course Mth271 ???}????????assignment #011
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17:49:06 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 --> This function is continuos at all real numbers because it is just a polynomial equation. confidence assessment: 1
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17:50:46 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 --> My number 16 problem was different than this I believe. My equation was 3 - 2x - x^2 I believe this is problem 19 maybe? If so I can see that the function is undefined when x^2 - 9 = 0. self critique assessment: 2
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17:53:41 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 --> My number 24 Is the cont for [[x-2]] + x. My answer was that the function is continous at all 2x + x values. I don't believe this is right. I still am having a problem with the lowest integer function. I understand we are to round to the lowest integar, but how do we do this when we have something other than just a plain integer? confidence assessment: 0
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17:54:07 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 --> See previous note. self critique assessment: 1
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17:55:29 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 --> Because the model is linear, it would be a continuous model, however actual revenue probably would not be continuous because it is nearly impossible to expect revenues to be smooth, with no jumps, leaps, or gaps. confidence assessment: 2
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17:55:50 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 assessment: 3
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17:55:54 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 3
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