Query 1

#$&*

course Mth 272

June 3, 20127:00pm

•the use of second-derivative tests to do the same,

• interpretation 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.

NOTE THAT PROBLEM NUMBERS ARE GIVEN FOR THE DOCUMENT Chapter 4 problems . THE NUMBERING OF PROBLEMS IN YOUR TEXT IS NOT RELATED TO THE NUMBER OF THE QUESTIONS ON THIS DOCUMENT

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Question: `q 4.1.7 (previously 4.1.16 (was 4.1.14)): Solve for x the equation 4^2=(x+2)^2 (This problem might not appear in your edition of the textbook; however this is basic algebra and you should be able to do it. The solution takes only a few steps.)

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Your solution:

4^2=(x+2)^2

4^2=x^2+4

16=x^2+4

12=x^2

Take the square root of (12)=take the square root of x

3.46=x

confidence rating #$&*: 3

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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. **

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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.

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Self-critique Rating: 3

@&

(x + 2)^2 is not x^2 + 4.

(x + 2)^2 = (x + 2) * ( x + 2).

Multiply that out using the distributive law. You will end up with a quadratic equation, which you can solve for x.

Be sure you also understand the given solution completely.

If there's anything you're not absolutely certain about here, you're welcome to submit a revision with a complete self-critique.

*@

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Question: `q4.1.28 (was 4.1.32) Sketch a graph of y =- 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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Your solution:

I believe this graph has an asymptote at -4 and curves negatively along the x axis.

confidence rating #$&*: 2.5

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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.

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Self-critique (if necessary): OK. I understood this question and could even form a mental picture before illustrating/sketching it out.

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Self-critique Rating: 3

@&

The asymptote is not at -4.

There is no such thing as negative curvature. You could describe the graph in terms of increasing or decreasing at an increasing or decreasing rate. Or more appropriately, using the language of first-semester calculus, you could describe it as increasing or decreasing, and concave up or concave down.

*@

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Question: `qHow does this graph compare to that of 5^-x, and why does it compare as it does?

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Your solution:

I believe this graph would have an asymptote at 5 and would curve in negative along the y-axis.

confidence rating #$&*: 3

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confidence 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.

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Self-critique (if necessary): OK.

@&

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

*@

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Self-critique Rating: 3

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Question: `q 4.2.6 (previously 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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Your solution:

This graph would be concave up because it is raised to a positive power, the graph will have an asymptote at positive 2.

confidence rating #$&*: 2.5

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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. **

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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.

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Self-critique Rating: 3

@&

You appear to have a reasonably good picture, but you are asserting asymptotes where they do not exist. So there are aspects of these graphs that require a little more of your attention.

*@

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Question: `qThe 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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Your solution:

The coefficient changes the equations, making x increase at a more rapid rate.

confidence rating #$&*: 3

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confidence 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. **

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Self-critique (if necessary): OK. This question was just covered in the given solution in the previous question and is fairly basic.

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Self-critique Rating: 3

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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?

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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

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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. *&*&

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Self-critique (if necessary): OK.

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Self-critique Rating: 3

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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

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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

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Self-critique (if necessary): I remember learning this material, but I forgot the formula to solve it.

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Self-critique Rating: 3

@&

This is covered in your text.

The basic concept is that the interest is divided into n parts, giving you growth rate r / n and growth factor (1 + r / n). For each of the n periods in a year you multiply by the growth factor so that in t years you will have multiplied by 1 + r / n a total of n * t times.

*@

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Question: `qHow did you obtain your result for continuous compounding?

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Your solution:

I did this wrong, but I multiplied the annual interest rate by the initial investment put in the account.

confidence 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. **

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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.

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Self-critique Rating: 3

@&

To understand why we raise e to the .05 * 40 requires that you completely understand compound interest, and the definition of e as the limiting value of (1 + 1 / n) ^ n.

At this point make sure that you understand that the limiting function is A = P e^(r t). I encourage you to think about why, but be sure you understand that as the number of compoundings approaches infinity, the result approaches this function.

