Math 117, Spring 2003, Math for Business and Economics Final Examination

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1 Math 117, Spring 2003, Math for Business and Economics Final Eamination Instructions: Try all of the problems and show all of your work. Answers given with little or no indication of how they were obtained may receive no credit. You may not use a calculator or the tet or notes of any kind. Good luck! 1. Solve each equation or inequality. a) +8 3 b) 2 5 < 7 2. Consider the relation defined by the equation 2 +4y 2 = 16. a) Is a function of y? Eplain. b) Find the domain of the relation. 3. Shown below are graphs of the functions f and g. Notice that the domain of f is [ 4, 4] and the domain of g is [0, 6] (i.e. what you see is the complete graph). y y = f() y 4 y = g() a) Find the value of f(2). b) Find the value of f 1 ( 1). c) Find the value of f g(4). d) Find all solutions to the inequality g() 0. e) Find all values of z for which g(z 2 )= 2. ( ) f f) Find the domain of. g

2 4. a) The graph of the function f is shown below. Carefully sketch the graph of y = f( 2)+1 on the blank aes provided. (Sorry - no picture in electronic version!) b) The graph below was obtained by performing several transformations to y = 1. Find its equation. (Sorry - no picture in electronic version!) 5. An oil spill covers a circular region in the ocean and is spreading out. Let t denote the number of hours that have passed since the spill occurred, let r denote the radius of the region covered in miles, and let C denote the cost of cleaning up the spill. Since the radius of the spill depends on how many hours have passed we can write r = f(t). The cost to clean up a spill depends on its radius, so C = g(r). a) What does the quantity g f(5) measure? Be as precise as you can. b) What does the quantity f 1 (2) measure? Be as precise as you can. 6. When a ball is thrown into the air its height, h, in feet, t seconds after being thrown is given by the equation h = 16t 2 +48t. a) When does the ball come back to the ground again? b) How high does the ball go? 7. a) How many turning points does the graph of f() = 2 ( +4) 3 ( 1) have? Eplain. b) Give an eample of a polynomial of degree 5 that has eactly 2 turning points. c) Consider the function f() = holes in its graph. 8. Consider the function f() = (22 +1)( 2) ( +3) 2 ( 1). Find the -coordinates of all the a) Find any vertical asymptotes. If there are none then indicate that this is the case. b) Find any horizontal asymptotes. If there are none then indicate that this is the case. c) Find all -intercepts. If there are none then indicate that this is the case. d) Find all intervals where f is positive and negative (i.e. find the sign graph of f). e) Find the y-intercept of f. f) Sketch a graph that is consistent with all of the information that you found in parts a) to e). 9. Simplify each epression.

3 a) log b) log 5 15 c) 3 e 2ln d) 4 ln(e 2 4 ) ( 1 2) log log Solve each equation or inequality. Write your answers in the simplest form you can. a) 8 2 =4 3 b) (3)2 +1 =3 2+1 ( c) = 600 2) d) log 10 (5 3) =log 10 (3 )+log 10 (3 + ) 11. a) Between which two integers does log 10 ( ) lie? b) Suppose log 10 =3.49. Which of the following intervals does lie in? Circle your answer. (0, 1) (1, 10) (10, 100) (100, 1000) (1000, 10, 000) 12. In a certain country, prices have increased by a factor of 1.44 in the last two years. a) By what percent have prices increased in the last 2 years? b) What is the annual inflation rate (epressed as % per year)? ( ) Sketch the graph of f() =6 2. Clearly mark any asymptotes and and y 3 intercepts. 14. On the campus of a certain university, parking tickets are $15 for the first offense. The cost doubles for each additional offense (so the cost for the second offense is $30, for the third offense is $60 etc.). Write a formula for the cost C as a function of the number of tickets n. 15. The drug Lorazepam, used to relieve aniety and stress, has a half-life of 14 hours. If someone takes a 2.5 mg tablet, how long does it take until only 1 mg remains in their body?

4 Solutions 1. a) First we rewrite the inequality as a rational inequality with 0 on one side Then do a sign graph (0 and 4 are -values that are indicated on the sign graph) to get the answer (0, 4]. b) The inequality is equivalent to 7 < 2 5 < 7 2<2<12 1<<6. 2. a) No. For eample if y =0thencould be either 4 or 4. b) Solving for y gives y 2 =(1/4)(16 2 ). To be able to get a value for y we need to be able to take a square root i.e. we require that Thus the domain is all values of for which 2 16 i.e a) 3 b) 3 c) f(g(4)) = f( 4) = 2 d) [0, 2] 6 e) z 2 =3,5soz=± 3,± 5 f) [0,2) (2, 4] 4. a) First reflect the given graph in the y-ais to get the graph of f 1 () =f( ). Then take each point on this graph and, leaving its y-coordinate unchanged, multiply its -coordinate by (1/2) to get the graph of f 2 () =f 1 (2) =f( 2). Then shift this graph up one unit to get the graph of f 3 () =f 2 ()+1=f( 2) a) The cost to clean up the spill 5 hours after it occurred. b) The number of hours it takes for the radius to spread to 2 miles. 6. a) We want to find the positive value of t when h = 0. Factoring we get h = 16t(t 3) so h =0whent=3. b) The graph of h is an upside down parabola with verte halfway between its zeros. Thus, the highest point on the graph is attained when t =(3/2) and the corresponding height at this point is 16(3/2)((3/2) 3) = 36 feet.

