course Mth163 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aSTUDENT RESPONSE: If you multiply any number by zero and you get zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why, when the magnitude | f(x) | of f(x) is greater than 1 (i.e., when the graph of f(x) is more than one unit from the x axis), then the product function will be further from the x axis than the g(x) function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If f(x) is 1 then the product of f g =g but if f(x) > 1 then the product of f g would be greater than g , thus the graph of the product is further away from the x axis than the graph of g(x) confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: If you multiply a number by another number greater than 1, the result is greater than the original number. If you multiply a number by another number whose magnitude is greater than 1, the result will have greater magnitude that the original number. If | f(x) | > 1 then the magnitude of f(x) * g(x) will be greater than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude increases so does the distance from the x axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok remember it was in comparison with the magnitude of | f(x) | > 1 ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why, when the magnitude | f(x) | of f(x) is less than 1 (i.e., when the graph of f(x) is less than one unit from the x axis), then the product function will be closer to the x axis than the g(x) function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the magnitude of | f (x) | < 1 then f(x) < 1 you will have f(x) * g(x) giving you a product with a less magnitude than g(x) because a number times another number that is less then 1 will give you a lesser number then that number, This will give you a product that is closer to the x axis than the g(x). confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: If you multiply a number by another number less than 1, the result is less than the original number. If you multiply a number by another number whose magnitude is less than 1, the result will have a lesser magnitude that the original number. If | f(x) | < 1 then the magnitude of f(x) * g(x) will be less than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude decreases so does the distance from the x axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why, when f(x) and g(x) are either both positive or both negative, the product function is positive. When f(x) and g(x) have opposite signs the product function is negative. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because when you multiply a negative number by a negative number you will always get a possitive number and same as a positive * a positive: So the product of f(x) andg(x) will be possitive if it is a negative * a negative or a positive * a positive. If you have a negative times a positive number the product will always be a negative number : So if mf(x) * g(x) is a negative * a positive you have a negative product. confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: This is basic multiplication: + * + = +, - * - = -, + * - = -. The product of like signs is positive, the product of unlike signs is negative. Since the product function results from multiplication of the two functions, these rules apply. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why, when f(x) = 1, the graph of the product function coincides with the graph of g(x). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because any number time 1 will equal that number so if f(x) = 1 then f g = g, giving you a product that is coincides with g(x) confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: g(x) * 1 = g(x) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `q problem 4 Sketch graphs for y = f(x) = 2^x and y = g(x) = .5 x, for -2 < x < 2. Use your graphs to predict the shape of the y = g(x) * f(x) graph. Describe the graphs of the two functions, and explain how you used these graphs predict the shape of the graph of the product function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If g(x) < 0 and f(x) > 0 then the product will be negative and lie below the x axis: If g(x) >0 and f(x) >0 then the product will be possitive and lie above the x axis growing greatly as g(x) and f(x) grow: If g(x) <0 and f(x) >0 the product will be positive and lie above the x axis: If g(x) = 0 then g f = 0 confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: Where g(x) is - and f(x) is + graph will be -. where g(x) =0 graph will be at 0where both are + graph will be positive and rise more steeply. y=2^x asymptote negative x axis y intercept (0,1) y=.5x linear graph passing through (0,0) rising 1 unit for run of 2 units INSTRUCTOR COMMENT: It follows that since one function is negative for x < 0 while the other is always positive the product will be negative for x < 0, and since both functions are positive for x > 0 the product will be positive for x > 0. Since one function is zero at x=0 the product will be 0 at x = 0. For x > 0 the exponential rise of the one graph and the continuing rise of the other imply that the graph will rise more and more rapidly, without bound, for large positive x. For x < 0 one function is positive and the other is negative so the graph will be below the x axis. For large negative x, one graph approaches 0 while the other keeps increasing in magnitude; it's not immediatly clear which function 'wins'. However the exponential always 'beats' a fixed power so the graph will be asymptotic to the negative x axis. It will reach a minimum somewhere to the left of the x axis, before curving back toward the x axis and becoming asymptotic. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The exponential always beats a fixed power so the graph will be asymptote to the negative x axis. It will reach a minimum somewhere to the left of the x axis, before curving back towards the x axis and becoming asymptotic. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q problem 7 range(depth) = 2.9 `sqrt(depth) and depth(t) = t^2 - 40 t + 400. At what times is depth 0. How did you show that the vertex of the graph of depth vs. time coincides with these zeros? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Depth(t) = t^2 – 40t + 400 since this is a quadratic function you can factor it out to find the zero’s and get (t – 20)(t – 20) zero is (20,0) The vertex occures at -b / (2a) -(40) / (2 * 1) = 20 The graph will stay above the x and open upward. confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The depth function is quadratic. Its vertex occurs at t = - b / (2 a) = - (-40) / (2 * 1) = 20. Its zeros can be found either by factoring or by the quadratic formula. t^2 - 40 t + 400 factors into (t - 20)(t - 20), so the only zero is at t = 20. This point (20, 0) happens to be the vertex, and the graph opens upward, so the graph never goes below the x axis. STUDENT QUESTION: When I simplified range(depth(t) = 2.9*'sqrt(t^2 - 40t +400) I got 2.9t - 58 which gives a negative range so I reversed it and got the correct results, what have I done wrong? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2.9*`sqrt(t^2-40t +400) 2.9 * `sqrt[(t-20)^2] 2.9(t-20) 2.9t-58 If I reverse it as the questions says I get -2.9 + 58 but why? confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `arange(depth(t) = 2.9*'sqrt(t^2 - 40t +400) = 2.9*'sqrt( (t-20)^2) ) = 2.9( t - 20) = 2.9t - 58 I know it should be -2.9 + 58 I just don't understand how to get there. Thanks INSTRUCTOR RESPONSE TO QUESTION: This is a great question. What is sqrt( (-5) ^ 2)? `sqrt( (-5)^2 ) isn't -5, it's 5, since `sqrt(25) = 5. This shows that you have to be careful about possible negative values of t - 20. This is equivalent to saying that `sqrt( (-5)^2 ) = | -5 = 5|. `sqrt( (t-20) * (t-20) ) has to be positive. So `sqrt( (t-20) * (t-20) ) = | t - 20 |. If t < 20 then t - 20 is negative so that | t - 20 | = -(t - 20) = 20 - t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If you can say `sqrt((-5)^2) = |-5=5| Then you can say the same for `sqrt((t-20)*(t-20))=|t – 20|. If t <20 then t-20 is negative so that |t-20| - -(t-20) = 20-t ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat two linear factors represent the depth function as their product? depth(t) = (t-20)(t-20) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qFor t = 5, 10 and 15, what are the ranges? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If depth(t) = (t-20)(t-20) or (t-20)^2 Depth(5) = (5-20)^2 = 225 Depth(10) = (10-20)^2 = 100 Depth(15) = (15-20)^2 = 25 So if the range(depth) = 2.9 * `sqrt(depth) then: Range(depth(5) = 2.9 * `sqrt(225) = 43.5 Range(depth(10) = 2.9 * `sqrt(100) = 29 Range(depth(15) = 2.9 * `sqrt(25) = 14.5 confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** depth(t) = t^2 - 40 t + 400 = (t-20)^2 so depth(5) = (5-20)^2 = (-15)^2 = 225 depth(10) = (10-20)^2 = (-10)^2 = 100 depth(15) = (15-20)^2 = (-5)^2 = 25. It follows that the ranges are range(depth(5)) = 2.9 sqrt(225) = 43.5 range(depth(10) = 2.9 sqrt(100) = 29 and range(depth(15) = 2.9 sqrt(25) = 14.5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the function range(depth(t))? Show that its simplified form is linear in time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As showed in the previous problems the function of range(depth(t) = 2.9 * `sqrt(t-20)^2 or 2.9 (t – 20) confidence rating: possitive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** range(depth(t) ) = 2.9 sqrt(depth(t)) = 2.9 sqrt(t^2 - 40 t + 400) = 2.9 sqrt( (t - 20)^2 ) = 2.9 | t - 20 |. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, range(depth(t)) = 2.9 |t-20| ------------------------------------------------ Self-critique rating: ********************************************* Question: `qproblem 8 Illumination(r) = 40 / r^2; distance = 400 - .04 t^2. What is the composite function illumination(distance(t))? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If illumination(r) = 40 / r^2 and distance = 400 - .04t^2 Then the function for illumination(distance(t)) = 40 / (400 - .04 t^2)^2 confidence rating:mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Illumination(r) = 40 / r^2 so Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qGive the illumination at t = 25, t = 50 and t = 75. At what average rate is illumination changing during the time interval from t = 25 to t = 50, and from t = 50 to t = 75? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Illumination(distance(25)) = 40 / (400 - .04*25^2)^2 = my calculator gives me 2.844444444E-4 which is ..000284 approx. Illumination(distance(50)) = 40 / (400 - .04*50^2))^2 = my calculator gives me 4.444444444 E-4 which is .000444 approx Illumination(distance(75)) = 40 / (400 - .04 * 75^2)^2 = .001361 The average rate change is: (.000444 - .0002840 / (50-25) =6.4E-6 which is.0000064 (.001361 - .000444) / (75-25) = 1.843E-5 which is .00001834 confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** illumination(distance(t)) = 40 / (400 - .04 t^2)^2 so illumination(distance(25)) = 40 / (400 - .04 * 25^2)^2 = .000284 illumination(distance(50)) = 40 / (400 - .04 * 50^2)^2 = .000444 illumination(distance(75)) = 40 / (400 - .04 * 75^2)^2 = .001306. from 25 to 50 change is .000444 - .000284 = .000160 so ave rate is .000160 / 25 = .0000064 from 50 to 75 change is .001306 - .000444 = .00086 so ave rate is .00086 / 25 = .000034 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???? Mr. smith, I don’t have the same adverage rate of change as you did, can you explain what I did wrong ????
