Math 105 Final Eam December 17, 013 Name: EXAM SOLUTIONS Instructor: Section: 1. Do not open this eam until ou are told to do so.. This eam has 1 pages including this cover. There are 1 problems. Note that the problems are not of equal difficult, so ou ma want to skip over and return to a problem on which ou are stuck. 3. Do not separate the pages of this eam. If the do become separated, write our name on ever page and point this out to our instructor when ou hand in the eam.. Please read the instructions for each individual problem carefull. One of the skills being tested on this eam is our abilit to interpret mathematical questions, so instructors will not answer questions about eam problems during the eam. 5. Show an appropriate amount of work including appropriate eplanation) for each problem, so that graders can see not onl our answer but how ou obtained it. Include units in our answer where that is appropriate. 6. You ma use an calculator ecept a TI-9 or other calculator with a full alphanumeric kepad). However, ou must show work for an calculation which we have learned how to do in this course. 7. If ou use graphs or tables to find an answer, be sure to include an eplanation and sketch of the graph, and to write out the entries of the table that ou use. 8. Turn off all cell phones, smartphones, and other electronic devices, and remove all headphones. 9. You must use the methods learned in this course to solve all problems. Problem Points Score LA Post-Test 5 1 10 7 3 10 13 5 5 6 5 7 8 8 11 9 5 10 11 7 1 10 Total 100
Math 105 / Final December 17, 013) page 1. [10 points] Foghorn is a chicken that is learning how to fl. In fact, he trains ever da b jumping off the top of his coop and flapping his wings. Toda, his height above the ground, in feet, t seconds after jumping is given b the function ht) = 16t + 0t + 6. Note that once he lands on the ground, he stas on the ground. a. [ points] How long after Foghorn jumps off his coop does he hit the ground? Be sure to show our work and give our final answer in eact form. To find when Foghorn hits the ground, we set ht) = 0 and solve for t. Using the quadratic formula, we find t = 0 ± 0 16)6) 16) = 0 ± 78 3 = 0 ± 8 3 So, t = 0.5 and t = 1.5. Because he starts on top of the coop, t = 0.5 does not make an sense as an answer. So, Foghorn hits the ground 1.5 seconds after jumping off the coop. Answer: 1.5 seconds b. [ points] Use the method of completing the square to put the formula for ht) into verte form. Carefull show our algebraic work step-b-step. ht) = 16t + 0t + 6 = 16 t 5 ) t + 6 = 16 t 5 5 8) t + = 16 t 5 ) 8 ) ) 5 + 6 = 16 t 5 + 16 8 8) ) ) 5 + 6 8 ) 5 + 6 = 16 t 5 + 1.5 8 8) Answer: ht) = 16 t 5 + 1.5 8) c. [ points] What is the maimum height Foghorn reaches? Answer: 1.5 feet When does he reach his maimum height? Answer: 5/8 seconds after he jumps The verte form gives a maimum height of 1.5 feet 5/8 seconds after he jumps. d. [ points] What are the domain and range of ht) in the contet of this problem? Use either interval notation or inequalities to give our answers. In the contet of this problem, the domain is from when he jumps to when he lands. The domain is then [0, 1.5]. The range is from the ground, 0 feet, to the maimum height he reaches, which is 1.5 feet. The range is then [0, 1.5]. Answers: Domain: [0, 1.5] Range: [0, 1.5]
Math 105 / Final December 17, 013) page 3. [7 points] Invertible functions f and g and a function h are described b the table, formula, and graph below. Use this information to answer the questions that follow. 3 1 0 1 3 f) 3 1 3 0 1 gt) = { + if < 3 if 3 Graph of = h) Evaluate each of the following quantities, if possible. If the specified quantit is undefined, write undefined. You do not have to show our work. However, an work ou show ma be worth partial credit. a. [1 point] f0)h ) Answer: f0) = 3 and h ) =. So, f0)h ) = 3)) = 6. b. [1 point] 3fg )) 6 g ) = + ) =. So 3fg )) = 3f) = 3 ) = 1. d. [1 point] g 1 ) g is invertible and piecewisedefined, so we must figure out which part of the piecewise formula has an output of. is defined for 3, so this piece gives outputs in the interval [8, ). So, we must solve + =. The solution is = 0. So g 1 ) = 0. Answer: 0 e. [1 point] gg 1)) Answer: 1 c. [1 point] f 1 h1) ) h1) = so h1) = =. Hence f 1 h1) ) = f 1 ) = 3. g 1) = + 1) = 3. So gg 1)) = g3) = 3 = 8. Answer: 8 f. [1 point] k 1) if k) = 1 3 h3) k 1) = 1 3 h3 1)) = 1 3 h 3) = 1 3 ) = 3 Answer: 3 Answer: /3 g. [1 point] Find the average rate of change of h) between = 1 and =. This average rate of change is given b h) h 1) 1) = ) 5 = 8 5. Answer: 8/5
