2 3 x = 6 4. (x 1) 6

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1 Solutions to Math 201 Final Eam from spring 2007 p. 1 of 16 (some of these problem solutions are out of order, in the interest of saving paper) 1. given equation: 1 2 ( 1) 1 3 = 4 both sides 6: ( 1) = 6 4 multiply numbers to cancel denominators: 3( 1) 2 = 24 distribute: = 24 combine like terms: 3 = 24 both sides +3: = given equation: both sides 2: cancel denominators: 2 = = = clean up: 2 = 14 5 both sides 14 and +5: = 0 factor: ( + 7)( 2) = 0 answers: = 7 or = 2. So the sum of the two answers is = Note that this equation can be written in the form 2u 2 + 3u = 2, where the u represents the quantity 1. This is sometimes called a quadratic form equation. The strategy for solving them is to solve the equation for u first, and then when the u-values are known, solve for from there: equation: 2u 2 + 3u = 2 both sides 2: 2u 2 + 3u 2 = 0 factor: (2u 1)(u + 2) = 0 finish solving for u: 2u 1 = 0 or u + 2 = 0 = u = 1 2 or u = 2. Replace u by 1 : 1 = 1 2 or 1 = 2 rewrite negative eponent: 1 = 1 2 or 1 = 2 1 reciprocate ( flip ) both sides of each equation: = 2 or = 1 2. Note: reciprocating both sides of an equation is only valid when you have (single fraction) = (single fraction).

2 4. This problem is just asking you to determine the smallest and largest allowable values that h can have, based on the given inequality. If we solve the inequality, the answers are revealed: inequality: 4h rewrite without the : 30 4h all sides + 50: 20 4h 80 all sides 4: 5 h 20. So minimum h value is 5 and maimum is a) (f + g)() = f() + g() = b) (fg)() = f()g() = ( + 1) (). c) (f g)() = f(g()) = f( ) = + 1. d) (g f)() = g(f()) = g( + 1) = + 1. e) From part (d), (g f)() = + 1. The only restriction on in this epression is caused by the squareroot. For a squareroot to be defined, the inside epression must be 0. So the domain here is + 1 0, which is the same [ as 1, which is written as the interval 1, ). 7. To find the formula for f 1 () when you are given the formula for f(): 1) write the relation down with y as the output (rather than f(), if the relation was given that way); 2) echange and y in the relation; 3) solve the new relation for y; 4) replace y by f 1 (). Here goes: original function: f() = 2 1 put in the y: y = 2 1 echange and y: = 2y 1 solve for y: both sides + 1: both sides 2: + 1 = 2y = y replace y by f 1 (): f 1 () =

3 5. This type of function is referred to in some tets as a case-defined function. The graph of such a function is made up of pieces of the graphs of the individual functions from each case. So we need to know what the graphs of the functions y =, y = 2, and y = 2 look like. They are all linear, and the slope and y-intercept of each equation can be found by thinking of it in the form y = m + b: y = is the same as y = 1 + 0, so this is a line with slope 1 and y-intercept 0. y = 2 is the same as y = 0 + 2, so it s a line with slope 0 (horizontal) and y-intercept 2. y = 2 is the same as y = 2 + 0, so it s a line with slope 2 and y-intercept 0. So the graphs of these three functions look like this: y = y = 2 y = 2 The graph of f agrees with that of y = only when < 0. The graph of f agrees with that of y = 2 only when 0 3. The graph of f agrees with that of y = 2 only when > 3. So we take the following pieces of the above graphs, and put them together to form the graph of f: y = y = 2 y = 2 Graph of f: y = f()

