course Mth 272
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RESPONSE -->
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16:49:06 You should understand the basic ideas, which include but are not limited to the following: rules of differentiation including product, quotient and chain rules, the use of first-derivative tests to find relative maxima and minima, the use of second-derivative tests to do the same, interpreation of the derivative, implicit differentiation and the complete analysis of graphs by analytically finding zeros, intervals on which the function is positive and negative, intervals on which the function is increasing or decreasing and intervals on which concavity is upward and downward. Comment once more on your level of preparedness for this course.
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16:49:23 4.1.16 (was 4.1.14): Solve for x the equation 4^2=(x+2)^2
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RESPONSE --> 4^2=(x+2)^2 4 = x + 2 2 = x confidence assessment: 3
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16:49:28
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RESPONSE --> confidence assessment:
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16:49:32 The steps in the solution: 4^2 = (x+2)^2. The solution of a^2 = b is a = +- sqrt(b). So we have x+2 = +- sqrt(4^2) or x+2 = +- 4. This gives us two equations, one for the + and one for the -: x+2 = 4 has solution x = 2 x+2 = -4 has solution x = -6. **
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RESPONSE --> self critique assessment:
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16:49:44 4.1.28 (was 4.1.32) graph 4^(-x). Describe your graph by telling where it is increasing, where it is decreasing, where it is concave up, where it is concave down, and what if any lines it has as asymptotes.
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RESPONSE --> The graph decreases at a decreasing rate moving left to right. The concave is up and with an asymptote of y = 0 confidence assessment: 3
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16:49:47 Many students graph this equation by plugging in numbers. That is a start, but you can only plug in so many numbers. In any case plugging in numbers is not a calculus-level skill. It is necessary to to reason out and include detailed reasons for the behavior, based ultimately on knowledge of derivatives and the related behavior of functions. A documented description of this graph will give a description and will explain the reasons for the major characteristics of the graph. The function y = 4^-x = 1 / 4^x has the following important characteristics: For increasing positive x the denominator increases very rapidly, resulting in a y value rapidly approaching zero. For x = 0 we have y = 1 / 4^0 = 1. For decreasing negative values of x the values of the function increase very rapidly. For example for x = -5 we get y = 1 / 4^-5 = 1 / (1/4^5) = 4^5 = 1024. Decreasing x by 1 to x = -6 we get 1 / 4^-6 = 4096. The values of y more and more rapidly approach infinity as x continues to decrease. This results in a graph which for increasing x decreases at a decreasing rate, passing through the y axis at (0, 1) and asymptotic to the positive x axis. The graph is decreasing and concave up. When we develop formulas for the derivatives of exponential functions we will be able to see that the derivative of this function is always negative and increasing toward 0, which will further explain many of the characteristics of the graph. **
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RESPONSE --> self critique assessment: 3
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16:50:01 How does this graph compare to that of 5^-x, and why does it compare as it does?
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RESPONSE --> confidence assessment:
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16:50:04 the graphs meet at the y axis; to the left of the y axis the graph of y = 5^-x is higher than that of y = 4^-x and to the right it is lower. This is because a higher positive power of a larger number will be larger, but applying a negative exponent will give a smaller results for the larger number. **
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RESPONSE --> self critique assessment:
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16:50:17 4.2.20 (was 4.1 #40) graph e^(2x) Describe your graph by telling where it is increasing, where it is decreasing, where it is concave up, where it is concave down, and what if any lines it has as asymptotes.
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RESPONSE --> Moving left to right the graph is increasing at an increasing rate and concaves up. confidence assessment: 3
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16:50:20 For large numbers x you have e raised to a large power, which gets extremely large. At x = 0 we have y = e^0 = 1. For large negative numbers e is raised to a large negative power, and since e^-a = 1 / e^a, the values of the function approach zero. } Thus the graph approaches the negative x axis as an asymptote and grows beyond all bounds as x gets large, passing thru the y axis as (0, 1). Since every time x increases by 1 the value of the function increases by factor e, becoming almost 3 times as great, the function will increase at a rapidly increasing rate. This will make the graph concave up. **
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RESPONSE --> self critique assessment: 3
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16:50:33 The entire description given above would apply to both e^x and e^(2x). So what are the differences between the graphs of these functions?
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RESPONSE --> e^(2x) will be twice as much as e^x. confidence assessment: 3
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16:50:36 Note that the graphing calculator can be useful for seeing the difference between the graphs, but you need to explain the properties of the functions. For example, on a test, a graph copied from a graphing calculator is not worth even a point; it is the explanation of the behavior of the function that counts. By the laws of exponents e^(2x) = (e^x)^2, so for every x the y value of e^(2x) is the square of the y value of e^x. For x > 1, this makes e^(2x) greater than e^x; for large x it is very much greater. For x < 1, the opposite is true. You will also be using derivatives and other techniques from first-semester calculus to analyze these functions. As you might already know, the derivative of e^x is e^x; by the Chain Rule the derivative of e^(2x) is 2 e^(2x). Thus at every point of the e^(2x) graph the slope is twice as great at the value of the function. In particular at x = 0, the slope of the e^x graph is 1, while that of the e^(2x) graph is 2. **
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RESPONSE --> self critique assessment:
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16:50:50 How did you obtain your graph, and what reasoning convinces you that the graph is as you described it? What happens to the value of the function as x increases into very large numbers? What is the limiting value of the function as x approaches infinity?
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RESPONSE --> As x increases, y get very large, so the graph will get real steep as x gets larger. as x approaches infinity the limit is infinity. As x approaches - infinity the limit is 0. confidence assessment: 3
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16:50:52 *& These questions are answered in the solutions given above. From those solutions you will ideally have been able to answer this question. *&*&
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RESPONSE --> self critique assessment: 3
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16:51:03 4.2.32 (formerly 4.2.43) (was 4.1 #48) $2500 at 5% for 40 years, 1, 2, 4, 12, 365 compoundings and continuous compounding
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RESPONSE --> A = P[1 + (r/n)]^nt A = 2500[1 + (0.05/1]^(1)(40) = 17,599.97 A = 2500[1 + (0.05/2]^(2)(40) = 18,023.92 A = 2500[1 + (0.05/4]^(4)(40) = 18,245.05 A = 2500[1 + (0.05/12]^(12)(40) = 18,396.04 A = 2500[1 + (0.05/365]^(365)(40) = 18,470.11 confidence assessment: 3
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16:51:06 A = P[1 + (r/n)]^nt A = 2500[1 + (0.05/1]^(1)(40) = 17599.97 A = 2500[1 + (0.05/2]^(2)(40) = 18023.92 A = 2500[1 + (0.05/4]^(4)(40) = 18245.05 A = 2500[1 + (0.05/12]^(12)(40) = 18396.04 A = 2500[1 + (0.05/365]^(365)(40) = 18470.11
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RESPONSE --> confidence assessment: 3
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16:51:17 How did you obtain your result for continuous compounding?
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RESPONSE --> A = Pe^rt A = 2500e^(.05)(40) A = 18,472.64 confidence assessment: 3
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16:51:19 For continuous compounding you have A = Pe^rt. For interest rate r = .05 and t = 40 years we have A = 2500e^(.05)(40). Evaluating we get A = 18472.64 The pattern of the results you obtained previously is to approach this value as a limit. **
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RESPONSE --> self critique assessment: 3
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16:51:32 4.2.40 (was 4.1 #60) typing rate N = 95 / (1 + 8.5 e^(-.12 t)) What is the limiting value of the typing rate?
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RESPONSE --> as t approaches 0 lim = 95 confidence assessment: 3
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16:51:34 As t increases e^(-.12 t) decreases exponentially, meaning that as an exponential function with a negative growth rate it approaches zero. The rate therefore approaches N = 95 / (1 + 8.5 * 0) = 95 / 1 = 95. *&*&
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RESPONSE --> self critique assessment: 3
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16:51:46 How long did it take to average 70 words / minute?
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RESPONSE --> 70 = 95 / (1 + 8.5 e^(-.12 t)) 70(1 + 8.5 e^(-.12 t) = 95 70+595e^(-.12t) = 95 595e^(-.12t) = 25 e^(-.12t) = 25/595 -.12t = ln(25/595) t = ln(25/595)/-.12 = 26.4 confidence assessment: 3
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16:51:49 *& According to the graph of the calculator it takes about 26.4 weeks to get to 70 words per min. This result was requested from a calculator, but you should also understand the analytical techniques for obtaining this result. The calculator isn't the authority, except for basic arithmetic and evaluating functions, though it can be useful to confirm the results of actual analysis. You should also know how to solve the equation. We want N to be 70. So we get the equation 70=95 / (1+8.5e^(-0.12t)). Gotta isolate t. Note the division. You first multiply both sides by the denominator to get 95=70(1+8.5e^(-0.12t)). Distribute the multiplication: 95 = 70 + 595 e^(-.12 t). Subtract 70 and divide by 595: e^(-.12 t) = 25/595. Take the natural log of both sides: -.12 t = ln(25/595). Divide by .12: t = ln(25/595) / (-.12). Approximate using your calculator. t is around 26.4. **
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RESPONSE --> self critique assessment: 3
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16:52:09 How many words per minute were being typed after 10 weeks?
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RESPONSE --> N = 95 / (1 + 8.5 e^(-.12 (10)) = 26.68 confidence assessment: 3
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16:52:13 *& According to the calculator 26.6 words per min was being typed after 10 weeks. Straightforward substitution confirms this result: N(10) = 95 / (1+8.5e^(-0.12* 10)) = 26.68 approx. **
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RESPONSE --> self critique assessment: 3
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16:52:23 Find the exact rate at which the model predicts words will be typed after 10 weeks (not time limit here).
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RESPONSE --> confidence assessment: 3
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16:52:26 The rate is 26.6 words / minute, as you found before. Expanding a bit we can find the rate at which the number of words being typed will be changing at t = 10 weeks. This would require that you take the derivative of the function, obtaining dN / dt. This question provides a good example of an application of the Chain Rule, which might be useful for review: Recall that the derivative of e^t is d^t. N = 95 / (1 + 8.5 e^(-.12 t)), which is a composite of f(z) = 1/z with g(t) = (1 + 8.5 e^(-.12 t)). The derivative, by the Chain Rule, is N' = g'(t) * f'(g(t)) = (1 + 8.5 e^(-.12 t)) ' * (-1 / (1 + 8.5 e^(-.12 t))^2 ) = -.12 * 8.5 e^(-.12 t)) * (-1 / (1 + 8.5 e^(-.12 t))^2 ) = 1.02 / (1 + 8.5 e^(-.12 t))^2 ). **
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RESPONSE --> self critique assessment: 3
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16:52:40 4.3.8 (was 4.2 #8) derivative of e^(1/x)
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RESPONSE --> f'(x) = 1/x = -1/x^2 derivative is e^(1/x)(-1/x^2) confidence assessment: 3
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16:52:43 There are two ways to look at the function: This is a composite of f(z) = e^z with g(x) = 1/x. f'(z) = e^z, g'(x) = -1/x^2 so the derivative is g'(x) * f'(g(x)) = -1/x^2 e^(1/x). Alternatively, and equivalently, using the text's General Exponential Rule: You let u = 1/x du/dx = -1/x^2 f'(x) = e^u (du/dx) = e^(1/x) * -1 / x^2. dy/dx = -1 /x^2 e^(1/x) **
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RESPONSE --> self critique assessment: 3
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16:52:54 Extra Question: What is the derivative of (e^-x + e^x)^3?
