at = 4.686 thousand units (4,686 units) and = 27.314 thousand units (27,314 units) R < C for 1 < 4.686 or 27.314 < 40 R > C for 4.686 < < 27.314 (C) Ma Rev: 50-1.25 2 = R -1.25( 2-40 + 400) + 500 = R -1.25( - 20) 2 + 500 = R Verte at (20, 500) Ma. Rev. = 500 thousand ($500,000) occurs when output is 20 thousand (20,000 units) Wholesale price at this output: p() = 50-1.25 p(20) = 50-1.25(20) = $25 (2-3) 98. (A) P() = R() - C() = (50-1.25) - (160 + 10) = -1.25 2 + 40-160 200 1 40-600 (B) P = 0 for = 4.686 thousand units (4,686 units) and = 27.314 thousand units (27,314 units) P < 0 for 1 < 4.686 or 27.314 < 40 P > 0 for 4.686 < < 27.314 (C) Maimum profit is 160 thousand dollars ($160,000), and this occurs at = 16 thousand units (16,000 units). The wholesale price at this output is p(16) = 50-1.25(16) = $30, which is $5 greater than the $25 found in 46(C). CHAPTER 2 REVIEW 77

99. (A) The area enclosed b the pens is given b A = (2) Now, 3 + 4 = 840 so Thus = 210-3 4 # A() = 2% 210 3 $ 4 & ( ' = 420-3 2 2 (B) Clearl and must be nonnegative; the fact that 0 implies and 210-3 4 0 210 3 4 840 3 280 Thus, domain A: 0 280 (C) 30,000 A 280 (D) Graph A() = 420-3 2 2 and = 25,000 together. There are two values of that will produce storage areas with a combined area of 25,000 square feet, one near = 90 and the other near = 190. (E) = 86, = 194 30,000 0 300 0 (F) A() = 420-3 2 2 = - 3 2 (2-280) Completing the square, we have A() = - 3 2 (2-280 + 19,600-19,600) = - 3 2 [( - 140)2-19,600] = - 3 2 ( - 140)2 + 29,400 The dimensions that will produce the maimum combined area are: = 140 ft, = 105 ft. The maimum area is 29,400 sq. ft. (2-5) 78 CHAPTER 2 FUNCTIONS AND GRAPHS

100. (A) Quadratic regression model, Table 1: (B) Linear regression model, Table 2: To estimate the demand at price level of $180, we solve the equation a 2 + b + c = 180 for. The result is 2,833 sets. To estimate the suppl at a price level of $180, we solve the equation a + b = 180 for. The result is 4,836 sets. (C) The condition is not stable; the price is likel to decrease since the suppl at the price level of $180 eceeds the demand at this level. (D) Equilibrium price: $131.59 Equilibrium quantit: 3,587 cookware sets (2-1) 101. (A) Eponential regression model Year 2015 corresponds to = 25; (25) 3,519 million subscribers. (B) Year 2018 corresponds to = 28. the model predicts that there will be (28) 7,594 million subscribers in 2018. (2-4) 102. (A) N(0) = 1 (B) We need to solve: N$ 1% ' = 2 # 2& 2 2t = 10 9 N(1) = 4 = 2 2 log 2 2t = log 10 9 = 9 N$ 3% ' = 8 = 2 3 # 2& 2t log 2 = 9 N(2) = 16 = 2 4 9 t = 2 log 2 14.95 M Thus, the mouse will die in 15 das. Thus, we conclude that N(t) = 2 2t or N = 4 t. (2-4, 2-5) CHAPTER 2 REVIEW 79

103. Given I = I 0 e -kd. When d = 73.6, I = 1 2 I. Thus, we have: 0 1 2 I 0 = I 0 e-k(73.6) e -k(73.6) = 1 2 -k(73.6) = ln 1 2 k = ln(0.5) 73.6 0.00942 Thus, k 0.00942. To find the depth at which 1% of the surface light remains, we set I = 0.01I 0 and solve 0.01I 0 = I 0 e -0.00942d for d: 0.01 = e -0.00942d -0.00942d = ln 0.01 d = ln 0.01 0.00942 488.87 Thus, 1% of the surface light remains at approimatel 489 feet. (2-4, 2-5) 104. Logarithmic regression model: Year 2015 corresponds to = 75; (75) 6,139,000 cows. (2-5) 105. Using the continuous compounding model, we have: 2P 0 = P 0 e 0.03t 2 = e 0.03t 0.03t = ln 2 t = ln 2 0.03 23.1 Thus, the model predicts that the population will double in approimatel 23.1 ears. (2-4, 2-5) 80 CHAPTER 2 FUNCTIONS AND GRAPHS

106. (A) Eponential regression model Year 2015 corresponds to = 35; (35) $789 billion. (B) To find when the ependitures will reach one trillion, solve ab = 1,000 for. The result is 38 ears; that is, in 2018. (2-4) CHAPTER 2 REVIEW 81