course Mth 271 011. `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 Rating: I feel fairly confident about this problem. ********************************************* 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: For the function |x-2| + 3, the vortex is (2,3). By inserting 0 in for x, I got the solution: |0-2| + 3 2 + 3 = 5 For the function x+5, the x intercept is (0,5) and the y intercept is (-5,0) confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Again, I don’t know what is meant by the intervals of continuity
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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 Rating: I feel fairly confident about this problem. ********************************************* 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: What model are you referring to?
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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. **