#$&* course MTH271 8-5-10 011. `query 11*********************************************
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Given Solution: `a The function is undefined where x^2 - 9 = 0, since division by zero is undefined. x^2 - 9 = 0 when x^2 = 9, i.e., when x = +-3. So the function is continuous on the intervals (-infinity, -3), (-3, 3) and (3, infinity). The expression (x - 3) / (x^2 - 9) can be simplified. Factoring the denominator we get (x - 3) / [ (x - 3) ( x + 3) ] = 1 / (x + 3). This 'removes' the discontinuity at x = +3. However in the given fom (x-3) / (x^2 + 9) there is a discontinuity at x = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):okay ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q1.6.24 (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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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). confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The graph of y = |x-2|+3 is translated 2 units in the x direction and 3 in the y direction from the graph of y = |x|. It forms a V with vertex at (2, 3). The given function follows this graph up to x = 0. It has slope -1 and y-intercept at y = | 0 - 2 | + 3 = 5. The graph then follows y = x + 5 for all positive x. y = x + 5 has y-intercept at y = 5. From that point the graph increases along a straight line with slope 1. So the graph of the given function also forms a V with vertex at (0, 5). Both functions are continuous up to that point, and both continuously approach that point. So the function is everywhere continuous. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):okay ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q1.6.66 (was 1.6.54 lin model of revenue for franchise Is your model continuous? Is actual revenue continuous? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a revenue comes in 'chunks'; everytime someone pays. So the actual revenue 'jumps' with every payment and isn't continuous. However for a franchise the jumps are small compared to the total revenue and occur often so that a continuous model isn't inappropriate for most purposes. ** GOOD STUDENT EXAMPLE The model would not be continuous. The VT vs. Boston College football game I watched actually inspired me on this one. A football franchise makes most, if not all, of its revenue on one day out of the week and that’s game day. Attendance and concessions will differ from each game depending on the anticipation beforehand so you may make a fortune on a game against a rival team and you may make less than average on a game that everyone knows the outcome of. Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):okay ------------------------------------------------ Self-critique rating #$&*3 Additional student questions and Instructor Responses for your reference (you are not required to answer these or ask questions about them, but if you wish you are welcome to do so; if you do insert solutions, questions, etc., mark you insertions with **** so the instructor can quickly recognize them): Student question: What are the intervals of continuity for the function f(x) = [[x-2]] + x? Instructor Response: [[x-2]] indicates 'the greatest integer in x - 2'. For example, if x = 3.9 we have [ [ x - 2 ] ] = [ [3.9 - 2]] = [ [ 1.9] ] = 1. [[ 1.9 ]] is the greatest integer less than or equal to 1.9, which is 1. if x = 3.99 we have [ [ x - 2 ] ] = [ [3.99 - 2]] = [ [ 1.99] ] = 1 [[1.99]] is still 1. if x = 3.99999 we have [ [ x - 2 ] ] = [ [3.99999 - 2]] = [ [ 1.99999] ] = 1 [[1.99999]] is still 1. However as soon a x = 4 we get [[4-2]] = [[2]] = 2. [[ 2 ]] is the greatest integer less than or equal to 2, which since 2 is an integer is just 2. Now if x = 3.9 we have [[x-2]] + x = [[3.9 - 2]] + 3.9 = 1 + 3.9 = 4.9. If x = 4 we get [[x-2]] + x = 6. What happens for values of x between 3.9 and 4, especially for values of x that approach 4 as a limit? How can you use these examples to understand this function and determine its intervals of continuity? "