course Mth 271 In my assignment I was supposed to do three randomized problems from week 5. When I followed the link I couldn't find any problems for week 5. Is there somewhere else I need to look?
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06:48:31 What is a polynomial with zeros at -3, 4 and 9? Describe the graph of your polynomial.
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RESPONSE --> A polynomial with zeros at -3, 4 and 9 is the same as (x+3)(x-4)(x-9). The graph travels up through the x axis at -3, comes back down through at x = 4 and goes back up at x = 9 confidence assessment: 3
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06:50:43 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 --> I see that a polynomial with -3, 4 and 9 as zeros is not limited to just (x+3)(x-4)(x-9). It can also be multiplied or divided by a constant and it can have other irreducible factors. self critique assessment: 2
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06:54:20 1.5.18 (was 1.5.16 right-, left-hand limits and limit (sin fn)
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RESPONSE --> In the graph, as x approaches c from the left, the limit is -2. The same is true as x approaches c from the right. Therefore the lim f(x) does exist because both limits are the same. confidence assessment: 2
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06:56:26 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 x->c(+) exists at -2. The limit x->c(-) also exists at -2. Therefore the limit x->c does exist and is at -2. confidence assessment: 2
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06:56:56 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: 3
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06:59:43 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 --> The limit as x approaches c from the right is 0. The limit as x approaches c from the left is 2. That means that both limits are one-sided. The limit x->c does not exist because the graph approaches different points from the two sides. confidence assessment: 3
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07:02:54 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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07:04:46 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 lim of (x+4)^(1/3) as x -> 4 is 2. I find this by using direct substitution. I substitute 4 for x. (4 + 4)^(1/3) = 8^(1/3) = 2. confidence assessment: 3
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07:04:56 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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07:11:28 1.5.48 (was 1.5.38 lim of (x^3-1)/(x-1) as x -> 1
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RESPONSE --> When I substitute 1 into the equation, the frachtion is undefined. I can divide the numerator and the denominator by (x-1), thought. That gives me [(x-1)(x^2+x+1)]/(x-1) = x^2 + x + 1. Now when I substitute 1, I find 1^2 + 1 + 1 = 3. Using the replacement theorem, I know that in the original fraction, the limit is 3 as x-> 1. confidence assessment: 3
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07:12:05 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 --> self critique assessment: 3
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07:17:27 1.5.70 (was 1.5.56 lim of 1000(1+r/40)^40 as r -> 6%
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RESPONSE --> When I substitute .06 for r, I find the lim is 1814.02. confidence assessment: 3
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07:20:46 $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 --> I see my mistake. When I typed the function in my calculator I put .06/4 instead of .06/40. self critique assessment: 2
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07:23:04 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It is interesting that a function can have two different limits depending on which side x approaches c from. self critique assessment: 3
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