Math 111 Final Eam Review KEY 1. Use the graph of = f in Figure 1 to answer the following. Approimate where necessar. a b Evaluate f 1. f 1 = 0 Evaluate f0. f0 = 6 c Solve f = 0. =, = 1, =, or = 3 Solution Set: {, 1,, 3} d Solve f = 7..5, 0.5, or 3.5 Solution Set: {.5, 0.5, 3.5} e f Determine if f is even, odd, or neither from its graph. Neither. The function is not smmetric about the -ais and is therefore not even. The function is not smmetric about the origin and is therefore not odd. State an local maimums or local minimums. There is a local maimum of at about -1.5 and a local maimum of at about.5. There is a local minimum of -7 at about 0.5. g State the domain and range of f. Domain:, Range:, ] h i j k Over what intervals is the function increasing?, 1.5 0.5,.5 Over what intervals is the function decreasing? 1.5, 0.5.5, Over what intervals is the function concave up? Concave up: 0.5, 1.5 Over what intervals is the function concave down? Concave down:, 0.5 1.5, Figure 1 6 6 l Find the zeros of f. The zeros are -, -1,, and 3. m Find a possible formula for this polnomial function. To find a formula, we note that since the zeros of the function are -, -1,, and 3, the function will have +, + 1, and 3 as factors. Since the graph of the function goes straight through each of its zeros, none of these factors repeat. Therefore a possible function is f = k + + 1 3. As the graph contains the point 0, 6, we know that f0 = 6. We will use this to find k: 6 = k0 + 0 + 10 0 3 1 = k Therefore a possible formula for this polnomial function is f = 1 + + 1 3. 6 6 1
. Let f = 1 +. a Find f 1. One Method: = 1 + Finding the inverse : = 1 + + = 1 + = 1 + = 1 = 1 = 1 1 = = 1 A Second Method: = 1 + + = 1 + = 1 = 1 = 1 = 1 f 1 = 1 Therefore f 1 = 1. This can be simplified to f 1 = + 1. b Confirm the inverse b computing f 1 f and f f 1. f 1 f = f 1 f 1 = f 1 + 1 + + 1 = 1 + 1 + + 1 = 1 + + + 1 + 1 + = 1 + = 5 5 = f f 1 = f f 1 = f = = = + 1 +1 +1 +1 +1 1 + 1 + + 1 1 + 1 + = 5 5 = Instructor: A.E.Car Math 111 Final Eam Review Ke Page of 19
c State the domain and range of f and f 1. Domain of f: { } Range of f: { } Domain of f 1 : { } Range of f 1 : { } d Evaluate f0. f0 = 1 e Evaluate f 1 0. f 1 0 = 1 f Solve f = 3. g The solution is -7. 1 + = 3 1 = 3 + 1 = 3 + 6 7 = Algebraicall determine if f is even, odd, or neither. To show that f is even, it must be shown that f = f. To show that f is odd, it must be shown that f = f. As f = 1 + = 1 + h and thus f f and f f, it holds that f is neither even or odd. State an horizontal and vertical asmptotes. There is a horizontal asmptote of = since the ratio of leading terms is. There is a vertical asmptote of = since the factor + appears in the denominator. i State an horizontal and vertical intercepts of f. The horizontal intercept occurs where f = 0 and is 1, 0. The vertical intercept occurs where = 0 and is 0, 1. j Sketch a graph of = f in Figure. The vertical asmptote is =. The horizontal asmptote is =. The -intercept is 0.5, 0. The onl zero is 0.5. The -intercept is 0, 0.5. Table 1. Behavior of R Zeros/ VA Interval,, 0.5 0.5, -3 0 3 R R 3 = 7 R0 = 0.5 R3 = 1 +/ + above - below + above Point 3, 7 0, 0.5 3, 1 Figure = - 8 6 = 8 6 6 8 6 8 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 3 of 19
3. Let f =. For each of the following, sketch a graph of the tranformation in Figure and write the simplified formula for the function. Describe the order of transformations, being as specific as possible and listing them in an appropriate order. a = f Reflect across the -ais. b = f + 1 Shift 1 unit left. c = f Stretch verticall b a factor of. d e f = f + 3 Shift up 3 units. = f + 1 + 3 Shift left 1 unit, stretch verticall b a factor of, shift up 3 units. = f3 Compress horizontall b a factor of 1 3. Figure 3. Graph of = a = b = + 1 c = d = + 3 e = + 1 + 3 f = 3 Figure. Complete Table below using the given values in the table. If an value is undefined, write undefined. Table - -1 0 1 f 1 0 1 g 0 - - g f - - 0 - - g f 8 0 - -8 f + g 6 3 0-1 - f g 1 1 und. 1 1 g 1 1 und. 0 und. -1 Instructor: A.E.Car Math 111 Final Eam Review Ke Page of 19