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Given Solution: `a** gradeAverage = -.5 + t / 10 = -.5 + 50 ( 1 - e^(-.02 (Q - 70) ) ) / 10 = -.5 + 5 ( 1 - e^(-.02 (Q - 70) ) ) So gradeAverage(t(100)) = -.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = -.5 + 5( 1 - .5488 ) = -.5 + 5 ( .4522 ) = -.5 + 2.26 = 1.76. gradeAverage(t(110)) = -.5 + 5 ( 1 - e^(-.02 ( 110 - 70) ) = -.5 + 5( 1 - .4493 ) = -.5 + 5 ( .5517 ) = -.5 + 2.76 = 2.26. gradeAverage(t(120)) = -.5 + 5 ( 1 - e^(-.02 ( 120 - 70) ) = -.5 + 5( 1 - .3678 ) = -.5 + 5 ( .6322 ) = -.5 + 3.16 = 2.66. gradeAverage(t(130)) = -.5 + 5 ( 1 - e^(-.02 ( 130 - 70) ) = -.5 + 5( 1 - .3012 ) = -.5 + 5 ( .6988 ) = -.5 + 3.49 = 2.99. As Q gets larger and larger Q - 70 will get larger and larger, so -.02 ( Q - 70) will be a negative number with increasing magnitude; its magnitude increases without limit. It follows that e^(-.02 ( Q - 70) ) = will consist of e raised to a negative number whose magnitude increases without limit. As the magnitude of the negative exponent increases the result will be closer and closer to zero. So -.5 + 5 ( 1 - e^(-.02 ( Q - 70) ) ) will approach -.5 + 5 ( 1 - 0) = -.5 + 5 = 4.5. Side note: For Q = 100, 200 and 300 we would have grade averages 1.755941819, 4.128632108, 4.449740821. To get a 4-point Q would have to be close to 200. Pretty tough course &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, but I could have made one problem like this: -.5 + [50(1-e^(-.02(100 – 70))) ] / 10
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Given Solution: `a110...2.2534, 120...2.66, 130...2.99 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the upper limit on the expected grade average that can be achieved by this student? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: when q is a very large number the bigger the bottom of the fraction involving e will get and the closer it will come to 0. if the fraction was 0, then t = 50 confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If Q is very, very large, e^-( .02(Q-70) ) would have a negative exponent with a very large magnitude and so would be very close to 0. In this case 50 ( 1-e^(-.02 (Q-70)) would be close to 50(1-0) = 50. Then the grade average would be -.5 + 50 / 10 = -.5 + 5 = 4.5 . DER [0.5488116360, 0.4493289641, 0.3678794411, 0.3011942119][1.755941819, 2.253355179, 2.660602794, 2.994028940] &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the composite function gradeAverage( t(Q) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: grade average (t(Q)) = -.5+[50(1-e^(-.02(Q-70))) ] / 10 confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a-.5+(50(1-e^(-.02(Q-70))/10 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat do you get when you evaluate your composite function at t = 100, 110, 120 and 130? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t(100) = -.5 + [ 50(1-e^(-.02(100-70))) ] / 10 = 1.756 approx t(110) = -.5 + [50(1-e^(-.02(110-70))) ] / 10 = 2.253 approx. t(120) = -.5 + [50(1-e^(-.02(120-70))) ] / 10 = 2.661 approx t(130) = -.5 + [50(1-e^(-.02(130-70))) ] /10 = 2.994 approx. confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `at=100...9.5 t=110...10.5 t=120...11.5 t=130...12.5 ???? Mr. Smith I don’t understand how you got this answer ???? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** you should get the same values you got before for these Q values. For example an approximate calculation for t = 130 is -.5 + 50(1-e^(-.02(130-70) ) / 10 = -.5 + 50 (1-e^-1.2) / 10 = -.5 + 50 (1 - .3) / 10 = -.5 + 35/10 = -.5 + 3.5 = 3, approx., pretty close to your 2.99 ** Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ???? Ok. but what is the question???? ------------------------------------------------ Self-critique rating: