course Mth 272 Applied Calculus II 06-19-2008
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00:45:13 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 --> answered previously self critique assessment: 3
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00:46:12 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 --> answered previously confidence assessment: 3
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00:46:24 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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00:46:45 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 --> answered previously confidence assessment: 3
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00:46:53 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 --> ok self critique assessment: 3
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00:47:02 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 --> answered previously confidence assessment: 3
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00:47:16 *& 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 --> ok self critique assessment: 3
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00:47:27 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 --> answered previously confidence assessment: 3
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00:47:34 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 --> ok confidence assessment: 3
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00:47:43 How did you obtain your result for continuous compounding?
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RESPONSE --> answered previously confidence assessment: 3
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00:47:52 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 --> ok self critique assessment: 3
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00:50:17 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 --> answered previously confidence assessment: 3
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00:50:24 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 --> ok self critique assessment: 3
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01:01:29 How long did it take to average 70 words / minute?
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RESPONSE --> Substitute 70 in place of N. ...Then solve for t. 70=95/(1+8.5e^-.12t) Multiply by denominator 70(1+8.5e^-.12t) = 95 Distribute 70 + 595e^-.12t = 95 Subt 70 595e^-.12t = 25 Divide by 595 e^-.12t = 25/595 Natural log -.12t = ln (25/595) Divide by -.12 t = (ln (25/595)) / (-.12) t = 26.4 confidence assessment: 2
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01:01:54 *& 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 --> ok self critique assessment: 3
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01:02:42 How many words per minute were being typed after 10 weeks?
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RESPONSE --> previously answered confidence assessment: 3
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01:02:52 *& 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 --> ok self critique assessment: 3
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01:03:13 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 --> answered previously confidence assessment: 3
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01:03:42 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 --> ok self critique assessment: 3
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01:11:55 4.3.8 (was 4.2 #8) derivative of e^(1/x)
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RESPONSE --> take derivative of the exponent and place in front of the term. Derivative of 1/x... Write 1/x as x^-1 ...-x^-2 so the derivative is: f""(x) = (-x^-2) e^(x^-1) or as: f""(x) = (-1/x^2) e^(1/x) confidence assessment: 3
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01:13:03 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 --> I did not write out what u=...but answered correctly. self critique assessment: 2
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01:25:41 Extra Question: What is the derivative of (e^-x + e^x)^3?
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RESPONSE --> 3(-e^-x + e^x) (e^-x + e^x)^2 Multiply by exponent...and subtract one from original exponent...then take derivative of inside. confidence assessment: 1
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01:27:20 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 --> I did not use the g and z but I think I understand the concept. self critique assessment: 2
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01:40:07 4.3.22. What is the tangent line to e^(4x-2)^2 at (0, 1)?
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RESPONSE --> The Parenthese is in the wrong place....for #22 in lesson 4.3. The left ( ) should be before the e. I'm confused whether to answer like the book or like it is shown to the left. Book: Find derivative and substitute 0 for x...to find slope...then use the y intercept to finish the equation...(0,1). The derivative is 8e^(4x)([e^(4x) - 2] Substitue 0 in for x and you get -8 so the equation using the slope and the (0,1) y intercept is: y=-8x+1 confidence assessment: 2
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01:44:45 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 --> This is not clear. The question in the book that you are referring to is not written the same on this program. The e is included in ( ) in the book. Am I missing something? self critique assessment: 1
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01:48:07 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 --> Unable to figure this one out. confidence assessment: 0
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01:57:30 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 --> I will definitley need to go back to the site referred to and review this concept. self critique assessment: 1
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02:05:10 4.3.34 (formerly 4.3.32) (was 4.2 #30) extrema of x e^(-x)
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RESPONSE --> (1, 0.368) I took Derivative which is = e^(-x) (-x+1) Set (-x+1)=0 and solve -x = -1 x = 1 which yeilds (1, 0.368) confidence assessment: 2
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02:07:56 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 --> e^-1 is equal to 0.368....is the decimal sufficient???? self critique assessment: 2
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02:11:36 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 --> Using given formula, substitute values a and b into equation....also the t values given 1 and 3 and work problem . 1week = 68.52 3 weeks = 37.85 confidence assessment: 2
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02:12:11 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 --> ok self critique assessment: 3
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02:14:37 ** At what rate is memory being lost at 3 weeks (no time limit here)?
