#$&* course mth 271 011. `query 11*********************************************
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Given Solution: `a The function is undefined where x^2 - 9 = 0, since division by zero is undefined. x^2 - 9 = 0 when x^2 = 9, i.e., when x = +-3. So the function is continuous on the intervals (-infinity, -3), (-3, 3) and (3, infinity). The expression (x - 3) / (x^2 - 9) can be simplified. Factoring the denominator we get (x - 3) / [ (x - 3) ( x + 3) ] = 1 / (x + 3). This 'removes' the discontinuity at x = +3. However in the given fom (x-3) / (x^2 + 9) there is a discontinuity at x = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q1.6.24 (was 1.6.22 intervals of cont for |x-2|+3, x<0; x+5, x>=0 What are the intervals of continuity for the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = x + 5 has y-intercept at y = 5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: 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? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 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? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!