course mth173 ?x???ú??????yh??assignment #011?????D????????Calculus I
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00:25:46 problem 1.7.6 (was 1.11.4) continuity of x / (x^2+2) on (-2,2)is the function continuous on the given interval and if so, why?
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RESPONSE --> Plugging the following terms into the equation we get: x y -1 -1 -1.99 -.33 -2.001 -.33 -2.01 -.33 0 0 1 .33 2 .33 This would make a graph of a near horizontal line in quadrant 3 that then goes through the origin and continues at the same near horizontal line in quadrant 1. The function is continuous
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00:27:23 ** The denominator would never be 0, since x^2 must always be positive. So you could never have division by zero, and the function is therefore defined for every value of x. The function also has a smooth graph on this interval and is therefore continuous. The same is true of the correct Problem 4, which is 1 / `sqrt(2x-5) on [3,4]. On this interval 2x-5 ranges continuously from 2*3-5=1 to 2*4-5=3, so the denominator ranges continuously from 1 to `sqrt(3) and the function itself ranges continuously from 1 / 1 to 1 / `sqrt(3). **
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RESPONSE --> okay
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00:41:49 query problem 1.7.16 (was 1.11.9) continuity of sin(x) / x, x<>0; 1/2 for x = 0. Where is the function continuous and where is it not continuous?
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RESPONSE --> x cannot = 0 because you cannot divide be zero and the finction cannot have a y intercept however it can get infinatly close to the y axis.
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00:48:11 ** Division by zero is not defined, so sin(x) / x cannot exist at x = 0. The function is, however, not defined at x = 0 by sin(x) / x; the definition says that at x = 0, the function is equal to 1/2. It remains to see what happens to sin(x) / x as x approaches zero. Does the function approach its defined value 1/2, in which case the value of the function at x = 0 would equal its limiting value x = 0 and the function would be continuous; does it approach some other number, in which case the limiting value and the function value at x = 0 would not the equal and the function would not be continuous; or does the limit at x = 0 perhaps not exist, in which case we could not have continuity. Substituting small nonzero values of x into sin(x) / x will yield results close to 1, and the closer x gets to 0 the closer the result gets to 1. So we expect that the limiting value of the function at x = 0 is 1, not 1/2. It follows that the function is not continuous. **
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RESPONSE --> I didn't understand what was ment by the part of the question that said at x = 0, the function is equal to 1/2. The closer x gets to 0 the closer y gets to 1 but at x = zero the value should be 1/2 so there would be a jump which could not happen mathematically at the near zero point which makes the function non continous.
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00:51:19 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> The basic idea is to give an very educated guess about where a point would land that is undefined by the function by getting several points extremely close to the undifined point and giving you an idea of where the undifined point would be, however if there is no undifined point then the function is continuous.
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