5. Find a formula for the piecewise-defined function graphed in Figure 5 below. + 5, 1 f =, 1 < < 1, 6. In Figure 6, graph the piecewise function defined b, 3 < 0 f =, 0 < < 1 1 +, 1 Figure 5. Graph of = f Figure 6. Graph of = f 7. The volume, V r in cubic centimeters of a circular balloon of radius r in centimeters is given b V r = 3 πr3. As someone blows air into the balloon, the radius of the balloon as a function of time t in seconds is given b r = gt = t. a Find and interpret V 3. V 3 = 3 π33 = 36π 113.1 b The volume of a balloon with a radius of 3 centimeters is approimatel 113.1 cm 3. Find and interpret g3. g3 = 3 = 6 c A balloon that has air blown into it for 3 seconds will have a radius of 6 cm. Find and interpret V g3. V g3 = V 6 = 3 π63 = 88π 90.8 The volume of a balloon that has been blown into for 3 seconds is about 90.8 cm 3. Instructor: A.E.Car Math 111 Final Eam Review Ke Page 5 of 19
d Find and interpret V gt. V gt = V t = 3 πt3 = 3 π 8t 3 = 3 3 πt3 e This function represents the volume of the balloon that has had air blown into it for t seconds. Eplain wh gv r in nonsense. The unit of the input of V is seconds. To input a variable whose unit is centimeters into the function V does not make sense. 8. Find the following for the functions f, g, and h defined b a f g f = 3 + 1 g = 3 + 1 h = 5 f g = f 3 + 1 = f 13 = 313 + 1 = 0 = 1 0 b h f 1 h f 1 = h f1 = h 31 + 1 1 = h 1 = 5 = c h + g 1 h + g 1 = h1 + g1 = 1 5 + 31 + 1 = 3 + = 1 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 6 of 19
d g g 0 g g 0 = g g0 = g 30 + 1 = g1 = 31 + 1 = e f g 0 f g 0 = f0 g0 = 30 + 1 30 + 1 = 1 f g g g g = g g = 3 + 1 3 + 1 = 9 + 6 + 1 g f g f g = f g = f 3 + 1 = 3 3 + 1 + 1 = 9 + 3 + 1 = 9 + h g h g h = g h = g 5 = 3 5 + 1 = 3 0 + 5 + 1 = 1 60 + 75 + 1 = 1 60 + 76 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 7 of 19
i h h h h = h h = h 5 = 5 5 = 10 5 = 15 9. Find the algebraic rule or formula of an eponential function that passes through each pair of points: a 1, 1 3 and 1, 1 We will find C and a for f = Ca. As f contains the points 1, 1 1 3 and 1, 1, we have 3 = Ca 1 and 1 = Ca 1. One method is to make a ratio in order to eliminate C: = Ca Ca 1 1 1 3 36 = a ±6 = a Since a must be positive, a = 6. Substituting to find C, we obtain C6 1 = 1 C = The eponential function passing through these points is defined b f = 6. An alternate method is to solve one equation for C and substitute. We will solve 1 3 = Ca 1 and substitute into 1 = Ca 1. Substituting to find a, we obtain 1 3 = Ca 1 1 3 a = C 1 = Ca 1 1 1 = a a 3 36 = a ±6 = a Since a must be positive, a = 6. The eponential function passing through these points is defined b f = 6. Instructor: A.E.Car Math 111 Final Eam Review Ke Page 8 of 19
b, 18 and 5, We will find C and a for f = Ca. As f contains the points, 18 and 5,, we have 18 = Ca and = Ca 5. Thus Substituting to find C: C 18 = Ca5 Ca 1 6 = a3 1 = a 1 = 18 C = 18 16 = 08 The eponential function passing through these points is defined b f = 08 1. 10. Write the following equations using eponents. a log 6 = 3 3 = 6 b ln e = 1 e 1/ = e 1 c log 10 = 100 10 = 1 100 11. Solve the following equations. Give the eact solution and then round accuratel to two decimal places. Clearl state each solution set. a 7 1 = 7 1 = 7 = 5 = log 7 5 0.83 The solution set is {log 7 5}. Equivalent solution sets are b e 5 = 10 { } ln5 amd ln7 { } log5. log7 { } ln10 The solution set is. 5 e 5 = 10 5 = ln10 = ln10 5 0.6 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 9 of 19
c 5e = 10 5e = 10 e = = ln 0.69 The solution set is {ln}. d 3 = 9 + 3 = 9 + 3 = 3 + 3 = 3 + = + = + 8 8 = 0 + = 0 = 0 or + = 0 = or = Both solutions check. The solution set is {, }. e log + 1 = log + 1 = Check: + 1 = + 1 = 16 = 15 = 15 log 15 + 1 = log 16 = The solution set is { } 15. Instructor: A.E.Car Math 111 Final Eam Review Ke Page 10 of 19
f log + log 3 = log Check: log 3 + log 3 = log The solution set is { 3}. g log log 3 = log log + log 3 = log log 3 = log 3 = = 3 Check: log 6 log 3 = log The solution set is {6}. log log 3 = log log = log 3 3 = = 6 h log 5 6 = log 5 log 5 6 = log 5 log 5 6 = log 5 6 = 1 + 36 = 13 + 36 = 0 9 = 0 = 9 or = Check = 9: Check = : log 5 6 is undefined. log 5 9 6 = log 5 3 = log 5 9 The solution set is {9}. i log 3 = 1 log 3 = 1 1/ = 3 = 3 = 9 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 11 of 19
Check: log 9 3 = log 9 3 1/ = 1 log 93 = 1 1 = 1 The solution set is {9}. j log1 = + log1 + Check: Left-hand side: log 1 99 Right-hand side: The solution set is log1 = + log1 + log1 log1 + = 1 log = 1 + 101 = log 00 101 10 = 1 1 + 100 = 1 1 + 1001 + = 1 100 + 100 = 1 99 = 101 99 101 = 0.98 + log 1 + 99 = + log 101 101 { 99 }. 101 = log 10 + log 101 = log 100 = log 00 101 101 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 1 of 19
1. The temperature of a cup of tea after it was brewed can be modeled b the function T t = 100e 0.1t + 68, where t is the number of minutes since the tea was brewed and T t is the temperature in degrees Fahrenheit at time t. a Find and interpret f0. b f0 = 100e 0.010 + 68 = 168 The initial temperature of the tea is 168 o F. Find and interpret f10. f10 = 100e 0.110 + 68 = 100e 1 + 68 10.8 After 10 minutes, the temperature of the tea is approimatel 10.8 o F. c Solve and interpret T t = 80. T t = 80 100e 0.1t + 68 = 80 100e 0.1t = 1 e 0.1t = 1 100 d e 0.1t = 3 5 3 0.1t = ln 5 t = ln 3 5 0.1 3 t = 10 ln 5 t 1. It takes approimatel 1. minutes for the tea to reach 80 o F. Graph = T t in our calculator. What is the horizontal asmptote? The horizontal asmptote is = 68. This happens to be the room temperature. Instructor: A.E.Car Math 111 Final Eam Review Ke Page 13 of 19
13. Tom and Jerr make separate investments at the same time. Their respective investments can be modeled b the functions T = ft = 50001.065 t and J = gt = 500 1 + 0.065 1t 1 where t is the number of ears since each investment began and T and J are their respective investment values in dollars. a b How much does Tom invest initiall? How much does Jerr invest initiall? Tom s initial investment was $5, 000. Jerr s initial investment was $, 500. What will the values of their respective investments be after 5 ears? Tom s investment after 5 ears: f5 = 50001.065 5 6850.3 Jerr s investment after 5 ears: g5 = 500 1 + 0.065 1 = 6.68 15 After 5 ears, Tom s investment will be worth $6850.3 and Jerr s investment will be worth $6.68. c How long will it take for Jerr s investment to double? As Jerr s initial investment is $, 500, we will solve gt = 9000: 9000 = 500 1 + 0.065 1t 1 = 1 + 0.065 1t 1 ln = ln 1 + 0.065 1t 1 ln = 1t ln 1 + 0.065 1 ln 1 ln 1 + 0.065 = t 1 t 10.69 Jerr s investment will double in approimatel 10.69 ears. d How long will it take for their investments to be worth the same amount? How much will their respective investments be worth at this time? Solve this problem graphicall using our calculator. To find these values, we will find the intersection point of ft and gt. Graphing 1 = 50001.065 and = 500 1 + 0.065 1, 1 the intersection point is 56.96, 180611.9. This means that in about 56.96 ears each investment will be worth about $180, 611.9. Instructor: A.E.Car Math 111 Final Eam Review Ke Page 1 of 19
1. The percentage of carbon 1, Q, remaining in a fossil t ears since deca began can be modeled b the function Q = ft = 100e 0.0001t a If a piece of cloth is thought to be 750 ears old. What percentage of carbon 1 is epected to remain in this sample? We will evaluate f750: b f750 = 100e 0.0001750 91.1 The cloth should have about 91.1% of its carbon 1 remaining. If a fossilized leaf contains 70% of its original carbon 1, how old is the fossil? We will solve ft = 70: The fossil is about 876.1 ears old. 70 = 100e 0.0001t 0.7 = e 0.0001t ln0.7 = 0.0001t ln0.7 0.0001 = t t 876.1 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 15 of 19
15. Find possible equations for each polnomial function in Figure 7. List the zeros and their multiplicit, the vertical intercept, and the long-run behavior. 8 8 6 6,. 8 6 6 8 8 6 6 8 6 6 8 8 a b Solution to Figure 7a Figure 7 Zeros The zero at 1, 0 has even multiplicit, so the function has + 1 as a factor. The zero at 5, 0 does not repeat, so the function has 5 as a factor. Vertical Intercept The vertical intercept is 0, 1. Long-run Behavior As, f As, f This is an odd-degree polnomial function with a positive leading coefficient. A possible formula for this function is of the form f = k + 1 5. We will use f0 = 1 to find k: 1 = k0 + 1 0 5 1 5 = k A possible formula for this polnomial function is f = 1 5 + 1 5. Solution to Figure 7b Zeros Four of the zeros are, 1, 0, and. None of these repeat as the graph goes straight through the horizontal ais at each zero. Thus the factors +, + 1,, and each appear once in the function. The zero at 5, 0 repeats an odd number of times since the graph of the function flattens or squishes there. Thus the factor 5 appears in the function an odd number of times. We will assume it appears three times. Vertical Intercept The vertical intercept is 0, 0. Long-run Behavior As, f Instructor: A.E.Car Math 111 Final Eam Review Ke Page 16 of 19
As, f This is an odd-degree polnomial function with a negative leading coefficient. A possible formula for this function is of the form f = k + + 1 5 3. We will use f =. to find k:. = k + + 1 5 3. = 0k 0.01 = k A possible formula for this polnomial function is f = 0.01 + + 1 5 3 or f = 1 100 + + 1 53. 16. Sketch a graph of = f for each polnomial function below. Also list the zeros and their multiplicit, the vertical intercept, and the long-run behavior. a f = + Zeros The zero at 0, 0 repeats twice. Thus the function will bounce at this point. The zero at, 0 does not repeat. Thus the function will go straight through the horizontal ais at this point. Vertical Intercept The vertical intercept is 0, 0. Long-run Behavior This is a third-degree polnomial function with a negative leading coefficient. As, f As, f The graph is shown in Figure 8. 8 6 8 6 6 8 6 8 Figure 8 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 17 of 19
b g = + 1 + Zeros The zero at, 0 does not repeat. Thus the function will go straight through the horizontal ais at this point. The zero at 1, 0 repeats twice. Thus the function will bounce at this point. The zero at, 0 does not repeat. Thus the function will go straight through the horizontal ais at this point. Vertical Intercept The vertical intercept is 0,. Long-run Behavior This is a fourth-degree polnomial function with a positive leading coefficient. As, f As, f The graph is shown in Figure 9. 1 1 10 8 6 3 1 1 3 6 8 10 1 1 Figure 9 Instructor: A.E.Car Math 111 Final Eam Review Ke Page 18 of 19
17. Find possible equations for each rational function in Figure 10 below. List the zeros and their multiplicit, an vertical asmptotes, an horizontal asmptotes, and the vertical intercept. 8 6 = 8 6 6 8 6 8 a = 1 Figure 10 Solution to Figure 10a Zeros At, 0, the graph of the function goes straight through the horizontal ais. Thus the factor + appears in the numerator and does not repeat. Vertical Intercept The vertical intercept is 0,. Vertical Asmptotes The vertical asmptote is = 1. As the function approaches positive infinit on one side of the asmptote and negative infinit on the other, the factor 1 appears in the denominator and either does not repeat or repeats an odd number of times. We will assume it does not repeat. Horizontal Asmptote/ Long-run Behavior The horizontal asmptote is =. Therefore the ratio of leading terms of this rational function is. k + A possible formula for this function is of the form f =. We will use f0 = to find k. This 1 should confirm the long-run behavior. k0 + = 0 1 = k A possible formula for this polnomial function is f = + 1 or f = + 1. Instructor: A.E.Car Math 111 Final Eam Review Ke Page 19 of 19