#$&* course Mth 163 4/26 6:00 24 query*********************************************
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Given Solution: `aSTUDENT RESPONSE: If you multiply any number by zero and you get zero. INSTRUCTOR NOTE: Right. If f(x) = 0 then f(x) * g(x) = 0; and the same is so if g(x) = 0. If one number in a product is zero, then the product is zero. ********************************************* Question: `qExplain why, when the magnitude | f(x) | of f(x) is greater than 1 (i.e., when the graph of f(x) is more than one unit from the x axis), then the product function will be further from the x axis than the g(x) function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When a number bigger than 1 is multiplied by another number the outcome is bigger than the number that you started with unless it is a negative and it will be smaller. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. INSTRUCTOR CLARIFICATION: Your response is correct if the original number is positive. However if it's negative, the result will be less than the original number (e.g.., if you multiply 3 by 2 you get 6, which is greater than the original number 3; however if you multiply -3 by 2 you get -6, which is less than -3). The general statement would be If you multiply a number by another number whose magnitude is greater than 1, the result will have greater magnitude than the original number.(e.g., the magnitude of -3 is | -3 | = 3; if you multiply | -3 | by 2 you get 6, which is greater than the magnitude 3 of the original number). Applying this to the present situation: 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 | g(x) |, and this represents its distance from the x axis. So when the magnitude increases so does the distance from the x axis. ********************************************* 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: With a number less than 1 is multiplied by a magnitude less than 1 the solution will be less than the beginning number. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aSTUDENT RESPONSE: If you multiply a number by another number less than 1, the result is less than the original number. If you multiply a number by another number whose magnitude is less than 1, the result will have a lesser magnitude that the original number. If | f(x) | < 1 then the magnitude of f(x) * g(x) will be less than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude decreases so does the distance from the x axis. ********************************************* Question: `qExplain why, when f(x) and g(x) are either both positive or both negative, the product function is positiv; and when f(x) and g(x) have opposite signs the product function is negative. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When 2 numbers are positive it will = a positive solution. The same with 2 negative numbers. If one number is positive and the other a negative it will = a negative number. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. INSTRUCTOR RESPONSE: Right. If f(x) and g(x) are both positive, then the product function f(x) * g(x) is positive. If f(x) and g(x) are both negative, then the product function f(x) * g(x) is positive. If f(x) and g(x) are unlike, then the product function f(x) * g(x) is negative. ********************************************* Question: `qExplain why, when f(x) = 1, the graph of the product function coincides with the graph of g(x). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Anything multiplied by 1 remains the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aSTUDENT RESPONSE: g(x) * 1 = g(x) ********************************************* Question: `q problem 4 Sketch graphs for y = f(x) = 2^x and y = g(x) = .5 x, for -2 < x < 2. Use your graphs to predict the shape of the y = g(x) * f(x) graph. Describe the graphs of the two functions, and explain how you used these graphs predict the shape of the graph of the product function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The function y = f(x) = 2^x is positive for any value of x because the power is postive. The function y = g(x) = .5 x is negative when x < 0, positive when x > 0 and zero when x = 0. When x < 0, then, f(x) is negative and g(x) is positive, so the product function f(x) g(x) must be negative. When x > 0, both functions are positive so the product function is positive. Since g(x) = 0 when x = 0, the product f(x) * g(x) will be 0 at x = 0. For x > 0 the exponential rise of the one graph and the continuing rise of the other imply that the graph will rise more and more rapidly, without bound, for large positive x. For x < 0, as we have seen, one function is positive and the other is negative so the graph will be below the x axis. For large negative values of x, one graph approaches 0 while the other keeps increasing in magnitude; it's not immediately clear which function 'wins'. However the exponential always 'beats' a fixed power so the graph will be asymptotic to the negative x axis. The graph will reach a minimum somewhere to the left of the x axis, before curving back toward the x axis and becoming asymptotic confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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: The function y = f(x) = 2^x is positive for any value of x (since any power of 2 is positive). The function y = g(x) = .5 x is negative when x < 0, positive when x > 0 and zero when x = 0. When x < 0, then, f(x) is negative and g(x) is positive, so the product function f(x) g(x) must be negative. When x > 0, both functions are positive so the product function is positive. Since g(x) = 0 when x = 0, the product f(x) * g(x) 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, as we have seen, one function is positive and the other is negative so the graph will be below the x axis. For large negative values of x, one graph approaches 0 while the other keeps increasing in magnitude; it's not immediately clear which function 'wins'. However the exponential always 'beats' a fixed power so the graph will be asymptotic to the negative x axis. The graph will reach a minimum somewhere to the left of the x axis, before curving back toward the x axis and becoming asymptotic. ** ********************************************* 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: Dept is 0 at t=20. The graph never goes below the x axis. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The depth function is quadratic. Its vertex occurs at t = - b / (2 a) = - (-40) / (2 * 1) = 20. Its zeros can be found either by factoring or by the quadratic formula. t^2 - 40 t + 400 factors into (t - 20)(t - 20), so its only zero is at t = 20. This point (20, 0) happens to be the vertex, and the graph opens upward, so the graph never goes below the x axis. ********************************************* Question: `qWhat two linear factors represent the depth function as their product? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Depth(t)=(t-20)(t-20) which is = to t-20 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: The quadratic function factors as depth(t) = (t-20)(t-20). This function consists of two identical linear factors, each equal to t - 20. ********************************************* Question: `qFor t = 5, 10 and 15, what are the ranges of the stream? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: depth(t) = t^2 - 40 t + 400 = (t-20)^2 depth(5) = (5-20)^2 = (-15)^2 = 225 depth(10) = (10-20)^2 = (-10)^2 = 100 depth(15) = (15-20)^2 = (-5)^2 = 25 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `a** depth(t) = t^2 - 40 t + 400 = (t-20)^2 so depth(5) = (5-20)^2 = (-15)^2 = 225 depth(10) = (10-20)^2 = (-10)^2 = 100 depth(15) = (15-20)^2 = (-5)^2 = 25. It follows that the ranges of the stream are range(depth(5)) = 2.9 sqrt(225) = 43.5 range(depth(10) = 2.9 sqrt(100) = 29 and range(depth(15) = 2.9 sqrt(25) = 14.5. ** ********************************************* Question: `qWhat is the function range(depth(t))? Show that its simplified form is linear in time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: range(depth(t) ) = 2.9 sqrt(depth(t)) = 2.9 sqrt(t^2 - 40 t + 400) = 2.9 sqrt( (t - 20)^2 ) = 2.9 [ t - 20] confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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 |. ** This function isn't actually linear. Its graph consists of two linear rays, the first of negative slope ending at t = 20 and the second of positive slope starting at t = 20. The graph forms a 'v' shape. In reality if the quadratic function represents the depth of water in a leaking container, the useful domain of the function ends when the container is empty (i.e., when the depth reaches its 'low point' corresponding to the vertex of the parabola), and the composite function would represent the stream range only up to this point. So the 'real-world' range(t) function would in fact be linear.** ********************************************* Question: `qproblem 8 Illumination(r) = 40 / r^2; distance = 400 - .04 t^2. What is the composite function illumination(distance(t))? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Illumination(r) = 40 / r^2 Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** Illumination(r) = 40 / r^2 so Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2. ** ********************************************* Question: `qGive the illumination at t = 25, t = 50 and t = 75. At what average rate is illumination changing during the time interval from t = 25 to t = 50, and from t = 50 to t = 75? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: illumination(distance(t)) = 40 / (400 - .04 t^2)^2 illumination(distance(25)) = 40 / (400 - .04 * 25^2)^2 = .000284 illumination(distance(50)) = 40 / (400 - .04 * 50^2)^2 = .000444 illumination(distance(75)) = 40 / (400 - .04 * 75^2)^2 = .001306. 25 to 50 change is .000444 - .000284 = .000160 rate is .000160 / 25 = .0000064 50 to 75 change is .001306 - .000444 = .00086 rate is .00086 / 25 = .000034 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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. ** ********************************************* Question: `q problem 10 gradeAverage = -.5 + t / 10. t(Q) = 50 (1 - e ^ (-.02 (Q - 70) ) ). If the student's mental health quotient is an average 100, then what grade average should the student expect? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Grade Average = -.5 + t / 10 = -.5 + 50 ( 1 - e^(-.02 (Q - 70) ) ) / 10 = -.5 + 5 ( 1 - e^(-.02 (Q - 70) ) ) So grade Average(t(100)) = -.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = -.5 + 5( 1 - .5488 ) = -.5 + 5 ( .4522 ) = -.5 + 2.26 = 1.76. grade Average(t(110)) = -.5 + 5 ( 1 - e^(-.02 ( 110 - 70) ) = -.5 + 5( 1 - .4493 ) = -.5 + 5 ( .5517 ) = -.5 + 2.76 = 2.26. grade Average(t(120)) = -.5 + 5 ( 1 - e^(-.02 ( 120 - 70) ) = -.5 + 5( 1 - .3678 ) = -.5 + 5 ( .6322 ) = -.5 + 3.16 = 2.66. grade Average(t(130)) = -.5 + 5 ( 1 - e^(-.02 ( 130 - 70) ) = -.5 + 5( 1 - .3012 ) = -.5 + 5 ( .6988 ) = -.5 + 3.49 = 2.99. -.5 + 5 ( 1 - e^(-.02 ( Q - 70) ) ) will approach -.5 + 5 ( 1 - 0) = -.5 + 5 = 4.5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** gradeAverage = -.5 + t / 10 = -.5 + 50 ( 1 - e^(-.02 (Q - 70) ) ) / 10 = -.5 + 5 ( 1 - e^(-.02 (Q - 70) ) ) So gradeAverage(t(100)) = -.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = -.5 + 5( 1 - .5488 ) = -.5 + 5 ( .4522 ) = -.5 + 2.26 = 1.76. gradeAverage(t(110)) = -.5 + 5 ( 1 - e^(-.02 ( 110 - 70) ) = -.5 + 5( 1 - .4493 ) = -.5 + 5 ( .5517 ) = -.5 + 2.76 = 2.26. gradeAverage(t(120)) = -.5 + 5 ( 1 - e^(-.02 ( 120 - 70) ) = -.5 + 5( 1 - .3678 ) = -.5 + 5 ( .6322 ) = -.5 + 3.16 = 2.66. gradeAverage(t(130)) = -.5 + 5 ( 1 - e^(-.02 ( 130 - 70) ) = -.5 + 5( 1 - .3012 ) = -.5 + 5 ( .6988 ) = -.5 + 3.49 = 2.99. As Q gets larger and larger Q - 70 will get larger and larger, so -.02 ( Q - 70) will be a negative number with increasing magnitude; its magnitude increases without limit. It follows that e^(-.02 ( Q - 70) ) = will consist of e raised to a negative number whose magnitude increases without limit. As the magnitude of the negative exponent increases the result will be closer and closer to zero. So -.5 + 5 ( 1 - e^(-.02 ( Q - 70) ) ) will approach -.5 + 5 ( 1 - 0) = -.5 + 5 = 4.5. Side note: For Q = 100, 200 and 300 we would have grade averages 1.755941819, 4.128632108, 4.449740821. To get a 4-point Q would have to be close to 200. Pretty tough course ********************************************* Question: `qWhat grade averages would be expected for mental health quotients of 110, 120 and 130? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 110=2.2534 120=2.66 130=2.99 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a110...2.2534, 120...2.66, 130...2.99 ********************************************* Question: `qWhat is the upper limit on the expected grade average that can be achieved by this student? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4.5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `a** If Q is very, very large, e^-( .02(Q-70) ) would have a negative exponent with a very large magnitude and so wouldbe very close to 0. In this case 50 ( 1-e^(-.02 (Q-70)) would be close to 50(1-0) = 50. Then the grade average would be -.5 + 50 / 10 = -.5 + 5 = 4.5. DER [0.5488116360, 0.4493289641, 0.3678794411, 0.3011942119][1.755941819, 2.253355179, 2.660602794, 2.994028940] ********************************************* Question: `qWhat is the composite function gradeAverage( t(Q) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a-.5+(50(1-e^(-.02(Q-70))/10 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a-.5+(50(1-e^(-.02(Q-70))/10 ********************************************* Question: `qWhat do you get when you evaluate your composite function at t = 100, 110, 120 and 130? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2.97 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: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. Evaluating the composite function more accurately, we get gradeAverage(100) = -.5 + ( 50 (1 - e ^ (-.02 ((100) - 70) ) ) ) / 10 = 1.76 gradeAverage(110) = -.5 + ( 50 (1 - e ^ (-.02 ((110) - 70) ) ) ) / 10 = 2.25 gradeAverage(120) = -.5 + ( 50 (1 - e ^ (-.02 ((120) - 70) ) ) ) / 10 = 2.66 gradeAverage(130) = -.5 + ( 50 (1 - e ^ (-.02 ((130) - 70) ) ) ) / 10 = 2.99 These values agree, as they must, with the values calculated previously. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: