course Mth 271 ‰æí¸ßê¸áï®’ã¡ú¢w®Ìîâ\ñÊÞÝÄ”assignment #011
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21:01:29 1.6.16 (was 1.6.14 intervals of cont for (x-3)/(x^2-9) What are the intervals of continuity for the given function?
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RESPONSE --> I set x^2 - 9 equal to zero and find the fraction is undefined at -3 and 3. The intervals of continuity, then, are (-inf., -3), (-3, 3), (3, inf.) confidence assessment: 1
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21:01:57 The function is undefined where x^2 - 9 = 0, since division by zero is undefined. x^2 - 9 = 0 when x^2 = 9, i.e., when x = +-3. So the function is continuous on the intervals (-infinity, -3), (-3, 3) and (3, infinity). The expression (x - 3) / (x^2 - 9) can be simplified. Factoring the denominator we get (x - 3) / [ (x - 3) ( x + 3) ] = 1 / (x + 3). This 'removes' the discontinuity at x = +3. However in the given fom (x-3) / (x^2 + 9) there is a discontinuity at x = -3. **
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RESPONSE --> self critique assessment: 3
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21:07:32 10-22-2008 21:07:32 1.6.24 (was 1.6.22 intervals of cont for |x-2|+3, x<0; x+5, x>=0 What are the intervals of continuity for the given function?
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21:08:18 1.6.24 (was 1.6.22 intervals of cont for |x-2|+3, x<0; x+5, x>=0 What are the intervals of continuity for the given function?
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RESPONSE --> I found that both graphs are continuous with each other. The interval of continuity is (-infinity, infinity) confidence assessment: 2
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21:08:47 The graph of y = |x-2|+3 is translated 2 units in the x direction and 3 in the y direction from the graph of y = |x|. It forms a V with vertex at (2, 3). The given function follows this graph up to x = 0. It has slope -1 and y-intercept at y = | 0 - 2 | + 3 = 5. The graph then follows y = x + 5 for all positive x. y = x + 5 has y-intercept at y = 5. From that point the graph increases along a straight line with slope 1. So the graph of the given function also forms a V with vertex at (0, 5). Both functions are continuous up to that point, and both continuously approach that point. So the function is everywhere continuous. **
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RESPONSE --> self critique assessment: 3
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21:12:16 1.6.66 (was 1.6.54 lin model of revenue for franchise Is your model continuous? Is actual revenue continuous?
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RESPONSE --> My model is continuous because revenue comes in daily instead of quarterly or yearly. The actual revenue is not a continuous function. I'm really just guessing here. I'm not sure at all. confidence assessment: 0
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21:14:41 revenue comes in 'chunks'; everytime someone pays. So the actual revenue 'jumps' with every payment and isn't continuous. However for a franchise the jumps are small compared to the total revenue and occur often so that a continuous model isn't inappropriate for most purposes. **
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RESPONSE --> I see that the jumps in revenue are small compared to the total revenue. Although the actual revenue isn't continuous, a model would likely be a continuous function because it would be too tedious to represent all the small jumps in revenue. self critique assessment: 2
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21:17:18 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I learned that a function that has one point that is undefined can sometimes be made continuous by redefining the function at that point. self critique assessment: 2
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