Math 105 / Final December 17, 013) page 3. [10 points] Let Gv) be the number of minutes it takes Goober the gorilla to eat a meal consisting of v pounds of vegetation. a. [ points] Suppose b and n are positive constants. Give a practical interpretation of the equation G 1 b) = n in the contet of this problem. Use a complete sentence and include units. It takes Goober b minutes to eat a meal consisting of n pounds of vegetation. b. [ points] Suppose that there are positive constants c and d so that a formula for Gv) is given b Gv) = cv d. If G) = 9 and G3) = 18, find the eact values of the constants c and d. G) = 9 and G3) = 18, so c) d = 9 and c3) d = 18. Taking ratios, we find c3) d c) d = 18 Using logarithms: ln1.5 d ) = ln) 9 3 d = d ln1.5) = ln) d ) d 3 = d = ln1.5) ln) 1.5 d = Substituting into the equation c) d = 9, we find c) ln1.5)/ ln) = 9 so c = 9 ln1.5)/ ln). Answers: c = 9 ln1.5)/ ln) and d = ln1.5) ln) c. [ points] Suppose that the number of minutes it takes Goober s friend Toober to eat a meal consisting of v pounds of vegetation is m = T v), which is given b the formula lnv + ) T v) = q + ln5) for some constant q. Find a formula for T 1 m). Show our work carefull. Note that our answer should be in eact form and be given in terms of m and q. To find T 1 m), we solve for v in the equation m = q + lnv + ) m = q + ln5) lnv + ) m q = ln5) lnv + ). ln5) Eponentiating: e ln5)m q) = v + e ln5)m q) = v ln5))m q) = lnv + ) Thus T 1 m) = e ln5)m q). Note that e ln5)m q) = e ln5)) m q = 5 m q so we can simplif to T 1 m) = 5 m q. Answer: T 1 m) = 5 m q or e ln5)m q) )
Math 105 / Final December 17, 013) page 5. [13 points] Severus Snake is slithering along the banks of a river. At noon, a scientist starts to track Severus s distance awa from the edge of the river. After a few minutes, the scientist realizes Severus s distance awa from the edge of the river is a sinusoidal function. Let Dt) be Severus s distance, in centimeters, awa from the edge of the river t seconds after noon. a. [5 points] At noon eactl, the scientist notes that Severus is 97 centimeters awa from the edge of the river, which is the farthest awa he ever gets. Three seconds after that, Severus s distance is 65 centimeters awa from the river, the closest he gets. Graph = Dt) for 0 t 1. Clearl label the aes and important points on our graph. Be ver careful with the shape and ke features of our graph.) cm) 97 6, 97) 65 3, 65) = Dt) 3 6 9 1 t sec) b. [6 points] Find the period, amplitude, equation of the midline, and a formula for the sinusoidal function Dt). Include units for the period and amplitude.) A maimum and succeeding minimum of a sinusoidal function occur half a period apart. In this case, the occur 3 seconds apart, so the period is 6 seconds. The midline is found b averaging the maimum and minimum values. This gives = 97+65 = 81. The amplitude is the distance between a maimum or minimum and the midline which is 97 81 = 16 centimeters. Because the maimum occurs at noon t = 0), the cosine function is a good candidate to use here. Let Dt) = A cosbt h)) + k. Then we have A = 16, B = π 6 = π 3, and k = 81. So, we π ) have Dt) = 16 cos 3 t h) + 81. Because the maimum occurs at t = 0, we do not need a ) πt horizontal or phase) shift. Thus, one possible formula for Dt) is Dt) = 16 cos + 81. 3 Period: Amplitude: 6 seconds 16 centimeters Midline: = 81 Formula: Dt) = 16 cos π 3 t) + 81 c. [ points] How far awa from the river is Severus 11 seconds after noon? Give our answer accurate to at least two decimal places. Severus is D11) centimeters awa ) from the river ) 11 seconds after noon. Using the 11π 1 formula from above, D11) = 16 cos + 81 = 16 + 81 = 8 + 81 = 89. So, Severus is 89 3 centimeters awa from the river 11 seconds after noon. Answer: 89 centimeters
Math 105 / Final December 17, 013) page 6 5. [5 points] Find a formula for one polnomial pz) that satisfies all of the following conditions: lim pz) = and lim pz) = z z The onl zeros of pz) are z =, z = 1, and z = 3. The point, 1) is on the graph of pz). The degree of pz) is at most 5. Show our work and reasoning carefull. You might find it helpful to first sketch a graph. There ma be more than one possible answer, but ou should give onl one answer. Because lim pz) = and lim pz) =, we know that the degree of pz) is z z even and the leading coefficient is negative. Because pz) has zeros at z =, z = 1, and z = 3, the degree of the polnomial must be at least 3. Since it has to be of even degree less than 5, the degree must be. Thus there must be eactl one double zero. Since the point, 1) is on the graph, we can see b sketching the graph that the double zero must be at z = 3. So, we have pz) = az + )z 1)z 3) for some negative constant a. Since p) = 1 we find 1 = p) = a + ) 1) 3) = a)1) 1) = a so a = 3. Thus, pz) = 3z + )z 1)z 3). Answer: pz) = 3z + )z 1)z 3) 6. [5 points] Find all solutions to the equation 5 tan + π ) 13 = 1 for between 0 and 5. Show our work carefull and give our answers) in eact form. 5 tan + π ) 13 = 1 so 5 tan + π ) = 5 and tan + π ) = 5. Using the inverse tangent function, we find that one solution to the equation is given b + π = arctan5) so = arctan5) π and = arctan5) π. Note that the period of 5 tan ) + π arctan5) π 13 is π/ and that the solution is not in the interval [0, 5]. Other solutions to the equation are obtained b adding integer multiples of the period π/ to arctan5) π arctan5) π. The resulting solutions in the interval [0, 5] are + π These can be simplified to arctan5), arctan5) π + π, arctan5) + π, and + 3π, and arctan5) π arctan5) + 3π. + 5π. Answer: = arctan5) + π, arctan5) + 3π, and arctan5) + 5π
Math 105 / Final December 17, 013) page 7 7. [8 points] Freckles and Comet are cats in the same household. Consider the functions F, C, and D which are defined as follows: F m) is the number of ounces of food that Freckles eats in month m. Cm) is the number of ounces of food that Comet eats in month m. Dq) is the cost of buing q cans of cat food at a time when there are no sale prices. Assume that D is invertible. For each of questions below, circle the one best answer from among the options provided. If none of the options are correct, circle none of these. Please note: To receive credit, ou must clearl circle our choices. Circle the entire answer. If there is an ambiguit in our answer, ou will not receive credit.) a. [1 point] What is the total number of ounces of food that Freckles and Comet eat in month m? Dm) + Cm) F m) + Cm) F Cm)) CF m)) none of these Freckles eats F m) ounces of food in month m. Comet eats Cm) ounces of food in month m. Combined, the eat F m) + Cm) ounces of food in month m. b. [ points] Suppose that there are 3 ounces of food per can. What is the total cost of the food Freckles eats in month? ) D) F ) 3D) D3F )) D none of these 3 3 If Freckles eats F ) ounces of food in month and there are 3 ounces ) of cat food per can, Freckles eats F ) 3 cans of cat food in month. The cost of this is D F ) 3. c. [1 point] Let Aq) be the average cost per can of buing q cans of cat food. Which of the following is a formula for Aq)? D 1 q) q Dq) F q) + Cq) Dq) q none of these The average cost per can of buing q cans of cat food is the cost of buing q cans of cat food, Dq), divided b the number of cans of cat food purchased, q. d. [1 point] When there are no sale prices, how man cans of cat food can be purchased at a time for $0? D0) D 1 0) 0D 1 1 q) none of these D0) If we want to spend $0 on cans of cat food, we want to solve for q in the equation Dq) = 0. Thus, we can bu D 1 0) cans of cat food for $0. e. [ points] Suppose that Comet eats at least twice as much food each month as Freckles eats. Which one of the following inequalities most accuratel describes this relationship? Cm) F m) Cm) F m) Cm) F m) Cm) F m) Twice as much food as Freckles eats in month m is F m). Comet eats Cm) ounces of food in month m, so the relationship is that Cm) F m) for all m. f. [1 point] If cat food goes on sale for 0% off its regular price, what is the cost of buing 0 cans of cat food at one time? 0.6D0) 1.D0) 0.D0) D8) none of these If cat food is at its regular price, the cost of buing 0 cans of cat food is D0). If we take 0% off the regular price, we have D0) 0.D0) = 0.6D0).
Math 105 / Final December 17, 013) page 8 8. [11 points] On the beaches of Meico, there is a population of pick snails that wait for special shells to wash up onto the shore. These snails can onl live in these particular shells, as the snails have become accustomed to the comfort in these shells. Suppose the number of hundreds of special shells on the beaches of Meico t ears after the beginning of 013 is ht) = t + 7)t 7) and the population, in hundreds, of pick snails t ears after the beginning of 013 is pt) = t ) 8t + 30). Throughout this problem, remember to clearl show our work and reasoning. a. [3 points] Find the leading term and an zeros of ht). If appropriate, write none in the answer blank provided. To find the zeros of ht), we set ht) = 0 and solve for t. Because t + 7 is alwas positive, ht) = 0 when t 7) = 0. So the zeros of ht) occur at t = 7/. To find the leading term, we take the product of the leading terms of its factors: leading term = t )t)t) = 16t Answers: Leading Term: 16t Zeros): t = 7/ b. [3 points] The number of shells per snail is Qt) = ht) pt). Find the equations of all vertical asmptotes V.A. ) and horizontal asmptotes H.A. ) of the graph of = Qt). If appropriate, write none in the answer blank provided. To find vertical asmptotes, since ht) and pt) have no common factors, we set pt) = 0 and solve for t. Because 8t + 30 is alwas positive, pt) = 0 onl when t ) = 0. So the onl zero of pt) is t =. So, since ht) 0, the onl vertical asmptote of Qt) is t =. To find horizontal asmptotes, we look at the long-run behavior of Qt). ht) lim t pt) = lim t + 7)t 7) t t ) 8t + 30) = lim 16t t 8t = ht) lim t pt) = lim t Thus, the horizontal asmptote of Qt) is =. t + 7)t 7) t ) 8t + 30) = lim 16t t 8t = Answers: V.A.: t = H.A.: =
Math 105 / Final December 17, 013) page 9 There is a competitive population of crabs that live on the same beaches. Suppose that there are 100 of these crabs at the beginning of 013, and that the population grows at a continuous annual rate of 35%. Let ct) be the population, in hundreds, of these crabs t ears after the beginning of 013. c. [ points] Find a formula for ct). If ct) = P e kt is the population, in hundreds, of these crabs t ears after the beginning of 013, then c0) = 1, so P = 1. With a continuous annual growth rate of 35%, we know that ct) = 1e.35t. Answer: ct) = 1e.35t d. [3 points] The crabs like the same special shells as the snails do. Write a formula for the ratio of the number of shells to the number of crabs t ears after the beginning of 013. Answer: ht) ct) = t + 7)t 7) 1e.35t In the long run, what happens to the ratio of the number of shells to the number of crabs? In other words, assuming the functions described in this problem continue to be accurate models, what happens to this ratio after man, man ears? You must clearl indicate our reasoning in order to receive an credit for this problem. The ratio of number of shells to the number of crabs t ears after the beginning of 013 is given b Rt) = ht) ct) To describe the ratio after man, man ears, we look at the long-run behavior. Because we are looking at ears after 013, we onl look at the limit of Rt) as t grows without bound, i.e. as t. Since eponential growth eventuall dominates polnomial growth, the limit is zero. That is ht) lim Rt) = lim t t ct) = lim t + 7)t 7) 16t t 1e.35t = lim t 1e.35t = 0 This means that in the long run, the ratio of the number of shells to the number of crabs approaches zero.
