course MTH 271 10/12, 1445 ԎYq{K`ݧޖassignment #011
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14:12:48 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 --> 1.6.18 (x - 3)/(x^2 - 9) (x-3) / (x-3)(x+3) 1 / (x+3) Continuous: (-infinity, -3)(-3, infinity) Discontinous @ -3 b/c if -3 is added to 3, 0 will be in the denominator making the function undefined. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:13:36 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 Rating:ent: 3
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14:19:00 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 --> (-infinity,infinity) confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:21:00 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 --> My method of solving the problem was reliant upon observing the combined graphs, rather than finding slopes and intercepts. *This problem does not seem to appear in 1.6. ------------------------------------------------ Self-critique Rating:ent: 3
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14:24:01 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 --> 1.6.65 Since the model developed for the revenue of the franchise is linear, it will be continuous. But, the actual revenue is unlikely to be continuous as there will be inevitable drops or jumps in revenue over time that will deviate from our linear model. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:24:09
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RESPONSE --> confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:24:29 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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14:24:43 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None noted. ------------------------------------------------ Self-critique Rating:ent: 3
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