course Mth 163 ??€?????????assignment #024??????????????Precalculus I
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00:30:14 code 496335
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00:30:17 assignment # 24
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00:30:19 properties of product functions
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00:31:10 Explain why, when either f(x) or g(x) is 0, then the product function also has a 0 for that value of x.
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RESPONSE --> When either f(x) or g(x) is 0, then the product function also has 0 for that x value. When any number is multiplied by zero, the product is zero.
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00:31:13 STUDENT RESPONSE: If you multiply any number by zero and you get zero.
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00:32:26 Explain 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.
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RESPONSE --> When the magnitude |f(x)| of f(x) is greater than 1, then the product function will be further from the x axis than the g(x) function. When a number greater than 1 is multiplied by another number greater than 1, the product is even larger than the original numbers.
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00:33:43 STUDENT 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.
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00:34:41 Explain 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.
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RESPONSE --> When the magnitude |f(x)| of f(x) is less than 1, the product function will be closer to the x axis than the g(x) function. When a number less than 1 is multiplied by another number less than 1, the product is even smaller than the original numbers.
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00:34:47 STUDENT 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.
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00:36:05 Explain 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.
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RESPONSE --> When f(x) and g(x) are either both positive or both negative, the product funtion is positive. When f(x) and g(x) have opposite signs the product function is negative. Multiplication rules say that pos*pos = pos, neg*neg=pos, neg*pos=neg. Since two functions are being multiplied, these basic rules apply.
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00:36:11 STUDENT 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.
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00:36:46 Explain why, when f(x) = 1, the graph of the product function coincides with the graph of g(x).
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RESPONSE --> When f(x)=1, the graph of the product function coincdes with the graph of g(x). Any number multiplied by 1 keeps the value of the original number.
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00:36:49 STUDENT RESPONSE: g(x) * 1 = g(x)
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00:42:47 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.
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RESPONSE --> y=g(x) is a straight line running through the origin with a slope of .5. y=f(x) is a graph that increases at an increasing rate passing through the y axis at y=1, the x axis seems to be an asymptote. the graph of y=fg(x) increases at an increasing rate, has pos y values when both graphs are positive (essentially when x values are pos) and the graph passes through the origin and has neg y values when the y=g(x) graph has neg y values (when there are neg x values)
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00:43:00 STUDENT 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. **
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00:46:36 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?
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RESPONSE --> depth = 0 at what times? using depth(t) = t^2-40t+400, I put these values into the quadratic funtion (40+-sqrt(1600-1600))/2 = (40+-sqrt 0 )/ 2 = 40/2 = 20, so there only one zero at (20,0). The point (20,0) is the vertex and only point passing through/on the x axis.
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00:46:50 ** 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?
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00:47:03 range(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. **
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00:47:26 What two linear factors represent the depth function as their product?
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RESPONSE --> depth(t)=(t-20)(t-20)
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00:47:27 depth(t) = (t-20)(t-20)
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00:50:01 For t = 5, 10 and 15, what are the ranges?
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RESPONSE --> depth(t)=t^2-40t+400 depth(5)=5^2-40(5)+400=225 range(225) = 2.9 sqrt225 = 43.5 depth(10)=10^2-40(10)+400=100 range(100) = 2.9 sqrt100 = 29 depth(15)=15^2-40(15)+400=25 range(25) = 2.9 sqrt25 = 14.5
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00:50:08 ** 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. **
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00:51:12 What is the function range(depth(t))? Show that its simplified form is linear in time.
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RESPONSE --> range(depth(t)) = 2.9 sqrt (depth(t)) = 2.9 sqrt(t^2-40t+400) = 2.9(t-20)^2 = 2.9 |t-20|
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00:51:18 ** 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 |. **
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00:52:25 problem 8 Illumination(r) = 40 / r^2; distance = 400 - .04 t^2. What is the composite function illumination(distance(t))?
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RESPONSE --> illumination(distance(t)) = 40/(distance(t))^2 = 40/(400-.04t^2)^2
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00:52:32 ** Illumination(r) = 40 / r^2 so Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2. **
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00:55:46 Give 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?
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RESPONSE --> 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 t=25-50 average rate .000444-.000284=.00016/25=.0000064 t=50-75 average rate .001306-.000444=.000862/25=.00003448
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00:56:28 ** 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. **
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00:58:00 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?
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RESPONSE --> grade average = -.5+5(1-e^(.02(Q-70)) Q=100 grade average = -.5+5(1-e^(.02(100-70))=1.76
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00:58:28 ** 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
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00:59:24 What grade averages would be expected for mental health quotients of 110, 120 and 130?
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RESPONSE --> grade average = -.5+5(1-e^(.02(110-70))=2.25 grade average = -.5+5(1-e^(.02(120-70))=2.66 grade average = -.5+5(1-e^(.02(130-70))=2.99
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00:59:27 110...2.2534, 120...2.66, 130...2.99
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01:02:37 What is the upper limit on the expected grade average that can be achieved by this student?
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RESPONSE --> Upper limit expected: When there is a large value of Q, e^(-.02(Q-70)) is negative but grows closer to equalling zero as Q grows. When e^(-.02(Q-70))=0, this would be the limit so 50(1-0)=50(1)=50 so -.5+50/10 = -.5+5 so a grade average of 4.5 is the upper limit.
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01:02:46 ** 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]
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01:03:46 What is the composite function gradeAverage( t(Q) )?
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RESPONSE --> gradeAverage(t(Q)) = -.5+(50(1-e^(-.02(Q-70)))/10 = -.5+5(1-e^(-.02(Q-70)))
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01:04:32 -.5+(50(1-e^(-.02(Q-70))/10
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RESPONSE --> I divided 50/10 to get 5 to distrubute to (1-e^(-.02(Q-70))
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01:05:39 What do you get when you evaluate your composite function at t = 100, 110, 120 and 130?
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RESPONSE --> I got the same values as found before approximately, (100,1.76), (110,2.25), (120,2.66), (130, 2.99)
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01:06:13 t=100...9.5 t=110...10.5 t=120...11.5 t=130...12.5
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01:06:22 ** 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 **
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01:06:24 Add comments on any surprises or insights you experienced as a result of this assignment.
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