Math Final Eam Review KEY. Use the graph of = f in Figure to answer the following. Approimate where necessar. a Evaluate f. f = 0 b Evaluate f0. f0 = 6 c Solve f = 0. =, =, =,or = 3 d Solve f = 7..5, 0.5, or 3.5 Figure e 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. f State an local maimums or local minimums. There is a local maimum of at about -.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 Over what intervals is the function increasing?,.5 0.5,.5 i Over what intervals is the function decreasing?.5,0.5.5, j Over what intervals is the function concave up? Concave up: 0.5,.5 k Over what intervals is the function concave down? Concave down:, 0.5.5, l Find the zeros of f. The zeros are, -,, 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, -,, and 3, the function will have +,+, 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++ 3. As the graph contains the point 0, 6, we know that f0 = 6. We will use this to find k: 6 = k0+0+0 0 3 = k Therefore a possible formula for this polnomial function is f = ++ 3.
. Let f = +. a Find f. = + + = + = = = = f = Therefore f =. This can be simplified to f = +. b Confirm the inverse b computing f f and f f. f f = f f = f + = = + + + + + + + + = ++ + = 5 5 = f f = f f = f = = + + + + + + + = + ++ = 5 5 = c State the domain and range of f and f. Domain of f: { } Range of f: { } Domain of f : { } Range of f : { } d Evaluate f0. f0 = e Evaluate f 0. f 0 = Instructor: A.E.Car Page of 0
f Solve f = 3. The solution is -7. + = 3 = 3+ = 3+6 7 = g Determine if f is even, odd, or neither from its formula. 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 = + = + and thus f f and f f, it holds that f is neither even or odd. h 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,0. The vertical intercept occurs where = 0 and is 0,. j Sketch a graph of = f in Figure. 8 8 Figure Instructor: A.E.Car Page 3 of 0
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 b f+ c f d f+3 e f++3 f f3 Figure 3. Graph of = a f = b f+ = + c f = d f+3 = +3 e f++3 = + +3 f f3 = 3 Figure. Complete Table 3 below using the given values in the table. If an value is undefined, write undefined. Table - 0 f 0 g 0 g f 0 g f 8 0 f + g 6 3 0 - f g und. g und. 0 und. - Instructor: A.E.Car Page of 0
5. Find a formula for the piecewise-defined function graphed in Figure 5 below. +5, f = 3, < <, 6. In Figure 6, graph the piecewise function defined b, < 0 f =, 0 < < +, -3 3-3 - -3-3 - 3-3 -3 Figure 5. Graph of = f Figure 6 7. The volume, V in cubic centimeters of a circular balloon of radius r in centimeters is given b V = fr = 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 f3. f3 = 3 π33 = 36π 3. The volume of a balloon with a radius of 3 centimeters is approimatel 3. cm 3. b Find and interpret g3. g3 = 3 = 6 A balloon that has air blown into it for 3 seconds will have a radius of 6 cm. c Find and interpret fg3. fg3 = f6 = 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 Page 5 of 0
d Find and interpret fgt. fgt = ft = 3 πt3 = 3 π 8t 3 = 3 3 πt3 This function represents the volume of the balloon that has had air blown into it for t seconds. e Eplain wh gfr in nonsense. The unit of the input of f is seconds. To input a variable whose unit is centimeters into the function f does not make sense. f Find and interpret r = f V. V = 3 πr3 3 π V = r3 3 3V π = r r = f V = 3 3V π The inverse function represents the radius of the balloon as a function of the volume. g Find and interpret f 0. f 0 = 3 3 0 π = 3 5 π.7 When the volume of the balloon is 0 cm 3, the radius is approimatel.7 cm. 8. Write the following using eponents. a log 6 = 3 3 = 6 b ln e = e / = e c log 0 = 00 0 = 00 9. Solve the following equations. Give the eact solution and then round accurate to two decimal places. Check our solutions for e through j. a 7 = 5 7 = 5 = log 7 5 0.83 The solution set is {log 7 5}. Equivalent solution sets are { } ln5 amd ln7 { } log5. log7 Instructor: A.E.Car Page 6 of 0
b e 5 = 0 { } ln0 The solution set is. 5 e 5 = 0 5 = ln0 = ln0 5 0.6 c 5e = 0 The solution set is {ln}. 5e = 0 e = = ln 0.69 d 3 = 7 5 One possible approach: 3 = 7 5 9 = 75 5 9 = 7 5 5 5 7 9 = 5 50 7 = 5 9 50 7 = 5 9 50 5 log = log 7 9 50 5 log = log 7 9 log 50 7 log 5 9 = 3.3 Instructor: A.E.Car Page 7 of 0
