Math 111 Final Exam Review KEY

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1 Math 111 Final Eam Review KEY 1. Use the graph of y = f in Figure 1 to answer the following. Approimate where necessary. a Evaluate f 1. f 1 = 0 b 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 Determine if f is even, odd, or neither from its graph. Neither. The function is not symmetric about the y-ais and is therefore not even. The function is not symmetric about the origin and is therefore not odd. f State any 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 Over what intervals is the function increasing?, ,.5 i Over what intervals is the function decreasing? 1.5,0.5.5, j Over what intervals is the function concave up? Concave up: 0.5, 1.5 k Over what intervals is the function concave down? Concave down:, , Figure 1 y l Find the zeros of f. The zeros are -, -1,, and 3. m Find a possible formula for this polynomial 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 As the graph contains the point 0, 6, we know that f0 = 6. We will use this to find k: 6 = k = k Therefore a possible formula for this polynomial function is f =

2 . Let f = 6 +. a Find f 1. y = 6 + Finding the inverse : = y 6 y + y + = y 6 y + = y 6 y + = y 6 y = y 6 y y = y 6 y y y = 6 y = 6 y = 6 Therefore f 1 = 6. This can be simplified to f 1 = +6. b Confirm the inverse by computing f 1 f and f f 1. f 1 f = f 1 f 6 = f = = = = = 1 1 = f f 1 = f f 1 = f = = = = = 1 1 = Instructor: A.E.Cary Math 111 Final Eam Review Key Page of

3 c State the domain and range of f and f 1. Domain of f: { } Range of f: {y y } Domain of f 1 : { } Range of f 1 : {y y } d Evaluate f0. f0 = 3 e Solve f = 3. The solution is -18. Solution Set: { 18}. 6 + = 3 6 = 3+ 6 = = f Algebraically 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 = 6 + = 6 + f = 6 + = +6 + and thus f f and f f, it holds that f is neither even or odd. g State any horizontal and vertical asymptotes of f. There is a horizontal asymptote of y = since the ratio of leading terms is. There is a vertical asymptote of = since the factor + appears in the denominator. h State any horizontal and vertical intercepts of f. The horizontal intercept occurs where f = 0 and is 3,0. The vertical intercept occurs where = 0 and is 0, 3. i Sketch a graph of y = f in Figure. The vertical asymptote is =. The horizontal asymptote is y =. The -intercept is 3,0. The only zero is 3. The y-intercept is 0, 1.5. Table 1. Behavior of R Zeros/ VA Interval,, 3 3, R R 6 = 9 R = 5 R10 = 1 +/ + above - below + above Point 6, 9, 5 10, 1 Figure =- 8 6 y= Instructor: A.E.Cary Math 111 Final Eam Review Key Page 3 of

4 3. Let f =. For each of the following, sketch a graph of the transformation 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 y = f Reflect across the -ais. b y = f Stretch vertically by a factor of. c y = f+3 Shift up 3 units. d y = f+1+3 Stretch vertically by a factor of, shift left 1 unit, and then shift up 3 units. e y = f3 Compress horizontally by a factor of 1 3. f y = 3f + 1 Stretch vertically by a factor of 3, reflect across the y-ais, compress horizontally by a factor of 1/, shift left units, and shift down 1 unit. Figure 3. Graph of y = y y y y a y = b y = c y = +3 y y y d y = e y = 3 f y = Figure. For each function below, identify the original or basic function and eplain how the graph is a transformation of the graph of the original function. State all steps to this transformation in an appropriate order. a g = The parent function is f = 3. Steps to the transformation: 1 A = 1 3 : Reflect across the -ais and compress vertically by a factor of 1 3. B = : Compress horizontally by a factor of 1. 3 h = 7: Shift right 7 units. k = : Shift down units. Note: Ordering BhAK also appropriate. Instructor: A.E.Cary Math 111 Final Eam Review Key Page of

5 b g = 7ln++5 The parent function is f = ln. Steps to the transformation: 1 A = 7: Stretch vertically by a factor of 7. h = : Shift left units. 3 k = 5: Shift up 5 units. c g = The parent function is f =. Steps to the transformation: 1 B = 1 : Reflect across the y-ais and stretch horizontally by a factor of. h = 1: Shift left 1 unit. 3 k = 3: Shift down 5 units. 5. The point,16 is on the graph of y = f. Determine the point on the graph of... a y = f h = 5: Shift 5 units right.,16 1,16 k = 7: Shift 7 units down. 1,16 1,9 The point 1,9 is on the graph of y = f 5 7. b y = f 1 A = 1: Reflect across the -ais.,16, 16 B = : Compress horizontally by a factor of 1., 16 1, 16 The point 1, 16 is on the graph of y = f. c y = 3f 1 A = 3: Stretch vertically by a factor of 3.,16,8 B = 1: Reflect across the y-ais.,8,8 The point,8 is on the graph of y = 3f. d y = 1 8 f A = 1 8 : Reflect across the -ais and compress vertically by a factor of 1 8.,16, 16, B = : Compress horizontally by a factor of 1.,, 3 h = 3: Shift left 3 units., 5,. k = 5: Shift up 8 units. 5, 5,3. The point 5,3 is on the graph of y = 1 8 f+3+5. Instructor: A.E.Cary Math 111 Final Eam Review Key Page 5 of

6 6. Complete Table below using the given values in the table. If any value is undefined, write undefined. Table f g g f g f -3 und. 3 und. f + g f g f 1 1 und. und Find a formula for the piecewise-defined function graphed in Figure 5 below. +5, 1 f =, 1 < < 1, 8. In Figure 6, graph the piecewise function defined by, 3 < 0 f =, 0 < < 1 1 +, 1 Figure 5. Graph of y = f y Figure 6. Graph of y = f y 9. The volume, Vr in cubic centimeters of a circular balloon of radius r in centimeters is given by Vr = 3 πr3. As someone blows air into the balloon, the radius of the balloon as a function of time t in seconds is given by r = gt = t. a Find and interpret V3. V3 = 3 π33 = 36π The volume of a balloon with a radius of 3 centimeters is approimately cm 3. Instructor: A.E.Cary Math 111 Final Eam Review Key Page 6 of

7 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 Vg3. Vg3 = V6 = 3 π63 = 88π 90.8 The volume of a balloon that has been blown into for 3 seconds is about 90.8 cm 3. d Find and interpret Vgt. Vgt = Vt = 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 why gvr 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. 10. Find the following for the functions f, g, and h defined by f = 3+1 g = 3 +1 h = 5 a f g f g = f 3 +1 = f 13 = = 0 = 1 0 Instructor: A.E.Cary Math 111 Final Eam Review Key Page 7 of

8 b h f1 h f1 = hf1 = h = h 1 = 5 = c h+g1 h+g1 = h1+g1 = = 3+ = 1 d g g0 g g0 = g g0 = g = g1 = = e f g0 f g0 = f0 g0 = = 1 f g g g g = g g = = Instructor: A.E.Cary Math 111 Final Eam Review Key Page 8 of

9 g f g h g h f g = f g g h = g h = f 3 +1 = = = 9 + = g 5 = = = = i h h h h = h h = h 5 = 5 5 = 10 5 = 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, 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 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 by f = 6. Instructor: A.E.Cary Math 111 Final Eam Review Key Page 9 of

10 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 18 = Ca5 Ca 1 6 = a3 1 = a Substituting to find C: C 1 = 18 C = = 08 The eponential function passing through these points is defined by f = Find the eact values of the epressions below. a log 6 = 3 b ln e = 1 1 c log 10 = Solve the following equations. Give the eact solution and then round accurately to two decimal places. Clearly state each solution set. a 7 1 = 7 1 = 7 = 5 = log The solution set is {log 7 5}. Equivalent solution sets are b e 5 = 10 { } ln5 and ln7 { } log5. log7 { } ln10 The solution set is. 5 e 5 = 10 5 = ln10 = ln Instructor: A.E.Cary Math 111 Final Eam Review Key Page 10 of

11 c 5e = 10 The solution set is {ln}. 5e = 10 e = = ln 0.69 d 3 = = = = 3 + = + = +8 8 = 0 + = 0 = 0 or + = 0 = or = Both solutions check. The solution set is {,}. e 3 +1 = 6 One Approach: 3 +1 = 6 An Alternate Approach: 3 +1 = 6 log 3 +1 = log6 +1log3 = log6 log 3 6 = +1 log = log3+log3 = log6 log3 = log6 log3 = log6 log3 log3 or = log log9 } The solution set is or { log6 log3 log3 { } log log9 or {log 9} or log { } log = = log 36 1 = Instructor: A.E.Cary Math 111 Final Eam Review Key Page 11 of

12 f 5 7 = = 3 7 log5 7 = log3 7 log5+log7 = log3+log 7 log5+log7 = log3+ 7log log5+log7 = log3+log 7log log5+log7 log7 log3+7log = log3+log 7log log7 log3+7log log5 log3+7log = log log7 log5 log3+7log = log log7 log5 log3+7log = log log7 log log = 7 log 5 3 log = 7 { } log The solution set is 5 3. This may be simplified in an appropriate equivalent form, such as log { } { } 7 log 5 3 ln or 5 3. log 7 ln 7 g log +1 = log +1 = +1 = +1 = 16 = 15 = 15 Check: The solution set is { } 15. log = log 16 = Instructor: A.E.Cary Math 111 Final Eam Review Key Page 1 of

13 h log +log 3 = log Check: log 3 +log 3 = log The solution set is { 3}. i 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 j log 5 6 = log 5 log 5 6 = log 5 log 5 6 = log 5 6 = 1+36 = = 0 9 = 0 = 9 or = Check = 9: log = log 5 3 = log 5 9 Check = : log 5 6 is undefined. The solution set is {9}. k log 3 = 1 log 3 = 1 1/ = 3 = 3 = 9 Instructor: A.E.Cary Math 111 Final Eam Review Key Page 13 of

14 Check: log 9 3 = log 9 3 1/ = 1 log 93 = 1 1 = 1 The solution set is {9}. l log1 = +log1+ Check: Left-hand side: log 1 99 Right-hand side: The solution set is log1 = +log1+ log1 log1+ = 1 log = = log = = = = 1 99 = = log = +log { 99 }. 101 = log 10 +log 101 = log 100 = log Instructor: A.E.Cary Math 111 Final Eam Review Key Page 1 of

15 m log 6 ++log 6 +3 = 1 log 6 ++log 6 +3 = 1 log = = = = = 0 +6 = 0 or +1 = 0 = 6 or = 1 Check: As log 6 6++log is undefined, -6 is a false solution. The solution set is { 1}. log 6 1++log 6 1+3? = 1 log 6 3+log 6? = 1 log 6 3? = log 6 6 = 1 1. The percentage of carbon 1, Q, remaining in a fossil t years since decay began can be modeled by the function Q = ft = 100e t a If a piece of cloth is thought to be 750 years old. What percentage of carbon 1 is epected to remain in this sample? We will evaluate f750: f750 = 100e The cloth should have about 91.1% of its carbon 1 remaining. b 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 years old. 70 = 100e t 0.7 = e t ln0.7 = t ln = t t Instructor: A.E.Cary Math 111 Final Eam Review Key Page 15 of

16 15. The temperature of a cup of tea after it was brewed can be modeled by the function Tt = 100e 0.1t +68, where t is the number of minutes since the tea was brewed and Tt is the temperature in degrees Fahrenheit at time t. a Find and interpret T0. T0 = 100e = 168 The initial temperature of the tea is 168 o F. b Find and interpret T10. T10 = 100e = 100e After 10 minutes, the temperature of the tea is approimately 10.8 o F. c Solve and interpret Tt = 80. Tt = e 0.1t +68 = e 0.1t = 1 e 0.1t = e 0.1t = t = ln 5 t = ln t = 10ln 5 t 1. It takes approimately 1. minutes for the tea to reach 80 o F. d Graph y = Tt in your calculator. What is the horizontal asymptote? The horizontal asymptote is y = 68. This happens to be the room temperature. Instructor: A.E.Cary Math 111 Final Eam Review Key Page 16 of

17 16. Tom and Jerry make separate investments at the same time. Their respective investments can be modeled by the functions T = ft = t and J = gt = t 1 where t is the number of years since each investment began and T and J are their respective investment values in dollars. a How much does Tom invest initially? How much does Jerry invest initially? Tom s initial investment was $5, 000. Jerry s initial investment was $, 500. b What will the values of their respective investments be after 5 years? Tom s investment after 5 years: f5 = Jerry s investment after 5 years: g5 = = After 5 years, Tom s investment will be worth $ and Jerry s investment will be worth $6.68. c How long will it take for Jerry s investment to double? As Jerry s initial investment is $,500, we will solve gt = 9000: 9000 = t 1 = t 1 ln = ln t 1 ln = 1tln ln 1ln = t 1 t Jerry s investment will double in approimately years. Instructor: A.E.Cary Math 111 Final Eam Review Key Page 17 of

18 17. The acidity of a solution is measured using the ph scale. Lucky for you, it s a logarithmic scale! The ph depends on the concentration, or molarity, in moles/liter of hydrogen ions. The formula for ph is given by ph = log [ H +], where [H + ] is the molarity of hydrogen ions. A low ph is very acidic like lemons and a high ph is very basic like bleach. a Coffee has a hydrogen ion concentration of about [ H +] = Find the ph of coffee. The ph of coffee is about.959. ph = log b Pure water is called neutral and has a ph of 7. Find the hydrogen ion concentration of pure water. 7 = log [ H +] 7 = log [ H +] 10 7 = [ H +] The hydrogen ion concentration of pure water is moles per liter. c The hydrogen ion concentration of bleach is about [ H +] = Find the ph of bleach. The ph of bleach is about ph = log d The ph of pure Hydrochloric acid HCl is 0. Find the hydrogen ion concentration of HCl. 0 = log [ H +] 0 = log [ H +] 10 0 = [ H +] 1 = [ H +] The hydrogen ion concentration of pure HCl is 1 mole per liter. 18. Let g = a Use your calculator to find the zeros of the function. The zeros of g are -1, 0 and. b Use your calculator to estimate the interval where the graph is concave down. Round accurately to the nearest tenth. The graph is concave down over the approimate interval 0, 1.5. c Use your calculator to estimate the local maimum value and where it occurs. The local maimum value is approimately.08 and occurs at approimately 0.8. d Use your calculator to estimate the absolute minimum value and where it occurs. The absolute minimum value is approimately -1.6 and occurs at approimately Instructor: A.E.Cary Math 111 Final Eam Review Key Page 18 of

