#$&* course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * What did the horizontal line test tell you for this function? For every horizontal below the 'peak' of this graph the graph will intersect the horizontal line in two points. This function is not one-to-one on the domain depicted here. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.43 / 7th edition 5.2.28 looks like cubic thru origin, (1,1), (-1,-1), sketch inverse. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In quadrant I the function is increasing at an increasing rate. The inverse function is just the opposite increasing at a decreasing rate in the first quadrant. In quadrant III the function is increasing at a decreasing rate. The inverse function is the just the opposite and is increasing at an increasing rate in the third quadrant confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Describe your sketch of the inverse function. The graph of the function passes through (0, 0), (1,1), and (-1,-1). The inverse function will reverse these coordinates, which will give the same three points. Between x = -1 and x = 1 the graph of the original function is closer to the x axis than to the y axis, and is horizontal at the origin. The graph of the inverse function will therefore be closer to the y axis than to the x axis for y values between -1 and 1, and will be vertical at the origin. For x < 1 and for x > 1 the graph lies closer to the y axis than to the x axis. The graph of the inverse function will therefore lie closer to the x axis than to the y axis for y < 1 and for y > 1. In the first quadrant the function is increasing at an increasing rate. The inverse function will therefore be increasing at a decreasing rate in the first quadrant. In the third quadrant the function is increasing at a decreasing rate. The inverse function will therefore be increasing at an increasing rate in the third quadrant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.32 / 7th edition 5.2.32 f = 2x + 6 inv to g = 1 / 2 * x - 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(g(x)) = 2 g(x) + 6 = 2 ( 1 / 2 * x - 3) + 6 = x - 6 + 6 = x g(f(x)) = 1 / 2 * f(x) - 3 = 1/2 ( 2 x + 6) - 3 = x + 3 - 3 = x Since f(g(x)) = g(f(x)) = x, the two functions are inverse confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Show that the functions f(x) and g(x) are indeed inverses. f(g(x)) = 2 g(x) + 6 = 2 ( 1 / 2 * x - 3) + 6 = x - 6 + 6 = x. g(f(x)) = 1 / 2 * f(x) - 3 = 1/2 ( 2 x + 6) - 3 = x + 3 - 3 = x. Since f(g(x)) = g(f(x)) = x, the two functions are inverse. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.52 / 7th edition 5.2.44. inv of x^3 + 1; domain range etc.. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = x^3 + 1 now we switch the x and y and solve x = y^3 + 1 x - 1 = y^3 y = (x-1)^1/3 both domain and ranger include all real numbers confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Give the inverse of the given function and the other requested information. The function is y = x^3 + 1. This function is defined for all real-number values of x and its range consists of all real numbers. If we switch the roles of x and y we get x = y^3 + 1. Solving for y we get y = (x - 1)^(1/3). This is the inverse function. We can take the 1/3 power of any real number, positive or negative, so the domain of the inverse function is all real numbers. Any real-number value of y can be obtained by using an appropriate value of x. So both the domain and range of the inverse function consist of all real numbers. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.64 / 7th edition 5.2.56. inv of f(x) = (3x+1)/(-x). Domain and using inv fn range of f. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = (3x + 1) / (-x) x = (3y + 1) / (-y) we know that y cannot be 0 because if we multiply both sides by 0 the se would be changed -xy = 3y + 1 -xy -3y = 1 now we factor out the y (-x -3)y = 1 now we get the y on a side by its self y = -1 / (x + 3) all real number except 3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * What is the domain of f? What is the inverse function? What does the inverse function tell you about the range of f? f(x) is defined for all x except x = 0, since division by 0 is not defined. If we switch x and y in the expression y = (3x + 1) / (-x) we get x = (3y + 1) / (-y). To solve for y we first multiply by -y, noting that this excludes y = 0 since multiplication of both sides by 0 would change the solution set. We get -y * x = -y * (3 y + 1) / (-y). The left-hand side simplifies to - x y and the right-hand side to 3 y + 1, giving us -x y = 3y + 1. Subtracting 3 y from both sides we get -x y - 3 y = 1. Factoring y out of the left-hand side (which becomes (-x - 3) * y; if you multiply this out you see that it is the same as -x y - 3 y) we get (-x - 3) y = 1, and dividing both sides by (-x - 3), which excludes x = -3, we get y = -1 / (x + 3). The domain of this function is the set of all real numbers except 3. Since the domain of the inverse function is the range of the original function, the range of the original function consists of all real numbers except 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.94 / 7th edition 5.2.74. T(L) = 2 pi sqrt ( L / 32.2). Find L(T). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = 2 pi sqrt( L / 32.2) First we square both sides to get rid of the sqrt T^2 = 4 pi^2 * L / 32.2 then we multiply both sides by 32.2 / ( 4 pi^2) L = T^2 * 32.2 / (4 pi^2) L(T) = 32.2 T^2 / (4 pi^2) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * We solve T = 2 pi sqrt( L / 32.2) for L. First squaring both sides we obtain T^2 = 4 pi^2 * L / 32.2. Multiplying both sides by 32.2 / ( 4 pi^2) we get L = T^2 * 32.2 / (4 pi^2). So our function L(T) is L(T) = 32.2 T^2 / (4 pi^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm. Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 036. * 36 ********************************************* Question: * 6.3.40 / 7th edition 5.3.38. Explain how you used transformations to graph f(x) = 2^(x+2) The function y = 2 ^ ( x+2) is a transformation of the basic function y = 2^x, with x replaced by x - 2. So the graph moves 2 units in the x direction. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All points on the graph of y = 2^(x+2) are shifted to the left of the original function y = 2^x td confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). The graph of y = 2^(x+2), being shifted in only the x direction, is also asymptotic to the negative x axis and passes through the points (0 - 2, 1) = (2, 1) and (1 - 2, 2) = (3, 2). The graph also increases at a rapidly increasing rate. All the points of the graph of y = 2^(x + 2) lie 2 units to the left of points on the graph of y = 2^x. To understand why the graph shifts to the left, and why we used the points (0 - 2, 1) = (2, 1) and (1 - 2, 2) = (3, 2) as a basis for the graph, consider the tables for y = 2^x and y = 2^(x + 2). The tables are given below: x y = 2^x X y = 2^(x+2) -3 1/8 -3 1/2 -2 1/4 -2 1 -1 1/2 -1 2 0 1 0 4 1 2 1 8 2 4 2 16 3 8 3 32 Observe that the y values 1/2, 1, 2, 4 and 8 in the y = 2^(x+2) column also occur in the y = 2^x column, but for different values of x: • The values of x for the y = 2^(x+2) function, corresponding to y values 1/2, 1, 2 and 4, are 2 units less than for the y = 2^x function. • This occurs because the exponent x + 2 is 2 units greater than the exponent x, so that x + 2 is always 2 units 'ahead' of the value of x. Thus y = 2^(x + 2) reaches its values 'earlier' than y = 2^x (for example y = 2^(x + 1) reaches the value y = 8 when x = 1, whereas y = 2^x doesn't reach y = 8 until x = 3). • This causes the y = 2^(x + 2) graph to be shifted 2 units to the left, relative to the graph of y = 2^x. The figure below depicts the graphs of the two functions: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * Extra Problem / 7th edition 5.3.42. Transformations to graph f(x) = 1 - 3 * 2^x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we start with our base which is 2^x which passes through (0,1), and (1,2) Then we will insert the -3 * 2^x and the graph will then shift and change the direction it is traveling in. it now travels through points (0,-3) and (1, -6) Now we will show our final graph inserting the 1 - 3 * 2^x this will shift the graph up making the points (0, -2) and (1, -5) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 1 - 3 * 2^x is obtained from y = 2^x by first vertically stretching the graph by factor -3, then by shifting this graph 1 unit in the vertical direction. The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). -3 * 2^x will approach zero as 2^x approaches zero, so the graph of y = -3 * 2^x will remain asymptotic to the negative x axis. The points (0, 1) and (1, 2) will be transformed to (0, -3* 1) = (0, -3) and (1, -3 * 2) = (1, -6). So the graph will decrease from its asymptote just below the x axis through the points (0, -3) and (1, -6), decreasing at a rapidly decreasing rate. To get the graph of 1 - 3 * 2^x the graph of -3 * 2^x will be vertically shifted 1 unit. This will raise the horizontal asymptote from the x axis (which is the line y = 0) to the line y = 1, and will also raise every other point by 1 unit. The points (0, -3) and (1, -6) will be transformed to (0, -3 + 1) = (0, -2) and (1, -6 + 5) = (1, -5), decreasing from its asymptote just below the line y = 1 through the points (0, -2) and (1, -5), decreasing at a rapidly decreasing rate. If you don't understand the above, then do as follows, without looking up at the solution given so far: • Plot the basic points (0, 1) and (1, 2) of the y = 2^x function. • Multiply your y values by 3 to get the basic points of the y = 3 * 2^x function, plot your points and sketch the graph (this is your 'vertical stretch'; you should see that it moves your original points 3 times as far from the x axis as before. The same thing happens to all the points of the original y = 2^x graph--they all move 3 times as far from the x axis.) • Multiply your y values by -1. This gives you the basic points of the y = - 3 * 2^x function, plot your points and sketch the graph.. • Add 1 to your y values, plot your points and sketch your graph. This gives you the basic points of the y = 1 -3 * 2^x function. (It should be clear that this 'shifts' the points of your y = - 3 * 2^x graph 1 unit in the vertical direction). Having done this, look again at the given solution. You might also consider the following table: x y = 2^x y = 3 * 2^x y = -3 * 2^x y = 1 -3 * 2^x 0 1 3 -3 -2 1 2 6 -6 -5 It should be clear how this table demonstrates the process described above (get y values of basic function, multiply by 3, multiply by -1, add 1), and how you transform the 'basic points' from (0, 1) and (1, 2) to get (0, 3) and (1, 6), then (0, -3) and (1, -6) and finally (0, -2) and (1, -5). You should identify these points on the graph depicted below, and having idenfied the basic points you should be able to identify which function is which. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * Extra Problem / 7th edition 5.3.60 Solve (1/2)^(1-x) = 4. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the law of logs we can get ln (1/2)^(1-x) =ln(4) (1 - x) ln(1/2) = 1n(4) 1 - x = ln(4)/1n(1/2) 1 - x = -2 x = 3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The reliable way to solve a problem of this nature is to recognize that the variable x occurs in an exponent, so we can 'get at it' by taking logs of both sides: (1/2)^(1-x) = 4. Taking the natural log of both sides, using the laws of logarithms, we get ln( (1/2)^(1 - x) = ln(4)) so that (1 - x) ln(1/2) = ln(4) and (1 - x) = ln(4) / ln(1/2). A calculator will reveal that ln(4) / ln(1/2) = -2, so that 1 - x = -2. We easily solve for x, obtaining x = 1. It is also possible to reason this problem out directly, and in this case our reasoning leads us to an exact solution: We first recognize one fact: • 4 is an integer power of 2, and 1/2 is an integer power of 2, so 4 must also be an integer power of 1/2. Since 4 = 2^2, and since 1/2 = 2^-1, we can recognize that 4 = 2^-2, and reason as follows: • (1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 4, so we know that the exponennt 1 - x must be -2 • If 1 - x = -2, it follows that -x = -2 - 1 = -3 and x = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.3.98 / 7th edition 5.3.78 A(n) = A0 e^(-.35 n), area of wound after n days. What is the area after 3 days and what is the area after 10 days? Init area is 100 mm^2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * If the initial area is 100 mm^2, then when n = 0 we have A(0) = A0 e^(-.35 * 0) = 100 mm^2. Since e^0 = 1 this tells us that A0 = 100 mm^2. So the function is A(n) = 100 mm^2 * e^(-.35 n). To get the area after 3 days we evaluate the function for n = 3, obtainind A(3) = 100 mm^2 * e^(-.35 * 3) = 100 mm^3 * .35 = 35 mm^2 approx.. After to days we find that the area is A(10) = 100 mm^2 * e^(-.35 * 10) 100 mm^3 * .0302 = 3.02 mm^2 approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.3.104 / 7th edition 5.3.84. Poisson probability 4^x e^-4 / x!, probability that x people will arrive in the next minute. What is the probability that 5 will arrive in the next minute, and what is the probability that 8 will arrive in the next minute? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The probability that 5 will arrive in the next minute is P(5) = 4^5 * e^-4 / (5 !) = 1024 * .0183 / (5 * 4 * 3 * 2 * 1) = 1024 * .0183 / 120 = .152 approx.. The probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-4 / (8 !) = .030 approx.. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm. Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 037. * 37 ********************************************* Question: * * * 7th edition only 5.4.14. 2.2^3 = N. Express in logarithmic notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: b^x = y is expressed in logarithmic notation as log{base b}(y) = x so we will substitute in the our equation log{base 2.2} * (N) = 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 2.2, x = 3 and y = N. So we write lob{base b}(y) = x as log{base 2.2}(N) = 3. 8th edition only 6.4.14. e^2.2 = N. Express in logarithmic notation. b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 3, x = 2.2 and y = N. So we write lob{base b}(y) = x as log{base e}(N) = 2.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * extra problem / 7th edition 5.4.18. x^pi = 3. Express in logarithmic notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base x} * (3) = pi confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: b^a = y is expressed in logarithmic notation as log{base b}(y) = a. In this case b = x, a = pi and y = 3. So we write lob{base b}(y) = a as log{base x}(3) = pi. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * extra problem / 7th edition 5.4.26.. log{base 2}?? = x. Express in exponential notation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2^x = ?? confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 2, a = x and y = ?? so the expression b^a = y is written as 2^x = ??. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.4.30 / 7th edition 5.4.36. Exact value of log{base 1/3}(9) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/3^a = 9 to solve this we must find what exponent will make both sides equal 1/3^2 = 1/9 so now we know that if we use -2 that will give us the recipical 1/3^-2 = 9 log{base 1/3}(9) = -2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 1/3, a is what we are trying to find and y = 9 so the expression b^a = y is written as (1/3)^a = 9. Noting that 9 is an integer power of 3 we expect that a will be an integer power of 1/3. Since 9 = 3^2 we might try (1/3)^2 = 9, but this doesn't work since (1/3)^2 = 1/9, not 9. We can correct this by using the -2 power, which is the reciprocal of the +2 power: (1/3)^-2 = 1/ ( 1/3)^2 = 1 / (1/9) = 9. So log base 1/3}(9) = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): oki ------------------------------------------------ Self-critique Rating: ********************************************* Question:. What is the domain of G(x) = log{base 1 / 2}(x^2-1) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For any positive value of b the domain of log{base b}(x) consists of all positive real numbers. It follows that log{base 1/2}(x^2-1) consists of all real numbers for which x^2 - 1 > 0. We solve x^2 - 1 > 0 for x by first finding the values of x for which x^2 - 1 = 0. We can factor the expression to get (x-1)(x+1) = 0, which is so if x-1 = 0 or if x + 1 = 0. Our solutions are therefore x = 1 and x = -1. It follows that x^2 - 1 is either positive or negative on each of the intervals (-infinity, -1), (-1, 1) and (1, infinity). We can determine which by substituting any value from each interval into x^2 - 1. On the interval (-1, 1) we can just choose x = 0, which substituted into x^2 - 1 gives us -1. We conclude that x^2 - 1 < 0 on this interval. On the interval (-infinity, -1) we could substitute x = -2, giving us x^2 - 1 = 4 - 1 = 3, which is > 0. So x^2 - 1 > 0 on (-infinity, -1). Substituting x = 2 to test the interval (1, infinity) we obtain the same result so that x^2 - 1 > 0 on (1, infinity). We conclude that the domain of this function is (-infinity, -1) U (1, infinity). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.4.58 / 7th edition 5.4.62. a such that graph of log{base a}(x) contains (1/2, -4). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: log{base a}(x) = y if a^y = x. The point (1/2, -4) will lie on the graph if y = -4 when x = 1/2, so we are looking for a value of a such that a ^ (-4) = 1/2. * * We easily solve for a by taking the -1/4 power of both sides: (a ^ (-4))^(-1/4) = 1/2 ^ (-1/4) so since (a^(-4))^(-1/4)) = a^(-4 * (-1/4)) = a^1 = a, and 1/2 ^(-1/4) = 16, we get a = (1/2)^-(1/4) = 16. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Use transformations to graph h(x) = ln(4-x). Give domain, range, asymptotes. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph of y = ln(x) is concave down, has a vertical asymptote at the negative y axis and passes through the points (1, 0) and (e, 1) (the latter is approximately (2.71828, 1) ). The graph of y = ln(x - 4), where x is replaced by x - 4, is shifted +4 units in the x direction so the vertical asymptote shifts 4 units right to the line x = 4. The points (1, 0) and (e, 1) shift to (1+4, 0) = (5, 0) and (e + 4, 1) (the latter being approximately (3.71828, 1). Since x - 4 = - (4 - x), the graph of y = ln(4 - x) is 'flipped' about the y axis relative to ln(x-4), so it the vertical asymptote becomes x = -4 and the graph passes through the points (-5, 0) and (-e-4, 1). The graph will still be concave down. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.4.98 / 7th edition 5.4.102. Solve log{base 6}(36) = 5x + 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6^(5x + 3) = 36 we have to solve for a which is 5x + 3 so we look at the 6 and know that 6^2 is 36 so now we can solve for x 5x + 3 = 2 5x = -1 x = -1/5 log{base 6}(36) = 2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * log{base b}(y) = a is expressed in exponential notation as b^a = y. In this case b = 6, a is what we are trying to find and y = 5x + 3 so the expression b^a = y is written as 6^(5x+3) = 36. We know that 6^2 = 36, so (5x + 3) = 2. We easily solve this equation to get x = -1 / 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.4.122 / 7th edition 6.3.102. F(t) = 1 - e^(-.15 t) prob of car arriving within t minute of 5:00. How long does it take for the probability to reach 50%? How long to reach 80%? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(t) = 1 - e^(-.15 t) = .5 -e^(-.15 t) = -.5 e^(-.15 t) = .5 ln( e^(-.15 t) ) = ln(.5) -.15 t = ln(.5) t = -ln(.5)/.15 = 4.62 approx. 50% t = -ln(.8)/.15= 1.49 approx. 80% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 3 * * The probability is F(t), so F(t) = 50% = .5 when 1 - e^(-.15 t) = .5. We subtract 1 from both sides the multiply by -1 to rearrange this equation to the form e^(-.15 t) = .5. Taking the natural log of both sides we get ln( e^(-.15 t) ) = ln(.5), which tells us that -.15 t = ln(.5) so that t = -ln(.5)/.15 = 4.6, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm. Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 038. * 38 ********************************************* Question: * 6.5.18 / 7th edition 5.5.18. Exact value of log{base 3}{(8) * log{base 8}(9). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: {base 3}{(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9) = 2 because three must be taken up to the second power to equal 9 {base 3}{(8) * log{base 8}(9) = 2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9). log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2. Thus log{base 3}(8) * log{base 8}(9) = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.5.24 / 7th edition 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln(2/3) = ln(2) - ln(3) = a - b confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ln(2/3) = ln(2) - ln(3) = a - b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Since ln(2) = a, and since ln(1) = 0, we have ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) - log{base 3}(v) as a single log. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 3}(u^2) - log{base 3}(v) = log{base 3}(u^2 / v) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * log{base 3}(u^2) - log{base 3}(v) = log{base 3}(u^2 / v). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln y = ln(x + C) = (x + c) = y^e confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * a = ln(b) means e^a = b, so y = ln(x+c) is translated to exponential form as (x+c) = e^y. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!