Another possible approach: 3 = 7 5 log 3 = log 7 5 log+log 3 = log7+log 5 log+log3 = log7+ log5 log+log3 = log7+log5 log5 log3 log5 = log7 log5 log log3 log5 = log7 log 5 log 7 log9 log5 = log log 5 9 7 log = log 5 5 log 9 5 7 = log 50 = log 7 50 log 9 5 3.3 { log 7 } { 50 log 50 } The solution set is log 7 9. Equivalentl, the solution set can be written as 5 log 5. Furthermore, if ou applied the natural logarithm function to each side, ou could obtain { 9 ln 50 } 7 ln 5 or { 9 ln 50 } 7 ln 5. 9 Instructor: A.E.Car Page 8 of 0
e log + = log + = + = + = 6 = 5 = 5 Check: log 5 + = log 6 = The solution set is { } 5. 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}. h log 5 6 = log 5 log log 3 = log log = log 3 3 = = 6 log 5 6 = log 5 log 5 6 = log 5 6 = +36 = 3+36 = 0 9 = 0 = 9 or = Check = 9: log 5 9 6 = log 5 3 = log 5 9 Check = : log 5 6 is undefined. The solution set is {9}. Instructor: A.E.Car Page 9 of 0
i log 3 = Check: log 3 = / = 3 = 3 log 9 3 = 9 = log 9 3 / = log 93 = = The solution set is {9}. j log = +log+ Check: Left-hand side: log 99 0 Right-hand side: The solution set is log = +log+ log log+ = log = + = log 00 0 0 = + 00 = + 00+ = 00+00 = 99 = 0 99 0 = 0.98 +log + 99 = +log 0 0 { 99 }. 0 = log 0 +log 0 = log 00 = log 00 0 0 Instructor: A.E.Car Page 0 of 0
0. Find the equation of an eponential function that passes through each pair of points: a, 3 and, We will find C and a for f = Ca. As f contains the points, 3 and,, we have 3 = Ca and = Ca. One method is to make a ratio in order to eliminate C: = Ca Ca 3 36 = a ±6 = a Since a must be positive, a = 6. Substituting to find C, we obtain C6 = 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 3 = Ca and substitute into = Ca. Substituting to find a, we obtain 3 = Ca 3 a = C = Ca = 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. b,8 and 5, We will find C and a for f = Ca. As f contains the points,8 and 5,, we have 8 = Ca and = Ca 5. Thus 8 = Ca5 Ca 6 = a3 = a Substituting to find C: C = 8 C = 8 6 = 08 The eponential function passing through these points is defined b f = 08. Instructor: A.E.Car Page of 0
. Mr. Oops invests $0,000 into an account. After ears, he has $,096. Let t be the number of ears after Mr. Oops investment began and let Q be the value of the investment at time t. a Write two ordered pairs of the form t, Q representing the information above. The two ordered pairs are 0,0000 and,096. b Assuming the investment decreases eponentiall, write an eponential function modeling the value of the investment, Q, at time t. Use the ordered pairs ou found in a to do so. We will find a function of the form Q = ft = Ca t. As the initial value is $0,000, we know that Ca 0 = 0000, and thus C = 0000. We will then solve Ca = 096 for a: 0000a = 096 a = 096 0000 a = 56 65 56 a = 65 = 5 = 0.8 An eponential function modeling this investment is Q = ft = 00000.8 t or Q = ft = 0000 5 t. c Assuming the investment decreases linearl, write a linear function modeling the value of the investment, Q, at time t. Use the ordered pairs ou found in a to do so. We will find a function of the form Q = gt = mt+b. As the initial value is $0,000, we know that 0000 = m0+b and thus b = 0000. We will find m using the slope formula: / m = 096 0000 0 = 590 = 76 Therefore a linear function modeling this investment is Q = gt = 76t+0000.. ThetemperatureofacupofteaafteritwasbrewedcanbemodeledbthefunctionT = ft = 00e 0.t +68, where t is the number of minutes since the tea was brewed and T is the temperature in degrees Fahrenheit. a Find and interpret f0. f0 = 00e 0.00 +68 = 68 The initial temperature of the tea is 68 o F. b Find and interpret f0. f0 = 00e 0.0 +68 = 00e +68 0.8 After 0 minutes, the temperature of the tea is approimatel 0.8 o F. Instructor: A.E.Car Page of 0
