#$&* course Mth 272 June 3, 20127:00pm •the use of second-derivative tests to do the same,
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I went about solving this the correct way using basic algebra, I’m not quite sure if it coincides with the answer in the given solution or not. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. ** STUDENT QUESTION To confirm, it has been a few months since working with first and second derivatives: In order to determine how this graph is portrayed without a graphing calculator, you would find the first derivative and set to solve for zero, giving x value? Additionally, use test values to determine increasing or decreasing. You would then do the same by finding the second derivative, solving for zero, in order to determine concavity and inflection points, by using text values? INSTRUCTOR RESPONSE You would set the first derivative equal to zero and solve the equation to find the critical points. You would then test each critical point. First-derivative test: • If the slope goes from negative to zero to positive as you move from left to right in the near neighborhood of the point, then the trend of graph goes from downward to level to upward and the critical point is a minimum. • If the slope goes from positive to zero to negative as you move from left to right in the near neighborhood of the point, then the trend of graph goes from upward to level to downward and the the critical point is a maximum. • If the slope goes from negative to zero to negative as you move from left to right in the near neighborhood of the point, or from positive to zero to positive, then the function goes either from downward to level to downward, or upward to level to upward and the critical point is a point of inflection. Second-derivative test: • If the second derivative takes a negative value at a critical point then the graph is level at that point but the trend of the graph is concave downward and the point is a max. If the second derivative is positive then the graph is concave upward and the point is a min. • If the second derivative is zero at the point and its sign changes as you move from left to right through the point, then you have a point of inflection. There are a couple more cases when the second derivative is zero, but this covers most applications. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. I understood this question and could even form a mental picture before illustrating/sketching it out. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a the graphs meet at the y axis, at the point (0, 1). To the left of the y axis the graph of y = 5^-x is 'higher' than that of y = 4^-x. To the right it is lower. The reasons for these behaviors: The zero power of any number is 1, so that both graphs pass through (0, 1). A given positive power of a larger number will be larger. However applying given negative exponent to a larger number result in a greater denominator than if the same exponent is applied to a smaller number, resulting in a smaller result. For example, if we apply the power 2 to both bases we obtain 4^2 = 16, but 5^2 = 25. The graph of 5^x is 'higher' than that of 4^x, when x = 2. If we apply the power -3 to both bases we obtain 4^-3 = 1 / 4^3 = 1 / 64, while 5^-3 = 1 / 5^3 = 1 / 125. The value of 5^-3 is less than the value of 4^-3. So the graph of 5^x is 'lower' than that of 4^x, when x = -3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK.
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I remember learning this information and being good at questions like this. I don’t think I need much review for this because I know how to do it with and without using a calculator to verify. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. This question was just covered in the given solution in the previous question and is fairly basic. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I plugged in numbers for x to get both the points for x and y then prepared my graph with my coordinates that I came up with. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a*& These questions are answered in the solutions given above. From those solutions you will ideally have been able to answer this question. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 4.2.8 (previously 4.2.32 (formerly 4.2.43) (was 4.1 #48)). If you invest $2500 at 5% for 40 years, how much do you have if you with 1, 2, 4, 12, 365 annual compoundings, and with continuous compounding YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Year 1- $2625 Year 2- 2756.25 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I remember learning this material, but I forgot the formula to solve it. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not quite sure why to raise the annual interest rate to the 40th power if the initial collection of interest is an independent events from the number of total years. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I remembered this rule from learning the equations to work these problems.
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Given Solution: `a*& 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Once I look at the problem given I understand what to do. I think my first instinct was to merely substitute the desired words per minute (70) for the 95 that’s included in the question. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qHow many words per minute were being typed after 10 weeks? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T= N / (1 + 8.5 e^(-.12 *10)) 95/(1+8.25e^(-1.2)) confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a*& 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The most confusing part of the problem for me was trying to figure out whether or not to include 95 in the equation like in the original since we were solving for N. This equation is related/ in the family with a previous one that I also forgot the initial formula to solve. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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) = 95 / 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)) ' * (-95 / (1 + 8.5 e^(-.12 t))^2 ) = -.12 * 8.5 e^(-.12 t)) * (-95 / (1 + 8.5 e^(-.12 t))^2 ) = 97 / (1 + 8.5 e^(-.12 t))^2 ), approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. I noticed when reading through the answer that the chain rule is an option in solving this problem. I had a very hard time learning the chain rule so I know that is something I will have to work on. ------------------------------------------------ Self-critique Rating:
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Given Solution: `a 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) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I remember my derivative rules and know whatever power e is raised to is the derivative of the equation. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t realize that the 3 that the entire expression is raised to is multiplied by the e’s in the expression rather than raised to those powers. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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 = 1.08 ( x - 0), or solving for y y = 1.08 x + 1. As previously noted, however, (0, 1) is not a point of the original graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I really need to review the chain rule. I knew this equation required it, but I wasn’t sure how to do it. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q 4.3.8 (formerly 4.3.24) (was 4.2.22) implicitly find y' for e^(xy) + x^2 - y^2 = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 ) . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 4.3.11 (previously 4.3.34 (formerly 4.3.32) (was 4.2 #30)) extrema of x e^(-x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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). ** STUDENT QUESTION So, I understand in order to solve for relative extrema, first find the derivative of the function. Apply product rule, set to solve for zero, and then to find critical points solving for x. However, didn’t quite understand how after factoring and ran into a problem. After e^-x + x(1-x) = 0; I see that e^-x can’t work, but we are still basically left with x(1 - x) = 0 ; What happened to the x outside of parenthesis, perhaps I am overlooking this? INSTRUCTOR RESPONSE e^-x + x(-e^-x) = 0. Factoring out e^-x we get e^(-x) (1-x) = 0. e^(-x) (1-x) could be written more explicitly as e^(-x) * (1-x), which means 'raise e to the power -x, then multiply the result by 1 - x'. That -x is the exponent of e. (1 - x) is not part of the exponent, since the exponentiation e^(-x) is done before the multiplication. Thus e^(-x) (1-x) is the product of two factors, e^(-x) and (1 - x). The product of two factors can be zero only if one of the two factors is zero. e^(-x) cannot be zero (because e^x is always positive e^(-x) = 1 / (e^x) is always positive). However 1 - x can be zero, when x = 1. Thus, since e^(-x) (1-x) is the derivative of the given function, the only critical point of that function occurs at x = 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I thought this problem was a matter of just taking the derivative of the expression and flipping the sign to find the positive and negative extremas. ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `a 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. ** YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: What is the question asking? After 3 weeks about 38% of memory is retained.
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Given Solution: `a** At what rate is memory being lost at 3 weeks (no time limit here)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: approximately 38% confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This answer was just given in the previous solution given. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q 4.3.13 (previously 4.2.48 (formerly 4.2.46) (was 4.2 #42)) effect of `mu on normal distribution YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe this averages the distribution. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!