course Mth 173 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q query problem 1.7.24 5th; 1.7.20 4th (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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For sin theta/theta theta=0 Sin(0)/0= undefined so it is non-continuous. For ½ for theta =0 I am not sure what to do in this case because the function does not incorporate theta so would it not just be a answer of .5???????
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am confused on the second part of this problem and how to do this without the theta being incorporated into the problem?????
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Given Solution: ** For h = .1, .01, .001 the values of (cos(h)-1 ) / h are -0.04995834722, -0.004999958472, -0.0005. The limit is zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have a fairly good idea of what is going on but I want to make sure I can explain it in my own words. The limit is the value that approached by a function but does not ever equal the function example being sinx/x I can see on the graph that the graph is approaching 1 but never makes it there, to further validate this I can solve the function sinx/x at point .1 to get .998334 at point .01 to get .999983334 at point .001 to get .99999998333 eventually this does make it to one but I am all but sure it is because my calculator eventually rounds the value to 1. Is my interpretation of the idea and process of this concept correct???????