*@

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Question: `q4.2.40 (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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Your solution: (-.12)

confidence rating #$&*: 2

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confidence rating #$&*: 2

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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. *&*&

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Self-critique (if necessary): I remembered this rule from learning the equations to work these problems.

@&

&#This also requires a self-critique.

&#

*@

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Self-critique Rating: 3

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Question: `qHow long did it take to average 70 words / minute?

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Your solution:

I’m not sure whether to plug in 70 words into n or place it as the limiting factor since you want to know that exact figure for words per minute.

confidence rating #$&*: 0

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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. **

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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.

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Self-critique Rating: 3

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Question: `qHow many words per minute were being typed after 10 weeks?

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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. **

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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.

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Self-critique Rating: 3

@&

All you need to do is evaluate your expression, which appears to be correct.

*@

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Question: `qFind the exact rate at which the model predicts words will be typed after 10 weeks.

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Your solution: Wouldn’t this question require he same steps as the previous one before it?

confidence rating #$&*: 2

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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.. **

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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.

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Self-critique Rating:

@&

The first thing you need to understand is that the rate is the derivative of the function.

Then you need to take the derivative, which will of course involve the chain rule.

The Recommended Review, to which a link is provided just prior to the actual table of assignments, might be helpful to you as it covers the rules of differentiation and important first-semester topics that also appear in the current chapter.

*@

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Question: `q 4.3.2 (previously 4.3.8 (was 4.2 #8)) derivative of e^(1/x)

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Your solution:

x/1

confidence rating #$&*: 2

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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) **

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Self-critique (if necessary): I remember my derivative rules and know whatever power e is raised to is the derivative of the equation.

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Self-critique Rating: 3

@&

Your statement isn't totally correct.

The key is to understand the chain rule, which I believe you need to review.

The recommended review, to which my previous note refers, might be helpful.

*@

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Question: `q 4.3.6. What is the derivative of (e^-x + e^x)^3?

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Your solution:

-3x+3x=0

confidence rating #$&*: 2

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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. **

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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.

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Self-critique Rating: 3

@&

Again this comes back to the rules of differentiation.

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Question: `q4.3.10 (previously 4.3.22). What is the tangent line to e^(4x-2)^2 at (0, 1)?

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Your solution:

confidence rating #$&*: 0

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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.

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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.

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Self-critique Rating: 2

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Question: `q 4.3.8 (formerly 4.3.24) (was 4.2.22) implicitly find y' for e^(xy) + x^2 - y^2 = 0

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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 ) . **

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `q 4.3.11 (previously 4.3.34 (formerly 4.3.32) (was 4.2 #30)) extrema of x e^(-x)

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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.

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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.

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Self-critique Rating: 2

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To find relative extrema you need to start by finding critical points, then apply a first- or second-derivative test.

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Question: `q 4.3.12 (previously 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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Your solution:

p = (100 - 20) e^(-.5*1) + 20

p=80e^.5+20

confidence rating #$&*: 3

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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. **

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Your solution:

What is the question asking? After 3 weeks about 38% of memory is retained.

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Had you evaluated your expression you would have had the answer to the first question.

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confidence rating #$&*: 3

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Given Solution:

`a** At what rate is memory being lost at 3 weeks (no time limit here)?

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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. **

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Self-critique (if necessary): This answer was just given in the previous solution given.

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Self-critique Rating: 3

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This question asks about the rate of memory loss at 3 weeks, not the memory retained at that point.

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Question: `q 4.3.13 (previously 4.2.48 (formerly 4.2.46) (was 4.2 #42)) effect of `mu on normal distribution

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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. **

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Self-critique (if necessary): OK.

"

Self-critique (if necessary):

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Self-critique rating:

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Question: `q 4.3.13 (previously 4.2.48 (formerly 4.2.46) (was 4.2 #42)) effect of `mu on normal distribution

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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. **

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Self-critique (if necessary): OK.

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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You do need a review of a number of first-semester topics.

See the notes I've inserted.

The page at

http://vhcc2.vhcc.edu/dsmith/Ac2/frames%20pages/review.htm

would constitute a good review of most of the first-semester topics covered in this assignment.

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