c Find and interpret f T. We will solve T = 00e 0.t +68 for t: T = 00e 0.t +68 T 68 = 00e 0.t T 68 = e 0.t 00 T 68 ln = 0.t 00 T 68 0ln = t 00 T 68 Therefore the inverse function is t = f T = 0ln. This function gives the time after 00 the tea was brewed as a function of the tea s temperature. d Find and interpret f 80. 80 68 f 80 = 0ln 00 3 = 0ln 5. It takes approimatel. minutes for the tea to reach 80 o F. e Graph the function f in our calculator. What is the horizontal asmptote? The horizontal asmptote is T = 68. This happens to be the room temperature. 3. Tom and Jerr make separate investments at the same time. Their respective investments can be modeled b the functions T = ft = 5000.065 t and J = gt = 500 + 0.065 t where t is the number of ears since each investment began and T and J are their respective investment values in dollars. a Who has the higher effective rate of interest? To find the effective rate of interest for each Jerr, we will use the formula r e = r e = + 0.065 0.067 Therefore Jerr s effective rate of interest is about 6.7%. + r n n : The effective rate of interest of Tom s investment is 6.5%. We can appl the above formula if necessar, which will give r e = +0.065 = 0.065 Therefore Jerr has the higher effective rate of interest. 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. Instructor: A.E.Car Page 3 of 0
c What will the values of their respective investments be after 5 ears? Tom s investment after 5 ears: f5 = 5000.065 5 6850.3 Jerr s investment after 5 ears: g5 = 500 + 0.065 = 6.68 5 After 5 ears, Tom s investment will be worth $6850.3 and Jerr s investment will be worth $6.68. d 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 + 0.065 t = + 0.065 t ln = ln + 0.065 t ln = tln + 0.065 ln ln + 0.065 = t t 0.69 Jerr s investment will double in approimatel 0.69 ears. e Howlongwillittakefortheirinvestmentstobeworththesameamount? Howmuchwilltheirrespective investments be worth at this time? Use our graphing calculator to answer these. Tofindthesevalues, wewillfindtheintersectionpointofftandgt. Graphing = 5000.065 and = 500, + 0.065 the intersection point is 56.96,806.9. This means that in about 56.96 ears each investment will be worth about $80, 6.9.. The percentage of carbon, Q, remaining in a fossil t ears since deca began can be modeled b the function Q = ft = 00e 0.000t a If a piece of cloth is thought to be 750 ears old. What percentage of carbon is epected to remain in this sample? We will evaluate f750: f750 = 00e 0.000750 9. The cloth should have about 9.% of its carbon remaining. b If a fossilized leaf contains 70% of its original carbon, how old is the fossil? We will solve ft = 70: The fossil is about 876. ears old. 70 = 00e 0.000t 0.7 = e 0.000t ln0.7 = 0.000t ln0.7 0.000 = t t 876. Instructor: A.E.Car Page of 0
5. 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,. 6 8 6 8 a b Figure 7 Solution to Figure 7a Zeros The zero at,0 has even multiplicit, so the function has + 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,. 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+ 5. We will use f0 = to find k: = k0+ 0 5 5 = k A possible formula for this polnomial function is f = 5 + 5. Solution to Figure 7b Zeros Four of the zeros are,,0, and. None of these repeat as the graph goes straight through the horizontal ais at each zero. Thus the factors +,+,, 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 As, f This is an odd-degree polnomial function with a negative leading coefficient. Instructor: A.E.Car Page 5 of 0
A possible formula for this function is of the form f = k++ 5 3. We will use f =. to find k:. = k++ 5 3. = 0k 0.0 = k A possible formula for this polnomial function is f = 0.0++ 5 3 or f = 00 ++ 53. 6. 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 6 8 Figure 8 b g = + + Zeros The zero at, 0 does not repeat. Thus the function will go straight through the horizontal ais at this point. The zero at,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 Instructor: A.E.Car Page 6 of 0
This is a fourth-degree polnomial function with a positive leading coefficient. As, f As, f The graph is shown in Figure 9. 8-3 - 3 - Figure 9 7. Find possible equations for each rational function in Figure 0 below. List the zeros and their multiplicit, an vertical asmptotes, an horizontal asmptotes, and the vertical intercept. 8 8 6 6,.6 6 8 6 8 a b Figure 0 Solution to Figure 0a 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 =. As the function approaches positive infinit on one side of the asmptote and negative infinit on the other, the factor appears in the denominator and either does not repeat or repeats an odd number of times. We will assume it does not repeat. Instructor: A.E.Car Page 7 of 0
Horizontal Asmptote/ Long-run Behavior The horizontal asmptote is =. Therefore the ratio of leading terms of this rational function is. A possible formula for this function is of the form f = k+. We will use f0 = to find k. This should confirm the long-run behavior. = k0+ 0 = k A possible formula for this polnomial function is f = + or f = +. Solution to Figure 0b Zeros At the point 0,0, the graph of the function flattens or squishes. Thus the factor appears in the numerator and repeats an odd number of times. We will assume it appears three times. Vertical Intercept The vertical intercept is 0, 0. Vertical Asmptotes For the vertical asmptote = 3, the function approaches positive infinit on one side of the asmptote and negative infinit on the other. Thus the factor +3 appears in the denominator once or an odd number of times. We will assume it appears once. For the vertical asmptote = 3, the function approaches positive infinit from both sides. Thus the factor 3 will appear in the denominator an even number of times. We will assume it appears twice. Horizontal Asmptote/ Long-run Behavior The horizontal asmptote is =. Therefore the ratio of leading terms of this rational function is. k 3 A possible formula for this function has the form f = +3 3. We will use f =.6 to find k. This should confirm the long-run behavior..6 =.6 = k 8 5 = k A possible formula for this polnomial function is f = k 3 +3 3 3 +3 3. 8. Sketch each rational function below. List the zeros and their multiplicit, an vertical asmptotes, an horizontal asmptotes, and the vertical intercept. a f = 3 +3+ We begin b factoring and simplifing this function as follows: f = 3 +3+ = 3+ ++ = 3 +, Thus the graph of f will be identical to that of the function, g = 3, but will have a hole at + =. Instructor: A.E.Car Page 8 of 0
Zeros The zero of this function is 3. As the factor +3 is in the numerator and does not repeat, the graph of the function will go straight through the horizontal ais at the point 3,0. Vertical Intercept The vertical intercept is 0, 3. Vertical Asmptotes Theverticalasmptoteis = sincethefactor isinthedenominator. Sincethisfactor does not repeat, the function will approach positive infinit on one side of the asmptote and negative infinit on the other. Horizontal Asmptote/ Long-run Behavior The ratio of leading terms is, and thus the horizontal asmptote is =. The last step in determining a graph for this function is to determine if the graph lies above or below the -ais on each side of its asmptote and zeros. Since the vertical asmptote occurs at = and the zero occurs at = 3, we will choose test points in the intervals,,,3 and 3,. The inputs chosen are = 3, = and = : Table Interval,, 3 3, Input chosen 3 Function value f 3 = 6 f = 3 f = 6 Location of graph below -ais above -ais below -ais Point on graph 3, 6, 3, 6 A graph of the function is shown in Figure. 8 6 6 8 Figure b g = + 6 First, it will be helpful to factor g: g = + +. Zeros The zero of this function is. As the factor + is in the numerator and does not repeat, the graph of the function will go straight through the horizontal ais at the point,0. Vertical Intercept The vertical intercept is 0,. Vertical Asmptotes Instructor: A.E.Car Page 9 of 0
The vertical asmptote are = and = since the factors and + appear in the denominator. Since these factor do not repeat, the function will approach positive infinit on one side of each asmptote and negative infinit on the other. Horizontal Asmptote/ Long-run Behavior The ratio of leading terms is, and thus the function will behave like g = in the long run. This means the function has the horizontal asmptote = 0. The last step in determining a graph for this function is to determine if the graph approaches positive or negative infinit from each side of its asmptotes. One option is to estimate the function s value near the asmptote. The other is to choose a specific input on either side of the asmptote and determine the function s value there. We will choose to evaluate f 5,f 3,f3 and f5: Table 3 Interval,,,, Input chosen 5 3 3 5 Function value f 5 = 3 f 3 = 7 f3 = 0 7 f5 = 9 Location of graph below -ais above -ais below -ais above -ais Point on graph 5, 3 3, 7 3, 0 7 5, 9 A graph of the function is shown in Figure. 8 6 6 8 Figure Instructor: A.E.Car Page 0 of 0