Solution manual for calculus and its applications 11th edition by bittinger

Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Chapter R Functions, Graphs, and Models Exercise Set R.1 Graph y = x + We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) −2 x−4 We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) Graph y = −3 −6 −4 −2 ( −2, ) (0, ) (3, ) Graph y = − x + Graph y = x − Graph y = −3 x We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) −1 0 −6 ( −1, 3) (0, 0) ( 2, −6) Graph y = − x Graph x + y = We solve for y first x+ y =5 y = 5− x subtract x from both sides y = −x + commutative property Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) −1 ( −3, −6) (0, −4 ) (3, −2 ) ( −1, 6) (0,5) ( 2, 3) Graph x − y = Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Chapter R Functions, Graphs, and Models Graph y − x = We solve for y first y − 2x = y = 2x + 4 y = x+ 8 1 y = x+ add 2x to both sides y −6 −7 −2 divide both sides by Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) −2 x ( x, y ) ( −6, ) (0, −2 ) (6, 3) 12 Graph x + y = 10 ( −2, 0) ( 2,1) (6, ) 10 Graph x + y = −9 11 Graph x − y = 12 We solve for y first x − y = 12 subtract x from both sides −6 y = 12 − x y= divide both sides by − (12 − x ) −6 y = −2 + x y = x−2 We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve 13 Graph y = x − We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) −2 −1 −1 −4 −5 −4 −1 ( −2, −1) ( −1, −4 ) (0, −5) (1, −4 ) ( 2, −1) Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Exercise Set R.1 14 Graph y = x − x y −2 −1 0 1 2 ( x, y ) ( −2, ) ( −1,1) (0, 0) (1,1) ( 2, ) 15 Graph x = − y Since x is expressed in terms of y we first choose values for y and then compute x Then we plot the points that are found and connect them with a smooth curve y x ( x, y ) −2 −2 −1 −1 −2 ( −2, −2 ) (1, −1) ( 2, ) ( −1,1) ( −2, ) 16 Graph x = y + 18 Graph y = − x 19 Graph y = − x We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) −2 −1 ( −2,3) ( −1, 6) (0, ) (1, 6) ( 2, 3) 17 Graph y = x We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Chapter R Functions, Graphs, and Models 20 Graph y = − x According to this model, the world record for the mile in 1954 is approximately 3.97532 minutes To convert this to traditional minutesseconds we multiply the decimal part by 60 seconds 0.97532 ( 60 ) = 58.5192 Therefore the world record for the mile in 1954 was 3:58.5 Likewise, we substitute 2008 in for x to get R = −0.00582 ( 2008) + 15.3476 21 Graph y + = x First we solve for y = 3.66104 According to the model, the world record for the mile in 2008 will be approximately 3.66104 minutes or 3:39.7 Finally, we substitute 2012 in for x to get R = −0.00582 ( 2012 ) + 15.3476 y + = x3 y = x3 − Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation Then we plot the points and connect them with a smooth curve y x ( x, y ) −2 −9 −1 −2 −1 ( −2, −9 ) ( −1, −2 ) ( 0, −1) (1, 0) ( 2, ) = 3.63776 According to the model, the world record for the mile in 2012 will be approximately 3.63776 minutes or 3:38.3 24 A = 0.5t + 3.45t − 96.65t + 347.7t , ≤ t ≤ We substitute t = A = 0.5 ( ) + 3.45 ( ) − 96.65 ( ) + 347.7 ( ) = 344.4 Approximately 344.4 milligrams of ibuprofen will remain in the blood stream hours after 400 mg have been swallowed 25 v ( t ) = 10.9t We substitute 2.5 in for t to get v ( 2.5) = 10.9 ( 2.5) = 27.25 White was traveling at 27.25 miles per hour when he reentered the half pipe 22 Graph y − = x 26 s ( t ) = 16t s ( t ) = 28 28 = 16t 23 R = −0.00582 x + 15.3476 We substitute 1954 in for x to get R = −0.00582 (1954 ) + 15.3476 28 =t 16 28 = t2 16 1.3228 ≈ t Danny took approximately 1.32 seconds to hit the ramp = 3.97532 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Exercise Set R.1 27 a) Locate 20 on the horizontal axis and go directly up to the graph Then move left to the vertical axis and read the value there We estimate the number of hearing-impaired Americans of age 20 is about 1.8 million Follow the same process for 40, 50, and 60 to determine the number of hearingimpaired Americans at each of those ages We estimate the number of hearing-impaired Americans of age 40 is about 3.7 million We estimate the number of hearing-impaired Americans of age 50 is about 4.4 million We estimate the number of hearing-impaired Americans of age 60 is about 4.5 million b) Locate on the vertical axis and move horizontally across to the graph There are two x-values that correspond to the y-value of They are 44 and 70, so there are approximately million Americans age 44 who are hearing-impaired and approximately million Americans age 70 who are hearing-impaired c) The highest point on the graph appears to correspond to the x-value of 58 Therefore, age 58 appears to be the age at which the greatest number of Americans are hearingimpaired d) Visually, we cannot tell precisely which point is the highest point on the graph or which x-value corresponds exactly to that point The graph is not detailed enough to make that determination 28 a) We estimate the incidence of breast cancer in 40-yr-old women is about 100 per 100,000 women b) We estimate that at age 67 and at age 88 the incidence of breast cancer is about 400 per 100,000 women c) The largest incidence of breast cancer occurs in woman who are approximately 79 years of age d) Visually, we cannot tell precisely which point is the highest point on the graph or which x-value corresponds exactly to that point The graph is not