19 19. Find possible formulas for each polynomial function in Figure 7. Clearly list the zeros and their multiplicities , a - b Figure 7 Solution to Figure 7a The zeros of f are -1 and 5. zero multiplicity factor -1 even: +1 5 odd: 1 5 A possible formula for this function is of the form f = k We will use the point 0, 1 to determine k. Note that any point eactly on the graph of y = f could be used. 1 = k = k A possible formula for this polynomial function is f = Solution to Figure 7b The zeros of f are -, -1, 0,, and 5. zero multiplicity factor - odd: odd: odd: 1 odd: 1 5 odd: A possible formula for this function is of the form f = k We will use the point 1, to find k: = k = k3 1 6 k = 1 96 A possible formula for this polynomial function is f = Instructor: A.E.Cary Math 111 Final Eam Review Key Page 19 of

20 0. Sketch a graph of y = f for each polynomial function below. Also list the zeros and their multiplicity, 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 polynomial function with a negative leading coefficient. As, f As, f The graph is shown in Figure Figure 8 Instructor: A.E.Cary Math 111 Final Eam Review Key Page 0 of

21 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 polynomial function with a positive leading coefficient. As, f As, f The graph is shown in Figure Figure 9 Instructor: A.E.Cary Math 111 Final Eam Review Key Page 1 of

22 1. Find possible formulas for each rational function in Figure 11 below. List the zeros and their multiplicity, any vertical asymptotes and any horizontal asymptotes. 6 6 y= 6 6 y= , 6 6 =1 =-1 = a b Figure 10 Solution to Figure 11a Table 3. Behavior of R Zeros/ VA Multiplicity Factor Numerator/denominator zero: - odd 1 + numerator VA: = 1 odd 1 1 denominator Thus a possible formula for R is of the form R = k+ 1 where k is a constant factor. The constant factor of k can be determined to be based upon the horizontal asymptote y =. Therefore a possible formula for R is R = + 1 Alternately, we could determine k by using one point on the graph that is not a horizontal intercept. Here, the point 0, is given: R0 = k = k = Instructor: A.E.Cary Math 111 Final Eam Review Key Page of

23 6 6 y= y= a =1 =-1 = b Figure 11 Solution to Figure 11b Table. Behavior of R Zeros/ VA Multiplicity Factor Numerator/denominator zero: 0 odd 1 numerator zero: 3 odd 1 3 numerator VA: = 1 even +1 denominator VA: = odd 1 denominator Thus a possible formula for R is of the form R = k 3 +1 where k is a constant factor. Since the horizontal asymptote is y = 0, we cannot use it to determine k. We can use the point 1, to determine it instead: The constant factor of k can be determined to be based upon the horizontal asymptote y =. Therefore a possible formula for R is = k = k 1 k = 1 R = Instructor: A.E.Cary Math 111 Final Eam Review Key Page 3 of

24 . Find a possible formula for the rational function in Figure 1 below. List the zeros and their multiplicity, any vertical asymptotes and any horizontal asymptotes. 10 =-3 =3 =6 8 6 y= 10, Figure 1 Solution to Figure 1 Table 5. Behavior of R Zeros/ VA Multiplicity Factor Numerator/denominator zero: - odd 1 + numerator zero: even numerator zero: 7 odd 1 7 numerator VA: = 3 even +3 denominator VA: = 3 odd 1 3 denominator VA: = 6 odd 1 6 denominator Thus a possible formula for R is of the form R = k where k is a constant factor. The constant factor of k can be determined to be based upon the horizontal asymptote y =. Therefore a possible formula for R is R = Instructor: A.E.Cary Math 111 Final Eam Review Key Page of

Math 111 Final Exam Review

Math 111 Final Exam Review Math 111 Final Eam Review With the eception of rounding irrational logarithmic epressions and problems that specif that a calculator should be used, ou should be prepared to do the entire problem without