sff09 - MATH 1111 SAMPLE FINAL EXAM 1. a- f(:€) = 3:52...

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Unformatted text preview: MATH 1111 SAMPLE FINAL EXAM 1. a- f(:€) = 3:52 —1— 1. 9(3) : a: — 1. Find (fog)(:r) and simplify b. Use the graphs of f and g to find (9 o f)(3]. l0 3. Subtract. Simplify as much as possible. 4. Simplify the expression. Your answer should contain no negative exponents. a (51341;)28-757} ' 153,!9 h. (9x3)%(m%)(6m)-2 3 O 5. Simplify this complex fraction. $4 LC :51 _|r_ :1 + l. :I‘ + 4 293+3 _3x—4 5 2 6. Solve for :t. 7. A total of $12,000 is invested in two corporate bonds that pay 7.5% and 9% simple interest. (Interest amount 2 interest rate times invested amount) The investor wants an annual interest incorne of $1005 from the investments. What amount should be invested in each bOnd? 8. On which intervaHs) is this function decreasing? What is f0)? On which intervaMs) is flat) > U? x/erE) 9. What is the domain of this function? f(:t) : m 3 10. Write the exponential equation through the points (33486) and (1",39366). You may use 3,: : Cam or 1 : Ce”, but the numbers come out nicer with y 2 Cam. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. Write the equation of the line which passes through the points (3,4) and (—6, —2). Write your answer in slope-intercept form. —2:r—7 fora". < —2 Sketch the ra h of = h g p fly) { :r+3 form>—2 What are the intercepts? _P_ q+r(l—q) Simplify the expression. ‘5/ 643273,:4 Solve for q S = 6:1: + 5 , . . = 7m 1 is aone—to—one function. Fmd f"1(:r:). fix) Solve for x using the quadratic formula: 3:2 — 6:1: + 2 = 0. You must show your work. Simplify if possible, but do not get a decimal apprOXimation. a. Find the vertex and intercepts of the parabola f (:r.) = 2m2 — 12:22 + 10. b. Write the ahk form of the equation of the parabola that has vertex (2, 3) and passes through the point (5,1). Solve for at. log3 1:2 = 2 + log3(52:) Solve for :12. iogg (:1: + 1) + logg + 3) = 1 How long (to the nearest tenth of a year) will it take for a $2,000 investment to grow to $10,000 at an interest rate of 8% compounded continuously? Show your work algebraically. Solve for x. It is not necessary to get a decimal approximation. (2X3-im—1-5) = 67x—8 3. Use the definition of logarithm to write 26 = 64 in logarithmic form. b. Evaluate log?] 62. a. Does the equation f0?) = 60(1035)t (with t in years) represent growth or decay? What‘s the growth/ decay factor? Rate? \Vill you have a doubling time or a half-life? What does the Rule of 70 predict? Use your calculator to get a more precise value. Write the equation in f(t) = Ce“ form. b. What’s the effective annual interest rate of an investment that has a nominal interest rate of 6% compounded continuously? flan) {233—7 forat>0 233—4 formfiO Find f(3). Given this table, find f(3). 26. 27. 28. 29. 30. 31. 32. f(33) : 10 — 3:2. Find f(—1). Records at a travel agency indicate that when the round-trip plane fare from Columbia to Chicago is $260, an average of 112 tickets are sold per week. and when the price is $180, an average of 176 tickets are sold. Assume that the relationship between price P and quantity sold Q is linear. Express P in terms of Q. What should the price he in order to sell 200 tickets? How many tickets will be sold if the price is $205? What are the units of the slope of the line? The speed V at which an autOmobile was traveling before the brakes were applied can sometimes be estimated from the length L of the skid marks. Assume that V is directly proportional to the square root of L. For a certain autornobile on a dry surface, L = 50 ft when V = 35 mi/hr. Estimate the initial speed of the automobile if the skid marks are 150 feet long. The half-life of radioactive actinium is 22 years. If the initial sample size is 100 grams, how much will remain after 40 years? The element decays according to the model 1; = yoekt. [Since this is decay, expect k to be negative] The population P of a town was 5000 people in 1990 (which correspOnds to t = U ), and 7200 people in 2003. Predict the population of this town in 2020. Bound to a whole number- How many years does it take for the populatiori to triple? Bound to a whole number. Use the continuous exponential growth model P = Fuel;t to answer these questions. Factor: a. xz—yg d. 3x3+2m2h12$—8 b. 3:3 — 3,:3 e. 4502 — 80!;2 c.. 33 + y3 f. 811:3 + n6 The first graph is y : f (at). Write the equation for the other graphs. 33. The population of a town quadruples every 15 years. What is the annual growth rate? Give your answer as a %. Round your answer to two numbers after the decimal point. 34. A credit card company offers a monthly interest rate of 2.2%. “that is the annual interest rate? Give your answer as a %. Round your answer to one number after the decimal point. 35. A health club has cost and revenue functions given by C = 115000 — 700:1: and R = 3000.15 — 203:2, where :1: dollars is the price of a one-year membership. Profit = Revenue — Cost, so at I R — C. What price should be charged to maximize profit? What is the maximum profit? 36. If any of the tables represent a linear function or an exponential function, write the equation for the function. If not, say “neither.” 37. Write each term as a number times 1:: to a power. 3 1 7 5 a. E?— b. c. 4Vm6+ W 38. Find the asymptotes and domain. Be sure to identity the asymptotes. (Review Q the rules for horizontal and slant asymptotes.) 17:3 ft”): (a:+3)(:r:—2) 39. Solve for 3: using the method of completing the square. 43:2 — 24:: — 8 = 0 You must show your work using this method. 40. Find the average rate of change of the number of farms in the U.S. between 1940 and 1990. number of farms (minions) 6 . 4 2 n _——| _.._- 1940 1960 1980 2000 ' Number of farms in the US (in millions) 41 43 44 who . Solve for m. 3(3: — T) . Solve for m. . Divide 153:3 — 141:2 m .53 + 11 =48' $+7=v13m+51 by 3:1: — 4. . Write the equations for these graphs. :1 '3 Solutions to the Sample Final Exam 1- a. (fogflw) =f(g(x))=f(m-1) =3($-1)2+1 = 3[(m—1)(:c—1)]+1=3[x2—2x+1]+1=3x2—6m+3+1=3x2—6x+4 b. Here, we will have to look at the pictures to get the function values. Remember that if, for example, the point (8, 9) is on the graph of y = f(x) we say that f(8) = 9. SO: (9 O f)(3) = 9(fl3ll = 9(2) = 2 We got f(3) = 2 from the point (3,2) on the f graph, and we got 9(2) = 2 from the point (2,2) on the g graph. 'l 2. To rationalize the denominator means to multiply the given expression by 1 in such a way that we end up with no radical in the denominator. There will still be a radical in the numerator when we finish. Here, the problem has a cube root. There are two factors under the radical, the 4 = 22 and the m. We would like each factor to have the power 3, because then we could simplify v3 233:3 = 23;. The trick is to multiply by *what we need* in order to make that happen. Once we choose the multiplier for the denominator, we have to put the same thing in the numerator, so that we are multiplying by 1. Here’s how it looks: 12 (6/2332) _ mid/25:2 _ 123/232 __ 6x73 2592 m 32332 22: a: 3. Get a common denominator, do the arithmetic in the numerator, remember to distribute properly, factor and cancel for the last step.- 655 _ 3 (m+2)_6$—3(x+2)_ 633—3.:c—6 T (:t—Q)(:L’+2) (x—Q) $+2 T(m—2)(m+2)_(m—2)(x+2)_ 3m—6 3(x—2) 3 (x—2)(m+2) (m—ZXm—l—Z) 3+2 4 a 25$8y2(3)$7 fl 7533153;2 _ 53315 ' ‘ 15y9 _ 153,9 ‘ y? b. 9%(x3)%(x%)6-2m-2 = 27(x%)(x%)(3lg)(x-2) = %m%+%-2 We have to get a common denominator to add those fractions in the exponent. _§ E+l_§_§ fl _4334 4 4—434 5. Multiply every term by the LCD, which is (a: + 4) (a: + 1). (:0+4)(33+1) + ($+4)(3:+1) 7(“4H12($+1) 3x+3—5x—20 _ —2..~;—17 7x+28+12$+12 _199.~+40 6. Cross-multiply. The h on the right can go with either the numerator or the denominator, but NOT with both! (23: + 3)(—2) 2 5(33: — 4) —4m—6=153:—20 —19m=—14 _ —14 _ 14 93- T r E 01‘ (2x + 3X2) = —5(3$ — 4) 4x+6= ~15x+20 1955' = 14 so m = 3!: 7. Let a: be the amount invested at 7.5%. Let y be the amount invested at 9%. m + y = 12000 { 075:5 + .093; = 990 55’ = 12000 — y 075(12000 — y) + .093; = 1005 900 — .0753] + .093; = 1005 .0153; = 105 y = 7000 ac = 12000 — 7000 = 5000 Invest $5000 at 7.5% and $7000 at 9%. 8. f is decreasing on the interval (1, 5). Remember that the numbers in the interval come from the x coordinates of the points where the function starts decreasing and stops decreasing. f (1) = 3 f(a:) > 0 on the intervals (01 4) and (6,00). 9. 6x — 5 Z 0 so 655 2 5 which means m 33 75 3 because the denominator cannot x2%,$#3 Gale: 2 be 0. All you need to write is 10. Start with the format of an exponential equation, 9 = Oct? Since the change in x is 4, we know that a is the fourth root of the ratio. 4 I 39366 a = — = 3 486 OK, now we have y = 0(3)? To find 0, use either point. 486 486 = 0(3)3 which gives 486 = C(27) so 0 = 3-7— : 18 and we get y = 18(3)? «— —-9 3 — (—6) 3 + 6 9 ‘3 9 3:950 Intercepts The x—intercept comes from the equation of the line that actually crosses the X—axis. 0 = —2m r T 23: = —7 53 = —% or a: = ~35 The x—intercept is (--3.5,0) The y—intercept comes from the equation of the line that actually crosses the y—axis. y = 0 + 3 = 3 The y-intercept is (0,3) 13. 14. 15. 16. There is more than one way to do this problem correctly. 3(q+p(1 - 9)) =10 501 + p - 139) = 29 Sq + 3p - Spq = p Sq-Spq=p-Sp' 918-829) =p-S'p _ p — Sp + 20(1 - S) 9‘ S—Sp 01" ‘1‘ 5(1-10) The rule is v5 3: = x, so we are looking for groups of 5 things. 5 (2)(2)(2)(2)(2)(2)$mmmyyyy = 258 v5 2332314 Call it y, change x and y, solve for y, call it the inverse. 6x + 5 6g + 5 y 753 — 1 en m 7y w 1 m(7y—1)=6y+5 7$y—$=6y+5 7xy—6y=m+5 y(7x—6)=$+5 __ 3+5 _ _1 y—7m_6—f (93)” a=1, b=*6, c=2 $=6i«/36—4(1)(2)=6im=6i2fi:2(3ifi)=3ifi 2(1) 2 2 2 .a. f(x)=2x2—12m+10 (1:2, b=—12, c=10 f: :5 =_‘(“12)=3 ” 2a 2(2) k = f(h) = 2(32) —12(3)+ 10 =18 — 36 + 10 2 —8 The vertex is (3, -—8). Set a: = 0 and solve for y to get the y-intercept (0, 10). Set y = 0 and solve for x to get the x—intercepts. 2(ch — 6:3 + 5) z 0 m2 — 63: + 5 = 0 (This is an equation, so I can divide each side by 2.) (x — 5)(sc — 1) = 0 so the X-intercepts are (5,0) and (1,0). 17. b. (2, 3) is (11,16) and (5,1) is (w, flag». Plug in all the numbers, solve for a, then write the equation. fix) 2 (1(3: — h)2 + k is a parabola with vertex ()1, k). 1 = a(5 — 2)2 + 3 18. 1033 (3:2) — 1083 (557) : (9)(5) = a," so a: z 45 19. 10g8((sc + 1)(sc + 3)) = 1 logs (1:2 +4m+3) = 1 81 = 392 +41: -1— 3 :32 + 4x — 5 = 0 (m+5)($—1) =0 93 = —5, :1: =1 BUT we can’t use :1: = —5, so :1: = 1 is the only answer. 20. 10000 = 200013-Ogt 5 : 6.08t 1115 = In 6081’ = .0815 m = t 2: 20.1 years 21. Use log or 111. In :1fl67x—8 1112 +11134m+5 =11167$_8 1112+ (4x+5)1n3 = (733—8)1I16 1n2+4xln3+5ln3 = 73:1n6—81n6 4321113 — 7.391116 = —81n6 —ln2—5ln3 x(4ln3~— 71116) = —81116-—1112— 51113 a? _ —81n6— 1n2 — 51113 _ 4m3—7m6 1 2 1 22.a.1og264==:6 b.10g562=: 319— (1R1 Og62 n 2564332773 1n 5 log 5 23. a. Growth because a. > 1. Growth factor is 1.035 Growth rate is 3.5%. Has a doubling time. 70 / 3.5 2 20 so doubling time is 20 years according to the Rule of 70. 2nd CALC intersect gives 20.148792 years. for) = 60 (eln(1.035))t so four) : 608.0344014267t b. A : Pe'06t so A = 130061836547)“: [do 6'06 on calculator.] r 2 a. — 1 = 061836547 so the effective annual interest rate is 6.1836547%. 24. f(3) = —1 25. f(3) = 19 26. f(—1)= 10 —[(-1)2] = 10 — 1 = 9 27. Points (Q, P) are (112, 260) and (176,180). I 180 - 260 —80 m : -— 2 -— 176 — 112 64 P — 260 = —1.25(Q — 112) P — 260 = -1.25Q + 140 P = —1.25Q + 400 = —1.25 Plug in Q = 200 to find the price. P : “125(200) + 400 = 150 Charge $150 to sell 200 tickets. Plug in P z 205 and solve for Q to find the number of tickets. 205 = —1.25Q + 400 205 — 400 = —1.25Q —195 = m1.25Q —195 . ._ . . —1§5 = Q : 156 If the price is $205, 156 tlckets Wlll be sold. AP . . To calculate slope, we found E, so the un1ts are $/t1cket. [Note: A lepe equal to —1.25 means that in order to sell one more ticket, we have to lower the price $1.25.] 28. V =WE 35 35 = lax/50 so — = k x/so 36 V = .— 150 a 6062177626 ( 730) (f ) The initial speed was about 60.6 miles per hour. 29. y = 1006‘“t Half remains after 22 years, so 50 = 100€k*22 or .5 = €22k 111(5) = lnezgk = 22k: in 22 Plug in 40 for t. After 40 years, 2835781305 grams remain. : k = —.03150669 and we have the equation y = 100€_'03150669t P = Poem 7200 = 50006M13) (t = 13 corresponds to 2003) 7200 _ 813}: 5000 _ ln = ln 613k =13k fl) = k = .0280494703 35 = 30 corresponds to 2020 P = 5OOU€('0280494703*30) : 1159903634 In 2020, there will be approximately 11599 people. Triple 5000 is 15000 15000 = 500019"?t Solve for t 3 = ekt 1113 = lnekt = kt ln3 T = t = 3916695317 The pOpulation triples in about 39 years. 31. 32. 33. 34. 35. a- (x - 0W + y) b- (w — 300:2 + my + 92) C- (a? + 0X36? — my + .112) d. Factor by grouping. 3m3 + 2:232 — 12x — 8 = 332(353 + 2) — 4(39: + 2) = (:1:2 — 4)(3:c + 2) = (as — 2)(.r + 2X33 + 2) [Keep going: difference of squaresl] e. 5(902 — 1602) 2 5((3a)2 — (4602) 2 5(30. — 4b)(30 +413) f. (2w)3+(n2)3 = (2w+n)((2w)2—(2w)(n)+(n2)2) = (2w+n)(4w2—2nw+n4) b. y = -f(3'3) P = P0(4)T where T is 15 years. New growth factor is a = 4T1? 13/1 2 109682498, so the annual growth rate is 9.68%. a.y=f(3:—2)+l c.y=f(—$)—3 A = A0(1.022)T where T is months. New growth factor is a = (1.022)12 = 1298406705. so the annual interest rate is 29.8%. 71' : (30003: — 2052) — (115000 — 7005) = 405:2 + 37005 — 115000 P ($3 Ver‘l’fiix P aPM x ($3 The first coordinate of the vertex is the price which maximizes profit. The second coordinate of the vertex is the maximum profit. You may find the vertex (Ink) by using your calculator, or by using the formulas h = and k z —3700 (QM-20) Charge $92.50 to maximize the profit. 11 = = 92.5 k = —20(92.5)2 + 3700(925) — 115000 = 55125 The maximum profit is $56.125.00. 36. 37. 38. 39. 40. 41. In the first table, 75 goes up by 1 and g(t) goes down by .25 every time. This one is linear. 9(t) —.25t + 7.60 In the second table, m goes up by 1 and there is a constant ratio of 1.5 when you look at the values of f(.’L'). This one is exponential. f(9:) : 63(15):” The third table is neither linear nor exponential. Remember to use the variable names given in the table! a. 3m‘7 b. gas—7 c. 433'?“ + 556—3 Vertical asymptotes: m = —3, a: = 2 Domain: m 75 —3, 93 # 2 The degree in the numerator is 1, and the degree in the denominator is 2. Horizontal asymptote: y = 0. 4932+2433—8z0 xg—fimh220 332—63322 b=—6 so%=—3 andthen(«f21)2=9 w2—6x+9=2+9 ($-3)2=11 x—32iVfi $=3:|:\/fi The average rate of change is the slope of the line between the points (1940, 6) and (1990, 2). 245 _ —4 19904940 * 50 The number of farms in the U.S. is decreasing at an average rate of .08 million —.08 million farms/year. farms per year, which is 80,000 farms per year. (33—7)% =16 ., ((m—7)%)2 =:t16% I We need :t because the denominator of the reciprocal power is an even number. a" — 7 = :|:1024 a: = 7 :t 1024, cc 2 1031, :12 = “1017 42. x+7=m ($+7)($+7)= (m2 $2+14x+49=13m+51 332+:c—2=0 (x+2)(33—1)=0 so = —2, m =1 and they both work. 43. \ 5x1 +l>< 4* I 3XFL'flISX3—1LWZ—5x + H IS‘LBWQ—le é); "'5'X-HI (ma-8% 3x+ll BX—‘f I? 6’ y:(m+3)2 f. y=1n(—m) l . :— l g y 33*ng ...
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This note was uploaded on 09/28/2011 for the course MATH 111 taught by Professor Hitchcock during the Fall '08 term at South Carolina.

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sff09 - MATH 1111 SAMPLE FINAL EXAM 1. a- f(:€) = 3:52...

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