detailed enough to make that determination 29 a) A = P (1 + i ) t A = 100, 000 (1 + 0.028) = 100, 000 (1.028) = 102,800 At the end of year, the investment is worth $102,800 i⎞ ⎛ b) A = P ⎜1 + ⎟ ⎝ n⎠ nt ⎛ 0.028 ⎞ A = 100, 000 ⎜1 + ⎟ ⎝ ⎠ = 100, 000 (1 + 0.014) = 100, 000 (1.014 ) 2⋅1 2 A = 100,000 (1.028196) = 102,819.60 At the end of year, the investment is worth $102,819.60 i⎞ ⎛ c) A = P ⎜1 + ⎟ ⎝ n⎠ nt ⎛ 0.028 ⎞ A = 100,000 ⎜1 + ⎝ ⎟⎠ = 100,000 (1 + 0.07 ) = 100,000 (1.07 ) 4⋅1 4 = 100,000 (1.0282953744 ) = 102,829.537 ≈ 102,829.54 At the end of year, the investment is worth $102,829.54 i⎞ ⎛ d) A = P ⎜1 + ⎟ ⎝ n⎠ nt ⎛ 0.028 ⎞ A = 100,000 ⎜1 + ⎝ 365 ⎟⎠ 365⋅1 = 100,000 (1 + 0.00076712329 ) 365 = 100,000 (1.00076712329) = 100,000 (1.02839458002) = 102,839.458002 ≈ 102,839.46 At the end of year, the investment is worth $102,839.46 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at 365 Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Chapter R Functions, Graphs, and Models e) There are 24 ⋅ 365 = 8760 hours in one year i⎞ ⎛ A = P ⎜1 + ⎟ ⎝ n⎠ nt e) There are 24 ⋅ 365 = 8760 hours in one year i⎞ ⎛ A = P ⎜1 + ⎟ ⎝ n⎠ ⎛ 0.028 ⎞ A = 100,000 ⎜1 + ⎝ 8760 ⎟⎠ 8760⋅1 nt ⎛ 0.022 ⎞ A = 300, 000 ⎜1 + ⎝ 8760 ⎟⎠ = 100,000 (1 + 0.00000196347 ) 8760 = 100,000 (1.00000196347 ) 8760 = 100,000 (1.02839563811) = 102,839.563811 ≈ 102,839.56 At the end of year, the investment is worth $102,839.56 8760⋅1 = 300, 000 (1.000002511416) 8760 ≈ 306, 673.13 At the end of year, the investment is worth $306,673.13 31 a) A = P (1 + i ) t A = 30, 000 (1 + 0.04 ) = 30, 000 (1.04 ) 30 a) A = P (1 + i ) t A = 300, 000 (1 + 0.022 ) = 300, 000 (1.022 ) = 306,600.00 At the end of year, the investment is worth $306,600.00 i⎞ ⎛ b) A = P ⎜1 + ⎟ ⎝ n⎠ nt 4⋅1 ≈ 306,654.65 At the end of year, the investment is worth $306,654.65 nt ⎛ 0.022 ⎞ A = 300, 000 ⎜1 + ⎟ ⎝ 365 ⎠ 365⋅1 = 300, 000 (1.000060273973) = 31, 212.00 At the end of year, the investment is worth $30,212.00 nt ⎛ 0.04 ⎞ A = 30,000 ⎜1 + ⎟ ⎝ ⎠ 2⋅1 = 30, 000 (1.0404 ) i⎞ ⎛ c) A = P ⎜1 + ⎟ ⎝ n⎠ ⎛ 0.022 ⎞ A = 300,000 ⎜1 + ⎟ ⎝ ⎠ i⎞ ⎛ d) A = P ⎜1 + ⎟ ⎝ n⎠ nt = 30, 000 (1.02 ) 2⋅1 = 306, 636.30 At the end of year, the investment is worth $306,636.30 = 300,000 (1.0055) i⎞ ⎛ b) A = P ⎜1 + ⎟ ⎝ n⎠ ⎛ 0.04 ⎞ A = 30, 000 ⎜1 + ⎝ ⎠⎟ nt ⎛ 0.022 ⎞ A = 300, 000 ⎜1 + ⎝ ⎠⎟ i⎞ ⎛ c) A = P ⎜1 + ⎟ ⎝ n⎠ = 31, 200.00 At the end of year, the investment is worth $30,200.00 = 30,000 (1.01) 4⋅1 = 30,000 (1.04060401) = 31, 218.1203 ≈ 31, 218.12 At the end of year, the investment is worth $30,218.12 365 ≈ 306, 672.93 At the end of year, the investment is worth $306,672.93 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Exercise Set R.1 i⎞ ⎛ d) A = P ⎜1 + ⎟ ⎝ n⎠ nt i⎞ ⎛ d) A = P ⎜1 + ⎟ ⎝ n⎠ ⎛ 0.04 ⎞ A = 30,000 ⎜1 + ⎟ ⎝ 365 ⎠ 365⋅1 ⎛ 0.05 ⎞ A = 1000 ⎜1 + ⎝ 365 ⎟⎠ = 30,000 (1.000109589 ) = 30,000 (1.04080847752 ) ≈ 31, 224.25 At the end of year, the investment is worth $30,224.25 e) There are 24 ⋅ 365 = 8760 hours in one year i⎞ ⎛ A = P ⎜1 + ⎟ ⎝ n⎠ nt 8760 ≈ 1051.27 At the end of year, the investment is worth $1051.27 = 30,000 (1.000004566210046 ) = 30,000 (1.04081067873) = 31, 224.3203618 ≈ 31, 224.32 At the end of year, the investment is worth $30,224.32 32 a) A = P (1 + i ) t A = 1000 (1 + 0.05) = 1050.00 At the end of year, the investment is worth $1050.00 nt ⎛ 0.05 ⎞ A = 1000 ⎜1 + ⎝ ⎟⎠ 2⋅1 = 1000 (1.050625) nt = 1000 (1.0125) 33 Using the formula: n ⎡ i ⎛ i ⎞ ⎤ ⎢ ⎜1 + ⎟ ⎥ 12 ⎝ 12 ⎠ ⎥ M = P⎢ n ⎢⎛ ⎥ i ⎞ ⎢ ⎜1 + ⎟ − ⎥ ⎢⎣ ⎝ 12 ⎠ ⎥⎦ We substitute 18,000 for P , 0.064 (6.4% = 0.064 ) for i, and 36 (3 ⋅ 12 = 36) for n Then we use a calculator to perform the computation ⎡ 0.064 ⎛ 0.064 ⎞ 36 ⎤ 1+ ⎢ ⎥ 12 ⎜⎝ 12 ⎟⎠ ⎥ M = 18, 000 ⎢ ⎢ ⎛ 0.064 ⎞ 36 ⎥ ⎢ ⎜1 + ⎥ − ⎟ 12 ⎠ ⎣⎢ ⎝ ⎦⎥ ≈ 550.86 The monthly payment on the loan will be approximately $550.86 ≈ 1050.63 At the end of year, the investment is worth $1050.63 ⎛ 0.05 ⎞ A = 1000 ⎜1 + ⎟ ⎝ ⎠ 8760⋅1 = 1000 (1.000005708) 8760⋅1 8760 i⎞ ⎛ c) A = P ⎜1 + ⎟ ⎝ n⎠ nt ⎛ 0.05 ⎞ A = 1000 ⎜1 + ⎝ 8760 ⎟⎠ ⎛ 0.04 ⎞ A = 30,000 ⎜1 + ⎝ 8760 ⎟⎠ 365 ≈ 1051.27 At the end of year, the investment is worth $1051.27 e) There are 24 ⋅ 365 = 8760 hours in one year = 31, 224.2543257 i⎞ ⎛ b) A = P ⎜1 + ⎟ ⎝ n⎠ 365⋅1 = 1000 (1.000136986) 365 i⎞ ⎛ A = P ⎜1 + ⎟ ⎝ n⎠ nt 4⋅1 ≈ 1050.95 At the end of year, the investment is worth $1050.95 34 30 years = 30 ⋅ 12 = 360 months P = 100, 000; i = 0.048 ⎡ 0.048 ⎛ 0.048 ⎞ 360 ⎤ 1+ ⎢ ⎥ 12 ⎜⎝ 12 ⎟⎠ ⎥ ⎢ M = 100, 000 ⎢ ⎛ 0.048 ⎞ 360 ⎥ ⎢ ⎜1 + −1 ⎥ ⎟ 12 ⎠ ⎣⎢ ⎝ ⎦⎥ ≈ 524.67 The monthly payment on the loan will be approximately $524.67 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Chapter R Functions, Graphs, and Models ⎡ (1 + i )n − ⎤ ⎥ 35 W = P ⎢ i ⎢⎣ ⎥⎦ We substitute 3000 for P, 0.0657 (6.57% = 0.0657 ) for i, and 18 for n ⎡ (1 + 0.0657 )18 − ⎤ ⎥ W = 3000 ⎢ 0.0657 ⎢⎣ ⎥⎦ ≈ 97,881.97 Rounded to the nearest cent, the annuity will be worth $97,881.97 after 18 years 36 We substitute 50,000 for W, 0.0725 ⎛ ⎞ ⎜⎝ % = 0.0725⎟⎠ in for i, and 20 for n Then we proceed to solve for P ⎡ (1 + 0.0725)20 − ⎤ ⎥ 50, 000 = P ⎢ 0.0725 ⎢⎣ ⎥⎦ 50,000 =P ⎡ (1 + 0.0725)20 − ⎤ ⎢ ⎥ 0.0725 ⎢⎣ ⎥⎦ 1186.74 ≈ P You will need to invest $1186.74 annually to reach a goal of $50,000 after 20 years 37 a) Locate 250,000 on the vertical axis and then think of horizontal lines extending across the graph from this point The years for which the graph lies above this line are the years for which the deer population was at or above 250,000 Those time periods are 1996 – 2000, and then a brief