#$&* course MTH 271 3/4 9:30 010. `query 10*********************************************
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Given Solution: `a A polynomial with zeros at -3, 4 and 9 must have factors (x + 3), ( x - 4) and (x - 9), and so must contain (x+3) (x-4) (x-9) as factors. These factors can be multiplied by any constant. For example 8 (x+3) (x-4) (x-9), -2(x+3) (x-4) (x-9) and (x+3) (x-4) (x-9) / 1872 are all polynomials with zeros at -3, 4 and 9. If -3, 4 and 9 are the only zeros then (x+3), (x-4) and (x-9) are the only possible linear factors. It is possible that the polynomial also has irreducible quadratic factors. For example x^2 + 3x + 10 has no zeros and is hence irreducible, so (x+3) (x-4) (x-9) (x^3 + 3x + 10) would also be a polynomial with its only zeros at -3, 4 and 9. The polynomial could have any number of irreducible quadratic factors. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q1.5.18 (was 1.5.16 right-, left-hand limits and limit (sin fn) I am not sure which question this should be, since I have the eigth edition of the book and all of the instructions so far have been for the sixth or seventh edition. However, question 18 shows a sine graph with (-2, -2) as a point, c = -2, and y = f(x) as the function. The question directs me to find the limit as x approaches c, c+ and c-. For c: the limit as x approaches -2 is -2. For c+: -2 For c-: -2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q1.5.22 (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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For problem 22, the limits are as follows: c+ = 0 c- = 2 c = This limit does not exist. That is because it is almost like two graphs on one page, and the limits of each piece are different because of the different places those two sides approach. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: The three limits are a. 0 b. 2 c. no limit The limit doesn't exist because while the function approaches left and right-handed limits, those limite are different. INSTRUCTOR COMMENT: That is correct. ADVICE TO ALL STUDENTS: Remember that the limit of a function at a point depends only on what happens near that point. What happens at the point iself is irrelevant to the limiting behavior of the function as we approach that point. STUDENT QUESTION Would it be ok to say that the two limits are different because the graph indicates -1 being undefined? INSTRUCTOR RESPONSE The limit as x approaches -1 isn't affected by the value of the function at x = -1, or by the lack of a value. You consider only what happens as you get close to -1. In considering the limit, you never actually get to -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q1.5.30 (was 1.5.26 lim of (x+4)^(1/3) as x → 4 What is the desired limit and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The limit should be at 2. This is true because the y value approaches 2 as x approaches 4. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The limit of a function at a point is not defined by its value at that point, but rather by its values near the point. A continuous function is one for which the values near a given point happen to approach the value of the function at that point. This is the case for most familiar functions at most points, so it can be difficult to appreciate why this is an important definition. The present function f(x) = (x + 4)^(1/3) is defined for all values of x, and like most functions with which you are familiar, is continuous for all values of x. The present question asks about the limit of this function as x approaches 4. If x is near 4, then in this case f(x) is near f(4). Furthermore no matter how close we want f(x) to be to f(4), we can guarantee that it's this close by requiring that x be close enough to 4. It is for this reason and in this sense that the limiting value of f(x) = (x + 4)^(1/3), as x approaches 4, is (4 + 4)^(1/3) = 2. For example, if we want f(x) to be within .001 of 2, this can certainly be guaranteed by letting x be within .00005 of 4. You might well ask how we came up with .0001 (answer: if y = (x+4)^(1/3) then x = y^3 - 4; the cube of a number that lies within delta of 2 is (2 - delta)^3 = 2^3 - 3 delta * 2^2 + 3 delta^2 * 2 - delta^3 = 8 - 12 delta + 6 delta^2 - delta^3; we use this to reason out how close x must be to 4 in order to guarantee that y be within .001 of 2. This boils down to the following: The function is continuous, so f(c) = c and in this case the limit is 8^(1/3), which is 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q1.5.48 (was 1.5.38 lim of (x^3-1)/(x-1) as x → 1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x^3-1)/(x-1) = x^2 + x + 1 When x = 1, y = 3 Therefore, the limit as x approaches 1 is 3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a As x -> 0 both numerator and denominator approach 0, so you can't tell just by plugging in numbers what the limit will be. The given function is not defined for x = 0, so it cannot be evaluated at x = 0. However 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. This expression is not the same as the original function, because this is defined at x = 0, whereas the original expression was not defined at x = 0. In any case the expression x^2 + x + 1 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). And it doesn't matter what happens to the function at the limiting point. The only thing that matters is what happens near the limiting point. So we can find the limit of (x^3-1)/(x-1) as x -> 1 by using the values of x^2 + x + 1 near x = 1. The expression x^2 + x + 1 is continuous, so as x -> 1, x^2 + x + 1 -> 3, and the limit of our original function, as x -> 1, is 3. It doesn't matter at all what the function does at x = 1, because the limiting value of x is not important when you take the limit--only x values approaching the limit matter. STUDENT COMMENT This is a little confusing. I don’t completely understand the steps that are written out here. INSTRUCTOR RESPONSE In order for me to clarify, you need to tell me as specifically as possible what you do and do not understand. I expect that you understand how factoring reduces the expression to x^2 + x + 1. I expect you understand that when x is near 1 the value of x^2 + x + 1 is near 3. The likely point of confusion is the rest of the explanation, which establishes why the expression x^2 + x + 1 applies only as long as x is not 1, and why this doesn't matter when finding the limit, which considers only values of the function when x is 'near' 1, what happens at x = 1. Let me know the specifics about what you do and do not understand, and I'll be glad to clarify further. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q1.5.70 (was 1.5.56 lim of 1000(1+r/40)^40 as r -> 6% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1000(1 + 0.06/40)^40 = 1061.788813 So, as x approaches 0.06, or 6%, the limit approaches 1061.788813. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a $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. ** Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!