course Mth 163
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15:53:23 Query problem 2. Describe the sum of the two graphs.
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RESPONSE --> one of th graphs shows the sum of a 'sine-wave' function which oscillates regularly above and below the x axis, and a the other function which is a parabolic function. at the x values where the sine wave passes through the a-axis, the sum function will be obtained by adding the y-value of the since function, which in this case is 0, to the value of the other function. the result will be a y value which is equal to the the one of the other function.
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15:55:28 ** The 'black' graph takes values 8, 3, 0, -1, 0, 3, 8 at x = -3, -2, -1, 0, 1, 2, 3. The 'blue' graph takes approximate values 1.7, .8, .2, -.1, -.4, -.6, -.8 at the same x values. The 'blue' graph takes value zero at approximately x = -.4. The sum of the two graphs will coincide with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The sum will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. **
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RESPONSE --> i was correct in my answer, but i didn't really give the values of the ""blue"" and ""black"" graphs as it related to the ones in the exercises. I do understand what is being said though.
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16:21:11 Where it is the sum graph higher than the 'black' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> I know that the 'sum graph' is represented by the 'red' graph and i can see the black graph. in the table 'Summing the y values of two functions' i can distinguish between the two 'graphs' and i can see just by looking that the 'red' graph is higher than the black graph. I also know that the 'red' graph is the 'sum graph' of the black and blue 'lines/graphs'. I can also see that one point of the black line is at (1, 1) and the blue point above it is at (1, 2), therefore that would make the red/sum graph would be at the point (1, 3) since we add the y values. Is this correct? Am i on the right track? The thing that i don't fullly understand is the exact intervals of the points. Also, i'm not sure i understand how the 'red/sum' graph can be lower than the 'black' graph, since just by looking, it's above the black line. Although in the other graph of y = x^2, Y= -x^3 and y = x^2 + (-x^3), i can see how the red graph is higher than the blue graph, when the green graph is positive, and lower when the green graph becomes negative and how the red 'squares' graph is never lower than the green 'triangles' graph at any point because the blue graph is numbers negative and how the red graph is higher than the green graph by exactly the height of the blue graph...but these tell of the red, blue and green graphs, not the black one. ??
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16:23:14 ** The sum of the graphs is higher than the 'black' graph where the 'blue' graph is positive, lower where the 'blue' graph is negative. The 'blue' graph is positive on the interval from x = -3 to x = -.4, approx.. This interval can be written [-3, -.4), or -3 <= x < -.4. **
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RESPONSE --> i understand a little more how you got it. i do understand the concept of when ""The sum of the graphs is higher than the 'black' graph where the 'blue' graph is positive, lower where the 'blue' graph is negative"". i can see this.
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16:45:56 Where it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> The sum of the graph is higher than the 'blue' graph where the 'black' graph is negative, lower where the 'black' graph is positive ?? I'm still not sure how to determine the intervals for the 'black' graph and where the intervals are positive from. ??
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16:49:51 ** The sum of the graphs is higher than the 'blue' graph where the 'black' graph is positive, lower where the 'black' graph is negative. The 'black' graph is positive on the interval from x = -1 to x = 1, not including the endpoints of the interval. This interval can be written (-1, 1) or -1 < x < 1. **
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RESPONSE --> i wasn't correct. i just thought that since the black graph is higher when the blue graph is positive, that it would be the opposite for the blue graph and that it would be higher where the black graph is negative. it just seemed logical to think this, but i'm not exactly confident in my understanding of the concept that the blue graph is higher where the black graph is positive and lower when the black graph is negative. ?
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16:56:47 Where does thus sum graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> The sum graph will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. This is because that when one graph is equal to zero, the other one is higher and vice versa. ??
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17:00:06 ** The sum coincides with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. The coordinates would be about (-.4, -.7), on the 'black' graph. **
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RESPONSE --> i was correct...maybe i am getting the hang of this more than i thought. The coordinates would be about (-.4, -.7) because when the black graph is -.4, the blue graph is zero and the when the blue graph is at -.7, the black graph is at zero??
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17:00:51 Where does thus sum graph coincide with the 'blue' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> The sum of the two graphs will coincide with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1.
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17:01:27 ** The sum coincides with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The coordinates would be about (-1, .2) and (1, -.4), on the 'blue' graph. **
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RESPONSE --> i do believe i'm getting the hang of this
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17:46:05 Query problem 3 Describe the quotient graph obtained by dividing the 'black' graph by the 'blue' graph. You should answer the following questions: Where it is the quotient graph further from the x axis than the 'black' graph, and where is it closer? Answer by giving specific intervals, and explaining why you believe these to be the correct intervals. Where it is the quotient graph on the same side of the x axis as the 'black' graph, and where is it on the opposite side, and why? Answer by giving specific intervals. Where does thus quotient graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs. Where does the quotient graph have vertical asymptote(s), and why? Describe the graph at each vertical asymptote.
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RESPONSE --> the quotient graph is further from the x-axis than the black graph when the parabolic graph line is below y = 1 (less than 1) or when the values are less. The quotient graph is closer to the x-axis than the black graph when the parabolic graph line is one or greater, or when the values are greater. I'm still not sure exactly how to find the exact intervals. ?? the quotient graph is on the same side of the x-axis as the black when dividing the sine values by greater parabolc values, or when the parabolic values are less than 1. It's on the opposite side of the x-axis when dividing values of sine function by values of the parabolic graph, or when it drops below y = 1 and passes through y = 0 the quotient vertical asymptotes are at every point where the parabolic graph is zero (0). what exactly does this mean? I couldn't understand and find the information for when the quotient graph coincides with the black graph and its specific coordinates. ??
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17:47:43 ** The 'black' graph is periodic, passing through 0 at approximately x = -3.1, 0, 3.1, 6.3. This graph has peaks with y = 1.5, approx., at x = 1.6 and 7.8, approx., and valleys with y = -1.5 at x = -1.6 and x = 4.7 approx. The 'blue' graph appears to be parabolic, passing thru the y axis at x = -1 and reaching a minimum value around y = -1.1 somewhere near x = 1. This graph passes thru the x axis at x = 5.5, approx., and first exceeds y = 1 around x = 7.5. The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the origin, since division by a number whose magnitude is less than 1 gives a result whose magnitude is greater than the number being divided. This will occur to the left of x = 1, and between about x = 2 and x = 7.5. Between about x = 0 and x = 1 the 'blue' graph is more than 1 unit from the x axis and the quotient graph will be closer to the x axis than the 'black' graph. The same is true for x > 7.5, approx.. The 'black' graph is zero at or near x = -3.1, 0, 3.1, 6.3. At both of these points the 'blue' graph is nonzero so the quotient will be zero. The 'blue' graph is negative for x < 5.5, approx.. Since division by a negative number gives us the opposite sign as the number being divided, on this interval the quotient graph will be on the opposite side of the x axis from the 'black' graph. The 'blue' graph is positive for x > 5.5, approx.. Since division by a positive number gives us the same sign as the number being divided, on this interval the quotient graph will be on the same side of the x axis as the 'black' graph. The quotient graph will therefore start at the left with positive y values, about 3 times as far from the x axis as the 'black' graph (this since the value of the 'blue' graph is about -1/3, and division by -1/3 reverses the sign and gives us a result with 3 times the magnitude of the divisor). The quotient graph will have y value about 2.5 at x = -1.6, where the 'black' graph 'peaks', but the quotient graph will 'peak' slightly to the left of this point due to the increasing magnitude of the 'blue' graph. The quotient graph will then reach y = 0 / (-1) = 0 at x = 0 and, since the 'black' graph then becomes positive while the 'blue' graph remains negative, the quotient graph will become negative. Between x = 0 and x = 2 the magnitude of the 'blue' graph is a little greater than 1, so the quotient graph will be a little closer to the x axis than the 'black' graph (while remaining on the other side of the x axis). At x = 3.1 approx. the 'black graph is again zero, so the quotient graph will meet the x axis at this point. Past x = 3.1 the quotient graph will become positive, since the signs of both graphs are negative. As we approach x = 5.5, where the value of the 'blue' graph is zero, the quotient will increase more and more rapidly in magnitude (this since the result of dividing a negative number by a negative number near zero is a large positive number, larger the closer the divisor is to zero). The result will be a vertical asymptote at x = 5.5, with the y value approaching +infinity as x approaches 5.5 from the left. Just past x = 5.5 the 'blue' values become positive. Dividing a negative number by a positive number near zero results in a very large negative value, so that on this side of x = 5.5 the asymptote will rise up from -infinity. The quotient graph passes through the x axis near x = 6.3, where the 'black' graph is again zero. To the right of this point both graphs have positive values and the quotient graph will be positive. Around x = 7.5, where the 'blue' value is 1, the graph will coincide with the 'black' graph, giving us a point near (7.5, 1.3). Past this point the 'blue' value is greater than 1 so that the quotient graph will become nearer the x axis than the 'black' graph, increasingly so as x (and hence the 'blue' value) increases. This will result in a 'peak' of the quotient graph somewhere around x = 7.5, a bit to the left of the peak of the 'black' graph. **
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RESPONSE --> i think i was partly correct, except i wasn't specific enough. i think i understand more about when the two graphs coincide.
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18:10:14 Query problems 7-8 Sketch the graph of y = x^2 - 2 x^4 by first sketching the graphs of y = x^2 and y = -2 x^4. How does the result compare to the graph of y = x^2 - x^4, and how do you explain the difference?
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RESPONSE --> i can see that the graph of y = x^2 - x^4 is a little distorted near the origin since it has the points of (-1, 0), (0, 0), and (1, 0) and then shoots down to (2, -12) and (-2, -12). should it be distorted?
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18:12:39 ** At x = 0, 1/2, 1 and 2 we have x^2 values 0, 1/4, 1 and 4, while -x^4 takes values 0, -1/16, -1 and -16, and -2 x^4 takes values 0, -1/8, -2 and -32. All graphs clearly pass through the origin. The graphs of y = x^2 - x^4 and y = x^4 - 2 x^4 are both increasingly negative at far right and far left. Graphical addition will show that y = x^2 - x^4 takes value 0 and hence passes thru the x axis when the graphs have equal but opposite y values, which occurs at x = 1 and x = -1. To the left of x = -1 and to the right of x = 1 the negative values of -x^4 overwhelm the positive values of x^2 and the sum graph will be increasingly negative, with values dominated by -x^4. Near x = 0 the graph of y = -x^4 is 'flatter' than that of y = x^2 and the x^2 values win out, making the sum graph positive. y = x^2 - 2 x^4 will take value 0 where the graphs are equal and opposite in value; this occurs somewhere between x = .8 and x = .9, and also between x = -.9 and x = -.8, which places the zeros closer to the y axis than those of the graph of y = x^2 - x^4. The graph of y = -2 x^4 is still flatter near x = 0 than the graph of y = x^2, but not as flat as the graph of y = -x^4, so while the sum graph will be positive between the zeros the values won't be as great. Outside the zeros the sum graph will be increasingly negative, with values dominated by -2x^4. **
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RESPONSE --> by saying that the graph of y = x^2 - x^4 was distorted, i meant that its vertex was a little 'flatter' at the origin than the other graphs. i couldn't think of the word and how to use it, therefore I said it was distorted. I didn't mean that it had peaks at the origin.
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18:17:09 How does the shape of the graph change when you add x to get y = -2 x^4 + x^2 + x, and how do you explain this change?
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RESPONSE --> the shape of the two graphs are similar in that they are both distorted (having peaks and valleys) at its vertex and the origin. except, the graph with x added to it, has a higher peak on one side than the other graph. when you add x to the original equation of y = -2x^4 + x^2, then you are adding another variable and it causes the line to increase 1 unit at the right peak.
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18:18:03 ** At x = 0 there is no change in the y value, so the graph still passes through (0, 0). As x increases through positive numbers we will have to increase the y values of y = x^2 -2 x^4 by greater and greater amounts. So it will take a little longer for the negative values of -2 x^4 to 'overwhelm' the positive values of x^2 + x than to overcome the positive values of x^2 and the x intercept will shift a bit to the right. As we move away from x = 0 through negative values of x we will find that the positive effect of y = x^2 is immediately overcome by the negative values of y = x, so there is no x intercept to the left of x = 0. The graph in fact stays fairly close to the graph of y = x near (0, 0), gradually moving away from that graph as the values of x^2 and -2 x^4 become more and more significant. **
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RESPONSE -->
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18:18:40 ** At x = 0 there is no change in the y value, so the graph still passes through (0, 0). As x increases through positive numbers we will have to increase the y values of y = x^2 -2 x^4 by greater and greater amounts. So it will take a little longer for the negative values of -2 x^4 to 'overwhelm' the positive values of x^2 + x than to overcome the positive values of x^2 and the x intercept will shift a bit to the right. As we move away from x = 0 through negative values of x we will find that the positive effect of y = x^2 is immediately overcome by the negative values of y = x, so there is no x intercept to the left of x = 0. The graph in fact stays fairly close to the graph of y = x near (0, 0), gradually moving away from that graph as the values of x^2 and -2 x^4 become more and more significant. **
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RESPONSE --> i think i got the correct graphs for each, but i wasn't specific enough.
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18:21:05 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> i wasn't familiar at all with the 'sum line' and how you could get it and graph it by adding two functions. I think it's neat and i think i'm getting the hang of it. Also, i never really heard of a graph having 'distortion' at the vertex, so this was new, but i feel that I am becoming more accustomed to it.
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18:32:39 163
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RESPONSE --> nothing is coming up.....i pushed 'Next question/answer' to start the assignment, but nothing come up but ************** and (code 496335) what is this?
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18:38:32 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 a 0 for that value of x, because anytime you multiply something my zero, the product is always zero.
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18:38:46 STUDENT RESPONSE: If you multiply any number by zero and you get zero.
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RESPONSE --> correct
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18:54:54 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 --> i'm not exactly certain of how to explain this concept, because i don't exactly understand the concept of the 'magnitude I f(x) I of f(x).
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18:58:07 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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RESPONSE --> ok, i understand what this student is saying. it's basic math knowledge that whenever you multiply a number by another number greater than 1, then the result is greater than the original number. I understand this. i just didn't think that that complex question had such a simple answer. i believe i'm making it harder on myself than what it really is on some things.
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19:00:37 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 --> because 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 less magnitude than the original number ??
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19:01:06 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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RESPONSE --> i believe i was correct
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19:04:38 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 --> because when you're multiplying, two negatives will always equal a positive as the same with two postives, but when you have opposite signs, as like a positive and a negative, it will always equal a negative. therefore when two functions are being multiplied together and have the same signs, then the product function will be positive. If the two functions that are being multiplied have opposite signs, then the product function will be negative.
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19:05:02 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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RESPONSE --> i was correct
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19:07:27 Explain why, when f(x) = 1, the graph of the product function coincides with the graph of g(x).
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RESPONSE --> because anything multiplied by one, is itself. therefore, when f(x) = 1, then, the product will be that of g(x), unless both f(x) and g(x) are one, which in this case, the product would just be one.
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19:08:11 STUDENT RESPONSE: g(x) * 1 = g(x)
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RESPONSE --> i was correct, except this is the function example of my answer.
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19:22:24 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 --> the graph for y = f(x) = 2^x, for -2 < x < 2, from left to right, starts out almost being parallel to the x-axis, but slightly increases until about y = 1, then it increases at a much rapid rate, butnot in a straight line. it curves to form what looks like half of a parabala. it continues to increase and become steeper. the graph for y = g(x) = .5x, for -2 < x < 2, from left to right, is a straight line that increases at a steady rate. it passes trhough the origin and continues. the graph for y = g(x) * f(x), for -2 < x < 2, from left to right, increases steadily and looks as though it's going in a straight line, until it hits about (-1, 0) in which it starts to slightly curve at an angle and keeps curving until about (2, 4) when it just increases upward, as like that of half a parabala.
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19:24:31 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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RESPONSE --> i saw that both g(x) and f(x) were opposite signs, therefore i knew that the product function would be negative, but i forgot to note it. i'm not sure if i got the same graph or not as you. i think i did...
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20:30:55 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 --> i'm not exactly sure how to determine the time of the depth and range. i know how to find range, range(depth) = 2.9 'sqrt(depth) 2.9 'sqrt(0) = 0 is this correct? I'm just not sure how to find the times with the function, depth(t) = t^2 - 40 t + 400. would i just substitue depth = 0 into the equation, but since it's not t then i wouldn't work.
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20:33:00 ** 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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RESPONSE --> I understand how to find the vertex using the -b/(2a) equation, but i didn't think to use it on this problem. I can understand why it works and the answer that was received.
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20:33:35 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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RESPONSE --> ok
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20:35:46 What two linear factors represent the depth function as their product?