time between 2001 and 2002 b) Locate 200,000 on the vertical axis and then think of a horizontal line extending across the graph from this point The years for which the graph touches this line are the years for which the population was at 200,000 Those time periods are 1987 and 1990 c) Locate the highest point on the graph and extend a line vertically to the horizontal axis The year which the deer population was the highest was 1999 d) Locate the lowest point on the graph and extend a line vertically to the horizontal axis In this case there are two points that are exactly at 200,000 The years when the deer population was the lowest are 1987 and 1990 38 Answers will vary, but might include references to milder winters, increase food sources, reduced number of predators etc… ⎡ (1 + i )n − ⎤ ⎥ we 39 a) Using the formula W = P ⎢ i ⎢⎣ ⎥⎦ substitute 1200 for P, 0.08 (8% = 0.08) for i and 35 for n ⎡ (1 + 0.08)35 − ⎤ ⎥ W = 1200 ⎢ 0.08 ⎢⎣ ⎥⎦ ≈ 206, 780.16 Sally will have approximately $206,780.16 in her account when she retires b) Sally invested $1200 per year for 35 years Therefore, the total amount of her original payments is: $1200i35 = $42, 000 Since the total amount in the account was $206,780, the interest earned over the 35 years is: $206, 780.16 − $42, 000 = $164, 780.16 Therefore, $42,000 was the total amount of Sally’s payments and $164,780.16 was the total amount of her interest 40 a) Using the formula: n ⎡ i ⎛ i ⎞ ⎤ ⎢ ⎜1 + ⎟ ⎥ 12 ⎝ 12 ⎠ ⎥ M = P⎢ n ⎢⎛ ⎥ i ⎞ ⎢ ⎜1 + ⎟ − ⎥ ⎣⎢ ⎝ 12 ⎠ ⎦⎥ We substitute 206,780.16 for P , 0.08 (8% = 0.08) for i, and 15 (80 − 65 = 15) for n Then we use a calculator to perform the computation ⎡ 0.08 ⎛ 0.08 ⎞180 ⎤ 1+ ⎢ ⎥ 12 ⎜⎝ 12 ⎟⎠ ⎥ M = 206, 780.16 ⎢ ⎢ ⎛ 0.08 ⎞180 ⎥ ⎢ ⎜1 + −1 ⎥ ⎟ 12 ⎠ ⎢⎣ ⎝ ⎥⎦ ≈ 1976.10 Sally should take a monthly payment of $1976.10 b) Sally received 180 payments of $1976.10, therefore she received a total of $1976.10i180 = $355, 698 during the 15 years Of that 42,000 was what she originally contributed, leaving $313,698 in interest Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at Exercise Set R.1 41 Graph y = x − 150 We use the following window Next, we type the equation in to the calculator The resulting graph is: Next, we type the equation into the calculator The resulting graph is: 44 Graph y = 23 − x Using a standard window 42 Graph y = 25 − x Using the following window: Results in the graph: The resulting graph is 45 Graph 9.6 x + 4.2 y = −100 First, we solve for y 9.6 x + 4.2 y = −100 4.2 y = −100 − 9.6 x subtract 9.6 x from both sides −9.6 x − 100 4.2 Next, we set the window to be: y= 43 Graph y = x + x − x − 13 We use the following window: Next, we type the equation into the calculator at the top of the next page Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at 10 Chapter R Functions, Graphs, and Models The resulting graph is: 46 Graph y = −2.3 x + 4.8 x − We use the following window: This resulting graph is: 48 Graph x = − y Solving for y yields: y = ± 8− x We use the standard window The resulting graph is: The resulting graph is: 47 Graph x = + y First we solve for y x = + y2 x − = y2 subtracting from both sides taking the square root of both sides ± x−4 = y Next, we set the window to the standard window: It is important to remember that we must graph both the positive root and the negative root We type both equations into the calculator at the top of the next column Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.5 36 Graph g ( x ) = 51 39 Solve x + x = First, we put the equation in standard form x2 + x − = subtract from both sides The equation is in standard form with a = 1, b = 6, c = −1 Next, we apply the quadratic formula x − 25 x−5 − b ± b2 − 4ac 2a Substituting the values for a, b, and c, we get: x= 37 Solve x − x = First, we put the equation in standard form x2 − x − = subtract from both sides The equation is in standard form with a = 1, b = −2, c = − Next, we apply the quadratic formula − b ± b2 − 4ac x= 2a Substituting the values for a, b, and c, we get: x= − ( −2 ) ± ( −2)2 − (1)(−2) (1) ± + ± 12 = = 2 2±2 = 12 = ⋅ = = ( 1± ( ) ) = 1± ⋅1 The solutions are + and − 38 Solve x − x + = First, we put the equation into standard form x2 − x − = Next we apply the quadratic formula x= = = − ( −2 ) ± ( −2)2 − (1)( −4) (1) ± + 16 ± 20 ± = = 2 ( 1± ) = 1± The solutions are + and − x= − ( 6) ± (6)2 − (1)( −1) (1) −6 ± 36 + −6 ± 40 = 2 −6 ± 10 = 40 = ⋅ 10 = 10 = = ( −3 ± 10 ) = −3 ± ( 10 ⋅1 The solutions are −3 + 10 and − − 10 40 Solve x + x = First, we put the equation into standard form x2 + 4x − = Next we apply the quadratic formula x= = = − (4) ± ( 4)2 − (1)( −3) (1) −4 ± 16 + 12 −4 ± 28 −4 ± = = 2 ( −2 ± ) = −2 ± The solutions are −2 + and − − 41 Solve x = x + First, we put the equation in standard form x − x − = subtract x and from both sides The equation is in standard form with a = 4, b = −4, c = − Next, we apply the quadratic formula − b ± b2 − 4ac 2a Substituting the values for a, b, and c, we apply the quadratic formula on the next page x= Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ ) Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ 52 Chapter R Functions, Graphs, and Models x= − ( − 4) ± 44 Solve p − p = First, we put the equation into standard form p2 − p − = Next we apply the quadratic formula ( − )2 − ( )( −1) (4) ± 16 + 16 ± 32 = 4±4 x= = = ( 1± 4⋅2 p= ) = 1± ( 32 = 16 ⋅ = ) 2 1+ 1− The solutions are and 2 = = − (4) ± ( 4)2 − ( 4)( −1) (4) ( 2⋅4 ) = −1 ± x= −1 + −1 − and 2 −b ± b2 − 4ac 2a Substituting the values for a, b, and c, we get: (8)2 − ( 3)( ) ( 3) −8 ± 64 − 24 −8 ± 40 = 6 −8 ± 10 = 40 = ⋅ 10 = 10 ( − ± 10 2⋅3 ) = −4 ± −7 ± 49 − 36 −7 ± 13 = 2 −7 + 13 −7 − 13 The solutions are and 2 1 46 Solve − = w w Multiplying both sides by w2 we get: 1⎞ ⎛ w2 ⎜1 − ⎟ = ⋅ w ⎝ w⎠ w w2 − w − = This is a quadratic equation in standard form Next, we apply the quadratic formula ) 10 −4 + 10 −4 − 10 The solutions are and 3 = (7)2 − (1)( 9) (1) w2 − w = = ( − (7) ± = y= − (8) ± x + x + = This is a quadratic equation in standard form with a = 1, b = 7, and c = Next, we apply the quadratic formula 43 Solve y + y + = The equation is in standard form with a = 3, b = 8, c = Next, we apply the quadratic formula y= =0 x Multiplying both sides by x, we get: 9⎞ ⎛ x ⋅ ⎜ x + + ⎟ = 0⋅ x ⎝ x⎠ x= The solutions are ± 25 + ± 33 = 4 + 33 − 33 The solutions are and 4 = − b ± b2 − 4ac 2a Substituting the values for a, b, and c, we get: − ± 16 + 16 − ± 32 − ± = = 8 −1 ± ( −5)2 − ( 2)(−1) ( 2) 45 Solve x + + 42 Solve − x = x − First, we put the equation into standard form x2 + x − = Next we apply the quadratic formula x= − ( − 5) ± w= − ( −1) ± ( −1)2 − (1)( −1) (1) 1± 1+ 1± = 2 1+ 1− The solutions are and 2 = Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.5 x3 = x 47 48 49 x5 = x a3 = a ( m n ) 63 ( b2 = b = b 51 t =t 52 c =c 53 x12 = x n am = a m n ) 64 t =t t = 12 = x3 =t 66 =t 56 m m x2 + = ( x + 7) = ( x + 7) ( 59 x = x1 = x 60 t 61 y3= 62 t m n −2 −1 b y b −1 ( ) −1 n ) = n am n y2 ⎛ −n ⎞ ⎜⎝ a = n ⎟⎠ a (a = b = m ) = n am n b −17 ⎛ −n ⎞ ⎜⎝ a = n ⎟⎠ a = 17 e (a 17 e 68 a =a m m ⎛ −n ⎞ ⎜⎝ n = a ⎟⎠ a e n m −19 (x = 19 m ) −3 −1 = ) = (x 70 (y +7 ) −1 = m n = n am ) m19 ) −3 1 ⎞ ⎛ −n ⎜⎝ a = n ⎠⎟ a (a = x −3 (y +7 ) = m n = n am y +7 2 (a m n = n am ) 71 t w (a m n = n am ) = t 73 −4 3 t n = n am ( 9) ( ) ⎛ = ⋅ 3; a m⋅n = a m ⎜⎝ 2 (a n =na ) = ( 3) = 27 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ ) ( ) = m = w = w4 = (a = 72 y2 m = = = =7t (a = ) 69 x3 + = x3 + am = a = m −2 m2 = 58 n ⎛ −n ⎞ ⎜⎝ n = a ⎟⎠ a 1 57 ( −5 = t t b = b 67 y 2 65 ⎛ −n ⎞ ⎜⎝ a = n ⎟⎠ a = −2 t 55 t = 50 54 am = a n The index is 2; 2 53 n⎞ ⎠⎟ ) Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ 54 Chapter R Functions, Graphs, and Models ( ) = ( 16 ) 74 16 = 16 75 81 The domain of a rational function is restricted to those input values that not result in division by To determine the domain of = ( ) = 1024 5 2 64 ( ) = 64 = ( 64 = (4) ) x3 x − 5x + we set the denominator equal to zero and solve: f ( x) = ( ) ⎟⎠ ⎛ = ⋅ 2; a m⋅n = a m ⎜⎝ 3 (a ( n =na 64 = ) n⎞ ) x − 5x + = ( x − 2)( x − 3) = = 16 76 ( ) = ( 8) 83= 77 16 3 = ( 2) = 4 ( ) = 16 = ( ) ⎠⎟ ⎛ = ⋅ 3; a m⋅n = a ⎝⎜ 4 (a ( 16 ) ( = (2) n = a n 16 = ) m n⎞ 82 ( ) = ( 25 ) 25 = 25 5 ) ( x + 5)( x + 1) = x = −5 or x = −1 Therefore, the numbers −5 and − are not in the domain The domain of f consists of all real numbers except −5 and − = (5) = 3125 83 The domain of the radical function f ( x ) = x + is restricted to those input values that result in the value of the radicand being greater than or equal to In other words, the domain will be the set of real numbers that satisfy the inequality x + ≥ To find the domain, we solve the inequality: 5x + ≥ x ≥ −4 x≥− Therefore, the domain of f consists of all real numbers greater than or equal to − , or in ⎡ ⎞ interval notation ⎢ − , ∞ ⎟ ⎣ ⎠ x = Therefore, is not in the domain The domain of f consists of all real numbers except f ( x) = x4 + x2 + 6x + x2 + 6x + = 79 The domain of a rational function is restricted to those input values that not result in division by To determine the domain of x − 25 f ( x) = x−5 we set the denominator equal to zero and solve: x−5= 80 f ( x) = Solve: =8 78 factoring x − = or x − = Principle of Zero Products x = or x = Therefore, and are not in the domain The domain of f consists of all real numbers except and x2 − x+2 Solve: x+2=0 x = −2 Therefore, −2 is not in the domain The domain of f consists of all real numbers except −2 84 f ( x) = 2x − Solve: 2x − ≥ 2x ≥ x≥3 The domain is the set of all real numbers greater than or equal to 3, or in interval notation [ 3, ∞ ) Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.5 85 The domain of the radical function f ( x ) = − x is restricted to those input values that result in the value of the radicand being greater than or equal to In other words, the domain will be the set of real numbers that satisfy the inequality − x ≥ To find the domain, we solve the inequality: 7− x ≥ 7≥ x Therefore, the domain of f consists of all real numbers less than or equal to 7, or in interval notation ( −∞,7] 86 f ( x) = − x Solve: 5− x ≥ 5≥ x Therefore, the domain of f consists of all real numbers less than or equal to 5, or in interval notation ( −∞, 5] 87 We set the demand equation equal to the supply equation and solve for x 1000 − 10 x = 250 + x 55 89 We set the demand equation equal to the supply equation and solve for x x = x 25 = x multiply both sides by x 25 = x take the square root of both sides ±5 = x Since it is not appropriate to have a negative price, the equilibrium price is hundred dollars or $500 To find the equilibrium quantity, we substitute for x into either the demand equation or supply equation We use the demand equation q= =1 ( 5) The equilibrium quantity is thousand units or 1000 units The equilibrium point is (5,1) 90 We set the demand equation equal to the supply equation and solve for x x = x 16 = x q = 1000 − 500 ±4 = x Since it is not appropriate to have a negative price, the equilibrium price is hundred dollars or $400 To find the equilibrium quantity, we substitute for x into either the demand equation or supply equation We use the demand equation q= ( 4) q = 500 The equilibrium quantity is 500 units The equilibrium point is (50,500 ) q =1 The equilibrium quantity is thousand units or 1000 units The equilibrium point is ( 4,1) 750 = 15 x 50 = x Thus, the equilibrium price is $50 To find the equilibrium quantity, we substitute 50 for x into either the demand equation or supply equation We use the demand equation q = 1000 − 10 (50 ) 88 We set demand equal to supply and solve to find equilibrium price 8800 − 30 x = 7000 + 15 x 1800 = 45 x 40 = x Substitute into the demand equation to find equilibrium quantity q = 8800 − 30 ( 40 ) = 7600 The equilibrium point is ( 40,7600) 91 We set the demand equation equal to the supply equation and solve for x ( x − 3)2 = x + x + x2 − x + = x2 + x + −6 x + = x + subtracting x from both sides = 8x 1= x The equilibrium price is $1 The solution is continued on the next page Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ 56 Chapter R Functions, Graphs, and Models To find the equilibrium quantity, we substitute for x into either the demand equation or supply equation We use the demand equation q = (1 − 3) 94 We set the demand equation equal to the supply equation and solve for x − x = x +1 0≤ x≤7 ( − x )2 = ( = ( −2 ) = ( x − )2 = x + x + = ( −3) = The equilibrium quantity is hundred units or 900 units The equilibrium point is (1,9 ) 93 We set the demand equation equal to the supply equation and solve for x 5− x = x +7 0≤ x≤5 ( − x )2 = ( x+7 ) Squaring both sides 25 − 10 x + x = x + 18 − 11x + x = (9 − x )( − x ) = 95 If the price per share S is inversely proportional to the prime rate R, then we have: k S= R We find the constant of variation by substituting 205.93 for S and 0.0325 (3.25%) for R k 205.93 = 0.0325 6.692725 = k 6.692725 The equation of variation is S = R If the prime rate rose to 4.75% we find S by substituting 0.0475 in for R 6.692725 S= 0.0475 = 140.899 ≈ 140.90 The price per share would be approximately $140.90 if the assumption of inverse proportionality is correct Factoring the quadratic − x = or − x = Principle of Zero Products = x or 2= x Since is not in the domain of the demand function, the equilibrium price is thousand dollars or $2000 To find the equilibrium quantity, we substitute for x into either the demand equation or supply equation We use the demand equation q = − (2) = The equilibrium quantity is thousand units or 3000 units The equilibrium point is ( 2, 3) 45 − 18 x + x = 15 − x = or − x = 15 = x or 3= x Since 15 is not in the domain of the demand function, the equilibrium price is thousand dollars or $3000 To find the equilibrium quantity, we substitute for x into either the demand equation or supply equation We use the demand equation q = − ( 3) = The equilibrium quantity is thousand units or 4000 units The equilibrium point is ( 3, ) 92 We set the demand equation equal to the supply equation and solve for x q = (1 − ) ) 49 − 14 x + x = x + The equilibrium quantity is hundred units or 400 units The equilibrium point is (1, ) x − x + 16 = x + x + −8 x + 16 = x + 10 = 10 x 1= x The equilibrium price is $1 To find the equilibrium quantity, we substitute for x into either the demand equation or supply equation We use the demand equation x +1 96 k p Find the constant of variation by making the following substitutions k 85, 000 = 2900 246,500, 000 = k Therefore, the equation of variation is: 246, 500, 000 x= p The solution is continued on the next page x= Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.5 57 If the price drops to $850, then 246,500,000 x= = 290,000 850 290,000 plasma TVs will be sold if the price is $850 97 a) R ( x ) = 11.74 x 0.25 R ( 40,000) = 11.74 ( 40, 000) w 0 1000 34.0 2000 64.2 3000 93.3 w 4000 121.5 5000 149.2 6000 176.5 7000 203.4 H ( w) H ( w) We plot the points in the table and graph the function 0.25 = 11.74 (14.14213562 ) = 166.0286722 ≈ 166 The maximum range will be approximately 166 miles when the peak power is 40,000 watts R (50,000) = 11.74 (50,000) 0.25 99 a) In 2005, t = 2005 − 1970 = 35 P = 1000 ( 35) + 14, 000 ≈ 99,130 = 11.74 (14.95348781) = 175.5539469 In 2005, average pollution was approximately 99,130 particles per cubic centimeter In 2008, t = 2008 − 1970 = 38 ≈ 176 The maximum range will be approximately 176 miles when the peak power is 50,000 watts R ( 60, 000) = 11.74 (60,000) P = 1000 ( 38) + 14, 000 ≈ 108, 347 0.25 In 2008, average pollution will be approximately 108,347 particles per cubic centimeter In 2014, t = 2014 − 1970 = 44 = 11.74 (15.6508458) = 183.7409297 ≈ 184 The maximum range will be approximately 184 miles when the peak power is 60,000 watts b) Plotting the points found in part (a) and connecting them with a smooth curve we see: 98 Substituting each value given for w in H ( w) = 0.059w0.92 and using a calculator to compute the values for H we create the following table in the next column where the values for H are rounded off to one decimal P = 1000 ( 44 ) + 14, 000 ≈ 127, 322 In 2014, average pollution will be approximately 127,322 particles per cubic centimeter b) Plot the points above and others, if necessary, and draw the graph 100 a) f (180 ) = 0.144 (180 ) ≈ 1.93 The surface area of a person whose mass is 75 kg and height