4 8. This function should be recognized as a quadratic function, because it fits the form of a quadratic function: y = f() = a 2 + b + c. This means its graph is a parabola. To find -intercept(s), set y = 0 and solve: 0 = = 0 = ( + 1)( + 1) = = 1. so -intercept is ( 1, 0). To find y intercept(s), set = 0 and find y: y = (0) 2 + 2(0) + 1 = y = 1. So the y-intercept is (0, 1). For a quadratic function (and for any polynomial function), domain is (, ) (all real numbers). The range of a quadratic function is found by figuring out whether the graph (which is a parabola) opens up or down, and by finding the verte. This parabola opens up, because the value of a is positive (a is 1 here). The verte is found as follows: verte -coordinate = h = b 2a = 2 2(1) = 1. verte y-coordinate = f(h) = f( 1) = ( 1) 2 + 2( 1) + 1 = 0. So the verte is at ( 1, 0), and the parabola opens up, so the range is [ 0, ). It was not asked here, but you should now be able to graph this function nicely, since the intercepts and verte are known: y = a) This question is asking us to find the equation of a line based on knowing two points that are on the line. We need to decide which variable (c or q) to treat as the output before we calculate the slope of the line, because slope must be calculated as (change in output)/(change in input). When the quantity variable q is involved, it is conventional to use q as input and the other variable (in this case c) as the output. The two points we ve been given are (q, c) = (10, 40) and (q, c) = (20, 70). So the slope of this linear relation is m = c 2 c = q 2 q = 3. Now that we know the slope of the line, we can write its equation down using the point-slope form: c 40 = 3(q 10), which can be written as c = 3q b) Using the formula derived in (a), when q = 35, the value of c is 3(35) + 10 = 115 (dollars).

5 9. To test for -ais symmetry, replace every y by ( y) and see if the relation can be legally simplified back to its original form: original relation: y 2 = 16 replace y by ( y): ( y) 2 = 16 simplify: y 2 = 16 (here, we use the fact that ( y) 2 = y 2 ) Since the negatives that were inserted simplified away and the relation ended up in its original form, this relation does have -ais symmetry. To test for y-ais symmetry, replace every by ( ) and see if the relation can be legally simplified back to its original form: original relation: y 2 = 16 replace by ( ): 4( ) 2 + y 2 = 16 simplify: y 2 = 16 (here, we use the fact that ( ) 2 = 2 ) Since the negatives that were inserted simplified away and the relation ended up in its original form, this relation does have y-ais symmetry. To test for origin symmetry, replace every by ( ) and every y by ( y) simultaneously, and see if the relation can be legally simplified back to its original form: original relation: y 2 = 16 replace by ( ) and y by ( y): 4( ) 2 + ( y) 2 = 16 simplify: y 2 = 16 (here, we use the facts that ( ) 2 = 2 and ( y) 2 = y 2 ) Since the negatives that were inserted simplified away and the relation ended up in its original form, this relation does have origin symmetry. As a check, remember that whenever a relation has any two of the -ais, y-ais, origin types of symmetry, it must also have the third type. So basically, when testing for these three types of symmetry, it is impossible to get two YES answers and one NO. 10. The given line s slope can be found by putting it in slope-intercept form: 3 2y + 7 = 0 = 3 7 = 2y = = y. So the slope of this line is 3 2. Any line which is perpendicular to this line must have a slope which is the negative reciprocal of 3 2, which is 2 3. Remember that parallel lines have the same slope, and perpendicular lines have negative-reciprocal slopes, like 3 2 and 2 3.

6 12. Area of a rectangle = A() = (length)(width) = (12 ) = Since this is a quadratic relation with a = 1, we know its graph is a parabola that opens down and lives in the, A plane, like this: A So to find out the value which will result in the maimum possible area value, we just need to find the value at verte of this parabola: verte -value = b 2a = 12 2( 1) = 6. If we had been asked to find the value of the maimum possible are, it would be A(6) = 12(6) (6) 2 = There are two methods of elimination that can be used to solve these systems alegbraically: substitution and addition. The method of substitution might go like this: 4 + 2y = 9 original system: 5 4y = 5 solve eqn A for y: 4 + 2y = 9 = 2y = 9 4 = y = C A B use eqn C to substitute for y in B : 5 4( 9 2 2) = 5 solve for : = 5 = = 5 = 13 = 13 = = 1. go back to C and solve for y: y = 9 2 2(1) = 5 2 = y = 3. So the solution to the system is = 1 and y = 5 2, or you can say (, y) = (1, 5 2 ). The method of addition might go like this: original system: 4 + 2y = 9 5 4y = 5 A B multiply eqn A by 2: 8 + 4y = 18 rewrite eqn B underneath it: 5 4y = 5 add the previous two eqns: y = 13 = 13 = 13 solve for : 13 = 13 = = 1. go back to A (or B ) to solve for y: 4(1) + 2y = 9 = 2y = 5 = y = 5 2. So the solution obtained is (again) = 1 and y = 5 2.

7 14. This is a nonlinear system of equations. Nonlinear systems are best solved by the substitution method. In this one, the easiest way to go is as follows. The first equation says that in the end, 2 will be the same as y So we can replace the 2 in the second equation by (y ). This eliminates and allows us to solve for y: y = (y ) 15 = 0 = y 2 y 2 = 0 = (y 2)(y + 1) = y = 2 or y = 1. Each y-value needs an -value to go with it: when y = 2, the first equation gives 2 = (2) = 2 = 17 = = ± 17. when y = 1, the first equation gives 2 = ( 1) = 2 = 14 = = ± 14. So the solution points for this system are (± 17, 2) and (± 14, 1). 15. The logarithmic functions are the inverses of the eponential functions. In particular, y = log 4 () is the inverse of y = 4. Since the inverse of any curve can be obtained by reflection about the line y =, we should know that the graph of y = log 4 () is just the reflection (about the line y = ) of the graph of y = 4. So the graph of y = log 4 () looks like the graph below. The graphs of y = 4 and the line y = are drawn as dashed curves so that the inverse relation between y = 4 and y = log 4 () can be seen. Again, three points are labeled with their coordinates, as requested. The lower y-ais is dashed to indicate that it serves as a vertical asymptote. y (4, 1) (1, 0) ( 1 4, 1) y = log 4 () Remember that for any function of the form f() = log b () with b > 1, its graph will have this same general appearance: eists only to the right of the y-ais, the negative y-ais serves as a vertical asymptote, goes through (1, 0), rises to the right, and is cupped down everywhere. Also remember that when generating input/output pairs for a logarithmic graph, it can be useful to remember that log b () = y is equivalent to b y = and that logarithms can be thought of like this: log b () = (the power you put on b to get ). So, for eample, log 4 (16) = (the power you put on 4 to get 16) = 2 ; log 4 ( 1 4 ) = (the power you put on 4 to get 1 4 ) = 1.

8 16. All of these questions can be answered by applying this property: (this is property 9 from below) log b () = y is equivalent to b y =. a) log 5 1/25 = = 5 = 1 25 = 5 = 5 2 = = 2. This one can also be done like this: (using property 2 from below) = log 5 1/25 = log 5 (1) log 5 (25) = 0 2 = 2. b) log 2 () = 2 = 2 2 = = 1 4 =. c) ln e 2 = = log e e 2 = = e = e 2 = = 2. d) log 4 = 1/2 = 1/2 = 4 = = 4 = = 16. There are other ways to think your way through these problems too. For eample, (c) can be done by remembering that in general, log b (b r ) = r, so = ln e 2 means = log e e 2 = 2. Here is a list of several useful logarithmic properties that you should be familiar with: 1) log b (AB) = log b (A) + log b (B) 2) log b (A/B) = log b (A) log b (B) 3) log b (A r ) = r log b (A) 4) log b () = log a() log a (b) 5) log b (1) = 0 ( change-of-base formula ) 7) log b (b r ) = r for any real number r 8) b log b (r) = r for any positive real number r 9) log b () = y is equivalent to b y = 10) log means log 10 and ln means log e. 6) log b (b) = Use property 2 mentioned in the last problem to combine the two log terms, and then rewrite the resulting equation using property 9: original equation: log(3 + 1) log( 2) = 1 combine log terms using property 2: ( ) log = 1 2 rewrite using property 9 (and prop. 10): 10 1 = finish: 10 = = 10( 2) = = = = 7 = 21 = = 3. This problem could also be done by checking the answer choices in the original equation to see which one works. But if it were not multiple choice, that would obviously not be an option. 18. Since ln e 2 is just 2 (by property 7), this equation becomes 2 =. Subtracting from both sides gives = 0.

9 19. Here s one way to get this one done: original: ln 3 y 5 rewrite as power: ln ( ( 3 y 5 ) 1/2) use property 3: 1 2 ln(3 y 5 ) use property 1: ( 1 2 ln( 3 ) + ln(y 5 ) ) use property 3 again: ( ) ln() + 5 ln(y) distribute: 3 2 ln() ln(y). 20. a) The size of a matri is given as (number of rows) (number of columns). So the size of matri C is 3 2. b) a 23 is the second-row, third-column entry of matri A. So a 23 = c) A + B T = T = = d) B I = B = e) A C = = R 1 + R R R 2 + R So the solution to the system is = 1, y = 2. a + by = k 23. The system c + dy = m So the given system can be rewritten as can always be rewritten as a c 4 7 = y 5 b = k. d y m 24. The idea here is that the answers to a linear system can always be caluluted as follows: [ ] [ ] inverse of answer answer = coefficient matri column matri So the answers here are calculated like this: y = =

10 R 1 R R 1 + R R R 2 + R At this point, the reduction is done. Notice that the equations in the final (reduced) matri give in terms of z and y in terms of z, and that s all. This indicates that z could be given any value, and then and y could be solved for based on the value of z. So this system has infinitely many solutions. In order to get full credit, we need to describe the form of all these solutions. This is done with the use of a parameter. Since z is free to take on any value, we set it equal to a parameter: z = r, where r can assume any real value. Then, solve for and y in terms of r, and we re done: eqn 1 = 5 3 z = 0 = = 5 3 z = = 5 3 r ; eqn 2 = y z = 0 = y = 2 3 z = y = 2 3 r. So the parametric family of solutions can be described as follows: = 5 3 r y = 2 3 r z = r, where r can be any real number. 26. a) 5 P 2 = b) 5 C 2 = 5! (5 2)! = 5! 3! = = 5 4 = ! (5 2)! 2! = 5! 3! 2! = (3 2)(2 1) = = The number of different groups of size 4 that can be formed, when choosing from among 10 different objects, is 10C 4 = 10! 6! 4! = = = = 210. ( )(4 3 2) This question is just asking how many different 3-person orderings are possible, when choosing from among 10. So the answer is 10P 3 = 10! (10 3)! = 10! = = = ! Answering the quiz is done in 5 stages (answering questions 1, 2, 3, 4 and 5). At each stage, there are 3 different possible ways to answer. So, thinking of a tree diagram and the Basic Counting Principle, the answer should be = 3 5 = 243.

11 30. This process can be done in the following stages: choose 2 of the 6 toys for one child, then choose two more of the remaining 4 toys for the net child, and (finally) give the last two toys to the third child. So the answer should be which is (# of ways to complete stage 1)(# of ways to complete stage 2)(# of ways to complete stage 3), ( 6 C 2 ) ( 4 C 2 ) ( 2 C 2 ) = 6! 4! 2! 4! 2! 2! 2! 2! 0!. Remebering that 0! = 1, the third fraction can be dropped completely, because it is equal to 1. Also, the two 4! epressions can be cancelled, which leaves 6! 2! 1 2! 2! = 6! 2! 2! 2! = = = a) The possible outcomes would look like this: HHHH, HHHT, HHTH, HTHH,... It seems like there might be too many to list quickly, so we should employ a counting technique to answer the question. This eperiment is done in 4 stages, and each stage has 2 possible outcomes (H or T), so the total number of different possible outcomes should be = 2 4 = 16. b) To answer a probability question, you have to understand the following idea: P(event E) = number of outcomes for which E occurs number of all possible outcomes Here, we already know that the number of all possible outcomes is 16. We just need to figure out how many of these outcomes there are for which no tails are observed. It should be clear that there is only one such outcome, namely HHHH. So: P(no tails) = c) Again, we know that there are 16 possible different outcomes. We just need to figure out how many of those 16 outcomes result in 3 heads and 1 tail. Listing them is probably the easiest way to see: HHHT, HHTH, HTHH, THHH. So P(3 heads and 1 tail) = 4 16 = 1 4.

12 32. We need to go back to the standard sample space fopr this eperiment, to make sure that all outcomes are equally likely. The standard sample space is just a an ordered list of the possible outcomes for the two dice: 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 Notice that these outcomes are all equally likely, and the ones that result in a sum of 7 lie on the diagonal from lower-left to upper-right: 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 Since there are 36 possible outcomes, and 6 of them result in a sum of 7, the answer to this question is P(sum is 7) = 6 36 = 1 6. b) The outcomes for which the sum is 2 or 3 are: 1,1 1,2 2,1. So P(sum is 2 or 3) = 3 36 = c) The outcomes for which the sum is greater than 9 are: 6,6 6,5 5,6 4,6 5,5 6,4. So P(sum > 9) = 6 36 = 1 6.

13 25. a.) Anyone who is solving a problem like this should understand the following two ideas, often refered to as the Corner Point Principle : If a linear programming problem has a closed and bounded feasible region, then the problem does have a solution, and the solution occurs at one of the feasible region s corner points. If a linear programming problem has a closed but unbounded feasible region, it may or may not have a solution. If it does have a solution, that solution will occur at one of the feasible region s corner points. Based on these ideas, linear programming problems like this one can be solved by graphing the feasible region and finding all its corner points, and then checking the value of the objective function at those corner points. So let s graph the feasible region: Solving 2 y 4: The points that barely work for this inequality are found by graphing the relation 2 y = 4. Since we know this relation has a straight-line graph, we can graph it by finding its intercepts: y = 0 = 2(0) y = 4 = y = 4; So 2 y = 4 graphs like this: y = 0 = 2 (0) = 4 = = 2. With a linear inequality like 2 y 4, the solution points will always lie to one side of the boundary line. To see which side, we can pick any point which is clearly on one side of the line, and test it in the inequality. Since (0, 0) clearly lies above the boundary line, and (0, 0) DOES satisfy the inequality [ (0) + (0) 4 is true], the points which solve 2 y 4 lie on or above the boundary line. So if we slightlty shade the SOLUTION side of the line, the solution set to 2 y 4 looks like this: 2 4 y 2 4 Solving + y 5 : The points that barely work for this inequality are found by graphing the relation + y = 5. Since we know this relation has a straight-line graph, we can graph it by finding its intercepts: y = 0 = (0) + y = 5 = y = 5; y = 0 = + (0) = 5 = = 5. So 2 + y = 10 graphs like this: 5 5 y Since (0)+(0) 5 is true, (0, 0) satisfies this inequality, which tells us that all points on or below the line satisfy this inequality. So, shading the SOLUTION side of the line, the solution set for + y 5 looks like this: 5 5

14 25. a.) (continued) p. 14 of 16 Solving 0, y 0 : These two inequalities simply say that neither nor y may be negative in the feasible region. This means that all allowable points must be in the first quadrant or on its boundary. So a graph of the points which solve 0 and y 0 looks like this (again slightly shading where the solution points are): y The feasible region for the entire problem is the region of points which satisfy all of the inequalites discussed earlier. So the graph of the feasible region will be a picture of the region where the the previously-shaded regions all overlap: y D The corner points of the feasible region must be found. Corner point A is clearly (0, 0), and we know B is (2, 0) and D is (0, 5) because these were intercepts in the earlier analysis. To find the eact coordinates of C (without guessing), note that C is the point where 2 y = 4 and + y = 5 intersect, so we solve the corresponding system: 2 y = y = 5 2 C FR add 1 and 2 to eliminate y: 3 = 9 solve for : 3 = 9 = = 3 A B use 2 to calculate y: (3) + y = 5 = y = 2. So the coordinates of corner point C are (3, 2). Now, we know (by the corner Point Principle) that the maimum value of Z will occur at one of these corner points. So we just need to check what the value of Z is at each corner point: At (0, 0): Z = 4(0) + 5(0) = 0 ; At (2, 0): Z = 4(2) + 5(0) = 8 ; At (0, 5): Z = 4(0) + 5(5) = 25 ; At (3, 2): Z = 4(3) + 5(2) = 22. So the maimum value of Z (keeping (, y) within the feasible region) is 25, which occurs at (, y) = (0, 5). p. 6 of 8

15 25. b.) When using the Simple Method, the inequalities 0 and y 0 are often referred to as trivial constraints. They must be there in order for the Simple Method to work, but they are not used when writing down the initial tableau. In order to write down the initial tableau, the non-trivial constraints and the objective equation must be rewritten like this: 2 y 4 2 y + s 1 = 4 + y 5 + y + s 2 = 5 Z = 4 + 5y 4 5y + Z = 0 The (non-trivial) constraints are always rewritten as equations instead of inequalities by introducing slack variables, and the objective equation is always rewritten by moving all terms to the left side of the equation. Now, the initial simple tableau can be written down. It s just the augmented matri for the system we just converted the problem to. Note that this system has 5 variables (, y, s 1, s 2, and Z) and 3 equations. Here is the initial tableau: s 1 s 2 Z y s 1 s 2 Z Tableau I As we perform the Simple method, this table will be transformed into other versions. With each version of the table, there is associated a guess (called a Basic Feasible Solution) as to what the solution to the problem might be. The variables written to the left of the tableau are meant to make it easy to read off the guess for each stage of the table. The values of the variables which appear to the left of the table can be read from the last column, and the variables not appearing to the left of the table are automatically assumed to be zero in the guess. So the guess (BFS) for the initial tableau above is this: If = 0 and y = 0, then s 1 = 4, s 2 = 5, and Z = 0. The numbers in the bottom row to the left of the Z column are called indicators. Because there are negative indicators in the table, we know that this guess is not the solution to the problem. That is, we should proceed with the Simple method. To do this, we first identify the most negative indicator, which is the 5. This identifies the second column as the pivot column. Then, for each row above the horizontal bar, we perform the calculation (answer column entry) (pivot column entry). The smallest positive result among these calculations identifies the pivot row. The pivot entry is then the entry that is in both the pivot row and the pivot column. All these calculations are shown below. In this case, the pivot entry turns out to be a 1 (boed below). s 1 s 2 Z y s 1 s 2 Z most negative indicator 4 ( 1) = 4 (ignore, because it s not positive) 5 1 = 5 (smallest positive result, because it s the only positive result)

16 25. b.) (continued) p. 16 of 16 Once the pivot entry is identified, the Simple method requires that we do this: (1) turn the pivot entry into a 1 by multiplying its row by an appropriate constant, and then (2) turn all the other pivot column entries into zeros by adding multiples of the pivot row to the other rows. The first step does not need to be done (pivot is already a 1). Here goes the rest: s 1 s 2 Z y s 1 s 2 Z R 2 + R 1 5R 2 + R 3 s 1 y Z y s 1 s 2 Z Tableau II Note how the variable above the pivot column (y) replaces the variable written to the left of the pivot row (s 2 ) in going from the first tableau to the net. The guess (BFS) corresponding to tableau II above is this: If = 0 and s 2 = 0, then s 1 = 9, y = 5, and Z = 25. If there were any negative indicators in tableau II, we would have to repeat the process all over again (find the pivot entry and reduce its column). But, since there are no negative indicators in tableau II, we are done, and the guess mentioned above is the solution to the problem. So: the maimum value of Z (keeping (, y) within the feasible region) is 25, which occurs when = 0 and y = 5. Since the slack variables were not in the original question, their values don t need to be mentioned in the final answer.