course Mth 271 ????????????assignment #010
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18:09:53 What is a polynomial with zeros at -3, 4 and 9? Describe the graph of your polynomial.
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RESPONSE --> (x-3)(x+4)(x+9) (x-3)(x+4) = x^2 - 3x + 4x - 12 (x + 9)= x^3 + 10x^2 - 3x - 108 This graph would have 3 different turning points at which it would cross the x-axis. confidence assessment: 3
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18:10:45 A polynomial with zeros at -3, 4 and 9 must have factors (x + 3), ( x - 4) and (x - 9), and so must contain (x+3) (x-4) (x-9) as factors. These factors can be multiplied by any constant. For example 8 (x+3) (x-4) (x-9), -2(x+3) (x-4) (x-9) and (x+3) (x-4) (x-9) / 1872 are all polynomizls with zeros at -3, 4 and 9. If -3, 4 and 9 are the only zeros then (x+3), (x-4) and (x-9) are the only possible linear factors. It is possible that the polynomial also has irreducible quadratic factors. For example x^2 + 3x + 10 has no zeros and is hence irreducible, so (x+3) (x-4) (x-9) (x^3 + 3x + 10) would also be a polynomial with its only zeros at -3, 4 and 9. The polynomial could have any number of irreducible quadratic factors. **
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RESPONSE --> My polynomial was just multiplying all factors together to come up with a completed polynomial. self critique assessment: 2
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18:11:20 1.5.18 (was 1.5.16 right-, left-hand limits and limit (sin fn)
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RESPONSE --> A) -2 B) -2 C)-2 confidence assessment: 2
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18:11:36 What are the three limits for your function (if a limit doesn't exist say so and tell why)?
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RESPONSE --> The limit of my function was -2 confidence assessment: 2
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18:11:58 Imagine walking along the graph from the right. As you approach the limiting x value, your y 'altitude' gets steadier and steadier, approaching closer and closer what value? You don't care what the function actually does at the limiting value of x, just how it behaves as you approach that limiting value. The same thing happens if you walk along the graph from the left. What does you y value approach? Is this the value that you approach as you 'walk in' from the right, as well as from the left? If so then it's your limit. **
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RESPONSE --> self critique assessment: 2
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18:12:21 1.5.22 (was 1.5.20 right-, left-hand limits and limit (discont at pt) What are the three limits for your function (if a limit doesn't exist say so and tell why)?
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RESPONSE --> A) 0 B) 2 C) Does not exist. confidence assessment: 2
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18:12:32 STUDENT RESPONSE: The three limits are a. 0 b. 2 c. no limit The limit doesn't exist because while the function approaches left and right-handed limits, those limite are different. INSTRUCTOR COMMENT: That is correct. ADVICE TO ALL STUDENTS: Remember that the limit of a function at a point depends only on what happens near that point. What happens at the point iself is irrelevant to the limiting behavior of the function as we approach that point.
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RESPONSE --> self critique assessment: 3
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18:14:01 1.5.30 (was 1.5.26 lim of (x+4)^(1/3) as x -> 4 What is the desired limit and why?
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RESPONSE --> The desired limit is 2 because the cubed root of 8 is 2. If it were a square root, it could either be 2 or -2 as an answer. confidence assessment: 2
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18:14:07 The function is continuous, so f(c) = c and in this case the limit is 8^(1/3), which is 2. **
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RESPONSE --> self critique assessment: 3
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18:16:08 1.5.48 (was 1.5.38 lim of (x^3-1)/(x-1) as x -> 1
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RESPONSE --> We must use the elimination process becasue 1 makes both sides equal zero. (x^3 - 1) / (x-1) = ((x-1) x^2 + x - 1) / (x - 1) = x^2 + x - 1 = 1^2 + 1 - 1 = 1 confidence assessment: 2
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18:17:32 As x -> 0 both numerator and denominator approach 0, so you can't tell just by plugging in numbers what the limit will be. If you factor x^3-1 into (x-1)(x^2+x+1) you can reduce the fraction to (x-1)(x^2+x+1) / (x-1) = x^2 + x + 1, which is equal to the original function for all x except 1 (we can't reduce for x = 1 because x-1 would be zero, and we can't divide by zero). As x -> 1, x^2 + x + 1 -> 3. It doesn't matter at all what the function does at x = 1, because the limiting value of x never occurs when you take the limit--only x values approaching the limit count. 3 is therefore the correct limit. **
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RESPONSE --> I had -1 instead of plus 1, then I would have came up with the right answer. I see my mistake now. self critique assessment: 2
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18:18:13 1.5.70 (was 1.5.56 lim of 1000(1+r/40)^40 as r -> 6%
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RESPONSE --> Yes, the limit is 1814.02 1000 (1 + (.06/4))^40 confidence assessment: 2
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18:18:53 $1000 *( 1+.06 / 40)^40 = 1061.788812. Since this function changes smoothly as you move through r = .06--i.e., since the function is continuous at r = .06--this value will be the limit. **
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RESPONSE --> I divided by 4 instead of 40, that was my problem on this question. self critique assessment: 2
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18:18:57 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 3
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