#$&* course MTH 163 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `aSTUDENT RESPONSE: If you multiply any number by zero and you get zero. INSTRUCTOR NOTE: Right. If f(x) = 0 then f(x) * g(x) = 0; and the same is so if g(x) = 0. If one number in a product is zero, then the product is zero. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why, when the magnitude | f(x) | of f(x) is greater than 1 (i.e., when the graph of f(x) is more than one unit from the x axis), then the product function will be further from the x axis than the g(x) function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For positive numbers, if you multiply by a number more than 1, than your result will be greater. confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ 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 you multiply by a number less than 1 than the result will be less than the original number. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: If you multiply a number by another number less than 1, the result is less than the original number. If you multiply a number by another number whose magnitude is less than 1, the result will have a lesser magnitude that the original number. If | f(x) | < 1 then the magnitude of f(x) * g(x) will be less than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude decreases so does the distance from the x axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain why, when f(x) and g(x) are either both positive or both negative, the product function is positiv; and when f(x) and g(x) have opposite signs the product function is negative. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because, two numbers that are positive will be positive, two numbers that are negative will be negative, and a negative times a positive will be negative. This is just due to simple math rules. confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ 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 if you times g(x) by 1 than it will be g(x). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: g(x) * 1 = g(x) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q problem 4 Sketch graphs for y = f(x) = 2^x and y = g(x) = .5 x, for -2 < x < 2. Use your graphs to predict the shape of the y = g(x) * f(x) graph. Describe the graphs of the two functions, and explain how you used these graphs predict the shape of the graph of the product function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When g(x) is negative and f(x) is positive, the result will be negative. When two are positive, it will be positive and it rises more. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 immediatly 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ 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: t^2 - 40 t + 400 is (t - 20)(t - 20) so it is t=20, so the zero is at t=20 20,0 is also the vertex. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. 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? `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. Let's first consider another 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. We conclude that sqrt(x^2) is not always equal to x. As in the above case, sqrt(x^2) can be equal to -x. We can settle the entire question by observing that • sqrt(x^2) = | x |. Applying this to the present problem: `sqrt( (t-20) ^2 ) cannot be negative. Using the observations made above, we see that `sqrt( (t-20)^2 ) = | t - 20 |. Thus the composite sqrt( (t - 20)^2 ) is equal to | t - 20 |. • If t > 20 then | t - 20 | = t - 20. • If t < 20 then t - 20 is negative so that | t - 20 | = -(t - 20) = 20 - t. STUDENT QUESTION I see how it's possible to get a positive value from solving the quadratic and a negative value if the quadratic factor, but it doesn't entirely make sense to me how that is possible or which form is truly correct? Dealing with a range in this circumstance, the positive value of course makes sense, but out of context, how would we know? INSTRUCTOR RESPONSE It is of course possible that the quadratic function would have been negative for some values of t, in which case the square root of the quadratic would not have been defined for those values of t. Those values of t would therefore have to be omitted from the domain of the composite function. There are two things that restrict the domain of a composite: • The function f(g(t)) is undefined if g(t) is undefined--i.e., if t is not in the domain of g, then it can't be in the domain of f(g(t)). • I addition the function f(g(t)) is undefined if z = g(t) is not in the domain of the function f(z). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat two linear factors represent the depth function as their product? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Depth =t-20 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qFor t = 5, 10 and 15, what are the ranges of the stream? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: depth(t) = t^2 - 40 t + 400 and = (t-20)^2 depth(5) = (5-20)^2= 225 depth(10) = (10-20)^2 = 100 depth(15) = (15-20)^2 = 25. range= 2.9 sqrt(225) = 43.5 range= 2.9 sqrt(100) = 29 range= 2.9 sqrt(25) = 14.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** depth(t) = t^2 - 40 t + 400 = (t-20)^2 so depth(5) = (5-20)^2 = (-15)^2 = 225 depth(10) = (10-20)^2 = (-10)^2 = 100 depth(15) = (15-20)^2 = (-5)^2 = 25. It follows that the ranges of the stream are range(depth(5)) = 2.9 sqrt(225) = 43.5 range(depth(10) = 2.9 sqrt(100) = 29 and range(depth(15) = 2.9 sqrt(25) = 14.5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the function range(depth(t))? Show that its simplified form is linear in time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t sureabout this one. But after looking at it, I think I understand pretty well. ------------------------------------------------ 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: = 40 / (400 - .04 t^2)^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Illumination(r) = 40 / r^2 so Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qGive the illumination at t = 25, t = 50 and t = 75. At what average rate is illumination changing during the time interval from t = 25 to t = 50, and from t = 50 to t = 75? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: illumination= 40 / (400 - .04 t^2)^2 = 40 / (400 - .04 * 25^2)^2 = .000284 = 40 / (400 - .04 * 50^2)^2 = .000444 = 40 / (400 - .04 * 75^2)^2 = .001306. confidence rating #$&*: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How exactly would I find the average rate of this? ------------------------------------------------ Self-critique rating:
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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): I definitely was very lost on this one. ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat grade averages would be expected for mental health quotients of 110, 120 and 130? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a110...2.2534, 120...2.66, 130...2.99 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the upper limit on the expected grade average that can be achieved by this student? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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] STUDENT COMMENT: I was able to determine the cap from trial and error, but I don't think I would've recognized it just from """"reading"""" the equation. I find it really interesting that no matter what number is plugged in higher than 1760 (approx) the result is always the same. INSTRUCTOR RESPONSE: An exponential function approaches its asymptote quickly, just as in the opposite direction it moves quickly away from its asymptote. It's useful to remember how and why an exponential function approaches zero as its exponent becomes increasingly negative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat is the composite function gradeAverage( t(Q) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -.5+(50(1-e^(-.02(Q-70))/10 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a-.5+(50(1-e^(-.02(Q-70))/10 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat do you get when you evaluate your composite function at t = 100, 110, 120 and 130? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: STUDENT COMMENT: Composite functions are fun. It's almost exactly the same as functions encountered in computer programming so I find it quite fascinating, helpful, and complementary to programming. I think this correlation helps me to better understand much of the precalculus I've seen thus far. Also, as I mentioned in one of the assignment's problems, I am really fascinated by the gradeAverage(t(Q)) function in that it has an upper limit that can't be breached no matter what number is plugged in for Q. It's a little disorienting when you're used to getting different output for different input. I don't quite comprehend how the function causes this to happen, though. INSTRUCTOR RESPONSE: We can go back to the statements '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 .' The key is that e^-( .02(Q-70) ) essentially 'disappears' for large values of Q, so that ( 1-e^(-.02 (Q-70)) will be pretty much equal to 1 and 50 ( 1-e^(-.02 (Q-70)) pretty much equal to 50. Plugging 50 into the grade average function we get -.5 + 50 / 10 = -.5 + 5 = 4.5, which is therefore the limiting value of the composite function. ********************************************* Question: Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Your response: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: STUDENT COMMENT: Composite functions are fun. It's almost exactly the same as functions encountered in computer programming so I find it quite fascinating, helpful, and complementary to programming. I think this correlation helps me to better understand much of the precalculus I've seen thus far. Also, as I mentioned in one of the assignment's problems, I am really fascinated by the gradeAverage(t(Q)) function in that it has an upper limit that can't be breached no matter what number is plugged in for Q. It's a little disorienting when you're used to getting different output for different input. I don't quite comprehend how the function causes this to happen, though. INSTRUCTOR RESPONSE: We can go back to the statements '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 .' The key is that e^-( .02(Q-70) ) essentially 'disappears' for large values of Q, so that ( 1-e^(-.02 (Q-70)) will be pretty much equal to 1 and 50 ( 1-e^(-.02 (Q-70)) pretty much equal to 50. Plugging 50 into the grade average function we get -.5 + 50 / 10 = -.5 + 5 = 4.5, which is therefore the limiting value of the composite function. ********************************************* Question: Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Your response: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!