5 7. a) This is a degree 6 polynomial so it has at most 5 turning points. It has one at = 0. It must have at least one between = 4 and= 0 and at least one between =0and=1. At= 4 there is no turning point but the graph is flat there and that flat part accounts for another two turning points. Thus, 3 turning points are forced on us and another two are accounted for. That makes 5 total that are accounted for so there can be no others. Thus, the graph has eactly 3 turning points. b) There are many possibilities. Here are two eamples that work: 3 ( 1)( 2) or 4 ( 1). c) Factoring we get f() = 2 (+1) bottom we get f() = (+1) 2. If we cancel common factors from top and +1. The graph of f doesn t have any holes because it it doesn t have any factors that are common to both the numerator and the denominator. It has a vertical asymptote at = 1. The function f is equal to f at all values of ecept possibly =0, 1wherefis not defined. This means that at all values of other than =0, 1they-value on the graph of f is the same as the y-value on the graph of f. Thus, the only places that f might have aholeisat=0or= 1. It doesn t have a hole at = 1 because there is a vertical asymptote there (try drawing a hole where there s a vertical asymptote!). Thus, f has eactly one hole at =0. 8. a) There are no factors common to the top and bottom so look for where the bottom is equal to 0. This is at = 3, 1. b) Both top and bottom have the same degree with leading coefficients 2 and 1 respectively. Thus, there is a horizontal asymptote at y = 2/1 = 2. c) Since there are no factors common to top and bottom look for values of where the numerator is equal to 0. Whatever is willbegreaterthanorequal to 1 so the only -intercept occurs at =2. d) You need to mark -3, 1 and 2 on your sign graph. You get that f is positive in (, 3), ( 3, 1) and (2, ) and negative in (1, 2). e) (1)( 2) 3 2 ( 1) = a) We are being asked to solve the equation 27 =1/9. Writing both sides with 3 as the base we get 3 3 =3 2 so 3 = 2 and= 2/3. ( ) (15)(20) b) Using properties of logs to combine the terms we get log 5 =log 5 25 = c) 3 e ln 2 = 3 2 = d) 4 ( 2 4) = a) Writing each side using 2 as the base gives us 2 3(2) =2 2( 3).Thus6=2 6 and = 3/2.

6 b) The variable is in the eponent but there is no one single easy base that we can write each side in terms of (as we did in a)). So, we ll need to take logs to get the variable downstairs. Before taking logs, it will simplify things if we divide both sides by 3. Thus, we get 2 +1 = 3 2 and then take logs to get (+1) ln2 = 2ln 3. Solving for (remember ln 2 and ln 3 are simply numbers) we ln 2 get ln 2+ln 2 = (2 ln 3) so ln 2 = (2 ln 3 ln 2) and = 2ln3 ln 2 = ln 2 ln(9/2). c) The variable is not in the eponent here so we won t need to take logs. Working to isolate the variable we get ( 1+ ) 4 = ) 1/4 1+ ( 6 2 = 5 ( (6 ) 1/4 = 2 1) 5 d) Using properties of logs (remember that we will need to check our answers at the end) we get: log 10 (5 3) = log 10 (9 2 ) 5 3 = = 0 ( 4)( +1) = 0 = 1,4 Substituting back into the original we see that = 4 is not a solution since the LHS (and RHS) are not defined when = 4. Thus, the only solution is = a) log 10 ( )=log ( 14). Since 3.24 is between 1 and 10, log is between 0 and 1. Thus log 10 ( ) is between -14 and -13. b) Rewriting in eponential form we see that = Thus, lies between 10 3 =1,000 and 10 4 =10, a) Since 1.44 = we see that the new prices are equal to the old prices plus % of the old prices. In other words, prices have increased by 44%. b) Every two years prices increase by a factor of 1.44, so every year they increase by a factor of 1.44 = 1.2. Since 1.2 =1+ 20 it follows that prices increase by a 100 factor of 20% per year, i.e. the annual inflation rate is 20%.

7 13. The graph of y =(1/3) is a decreasing eponential that has y-intercept 1 and as,y 0. The graph of y =2(1/3) looks the same on a sketch but its y- intercept is 2 instead of 1. Reflect this graph in the ais to get y = 2(1/3) and then shift this graph up 6 units to get y =6 2(1/3).Youendupwithagraphthat is increasing everywhere, concave down everywhere, has y-intercept 4, and as, y 6. To find the -intercept solve 0 = 6 2(1/3) i.e. 3 = (1/3) so = This is describing eponential growth where every time n increases by 1, C increases by a factor of 2. Thus, C = C 0 2 n. To find C 0 note that when n =1,C=15so 15 = C 0 2 i.e. C 0 =7.5. Thus C =(7.5)2 n. 15. Let A be the amount remaining in the body t hours after the tablet is taken. A decreases eponmntially so it has the form A = A 0 e kt. A 0 is the initial amount so it is equal to 2.5. The half-life is 14 hours so k =(ln2)/14. Thus A =2.5e ln 2 14 t.wewant ( ( ) to find t when A = 1. Thus we want to solve 1 = 2.5e ln 2 2 ln 2 14 t.wegetln = t 5) 14 so t = 14 ln( 5 2) hours. ln 2