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RESPONSE --> If one is retaining only 38%...I would assume he/she will have lost 62 %. Simply subtraction. self critique assessment: 1
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02:32:47 The average rate of change of y with respect to t is ave rate = change in y / change in t. This is taken to the limit, as t -> 0, to get the instantaneous rate dy/dt, which is the derivative of y with respect to t. This is the entire idea of the derivative--it's an instantaneous rate of change. The rate of memory loss is the derivative of the function with respect to t. dp/dt = d/dt [ (100 - a) e^(-bt) + a ] = (100-a) * -b e^-(bt). Evaluate at t = 3 to answer the question. The result is dp/dt = -8.93 approx.. This indicates about a 9% loss per week, at the 3-week point. Of course as we've seen you only have about 38% retention at t = 3, so a loss of almost 9 percentage points is a significant proportion of what you still remember. Note that between t = 1 and t = 3 the change in p is about -21 so the average rate of change is about -21 / 2 = -10.5. The rate is decreasing. This is consistent with the value -8.9 for the instantaneous rate at t = 3. **
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RESPONSE --> Way off....I see now. I thought this might be just any easy subtraction question...meant to be funny, but I guess the joke is on me. :) I went back and did work out the question and was able to get the right answer. self critique assessment: 2
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02:44:42 4.2.48 (formerly 4.2.46) (was 4.2 #42) effect of `mu on normal distribution
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RESPONSE --> Determines where the maximum points are on each graph. confidence assessment: 1
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02:45:35 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 --> ok...lots more detailed than my response. self critique assessment: 2
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02:47:21 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> My critique responses probably need inprovement. I look at things to try to make them as simple as possible. I need to be more detailed and look deeper into each question. confidence assessment: 2
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02:48:17 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 --> Good to know! confidence assessment: 2
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12:50:33 4.4.4 (was 4.3 #40 write ln(.056) = -2.8824 as an exponential equation
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RESPONSE --> e^-2.8824... = 0.056 By rules of exponential and logarithmic functions confidence assessment: 2
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12:51:29 y = ln x is the same as e^y = x, so in exponential form the equation should read e^-2.8824 = .056 **
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RESPONSE --> ok self critique assessment: 3
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12:53:55 4.4.8 (was 4.3 #8) write e^(.25) = 1.2840 as a logarithmic equation
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RESPONSE --> Using rules of exponential and logarithmic functions. e^x = y is equivalent to ln y=x So.... ln 1.2840 = 0.25 confidence assessment: 2
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12:54:48 e^x = y is the same as x = ln(y) so the equation is .25 = ln(1.2840). **
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RESPONSE --> ok self critique assessment: 3
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13:00:05 4.4.16 (was 4.3 #16) Sketch the graph of y = 5 + ln x.
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RESPONSE --> Used Graphing Calculator to make table of values to sketch graph on my paper. (1,5) (2,5.6) (3,6.099) (4,6.386) (5,6.61) Find x-intercept. Substitute 0 for y and solve for x. 0=5 + ln x -5 = ln x inverse e^-5 = e^(lnx) e^-5 = x .007 =x So...the x intercept is (0.007,0) confidence assessment: 2
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13:05:24 Plugging in values is a good start but we want to explain the graph and construct it without having to resort to much of that. The logarithmic function is the inverse of the exponential function, which is concave up, has asymptote at the negative x axis and passes thru (0, 1). Reversing the coordinates of the exponential graph gives you a graph which is concave down, asymptotic to the negative y axis and passes thru (1, 0). Adding 5 to get y = 5 + ln x raises this graph 5 units so that the graph passes thru (1, 5) and remains asymptotic to the negative y axis and also remains concave down. The graph is asymptotic to the negative y axis, and passes through (1,5). The function is increasing at a decreasing rate; another way of saying this is that it is increasing and concave down. STUDENT COMMENT: I had the calculator construct the graph. I can do it by hand but the calculator is much faster. INSTRUCTOR RESPONSE: The calculator is faster but you need to understand how different graphs are related, how each is constructed from one of a few basic functions, and how analysis reveals the shapes of graphs. The calculator doesn't teach you that, though it can be a nice reinforcing tool and it does give you details more precise than those you can imagine. Ideally you should be able to visualize these graphs without the use of the calculator. For example the logarithmic function is the inverse of the exponential function, which is concave up, has asymptote at the negative x axis and passes thru (0, 1). Reversing the coordinates of the exponential graph gives you a graph which is concave down, asymptotic to the negative y axis and passes thru (1, 0). Adding 5 to get y = 5 + ln x raises this graph 5 units so that the graph passes thru (1, 5) and remains asymptotic to the negative y axis and also remains concave down. **
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RESPONSE --> I did the common student response..Graphed in my calculator and used table to transfer to my graph paper. I understand that it is concave down and increasing. Is the x-intercept important? (0.007,0) self critique assessment: 2
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14:34:48 4.4.22 (was 4.3 #22) Show e^(x/3) and ln(x^3) inverse functions
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RESPONSE --> Take te inverse of f(x) to see if it equals g(x). f(x)=e^(x/3) reverse x and y. x = e^(y/3) ln of each side ln x = ln e^(y/3) simplify ln x = (y/3) Multiply both sides by 3 3 ln x = y Simplify the g(x) equation. y = ln x^3 y = 3 ln x This proves these two functions are inverses of each other. confidence assessment: 2
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14:46:17 GOOD STUDENT RESPONSE: Natural logarithmic functions and natural exponential functions are inverses of each other. f(x) = e^(x/3) y = e^(x/3) x = e^(y/3) y = lnx^3 f(x) = lnx^3 y = ln x^3 x = lny^3 y = e^(x/3) INSTRUCTOR RESPONSE: Good. f(x) = e^(x/3) so f(ln(x^3)) = e^( ln(x^3) / 3) = e^(3 ln(x) / 3) = e^(ln x) = x would also answer the question MORE ELABORATION You have to show that applying one function to the other gives the identity function. If f(x) = e^(x/3) and g(x) = ln(x^3) then f(g(x)) = e^(ln(x^3) / 3) = e^( 3 ln(x) / 3) = e^(ln(x)) = x. **
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RESPONSE --> Makes sense. I reworked the question the way you suggested and also went back in the book and tried #21 similarly....I was able to complete the concept. confidence assessment: 2
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14:51:05 4.4.46 (was query 4.3 #44) simplify 1/3 [ 2 ln(x+3) + ln x - ln(x^2-1) ]
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RESPONSE --> By the properties of Logarithms. ln xy = ln x + ln y ln x/y = ln x - ln y ln x^n = n ln x 1/3 [((x+3)^2 (x)) / (x^2-1)] [((x+3)^2 (x)) / (x^2-1)]^(1/3) confidence assessment: 2
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14:54:02 1/3 [ 2 ln(x+3) + ln x - ln(x^2-1) ] = 2/3 ln(x+3) + 1/3 ln x - 1/3 ln(x^2-1) = ln(x+3)^(2/3) + ln(x^(1/3)) - ln((x^2-1)^(1/3)) = ln [ (x+3)^(2/3) (x^(1/3) / (x^2-1)^(1/3) ] = ln [ {(x+3)^2 * x / (x^2-1)}^(1/3) ] **
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RESPONSE --> oops I forgot to put the ln in front. self critique assessment: 2
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14:58:20 4.4.58 (was 4.3 #58) solve 400 e^(-.0174 t) = 1000.
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RESPONSE --> Solve for t. 400 e^(-.0174 t) = 1000 Divide by 400 e^(-.0174 t) = 5/2 Take ln of each side ln e^(-.0174 t) = ln 5/2 -.0174 t = ln 5/2 Divide by -.0174 t = (ln (5/2)) / -0.0174 t = -52.66 confidence assessment: 2
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14:59:00 The equation can easily be arranged to the form e^(-.0174) = 2.5 We can convert the equation to logarithmic form: ln(2.5) = -.0174t. Thus t = ln(2.5) / -.0174 = 52.7 approx.. **
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RESPONSE --> ok self critique assessment: 3
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15:13:46 4.4.72 (was 4.3 #68) p = 250 - .8 e^(.005x), price and demand; find demand for price $200 and $125
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RESPONSE --> Substitute the values of P and solve for x. 200 = 250 - 0.8e^.005x Subt 250 -50 = -0.8e^.005x Div by -0.8 62.5 = e^.005x ln of both sides ln 62.5 = ln e^.005x ln 62.5 = .005x Divide by 0.005 (ln 62.5) / 0.005 = x Evaluate 827.03 = x 125 = 250 - 0.8e^.005x Subt 250 -125 = -0.8e^.005x Div by -0.8 156.25 = e^.005x ln of both sides ln 156.25 = ln e^.005x ln 156.25 = .005x Divide by 0.005 (ln 156.25) / 0.005 = x Evaluate 1010.29 = x confidence assessment: 2
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15:14:06 p = 250 - .8 e^(.005x) so p - 250 = - .8 e^(.005x) so e^(.005 x) = (p - 250) / (-.8) so e^(.005 x) = 312.5 - 1.25 p so .005 x = ln(312.5 - 1.25 p) and x = 200 ln(312.5 - 1.25 p) If p = 200 then x = 200 ln(312.5 - 1.25 * 200) = 200 ln(62.5) = 827.033. For p=125 the expression is easily evaluated to give x = 1010.29. **
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RESPONSE --> ok self critique assessment: 3
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