Math 105 / Final December 17, 013) page 10 9. [5 points] Note that throughout this problem, ou are not required to show our work. A portion of the graph of a sinusoidal function h) is shown below. = h) 6 6 a. [ points] Which, if an, of the figures below shows part of the graph of = 1 h)? Note that the scale is smaller than in the original graph above. Be sure to pa attention to the scale indicated on the aes. Option A) Option B) Option C) Circle our one final answer below. Onl the answer ou circle below will be graded.) First, we verticall compress the graph of h) b a factor of 1. Then, we reflect the resulting graph across the -ais. Option A Option B Option C none of these b. [3 points] Which, if an, of the figures below shows part of the graph of = h + )? Note that the scale is smaller than in the original graph above. Be sure to pa attention to the scale indicated on the aes. Option A) Option B) Option C) Circle our one final answer below. Onl the answer ou circle below will be graded.) Note that h + ) = h + 1)). So the graph of h + ) can be obtained from that of h) b first compressing horizontall b a factor of 1 and then shifting the resulting graph 1 unit to the left. Option A Option B Option C none of these
Math 105 / Final December 17, 013) page 11 10. [ points] Suppose that the number of acorns in Squish squirrel s nest is proportional to the cube of the number of squirrels currentl living there. If there are 113 acorns in his nest when there are two squirrels living there, how man acorns will there be in Squish s nest when there are four squirrels living there? Remember to show our work carefull. Let A be the number of acorns in Squish squirrel s nest. Let Q be the number of squirrels living there. Then we know that A = kq 3 where k is some constant. If there are 113 acorns in his nest when there are two squirrels living there, then 113 = k 3 ) so k = 113 8 and a general formula is A = 113 8 Q3. Therefore, when there are four squirrels living in the nest, A = 113 8 3 ) = 113 8 6) = 1138) = 90. So there are 90 acorns in Squish squirrel s nest when there are four squirrels living there. Answer: 90 acorns 11. [7 points] Wolfgang the wolf is on a 10-foot long leash that is tied to a post that is 0 feet west of a fence. Wolfgang θ Post Q 0 feet Fence Fence P0,0) Because he dislikes being on his leash, he stas 10 feet awa from the post at all times. a. [ points] Suppose we think of the origin at the point P as shown in the diagram and that the unit of measurement is feet so that the coordinates of the post are 0, 0). Find Wolfgang s coordinates when he is at the angle θ shown in the diagram. Your answer should be in terms of θ.) If the post were at the origin, his coordinates would be 10 cosθ), 10 sinθ)). Since the post is 0 feet to the left of the origin, his first coordinate will be 0 units less, i.e. 0+10 cosθ). This does not change the second coordinate. Answer: Wolfgang s coordinates are 0 + 10 cosθ), 10 sinθ) ). b. [3 points] Wolfgang starts walking counterclockwise from the point Q. The angle θ through which Wolfgang has walked is a function of the amount of time he has been walking. Let θ = zt) be the angle in radians) through which Wolfgang has walked after he has been walking for t minutes. Let At) be the distance Wolfgang has traveled along the circle in t minutes. Find a function ft) such that At) = fzt)). such that Using the arclength formula, At) = 10θ = 10zt). We want to find a function ft) fzt)) = At) i.e. so that fzt)) = 10zt). The above shows that if the function f is given an input of zt), the output is 10zt). Thus, f should be a function that takes its input and multiplies it input b 10. So ft) = 10t. Answer: ft) = 10t
Math 105 / Final December 17, 013) page 1 1. [10 points] Consider the functions f, g, and h defined as follows: f) = a + b g) = c d h) = w1 + r) for nonzero constants a, b, c, d, r, and w with r > 1. For each of the questions below, circle all the correct answers from among the choices provided, or circle none of these if appropriate. a. [ points] The graph of which functions) definitel has at least one horizontal intercept? f) g) h) none of these f) will alwas have a horizontal intercept since it is linear with nonzero slope. If d < 0, e.g. if g) = 10 1, then g) does not have a horizontal intercept. h) does not have a horizontal intercept because it is an eponential function and w is nonzero. b. [ points] The graph of which functions) definitel has at least one horizontal asmptote? f) g) h) none of these f) is a linear function with nonzero slope so does not have a horizontal asmptote. If d > 0, e.g. if g) = 10, then it does not have a horizontal asmptote. = h) is an eponential function so has horizontal asmptote = 0. c. [ points] Which functions) isare) definitel invertible? f) g) h) none of these The linear function with nonzero slope, f), and the eponential function, h) are definitel invertible. The pass the horizontal line test, for eample.) g) ma or ma not be invertible. For eample, if g) = 10, then it is not invertible. d. [ points] How man times could the graph of f) intersect the graph of h)? 0 1 3 more than A linear function can intersect an eponential function either 0, 1, or times. For eample, = + 1 and = 1 + 3) do not intersect, whereas = + 1 and = 1 + 3) intersect eactl once, and = + 5 and = 1 + 3) intersect eactl two times. The cannot intersect more than two times due to their long-run behavior.) e. [ points] Suppose the graph of h is concave up. Which of the following isare) definitel true? w > 0 w < 0 r > 0 r < 0 none of these An eponential function is concave up if and onl if its initial value is positive. That is, whether 1 < r < 0 or r > 0, if w > 0, then the graph of h will be concave up whereas if w < 0, the graph of h will be concave down.