course Mth163 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: `a** You can have as many as 6 linear and 3 irreducible quadratic factors for a polynomial of degree 6. For a polynomial of degree 6: If you have no irreducible quadratic factors then to have degree 6 you will need 6 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 4 linear factors. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 2 linear factors. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you can have no linear factors. For a polynomial of degree 7: If you have no irreducible quadratic factors then to have degree 7 you will need 7 linear factors. If you have exactly one irreducible quadratic factor then this factor is of degree 2 and you will need 5 linear factors to give you degree 7. If you have exactly two irreducible quadratic factors then the product of these factors is of degree 4 and you will need 3 linear factors to give you degree 7. If you have three irreducible quadratic factors then the product of these factors is of degree 6 and you will need 1 linear factor to give you degree 7. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, It didn’t ask for a seven degree but I understand the combinations of linear and irreducible quad. factors to get your required degree. Self-critique rating:ok ********************************************* Question: `qFor a degree 6 polynomial with one irreducible quadratic factor and four linear factors list the possible numbers of repeated and distinct zeros. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: An irreducible quadratic factor has no zeros and 4 linear factors has up to 4 possible zeros. You could have 1 linear repeated 4 times, 1 linear repeated 2 times * another 1 linear repeated 2 times., You could have 1 linear repeated 3 times * 1 linear , or 4 different linear equations Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** there could be 1 root repeated 4 times, 2 roots with 1 repeated 3 times and the other distinct from it, 2 distinct roots each repeated twice, three distinct roots with one of them repeated twice, or four distinct roots. Explanation: You have one zero for every linear factor, so there will be four zeros. Since the degree is even the far-left and far-right behaviors will be the same, either both increasing very rapidly toward +infinity or both decreasing very rapidly toward -infinity. You can have 4 distinct zeros, which will result in a graph passing straight thru the x axis at each zero, passing one way (up or down) through one zero and the opposite way (down or up) through the next. You can have 2 repeated and 2 distinct zeros. At the repeated zero the graph will just touch the x axis as does a parabola at its vertex. The graph will pass straight through the x axis at the two distinct zeros. You can have 3 repeated and 1 distinct zero. At the 3 repeated zeros the graph will level off at the instant it passes thru the x axis, in the same way the y = x^3 graph levels off at x = 0. The graph will pass through the x axis at the one distinct zero. You can have two pairs of 2 repeated zeros. At each repeated zero the graph will just touch the x axis as does a parabola at its vertex. Since there are no single zeros (or any other zeros repeated an odd number of times) the graph will not pass through the x axis, so will remain either entirely above or below the x axis except at these two points. You can have four repeated zeros. At the repeated zero the graph will just touch the x axis, much as does a parabola at its vertex except that just as the y = x^4 function is somewhat flatter near its 'vertex' than the y = x^2 function, the graph will be flatter near this zero than would be a parabolic graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Since the degree is even you will have behavior on the far left and the far right to be the same either both increasing or decreasing rapidly toward infinity. You could have 4 distinct zeros or 2 repeated and 2 distinct zeros, 3 repeated and 1 distinct zero. Remember a graph of the y = x^4 will be flatter lying closer to its zeros than the parabola function y = x^2. Self-critique rating:ok ********************************************* Question: `qDescribe a typical graph for each of these possibilities. Describe by specifying the shape of the graph at each of its zeros, and describe the far-left and far-right belavior of the graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: with 4 zeros the graph curves up above the x axis crossing at the f zero points and is the same to the right and to the left with 2 repeated and 2 zeros the graph curves up above the x axis with just touching the x axis and crossing it at the 2 zeros with 3 repeated and 1 zero the graph curves up leveling out and then crossing the x axis and then leveling out before increasing towards the opposite direction. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aERRONEOUS STUDENT SOLUTION AND INSTRUCTOR CORRESION for 4 distincts, the graph curves up above the x axis then it crosses it 4 timeswithe the line retreating from the direction it came from for say 2 distincts and 2 repeats, line curves above x axis then it kisses the axis then it crosses it twice, retreating to the side it came from for 1 distinct and 3 it curves above x axis then it crosses once and kisses, finnaly heading off to the opposite side it came from INSTRUCTOR COMMENTS: {The graph can't go off in th opposite direction. Since it is a product of four linear factors and any number of quadratic factors its degree is even so its large-x behavior is the same for large positive as for large negative x. It doesn't kiss at a degree-3 root, it acts like the y = x^3 polynomial, leveling off just for an instant as it passes through the zero and on to the other side of the axis. ** problems 3-5 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique rating: ********************************************* Question: `qIt doesn't matter if you don't have a graphing utility--you can answer these questions based on what you know about the shape of each power function. Why does a cubic polynomial, with is shape influenced by the y = x^3 power function, fit the first graph better than a quadratic or a linear polynomial? What can a cubic polynomial do with this data that a quadratic can't? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A linear is just a straight line, quadratics can either curve upwards or downwards and cubic graphs can change curves from upward to downwards. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the concavity (i.e., the direction of curvature) of a cubic can change. Linear graphs don't curve, quadratic graphs can be concave either upward or downward but not both on the same graph. Cubics can change concavity from upward to downward. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Concavity – means the direction of curvature. Note that quadratics can only do one direction at a time. Self-critique rating:ok ********************************************* Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The 4 degree polynomial is flatter and closer to it’s zeros of that graph and the 6 degree polynomial is even flatter becoming even closer than the other two graphs to it’s zeros. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** higher even degrees flatten out more near their 'vertices' ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique rating: ********************************************* Question: `qOn problem 5, how do the shapes of 4th-degree polynomials and 6th-degree polynomials progressively differ from the shape of a 2d-degree polynomial in such a way as to permit a better and better fit to the graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There were more changes with curvature with the 4 degree and even more with the 6 degree allowing a better fit to the graph. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT RESPONSE: progressively more flexing, because more curves, and fit graph better the that of a lesser degree INSTRUCTOR COMMENT: On a degree-2 polynomial there is only one change of direction, which occurs at the vertex. For degrees 4 and 6, respectively, there can be as many as 3 and 5 changes of direction, respectively. For higher degrees the graph has more ability to 'wobble around' to follow the data points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2 degree only changes one time For higher degrees there is more ability to ‘wobble around’ to follow data point. Self-critique rating:ok ********************************************* Question: `qWhat is the degree 2 Taylor approximation for f(t) = e^(2t), and what is your approximation to f(.5)? How close is your approximation to the actual value of e^(2t)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would obtain f(t) = 1 + 2t + (2t)^2 / 2 = 1 + 2t + (4t) ^2 / 2 For f(5) we have: f(.5) = 1 + 2*.5 + (2*.5)^2 / 2 = 1 +1 + .5 = 2.5 In F(.5) = e^(2 * .5) = 2.718 approx….this is only a difference of .218 between the two f(.5) Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The degree 2 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! = 1 + 2 t + 4 t^2 / 2 = 1 + 2 t + 2 t^2. Therefore we have T(.5) = 1 + 2 * .5 + 2 * .5^2 = 1 + 1 + 2 * .25 = 1 + 1 + .5 = 2.5. The actual value of e^(2t) at t = .5 is f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx.. The approximation is .218 less than the actual function. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, should have said .218 less than the actual function. Self-critique rating:ok ********************************************* Question: `qBy how much does your accuracy improve when you make the same estimate using the degree 3 Taylor approximation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the 3 degree Taylor fit is 1 + 2t + (2t)^2/2! + (2t)^3/3! = 1 + 2t + 4t^2/2 + 8t^3/6 = 1 + 2t + 2t^2 + 4t^3/3 Now to check our accuracy with t(.5) = 1 + 2*.5 + 2 *(.5)^2 + 4 * (.5)^3 / 3 = 2.667 approx. Using f(t) = e^(2t)…and t = .5..we have….f(.5) = e^(2 * .5) = 2.718 appox….which is .051 lees than the actual function. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The degree 3 Taylor polynomial is f(t) = 1 + 2t + (2t)^2 / 2! + (2t)^3 / 3! = 1 + 2 t + 4 t^2 / 2 + 8 t^3 / 6 = 1 + 2 t + 2 t^2 + 4 t^3 / 3. Therefore we have T(.5) = 1 + 2 * .5 + 2 * .5^2 + 4 * .5^3 / 3 = 1 + 1 + 2 * .25 + 4 * .125 / 3 = 1 + 1 + .5 + .167 = 2.667. The actual value of e^(2t) at t = .5 is f(.5) = e^(2 * .5) = e^1 = e = 2.718, approx.. The approximation is .051 less than the actual function. This is about 4 times closer than the approximation we obtained from the degree-2 polynomial. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Note that this is 4 times closer than the approx. we obtained from the 2 degree polynomial Self-critique rating:ok ********************************************* Question: `qDescribe your graph of the error vs. the degree of the approximation for degree 2, degree 3, degree 4 and degree 5 approximations to e^.5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For a 2 degree = 1 + .5 + .5^2/2 = 1.625…e^.5 = 1.649 approx…difference of .024 for a 3 degree = 1 + .5 + .5^2/2 + .5^3/6 = 1.646 approx….difference from e^.5 = .003 for a 4 degree = 1 + .5 + .5^2/2 + .5^3/6 + .5^4/24 = 1.648 approx…difference from e^.5 = .001
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Given Solution: `a** The errors, rounded to the nearest thousandth, are: degree-2 error: -.218 degree-3 error: -.051 degree-4 error: -.010 degree-5 error: -.002 A graph of error vs. degree decreases rapidly toward the horizontal axis, showing that the error decreases rapidly toward zero as the degree increases. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): our degrees in error did not match up showing that I must not have the correct way of figuring out the degrees. ???? Mr. Smith, could you please give me a step by step procedure of how you came by your error so I can see where I was incorrect, thank-you ????