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RESPONSE --> 3(e^-x + e^x)^2(-e^-x + e^x) confidence assessment: 3
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16:52:56 This function is the composite f(z) = z^3 with g(x) = e^-x + e^x. f ' (z) = 3 z^2 and g ' (x) = - e^-x + e^x. The derivative is therefore (f(g(x)) ' = g ' (x) * f ' (g(x)) = (-e^-x + e^x) * 3 ( e^-x + e^x) ^ 2 = 3 (-e^-x + e^x) * ( e^-x + e^x) ^ 2 Alternative the General Power Rule is (u^n) ' = n u^(n-1) * du/dx. Letting u = e^-x + e^x and n = 3 we find that du/dx = -e^-x + e^x so that [ (e^-x + e^x)^3 ] ' = (u^3) ' = 3 u^2 du/dx = 3 (e^-x + e^x)^2 * (-e^-x + e^x), as before. **
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RESPONSE --> self critique assessment: 3
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16:53:08 4.3.22. What is the tangent line to e^(4x-2)^2 at (0, 1)?
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RESPONSE --> Equation in book is y = (e^(4x)-2)^2 (0,1) y = (e^(4x)-2)^2 y' = 2(e^4x - 2)(4e^4x) y'(0) = -8 y - 1 = -8(x-0) y = -8x + 1 confidence assessment: 3
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16:53:10 FIrst note that at x = 0 we have e^(4x-2) = e^(4*0 - 2)^2 - e^(-2)^2, which is not 1. So the graph does not pass through (0, 1). The textbook is apparently in error. We will continue with the process anyway and note where we differ from the text. }The function is the composite f(g(x)) wheren g(x) = e^(4x-2) and f(z) = z^2, with f ' (z) = 2 z. The derivative of e^(4x-2) itself requires the Chain Rule, and gives us 4 e^(4x-2). So our derivative is (f(g(x))' = g ' (x) * f ' (g(x)) = 4 (e^(4x-2) ) * 2 ( e^(4x - 2)) = 8 ( e^(4x - 2)). Now at x = 0 our derivative is 8 ( e^(4 * 0 - 2)) = 8 e^-2 = 1.08 (approx). If (0, 1) was a graph point the tangent line would be the line through (0, 1) with slope 1.08. This line has equation y - 1 = .0297 ( x - 0), or solving for y y = .0297 x + 1. As previously noted, however, (0, 1) is not a point of the original graph.
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RESPONSE --> self critique assessment: 3
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16:53:22 4.3.26 (formerly 4.3.24) (was 4.2.22) implicitly find y' for e^(xy) + x^2 - y^2 = 0
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RESPONSE --> e^(xy) + x^2 - y^2 = 10 (y + x y' ) * e^(xy) + 2x - 2y y' = 0 y e^(xy) + x y ' e^(xy) + 2x - 2 y y' = 0 (x e^(xy) - 2y ) y' = - y e^(xy) - 2x y' = [- y e^(xy) - 2x] / (x e^(xy) - 2y ) confidence assessment: 3
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16:53:24 The the q_a_ program for assts 14-16 in calculus 1, located on the Supervised Study ... pages under Course Documents, Calculus I, has an introduction to implicit differentiation. I recommend it if you didn't learn implicit differentiation in your first-semester course, or if you're rusty and can't follow the introduction in your text. The derivative of y^2 is 2 y y'. y is itself a function of x, and the derivative is with respect to x so the y' comes from the Chain Rule. the derivative of e^(xy) is (xy)' e^(xy). (xy)' is x' y + x y' = y + x y '. the equation is thus (y + x y' ) * e^(xy) + 2x - 2y y' = 0. Multiply out to get y e^(xy) + x y ' e^(xy) + 2x - 2 y y' = 0, then collect all y ' terms on the left-hand side: x y ' e^(xy) - 2 y y ' = -y e^(xy) - 2x. Factor to get (x e^(xy) - 2y ) y' = - y e^(xy) - 2x, then divide to get y' = [- y e^(xy) - 2x] / (x e^(xy) - 2y ) . **
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RESPONSE --> self critique assessment: 3
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16:53:41 4.3.34 (formerly 4.3.32) (was 4.2 #30) extrema of x e^(-x)
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RESPONSE --> f(x) = x e^(-x) f'(x) = x' e^(-x) + x (e^-x)' = 0 f'(x) is zero at x = 1 extrema is (1, e^-1) confidence assessment: 3
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16:53:43 Again the calculator is useful but it doesn't replace analysis. You have to do the analysis for this problem and document it. Critical points occur when the derivative is 0. Applying the product rule you get x' e^(-x) + x (e^-x)' = 0. This gives you e^-x + x(-e^-x) = 0. Factoring out e^-x: e^(-x) (1-x) = 0 e^(-x) can't equal 0, so (1-x) = 0 and x = 1. Now, for 0 < x < 1 the derivative is positive because e^-x is positive and (1-x) is positive. For 1 < x the derivative is negative because e^-x is negative and (1-x) is negative. So at x = 1 the derivative goes from positive to negative, indicating the the original function goes from increasing to decreasing. Thus the critical point gives you a maximum. The y value is 1 * e^-1. The extremum is therefore a maximum, located at (1, e^-1). **
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RESPONSE --> self critique assessment: 3
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16:53:56 4.3.42 (formerly 4.3.40) (was 4.2 #38) memory model p = (100 - a) e^(-bt) + a, a=20 , b=.5, info retained after 1, 3 weeks.How much memory was maintained after each time interval?
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RESPONSE --> p = (100 - 20) e^(-.5(1)) + 20 = 68.52 p = (100 - 20) e^(-.5(3)) + 20 = 37.85 confidence assessment: 3
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16:53:58 Plugging in a = 20, b = .5 and t = 1 we get p = (100 - 20) e^(-.5 * 1) + 20 = 80 * e^-.5 + 20 = 68.52, approx., meaning about 69% retention after 1 week. A similar calculation with t = 3 gives us 37.85, approx., indicating about 38% retention after 3 weeks. **
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RESPONSE --> self critique assessment: 3
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16:54:09 ** At what rate is memory being lost at 3 weeks (no time limit here)?
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RESPONSE --> dp/dt = (100-a) * -b e^-(bt) dp/dt(3) = (100-20) * -.5 e^-(.5(3) = -8.93 % per week self critique assessment: 3
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16:54:29 4.2.48 (formerly 4.2.46) (was 4.2 #42) effect of `mu on normal distribution
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RESPONSE --> confidence assessment:
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16:54:31 The calculator should have showed you how the distribution varies with different values of `mu. The analytical explanation is as follows: The derivative of e^[ -(x-`mu)^2 / (2 `sigma) ] is -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma. Setting this equal to zero we get -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma = 0. Dividing both sides by e^[ -(x-`mu)^2 / 2 ] / `sigma we get -(x - `mu) = 0, which we easily solve for x to get x = `mu. The sign of the derivative -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma is the same as the sign of -(x - `mu) = `mu - x. To the left of x = `mu this quantity is positive, to the right it is negative, so the derivative goes from positive to negative at the critical point. By the first-derivative test the maximum therefore occurs at x = `mu. More detail: We look for the extreme values of the function. e^[ -(x-`mu)^2 / (2 `sigma) ] is a composite of f(z) = e^z with g(x) = -(x-`mu)^2 / (2 `sigma). g'(x) = -(x - `mu) / `sigma. Thus the derivative of e^[ -(x-`mu)^2 / (2 `sigma) ] with respect to x is -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma. Setting this equal to zero we get x = `mu. The maximum occurs at x = `mu. **
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RESPONSE --> self critique assessment:
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16:54:33 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> confidence assessment:
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16:54:36 Typical Comment so if you feel very rusty you'll know you aren't along: Good grief, lol where to start!!! Just kidding! I guess I really need to be refreshed on how to handle deriving the exponential function with e. 4.2 was the killer for me here with only minimum examples in the section I had to review my old text and notes. It's just been so long.
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RESPONSE --> confidence assessment:
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