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RESPONSE --> range and time ??
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20:36:46 depth(t) = (t-20)(t-20)
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RESPONSE --> how did you get this answer? I don't understand how these can be the two linear factors. ??
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20:39:41 For t = 5, 10 and 15, what are the ranges?
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RESPONSE --> for depth(t) = (t-20)(t-20), for t = 5, 10 and 15 depth(5) = (5-20)(5-20) = (-15)(-15) = 225 depth(10) = (10-20)(10-20) = (-10)(-10) = 100 depth(15) = (15-20)(15-20) = (-5)(-5) = 25
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20:40:45 ** 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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RESPONSE --> i got half of it right. i didn't finish it however, because i didn't the ranges. i understand how to get them though
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20:42:54 What is the function range(depth(t))? Show that its simplified form is linear in time.
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RESPONSE --> horizontal range is given as a function of depth by the function range(depth(t)) = range * (depth*t)
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20:43:40 ** 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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RESPONSE --> i didn't realize that the question was for the previous one. i thought you just wanted us to explain it.
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20:46:17 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 --> i'm not sure how to solve it without the values of t ??
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20:46:51 ** Illumination(r) = 40 / r^2 so Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2. **
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RESPONSE --> i believe i understand how it was put together.
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20:53:52 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 --> i'm not exactly sure which equation to plug t into ??
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20:56:19 ** 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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RESPONSE --> ok, i used this equation to plug t = 25, 50, and 75 into, but when i solved it for 25, i got 2.844e-4 and i couldn't get it to change into normal view. how do i do this? i know how to find the change between two of the rates.
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21:03:00 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 --> after solving the equation for t(Q) = 100, i got that t = 22.56, but this wouldn't seem right would it? how can you get a grade average of 22.56? i must have done something wrong while working it out.
t(Q) is the time spent studying; Q is the mental health quotient. So if mental health quotient is 100, you wouldn't say t(Q) = 100; t(Q) would be time spent studying, not mental health quotient. If mental health quotient in 100, then you plug in 100 for Q..................................................
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21:05:12 ** 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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RESPONSE --> i didn't get the answer you did, i realize that it must've been a mistake in working the problem. i understand how to find t(Q) when i have the 'Q' value.
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21:06:08 What grade averages would be expected for mental health quotients of 110, 120 and 130?
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RESPONSE --> 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. is this correct?
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21:06:32 110...2.2534, 120...2.66, 130...2.99
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RESPONSE --> i was correct
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21:08:11 What is the upper limit on the expected grade average that can be achieved by this student?
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RESPONSE --> wouldn't there not be a limit as to what the student could achieve, since there isn't a limit on a person's mental health quotient (or IQ) ??
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21:09:42 ** 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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RESPONSE --> i'm not sure i understand how the negatives come into play in solving this. if Q is very large, how can it be negative? wouldn't it be a positive number?
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21:11:50 What is the composite function gradeAverage( t(Q) )?
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RESPONSE --> i'm not sure how to answer this question. would it be gradeAverage(t(Q)) = -.5 + 5 ( 1 - e^(-.02 (Q - 70) ) ) ???
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21:12:57 -.5+(50(1-e^(-.02(Q-70))/10
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RESPONSE --> i was right except i left off the / 10 where did this come from? i don't remember seeing it in the previous problems that envolved solving gradeAverage(t(Q)) ??
The expression is -.5 + 5 ( 1 - e^(-.02 ( Q - 70) ) ) . Since 5 = 50 / 10, this can be written as 50 ( 1 - e^(-.02 ( Q - 70) ) ) / 10..................................................
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21:14:55 What do you get when you evaluate your composite function at t = 100, 110, 120 and 130?
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RESPONSE --> would you just plug in t = 100,110,120, and 130 into the equation gradeAverage( t(Q) ) = .5+(50(1-e^(-.02(Q-70))/10 i'm don't fully understand how to solve this problem.
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21:16:01 t=100...9.5 t=110...10.5 t=120...11.5 t=130...12.5
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RESPONSE --> ok, but how did you get the answers?
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21:16:21 ** 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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RESPONSE --> ok
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21:17:47 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> i'm still having a little difficulty in understanding the full concept of how to solve the composite function gradeAverage( t(Q) ). most of the other material covered, i feel i know how to do, and am understanding it.
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