course MTH 158 ºõ{ª¿ÉXÙâçæõ”áTœûÚ®{bëÂäþÎÏ—ôÕassignment #035
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09:08:30 5.2.18. Horiz line test, looks like log. What did the horizontal line test tell you for this function?
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RESPONSE --> That the function is one-to-one confidence assessment: 3
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09:08:40 There is no horizontal line that passes through this graph more than once. The function is strictly increasing, taking each y value only once. The function is therefore one-to-one on its domain.
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RESPONSE --> ok self critique assessment: 3
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09:09:13 5.2.20. Horiz line test, looks like inverted parabola or hyperbola. What did the horizontal line test tell you for this function?
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RESPONSE --> The horizontal line passes through the graph more than one time, so the graph is not one-to-one. confidence assessment: 3
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09:09:27 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.
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RESPONSE --> ok self critique assessment: 3
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09:39:26 5.2.28 looks like cubic thru origin, (1,1), (-1,-1), sketch inverse. Describe your sketch of the inverse function.
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RESPONSE --> The sketch of the inverse I drew looks like the original cubic graph seen in a mirror. The graph still goes through the points (1,1) and (-1, -1), but in quadrant III, istead of the original graph starting from -infinity below the y = x line, intersecting the line at (-1,-1), and rising above until it meets at the origin again, the inverse graph starts from -infinity above the y = x line, then curves below the line after the intersection at (-1,-1). Above the origin, the inverse graph curves above the y = x line, intersects it at (1,1), and continues to infinity below the line. confidence assessment: 3
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09:40:07 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.
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RESPONSE --> ok self critique assessment: 3
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09:52:59 5.2.32 f = 2x + 6 inv to g = 1 / 2 * x – 3. Show that the functions f(x) and g(x) are indeed inverses.
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RESPONSE --> g(f(x)) = 2 (1/2 x - 3) + 6 = x - 6 + 6 = x f(g(x)) = 1/2 (2x + 6) - 3 = x + 3 - 3 = x confidence assessment: 3
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09:53:06 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.
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RESPONSE --> ok self critique assessment: 3
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10:14:43 5.2.44. inv of x^3 + 1; domain range etc.. Give the inverse of the given function and the other requested information.
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RESPONSE --> y = x^3 + 1 , x = y^3 + 1 , y^3 = x - 1 , y = (x - 1)^1/3 Checking, f(f^-1(x)) = f ((x-1)^1/3) = ((x-1)^1/3)^3 + 1 = x - 1 + 1 = x Checking, f^-1(f(x)) = f^-1 (x^3 + 1) = ((x^3 + 1) - 1)^1/3 = (x^3)^1/3 =x Domain of f = range of f^-1 = (-infinity, infinity) Range of f = domain of f^-1 = (-infinity, infinity) confidence assessment: 3
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10:15:10 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.
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RESPONSE --> ok self critique assessment: 3
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10:46:15 5.2.56. inv of f(x) = (3x+1)/(-x). Domain and using inv fn range of f. What is the domain of f? What is the inverse function? What does the inverse function tell you about the range of f?
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RESPONSE --> x = (3y + 1) / -y is the inverse x(-y) = 3y + 1 -xy = 3y + 1 -xy - 1 = 3y y = (-xy - 1) / 3 Domain of f = range of f^-1 = all real numbers except 0 Range of f = domain of f^-1 = all real numbers confidence assessment: 2
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10:48:01 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 -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 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.
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RESPONSE --> I didn't see that I could subtract 3y, then factor y out of all but one side... That definitely helps. self critique assessment: 3
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11:02:52 5.2.74. T(L) = 2 pi sqrt ( L / g). Find L(T).
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RESPONSE --> T = 2pi sqrt(L/g) ( sqrt(g)) T = 2 pi sqrt(L/g)(sqrt(g)) T sqrt(g) = 2 pi sqrt(L) T (sqrt(g))(sqrt(L)) = 2 pi L L = T (sqrt(g))(sqrt(L)) / 2pi L(T) = T (sqrt(g))(sqrt(L)) / 2pi g > 0, L > 0 confidence assessment: 2
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11:03:38 We solve T = 2 pi sqrt( L / g) for L. First squaring both sides we obtain T^2 = 4 pi^2 * L / g. Multiplying both sides by 6 / ( 4 pi^2) we get L = T^2 g / (4 pi^2). So our function L(T) is L(T) = T^2 g / (4 pi^2).
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RESPONSE --> Oh my goodness, so you just had to square both sides... argh. Ok, I'm sorry. self critique assessment: 2
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»Ü†ïáÍ´d´Å¤£ÁÁýà‹¨ï™Šúš assignment #036 036. Query 36 College Algebra 07-29-2008
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13:01:52 5.3.38. Explain how you used transformations to graph f(x) = 2^(x+2)
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RESPONSE --> Using the graph of y = 2^x, shift it to the right 2 units confidence assessment: 2
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13:02:12 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. 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 right of points on the graph of y = 2^x.
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RESPONSE --> ok self critique assessment: 3
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13:17:38 5.3.42. Transformations to graph f(x) = 1 – 3 * 2^x
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RESPONSE --> Using the graph of y = 2^x, stretch the graph vertically by a factor of 3, reflect about the x-axis, and move graph up 1 unit Domain is the set of all real numbers Range is (1, -infinity) Horizontal asymptote is y = 0 confidence assessment: 2
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13:18:24 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.
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RESPONSE --> Was my answer correct?
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13:23:36 5.3.60 Solve (1/2)^(1-x) = 4.
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RESPONSE --> (2^-1)^1 - x = 2^2 2^x-1 = 2^2 x - 1 = 2 x = 3 confidence assessment: 2
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13:23:41 (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.
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RESPONSE --> ok self critique assessment: 3
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13:31:46 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.
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RESPONSE --> After 3 days the area is approx. 34.99 square mm After 10 days the area is approx. 3.02 square mm confidence assessment: 2
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13:32:02 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 mm63 * .0302 = 3.02 mm^2 approx..
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RESPONSE --> ok self critique assessment: 3
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13:46:28 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?
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RESPONSE --> P(x) = (4^x e^-4) / x! P(x) = (4^5 e^-4) / 5! P(x) = 0.1563 There is a 15.63% chance that 5 people will arrive in the next minute. P(x) = (4^8 e^-4) / 8! P(x) = .0298 There is a 2.98% chance that 8 people will arrive in the next minute. confidence assessment: 2
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13:47:06 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..
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RESPONSE --> ok, is that close enough to my answer?
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13:49:06 The probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-8 / (8 !) = .00055 approx..
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RESPONSE --> The formula was not P(x) = (4^x e^-x) / x!, it was P(x) = (4^x e^-4) / x! there should be no e^-8 in question self critique assessment: 3
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