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Given Solution: `a** The function is P2(x) = (x-1) - (x-1)^2/2. A table of values of ln(x), P2(x) and P2(x) - ln(x): x ln(x) P2(x) P2(x) - ln(x) .6 -0.5108256237 -0.48 0.03082562376 .8 -0.2231435513 -0.22 0.003143551314 1.2 0.1823215567 0.18 -0.002321556793 1.4 0.3364722366 0.32 -0.01647223662 At x = 1 we have ln(x) = ln(1) = 0, and P2(x) = P2(1) = (1-1) - (1-1)^2 / 2 = 0. There is no difference in values at x = 1. As we move away from x = 1 the approximation becomes less and less accurate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): P2(x) = (1-1) – (1-1)^2/2 So in ln(x), x = 1, we have ln(1) = 0, and P2(x) = P2(1) = (1-1) – (1-1)^2/2 = 0, showing no difference in x values. As we move away from the x = 1 the approx. becomes less and less accurate. Self-critique rating:ok ********************************************* Question: `qWhat is the error in the degree 2 approximation to ln(x) for x = .6, .8, 1.2 and 1.4? Why does the approximation get better as x approaches 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: for x = .6, we have ln(.6) = -.5108 approx and forP2 we have P2(.6) = (.6 – 1) – (.6-1)^2/2 = -.48 with an error of -.0308 for x = .8 we have ln(x) = ln(.8) = -.2231approx. and P2(x) = P2(.8) = (.8 – 1) – (.8 -1)^2/2 = -.22 with an error of -.0031 for x = 1.2 we have ln(x) = ln(1.2) = .1823 approx. and P2(x) = P2(1.2) = (1.2 – 1) – (1.2 – 1)^2/2 = .18 a difference of .0023 for x – 1.4 we have ln(x) = ln(1.4) = .3325 approx. and P2(x) = P2(1.4) = (1.4 – 1) – (1.4 – 1)^2/2 = .32 an error of .0165 the approximation gets better as we move closer to 1 because there is no error in 1 and the further we move away from 1 the less and less accurate it gets. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The respective errors are .03, .00314, .00232, .016472. There is no error at x = 1, since both the function and the approximation give us 0. As we move away from 1 the approximation becomes less and less accurate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, we have some differences with or approx. but that is only due to rounding Self-critique rating: ********************************************* Question: `qproblem 12. What does the 1/x graph do than no quadratic function can do? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the y = 1/x graph has a vertical and horizontal asymptote and the quadratic functions are more like parabolas. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The y = 1/x graph has vertical asymptotes at the y axis and horizontal asymptotes at the x axis. The parabolas we get from quadratic functions do have neither vertical nor horizontal asymptotes. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique rating: ********************************************* Question: `qWhat are the errors in the quadratic approximation to 1/x at x = .6, .8, 1, 1.2, and 1.4? Describe a graph of the approximation error vs. x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: looking back at some of the previous problems I think the way to find the errors is to take P2 – 1/x, remembering P2(x) = 1 – (x-1) + (x-1)^2/2 ???? Mr. Smith, what I am confused about is what is the equation to work out 1/x . I have looked at our notes and see the polynomial approximations and the chart with the degree of three but I’m not sure what to use. ????
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Given Solution: `a** The quadratic approximation to 1/x is the second-degree Taylor polynomial P2(x) = 1 - (x - 1) + (x - 1)^2. A table of values of 1/x, P2(x) and P2(x) - 1/x: x 1/x P2(x) P2(x) - 1/x .6 1.666666666 1.56 - 0.1066666666 .8 1.25 1.24 - 0.01 1.2 0.8333333333 0.84, 0.006666666666 1.4 0.7142857142 0.76 0.04571428571. A graph of appoximation error vs. x decreases at a decreasing rate to 0 at x = 1, then increases at an increasing rate for x > 1. This shows how the accuracy of the approximation decreases as we move away from x = 1. The graph of approximation error vs. x gets greater as we move away from x = 1.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: