Math 101 Final Exam Review Solutions Eric Schmutz Problem 1. Write an equation of the line passing through (,7) and (-1,1). Let (x 1, y 1 ) = (, 7) and (x, y ) = ( 1, 1). The slope is m = y y 1 x x 1 = 1 7 1 = 6 3 =. Therefore a point slope form equation for the line is y 7 = (x ) To get the slope-intercept for, solve for y: Answer : y = x + 3 y = (x ) + 7 = x + 7 = x + 3 Problem. Find an equation of the line that is perpendicular to x - 3y = 5 and passes through the point (-, 6). There are two lines in this problem: the given line x 3y = 5 whose slope is m given =, and the unknown line. Because the unknown line 3 is perpendicular to the given line, we know its slope: m unknown = 1 m given = 3. We also know that the unknown line passes through (-, 6). Since we know both the slope and a point, we can immediately write down a point-slope form equation for the line: y 6 = 3 (x ).
To get the slope-intercept for, solve for y: y = 3 (x + ) + 6 = 3 x 3 + 6 = 3 x + 3 Answer :y = 3 x + 3 Problem 3. Let (x 1, y 1 ) = (0, 3), and and (x, y ) = (100, 1). The slope is m = y y 1 x x 1 = 1 3 100 0 = 180 100 = 1.8 Therefore a point slope form equation for the line is y 3 = 1.8(x 0) To get the slope-intercept for, solve for y: Answer :y = 1.8x + 3 Problem. y = 1.8(x 0) + 3 = 1.8x 0 + 3 = 1.8x + 3 The graph shown fails the vertical line test. For example, there are two different points on the curve having x-coordinate1. What is f(1)? We can t say because, in fact, this is not the graph of a function. Problem 5. Find the domain for each of the following functions. (a) f(x) = x x is in the domain if and only if the expression under the -sign is not negative. x is not negative if and only if x is not positive. The domain consists of those real numbers that are 0. Answer :(, 0]
7 (b)f(x) = x x x is in the domain if and only if the expression under the - sign is positive. (This time it can t be zero because of the fraction. Notice that x x = (x )(x + 1). There are two ways that this expression can be positive: if both factors are positive, or if both factors are negative. Both factors are positive if x >. Both factors are negative if x < 1. Answer :(, 1) (, ) (c) h(x) = log 5 (x 1) By definition, log 5 (x 1) is the value of y for which 5 y = x 1. Since 5 to any power is positive, we must have x 1 > 0. Problem 6. Answer :(1, ) Compare figures 7 and 6 on page 87 (section 10.), and observe that ( 1 )x = x. More generally, if b > 1, then the graph of f(x) = b x looks like figure 6 (growing very quickly), and the graph of f(x) = b x = 1 b x looks like that in figure 7 (decaying to zero very quickly). If you think carefully about this, it might help you better understand the financial math formulas in chapter 5. Problem 7. Find the vertex of the graph of the function f(x) = x x 3. One way to do this is by completing the square as follows: x x 3 = (x x + 1) = (x 1). It is clear that (x 1) is smallest when x = 1, and y = (1 1) = 1. It is also acceptable to memorize a specific formula: for the parabola y = ax + bx + c, the vertex is at x = b. In this example a = 1, b = a, c = 3, so the vertex is at x = = 1. We find the y coordinate by plugging x = 1 back into the equation = x x 3 to get y =. Problem 8. Solve log 8 (x) + log 8 (x 1) =. 3
First note that the domain of log 8 (x 1) is x > 1. If x > 1, the we can use the fact that the log of a product is the sum of the logs : log 8 (x(x 1)) =. If the left side is equal to the right side, then certainly 8 left side = 8 right side ; 8 log 8 (x(x 1)) = 8 From the definition of a logarithm we have 8 log 8 (t) = t for any t > 0.In particular when t = x(x ). Therefore Now a bit of algebra: x(x 1) = 8 x(x 1) = 8 if and only if x 1x = 6 if and only if x 1x 6 = 0 if and only if (x 16)(x + ) = 0 x = 16 or x =. Recall that x = is ruled out because it is not in the domain of log 8 (x 1). We are therefore left with just one solution x = 16. Answer :x = 16 Problem 9 A company rents out cares for $ 5 day plus 1 cents a mile. Express the daily cost C as a function of x = the number of miles driven. Answer :C(x) = 5 +.1x dollars Problem 10 For a store that installs satellite receivers, the cost to the store is C(x) = 80x+1950 dollars to install x receivers, and that it revenue it gets back in payments from customers is R(x) = x + 0x dollars.
(a) Find the break even points. The store breaks even if the payments it receives are neither more or less than the costs incurred. So we set C(x) = R(x) and solve for x. C(x) = R(x) 80x + 1950 = x + 0x x + 80x + 1950 = 0x x 160x + 1950 = 0 x 80x + 975 = 0 (x 16)(x 5) = 0 x = 16 or x = 5. Answer :x = 16 or x = 5. (b) Find the number x of receivers that maximizes profit, and the maximum profit. The profit is P (x) = R(x) C(x), so P (x) = ( x + 0x ) (80x + 1950) = x + 160x 1950 The vertex of a parabola y = ax + bx + c always has x coordinate x = b. In this case a =, b = 160, c = 1950, so we get a x = 160 = 0 is the number of receivers that maximizes profit. To get the maximum profit, plug x = 0 back into P (x) : P (0) = (0 ) + 160(0) 1950 = 150 Answer :x = 0, P (0) = $150 Problem 11. Find the vertical and horizontal asymptotes of f(x) = 3x 1 1x+. When x =, the denominator is zero and the numerator is not, so x = is a vertical asymptote. The horizontal asymptote is y = 3 1 Problem 1. Simplify f(x) = ln(e 7x ) 5x ln(e) + ln(1). 5
ln(1) = 0 and ln(e) = 1 and ln(e 7x ) = 7x ln(e) = 7x. Therefore f(x) = 7x 5x ln(e) + 0 = x 1 Problem 13. How long will it take money to double if it is invested at 6%, compounded monthly. If an initial amount Ais invested for x months, the value of the investment at the end of this period is A(1 +.06 1 )x. The problem is to choose x so that A(1+.06 1 )x = A. If we divide by A it becomes clear that the answer does not depend on the amount invested: (1 +.06 1 )x = Taking the logarithm of both sides, we get log((1 +.06 1 )x ) = x log(1 +.065 1 ) = log Solving for x, we get log x = log(1 +.06 log = log(1.005) = 138.975... So the answer is 139 months, because of monthly compounding. Problem 1. The Jones family gets a 30-year $ 150,000 mortgage at 8.75%. However, after years 9 months of making monthly payments, they decide to pay off the mortgage early in order to relocate. How much do they have to pay? (This problem is very similar to Example, page 1 in the text.) First let s calculate the monthly payment before relocation. Using the Amortization Payments formula in the green box on page 11 we get the monthly payment R. R = 6 1 ) P i 1 (1 + i) n,
where P = $150, 000 is the amount of money they borrow from the bank in order to buy the house, at the beginning of the 30-year period; i =.0875, because the annual interest rate is given as.0875, 1 and there are 1 payment periods per year, and n = 1 360 is the number of monthly payments that would have been be made over 30 year period if they hadn t decided to pay off the loan early. R =.0875 $150, 000 1 1 (1 +.0875 = 1180.05. ) 360 1 After years 9 months, they have made 1 + 9 = 57 payments, and still have 360 57 = 303 payments on the loan. To calculate the amount that they need to pay we agin use the Present Value of an Ordinary Annuity formula at the top of page 10. 1 (1 + i) 303 P = R i.0875 1 (1 + 1 = $1180.05 ) 303.0875 1 $13, 97. Remark Although they have made more than $ 67,6.85 in payments, they still owe much more than 150, 000 67, 6.85 = 8, 737.15 dollars to the bank. This is because in the early years, most of the payment is interest, with only a small amount used to reduce the amount owed. Problem 15. If an account earns % 11.5 interest, how much should you deposit each month in order to have $ 150,000 at the end of 0 years. For this we use the Future Value of an Ordinary Annuity formula in the green box on page 03. Here R is the unknown monthly payment, to be determined, S = $150, 000 is the value at the end of 0 years, i =.115 1 because the annual interest rate is 11.5% and there 1 payment periods per year, and n = 1 0 = 0 is the total number of payment periods. ( ) (1 + i) n 1 $150, 000 = R i 7
Solving for R, we get = R = $150, 000i (1 + i) n 1 $150, 000.115 1 (1 +.115 1 )0 1 $167.63 Problem 16. What is the interest rate of a loan charging $ 18 simple interest on a principal of $ 150 after years. In the formula I = P rt, it is given that I = $18 and t = and P = $150. To find the annual interest rate r, we simply (sorry) solve 18 = 150 r for r: r = 18 150 =.06 Answer :6% Problem 17. Brandyswine Fund returned 19.6% compounded quarterly over a period of years. If $ 5000 was invested at the beginning of this period, what was its value of the investment the end of four years? For this we can use the Compund Amount formula in the green box on page 191: A = P (1 + i) n, where A is the unknown final value of the investment that we want to calculate P = $5000 is the amount invested i =.196 is the quarterly interest rate n = = 16 is the number of quarters in four years. So ( A = $5000 1 +.196 ) 16 $10, 79. Answer :$10, 79. Problem 18 Solve for x 8
(a)e x +5x+6 = 1 Taking the logarithm of both sides, we get ln(e x +5x+6 ) = ln(1) We know that ln(1) = 0 and that, for any y, ln(e y ) = y ln(e) = y 1 = y In particular, for y = x + 5x + 6, we get ln(e x +5x+6 ) = x + 5x + 6 = 0 It is easy to solve this quadratic formula by factoring: (x + )(x + 3) = 0 has solutions x = and x = 3. Answer :x =, 3 (b) 81 x = 7 x+3 Let s rewrite both sides using 3 as a common base so we wont have two different bases. 81 = 3 and 7 = 3 3. Hence the problem is to solve the following equation for x: (3 ) x = (3 3 ) x+3 (1) Recall that a basic rule for exponents is that (b a ) c = b ac. Hence equation (1) is equivalent to 3 16x = 3 3x+9 Taking base-3 logarithms of both sides, we get Therefore 13x = 9 and x = 9 13 16x = 3x + 9 Answer :x = 9 13 Problem 19 Solve each system using any method (a) x + 3y =63 x + 5y =10 9
(b) ( ) 1 3 63 The corresponding augmented matrix is Subtract R 1 from R to get Divide Row ( ) 5 10 1 3 63 ( ) 0 7 11 1 3 63 by -7 to get. Finally, subtract 3 R from R1 ( 0 1 ) 16 1 0 15 to get. 0 1 16 Answer :x = 15, y = 16 x + 3y =1 x + 6y = ( ) 3 1 The corresponding augmented matrix is Subtract ( ) 6 3 1 twice Row 1 from Row to get For this system, 0 0 0 every point on the line x + 3y = 1 is a solution. Remark One way to describe the solution set is to solve for x in terms of y and use y as a parameter like this: the points on the line x + 3y = 1 satisfy x = 1 3 y. Hence the line consists of all points of the form ( 1 3 y, y) (One could also solve for y in terms of x and use x as a parameter) Problem 0 Let x be the number of antibiotic prescriptions and y the number of cough suppressant prescriptions that a drugstore fills. The total number of prescriptions filled is 9, and there are 3 more antibiotic prescriptions than cough suppressant prescriptions. Find x and y. The given information is x + y = ( 9 and x y ) = 3. The cor- 1 1 9 responding augmented matrix is Subtract Row ( 1 1 ) 3 1 1 9 1 from Row to get Divide R by - to get ( ) 0 58 1 1 9 Finally, subtract Row from Row 1 to get 0 1 9 10 ( 1 0 63 0 1 9 )
Answer :x = 63, y = 9 Problem 1 Solve the system x 1 + x + x 3 = x 1 x + x 3 = 1 x 1 + 3x x 3 = 18 The corresponding augmented matrix is 1 1 1 1 1 1 3 18 Working from left to right, begin by clearing the first column. Subtract twice Row 1 from Row to get 1 1 1 0 1 8 1 3 18 Add Row 1 to Row 3 to get 1 1 1 0 1 8 0 1 0 Now we are done with the first column, and can work on the second column. Add Row to Row 3 to get 1 1 1 0 1 8 0 0 8 Divide Row by - to get 1 1 1 1 0 1 0 0 8 11
Subtract Row from Row 1 to get 3 1 0 1 0 1 0 0 8 Now we are done with the second column, and can work on the third. Divide Row 3 by - to get 1 0 3 0 1 1 0 0 1 1 Subtract 1 Row 3 from Row to get 1 0 3 0 1 0 3/ 0 0 1 1 Finally, subtract 3 Row 3 from Row 1 to get 1 0 0 9/ 0 1 0 3/ 0 0 1 1 Answer :x = 9/ = 1.5, y = 3/ = 1.5, z = 1 Problem For problems -5 it is given that A = ( ) 1 1, B = 1 ( ) 3 6 8, C = 7 5 1 1 1 1 For problem, Find AC. ( ) ( + ) ( 1 1) ( + ) AC = ( 1 + ) ( + + ) ( 1 + ) ( ) 6 = 7 6 7 1
Problem 3 Find AB This multiplication cannot be done because the number of columns of the left matix A is not equal to the number of rows of the second. Problem Find A + C A has only two rows, and C has three rows, so we cannot add them Problem 5 Find 3I C If I is a 3 3 identity matrix, then 3 0 0 1 1 1 3I C = 0 3 0 1 1 = 1 5 1 0 0 3 1 1 1 Problem 6 Let x, y, z respectively be the number 6 inch, 3 inch, and inch sets sold. Set up one equation for each of the inputs. The each column will correspond to one of the three sizes of sets. Assembly equation: x + y + 3z = 10 Soldering equation:3x + y + z = 30 Finishing equation: x + 3y + z = 60 1 3 Problem 7 Are these two matrices inverses of each other? A = 1 1 1, 0 1 3 3 1 B = 3 3. 1 1 1 Just multiply A times B and see if you get the identity matrix. ( + 3 + 0) ( + 3 + 1) ( 6 + 3 + 3) 1 0 0 AB = (3 3 + 0) (6 3 ) (9 3 6) = 0 1 0 ( 1 + 1 + 0) ( + 1 + 1) ( 3 + 1 + 3) 0 0 1 Yes, they are inverses of each other because their product is the identity matrix. ( ) 5 Problem 8 Find the inverse of A =. 1 3 13
If a sequence of row operations converts A into the identity matrix, then the same sequence of row operations converts an identity matrix into the inverse of A. We can do both simultaneously by adjoining and identity matrix to the right side of A: ( 5 1 ) 0 1 3 0 1 Swap the to rows to get ( 1 3 0 ) 1 5 1 0 Now subtract twice Row 1 from Row to get ( 1 3 0 ) 1 0 1 1 Ad three time Row to Row 1 to get ( 1 0 3 ) 5 0 1 1 Finally, multiply Row by -1 to get ( 1 0 3 ) 5 0 1 1 ( ) 3 5 The inverse of A is A 1 = 1 Problem 9 Rewrite the following system of equations as a matrix equation AX = B, and use the inverse to find the solution. 13x 1 + x = 33 3x 1 + x = 8 ( ) 13 We can write the system as AX = B, where A = and X = ( ) ( ) 3 1 x1 33 and B = If we can find A x 8 1, then we can multiply both sides of AX = B by A 1 to get A 1 AX = A 1 B 1
IX = A 1 B X = A 1 B To find A 1 use the method of problem 8. ( 13 1 ) 0 3 1 0 1 Thus X = A 1 B = ( 1 0 1 ) 3 1 0 1 ( 1 0 1 ) 0 1 3 13 ( ) ( ) ( ) 1 33 1 = 3 13 8 5 Problem 30 Let x be the number of children s tickets sold for $6 each, y the number adult tickets sold for $1 each. On Saturday we have x + y = 180 6x + 1y = 1560 On Sunday the numbers of tickets sold are different, but the prices are the same: x + y = 10 6x + 1y = 110 How many of each kind of ticket are sold on each day? We could solve these systems separately, but the row operations are the same so it is more efficient to solve them together like this ( 1 1 180 ) 10 6 1 1560 110 Subtract six times Row 1 from Row to get ( 1 1 180 ) 10 0 6 80 0 15
Divide Row by 6 to get ( 1 1 180 ) 10 0 1 80 70 Subtract Row from Row 1 to get ( 1 0 100 ) 50 0 1 80 70 Now we can read of the solutions: on Saturday, x = 100, y = 80; on Sunday x = 50, y = 70 16