course Mth 271 ªv¾¬úʤ¾¿NËÏÔ•fòYþ¸¬l¿Üêøassignment #010
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17:53:55 What is a polynomial with zeros at -3, 4 and 9? Describe the graph of your polynomial.
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RESPONSE --> x^3-10x^2-3x+108 intersects the x axis at -3, 4, and 9 confidence assessment: 3
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17:54:43 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 --> yup self critique assessment: 3
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17:55:46 1.5.18 (was 1.5.16 right-, left-hand limits and limit (sin fn)
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RESPONSE --> all three limits are -2 confidence assessment: 3
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17:55:57 What are the three limits for your function (if a limit doesn't exist say so and tell why)?
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RESPONSE --> -2 confidence assessment: 3
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17:56:22 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 --> yup self critique assessment: 3
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17:57:21 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 --> lim as x approaches -1 from the right is 0 lim as x approaches -1 from the left is 2 lim as x approaches -1 is undefined confidence assessment: 3
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17:57:39 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 --> yup self critique assessment: 3
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17:58:20 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 --> substitute in 4, lim= 2 confidence assessment: 3
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17:58:29 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 --> yup self critique assessment: 3
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17:59:31 1.5.48 (was 1.5.38 lim of (x^3-1)/(x-1) as x -> 1
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RESPONSE --> when you subst in 1, both the numerator and denominator are zero. x-1 is a factor of both the num and denom, so factor it out, which leaves x^2+x subst in 1, and you get 2 confidence assessment: 3
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18:01:55 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 --> okay i see where i made a mistake in factoring self critique assessment: 3
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18:04:55 1.5.70 (was 1.5.56 lim of 1000(1+r/40)^40 as r -> 6%
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RESPONSE --> if you plug in .06, you get A=$1814.02 i graphed it but not sure about the limit confidence assessment: 1
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18:05:46 $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 --> okay self critique assessment: 3
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18:05:54 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none self critique assessment: 3
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