is 180 cm is approximately 1.94m2 b) f (170 ) = 0.144 (170 ) ≈ 1.88 The surface area of a person whose mass is 75 kg and height is 170 cm is approximately 1.88m2 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ 58 Chapter R Functions, Graphs, and Models c) Plotting the points from the previous page and others, if necessary, the graph is: 103 Answers will vary A rational function is the quotient of two polynomial functions Every polynomial function can be expressed as a rational function with a denominator of 1, so every polynomial function is a rational function However, most rational functions are not polynomial functions 104 We graph f ( x ) = x − x on our calculator 101 The number of cities N with a population greater than S is inversely proportional to S k N= S k 52 = Substituting 350,000 18, 200, 000 = k The equation of variation is 18, 200, 000 N= S We find N when S is 500,000 18, 200, 000 N= = 36.4 ≈ 36 500, 000 Using the fact that there were 52 cities with a population greater than 350,000 and 36 cities with a population of 500,000 or greater, we estimate that there are 52 − 36 = 16 cities with a population between 350,000 and 500,000 Using the ZERO feature we find the zeros are -1, 0, Algebraically we find the zeros by solving the equation: x3 − x = ( ) x x2 − = x ( x − 1)( x + 1) = x = or x − = or x + = x = or x = or x = −1 The zeros are -1, 0, 105 f ( x ) = x − x − 14 x − 10 Enter the function into your calculator To estimate the number of cities with a population between 300,000 and 600,000 we find N when S is 300,000 and when S is 600,000 18, 200, 000 N= ≈ 60.667 ≈ 61 300, 000 18, 200, 000 N= ≈ 30.33 ≈ 30 600, 000 There are 61 cities with a population greater than 300,000 and 30 cities with a population greater than 600,000, so there are 61 − 30 = 31 cities with a population between 300,000 and 600,000 102 Answers will vary A function has at most one y-intercept If a graph has more than one yintercept, then the vertical line x = intersect the graph in more than one point Thus, the graph fails the vertical line test, and the graph does not represent a function Using the window: We see the graph: Now using the ZERO feature on the calculator, we approximate the zeros The zeros are -1.831, -0.856, 3.188 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.5 ( x −4 + x −7 )− Using the window: 106 f ( x ) = 59 We get the graph: The zeros are −2.449 and 2.449 We get the graph: 109 f ( x ) = x + + x − − Enter the function into your calculator The zeros are 1.5 and 9.5 107 f ( x ) = x + x − 36 x − 160 x + 300 Using the standard window: Enter the function into your calculator We see the graph: Using the window: We see the graph: Now using the ZERO feature on the calculator, we approximate the zeros The zeros are –2 and 110 f ( x ) = x + + x − Enter the function into your calculator Now using the ZERO feature on the calculator, we approximate the zeros The zeros are 1.489 and 5.673 Using the standard window: 108 f ( x ) = − x − Using the window: The solution is continued on the next page Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ 60 Chapter R Functions, Graphs, and Models We see the graph: This results in the following graph: We see that the graph does not cross the x-axis, there are no real zeros You can see that it difficult to see all of the zeros It is recommended to zoom in around each zero and then use the ZERO feature to approximate each zero The zeros are: –10.153, –1.871, –0.821, –0.303, 0.098, 0.535, 1.219, and 3.297 111 f ( x ) = x + + x − − Enter the function into your calculator 113 We enter the demand and supply equations into the graphing editor on the calculator Using the standard window: Using the window: We see the graph: We see the graph: We see that the graph intersects the x-axis between −1 ≤ x ≤ The zeros of this function are all real numbers in [ −1, ] 112 f ( x ) = x + x − 28 x − 56 x + 70 x + 56 x Using the intersect feature on the calculator we find the intersection to be: −28 x − x + It is difficult to choose a window that shows the entire graph and all of the function’s zeros We use the window: The equilibrium point for this market is (75.11, 7.893) In other words, 7893 units will be sold at a price of $75.11 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.6 Exercise Set R.6 The data is decreasing at a constant rate, so a linear function f ( x ) = mx + b could be used to model the data The data falls first and then rises over the domain, so a quadratic function f ( x ) = ax + bx + c, a > could be used to model the data The data rises first and then falls over the domain, so a quadratic function f ( x ) = ax + bx + c, a < could be used to model the data The data rises and falls several times over the domain This implies that a polynomial function that is neither quadratic nor linear would be best to model the data The data is increasing at a constant rate, so a linear function f ( x ) = mx + b could be used to model the data The data rises first and then falls over the domain, so a quadratic function f ( x ) = ax + bx + c, a < could be used to model the data The data rises and falls several times over the domain This implies that a polynomial function that is neither quadratic nor linear would be best to model the data The data rises first and then falls over the domain, so a quadratic function f ( x ) = ax + bx + c, a < could be used to model the data The data is decreasing at a constant rate, so a linear function f ( x ) = mx + b could be used to model the data 10 a) We find the linear function f ( x ) = mx + b that contains the data points (1, 6.98) and (12, 3.61) 61 We find the slope containing these two points first 3.61 − 6.98 −3.37 m= = ≈ −0.306 12 − 11 Next, we use the slope and one of the points and substitute into the point-slope equation to find the equation of the line y − y1 = m ( x − x1 ) y − 6.98 = −0.306 ( x − 1) y − 6.98 = −0.306 x + 0.306 y = −0.306 x + 7.286 The linear function that models the data is: f ( x ) = −0.306 x + 7.286 Alternatively, we substitute in to the equation y = mx + b to obtain a system of equations 6.98 = m ⋅ + b (1) 3.61 = m ⋅ 12 + b (2) Subtracting each side of Equation (1) from each side of Equation (2) we get: −3.37 = 11m −3.37 =m 11 −0.306 ≈ m Now substitute m = −0.306 in for either Equation (1) or (2) and solve for b We use Equation (1) 6.98 = ( −0.306)(1) + b 6.98 = −0.306 + b 7.286 = b Therefore, the linear function that fits the data is f ( x ) = −0.306 x + 7.286 b) In June of 2000, x = 18 f (18) = −0.306 (18) + 7.286 ≈ 1.78 The prime rate in June of 2009 is approximately 1.78% 11 a) We will use the points (0, 3.6 ) and (9,5.6 ) First, we find the slope: 5.6 − 3.6 = 9−0 Next, since we chose the y-intercept, we can substitute into the slope-intercept equation y = mx + b y = x + 3.6 m= Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ 62 Chapter R Functions, Graphs, and Models Alternatively, we could have found the equation on the previous page by substituting the data points in to the equation y = mx + b to obtain a system of equations 3.6 = m ⋅ + b (1) (2) 5.6 = m ⋅ + b Subtracting each side of Equation (1) from each side of Equation (2) we get: = 9m =m Now substitute m = in for either Equation (1) or (2) and solve for b We use Equation (1) ⎛2⎞ 3.6 = ⎜ ⎟ ( 0) + b ⎝9⎠ 3.6 = b Therefore, the linear function that fits the data is y = x + 3.6 b) In 2012, x = 2012 − 2000 = 12 Substituting 12 for x we have: y = (12 ) + 3.6 = 6.266 ≈ 6.3 The average salary of NBA players in 2012 will be approximately $6.3 million In 2020, x = 2020 − 2000 = 20 Substituting 12 for x we have: y = ( 20) + 3.6 = 8.044 ≈ 8.0 The average salary of NBA players in 2020 will be approximately $8.0 million c) Substituting 9.0 for y and solving for x we have: 9.0 = x + 3.6 5.4 = x 9 (5.4 ) = x 24.3 ≈ x 24.3 years after 2000 or approximately 2024 the average salary of NBA players will reach $9.0 million 12 a) Consider the general quadratic function y = ax + bx + c Using the points (0, 0) , ( 2, 200) , and (3,167 ) , we substitute each point into the general quadratic function to obtain the system of equations = a ⋅ 02 + b ⋅ + c 200 = a ⋅ 22 + b ⋅ + c 167 = a ⋅ 32 + b ⋅ + c Which gives us: 0=c 200 = 4a + 2b + c 167 = 9a + 3b + c From the first equation we see that c = We substitute this value for c into the other two equations and our system is reduced to: 200 = 4a + 2b 167 = 9a + 3b Solving this system of equations, we get: 133 566 a=− ,b = ,c = 3 This gives us the quadratic function: 133 566 f ( x) = − x + x+0 3 133 566 =− x + x 3 Writing the equation with proper fractions we have: f ( x ) = −44 13 x + 188 23 x b) f ( ) = −44 13 ( ) + 188 23 ( ) ≈ 45.3 Approximately 45.3 mg of albuterol will be in the bloodstream after hours c) Answers will vary It does not make sense to use this function for t = because albuterol will leave the blood stream entirely after about hours is not in the domain of this function 13 a) Consider the general quadratic function y = ax + bx + c , where y is the braking distance, in feet, and x is the speed, in miles per hour Using the points ( 20, 25) , ( 40,105) , and (60, 300) , we substitute each point into the general quadratic function to get the system of equations on the next page Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.6 63 Using information from the previous page, we obtain the following system of equations: 25 = a ⋅ 202 + b ⋅ 20 + c 105 = a ⋅ 402 + b ⋅ 40 + c 300 = a ⋅ 602 + b ⋅ 60 + c or, 25 = 400a + 20b + c 105 = 1600a + 40b + c 300 = 3600a + 60b + c Solving the system of equations, we get: a = 0.14375, b = −4.625, and c = 60 Therefore, the function is y = 0.144 x − 4.63 x + 60 b) We substitute 50 for x and compute the value of y Subtracting each side of Equation (1) from each side of Equation (2) we get: 51.6 = 40m 1.29 = m Now substitute m = 1.29 in for either Equation (1) or (2) and solve for b We use Equation (1) 1.4 = (1.29 )( 30) + b 1.4 = 38.7 + b −37.3 = b Therefore, the linear function that fits the data is y = 1.29 x − 37.3 b) We plot the points in the table in the text and then graph the function found in part (a) on the same set of axes y = 0.144 (50) − 4.63 (50 ) + 60 = 188.5 The breaking distance of a car traveling at 50 mph is about 188.5 ft c) No, the function does not make sense for speeds less than 15 mph The vertex of the parabola occurs around 16 mph Thus for speeds less than 16 mph the breaking distance starts increasing as the speed decreases 14 a) N ( x ) = ax + bx + c 16 a) Answers will vary depending on which points are used to find the function We will use the points ( 30, 7.3) and 100 = a ⋅ 602 + b ⋅ 60 + c 130 = a ⋅ 802 + b ⋅ 80 + c 200 = a ⋅ 1002 + b ⋅ 100 + c Solving this system we get: a = 0.05, b = −5.5, and c = 250 Therefore, the function is: N ( x ) = 0.05 x − 5.5 x + 250 b) N (50 ) = 0.05 (50 ) − 5.5 (50) + 250 = 100 At 50 km/h, 100 accidents occur per 200 million km driven 15 a) Answers will vary depending on which points are used to find the function We will use the points ( 30,1.4 ) and (70,53.0) We substitute in to the equation y = mx + b to obtain a system of equations 1.4 = m ⋅ 30 + b 53.0 = m ⋅ 70 + b c) Substituting in 55 for x we have: y = 1.29 (55) − 37.3 = 33.65 Approximately 33.65% of 55-yr-old women have high blood pressure (1) (2) ( 70, 34.9 ) We substitute in to the equation y = mx + b to obtain a system of equations 7.3 = m ⋅ 30 + b (1) (2) 34.9 = m ⋅ 70 + b Subtracting each side of Equation (1) from each side of Equation (2) we get: 27.6 = 40m 0.69 = m Now substitute m = 0.69 in for either Equation (1) or (2) and solve for b We use Equation (1) 7.3 = ( 0.69 )( 30) + b 7.3 = 20.7 + b −13.4 = b Therefore, the linear function that fits the data is y = 0.69 x − 13.4 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ 64 Chapter R Functions, Graphs, and Models b) We plot the points in the table in the text and then graph the function found in part (a) on the same set of axes c) Substituting in 55 for x we have: y = 0.69 (55) − 13.4 = 24.55 Approximately 24.55% of 55-yr-old men have high blood pressure 17 Answers will vary With the small amount of data, it is difficult to determine what trend the data is taking on Since linear functions are easier to find than other polynomial functions, we might want to trade off accuracy for simplicity 18 Answers will vary During the late spring and early summer, the days grow longer and thus the number of hours of daylight increases After the summer solstice, the hours of daylight begin to decrease Since the data values increase then decrease, a quadratic function with a < would be appropriate to model the number of hours of daylight for the dates April 22 to August 22 19 Answers will vary Since the data deals with the amount of Albuterol in the bloodstream, the domain should be restricted to nonnegative numbers because the amount of Albuterol in the bloodstream cannot be measured until the drug is ingested The upper restriction on the domain will be the amount of time it takes for Albuterol to exit the system For the data in Exercise 12, this domain should not extend beyond [ 0, 4] 20 Answers will vary The domain must be restricted to the nonnegative numbers since negative speed have no meaning A car traveling at miles per hour would have a breaking distance of feet, so could be included in the domain Realistically, there will also be an upper limit on the domain The upper limit would be based upon the maximum speed of the automobile 21 a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y Using the linear regression feature we get: The linear function that fits the data is y = −0.224 x + 6.5414 b) Substituting 18 in for x we get: y = −0.224 (18) + 6.5414 ≈ 2.509 ≈ 2.51 The prime rate in June 2009 will be approximately 2.51% c) The actual prime rate in June of 2009 was 3.25% So the linear regression on the calculator gave us a closer approximation to the actual prime rate for that month The regression answer seems more plausible; it uses all the data d) Using the cubic regression feature on the calculator results in: y = −0.009856 x + 0.1993 x − 1.3563 x + 8.103 Substituting in 18 for x we get: y = −0.009856 (18) + 1993 (18) − 1.3563 (18) + 8.103 ≈ −9.207 The cubic function estimated the prime rate to be -9.207% for June of 2005 Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ Solution Manual for Calculus And Its Applications 11th Edition by Bittinger Full file at https://TestbankDirect.eu/ Exercise Set R.6 e) 65 Answers will vary The linear model is more appropriate for this data even though the cubic model gave us a more accurate approximation The leading coefficient on the cubic model is approximately −0.0004856 The additional accuracy gained in the cubic model does not justify the amount of effort needed to create and to use the cubic model for calculations It is best to use the simpler linear model in this case Furthermore, the model is not valid very far into the future since we calculate a negative interest rate in June 2009 c) Answers will vary The data appears to be nonlinear, however it does appear to have a linear trend This means that even though the data increases and decreases, it is doing so along an upward sloping line Therefore if the fluctuations about the linear trend are not too large, we can fit the data with a linear model and sacrifice a little bit of accuracy for much greater simplicity Further more, we see that the leading coefficient for the cubic model is very close to zero This leads us to believe the linear model might be good enough to use 22 a) Letting x be the number of years after 1996, we enter the data into the statistics editor of the graphing calculator We enter the years since 1996 into L1 and the trade deficit into L2 Next, we use the cubic regression feature on the calculator to fit the data to the model The cubic function that fits the data is: y = −74.704 x + 834.04 x + 1730.3 x + 53831.3 b) To predict the trade deficit in 2012, we substitute 16 in for x and compute y y = −74.704 (16) + 834.04 (16) + 1730.3 (16 ) + 53831.3 ≈ −10, 954 The trade deficit in 2012 will be approximately negative 10,954 million dollars In other words, the model predicts a trade surplus of $10,954 million Copyright © 2014 Pearson Education, Inc Publishing as Addison-Wesley Full file at https://TestbankDirect.eu/ ... 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