query24

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course Mth163

July 16,20135:08 pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

024. `query 24

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Question: `qExplain why, when either f(x) or g(x) is 0, then the product function also has a 0 for that value of x.

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Your solution:

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In order to get zero as a product one of the factors must be zero

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confidence rating #$&*: 3

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Given Solution:

`aSTUDENT RESPONSE: If you multiply any number by zero and you get zero.

INSTRUCTOR NOTE: Right. If f(x) = 0 then f(x) * g(x) = 0; and the same is so if g(x) = 0. If one number in a product is zero, then the product is zero.

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Self-critique (if necessary):

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Question: `qExplain why, when the magnitude | f(x) | of f(x) is greater than 1 (i.e., when the graph of f(x) is more

than one unit from the x axis), then the product function will be further from the x axis than the g(x) function.

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Your solution:

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when the magnitude | f(x) | of f(x) is greater than 1,it will be further and further from the x axis because it will keep getting bigger in the positive direction that’s why when the sine peak was at a max value of 1, the linear reg is farther away and farther away from initial graph or product function is farther away from the initial .

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Given Solution:

`aSTUDENT RESPONSE: If you multiply a number by another number greater than 1, the result is greater than the

original number.

INSTRUCTOR CLARIFICATION:

Your response is correct if the original number is positive. However if it's negative, the result will be less than the original number (e.g.., if you multiply 3 by 2 you get 6, which is greater than the original number 3; however if you multiply -3 by 2 you get -6, which is less than -3).

The general statement would be

• If you multiply a number by another number whose magnitude is greater than 1, the result will have greater magnitude than the original number.

(e.g., the magnitude of -3 is | -3 | = 3; if you multiply | -3 | by 2 you get 6, which is greater than the magnitude 3 of the original number).

Applying this to the present situation:

• If | f(x) | > 1 then the magnitude of f(x) * g(x) will be greater than the magnitude of g(x).

The magnitude of g(x) at a given value of x is | g(x) |, and this represents its distance from the x axis. So when the magnitude increases so does the distance from the x axis.

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Self-critique (if necessary):

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Even when you multiply by a negative number your magnitude (distance traveled???) will still be larger than the starting point because you have yet to travel.

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Question: `qExplain why, when the magnitude | f(x) | of f(x) is less than 1 (i.e., when the graph of f(x) is less than one unit from the x axis), then the product function will be closer to the x axis than the g(x) function.

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Your solution:

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Your fractioning off until you reach an asymptote

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Multiplying by values less than 1, even if they approach zero, doesn't lead to vertical asymptotes. Unless the other function has a vertical asymptote, this would simply lead to values of the product function that approach zero.

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Given Solution:

`aSTUDENT RESPONSE: If you multiply a number by another number less than 1, the result is less than the original

number.

If you multiply a number by another number whose magnitude is less than 1, the result will have a lesser magnitude that the

original number.

If | f(x) | < 1 then the magnitude of f(x) * g(x) will be less than the magnitude of g(x).

The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude decreases so does the

distance from the x axis.

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Self-critique (if necessary):

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So would this indicate an asymptote???

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Question: `qExplain why, when f(x) and g(x) are either both positive or both negative, the product function is positiv; and when f(x) and g(x) have opposite signs the product function is negative.

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Your solution:

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Poss*poss=poss

Neg*neg=poss

Neg*poss=neg

Poss*neg=neg

Two signs times each other are always pos

When the amount of numbers being multiplied is odd and you have at least one neg factor your product will be neg

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Given Solution:

`aSTUDENT RESPONSE: This is basic multiplication: + * + = +, - * - = -, + * - = -. The product of like signs is

positive, the product of unlike signs is negative. Since the product function results from multiplication of the two functions,

these rules apply.

INSTRUCTOR RESPONSE:

Right.

If f(x) and g(x) are both positive, then the product function f(x) * g(x) is positive.

If f(x) and g(x) are both negative, then the product function f(x) * g(x) is positive.

If f(x) and g(x) are unlike, then the product function f(x) * g(x) is negative.

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Self-critique (if necessary):

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Question: `qExplain why, when f(x) = 1, the graph of the product function coincides with the graph of g(x).

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Your solution:

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Anything times one is its self

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Given Solution:

`aSTUDENT RESPONSE: g(x) * 1 = g(x)

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Question: `q problem 4 Sketch graphs for y = f(x) = 2^x and y = g(x) = .5 x, for -2 < x < 2. Use your graphs to predict the shape of the y = g(x) * f(x) graph.

Describe the graphs of the two functions, and explain how you used these graphs predict the shape of the graph of the

product function.

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Your solution:

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when using the table you just multiply the f(x) row by the g(x) row to get g(x)*f(x)

X

-2

-1.5

-1

0

1

1.5

5

. f(x)

.25

.35

.5

1

2

2.8

4

. g(x)

-1

-.75

-.5

0

.5

.75

1

. g(x)*f(x)

-.25

-.26

-.26

0

1

2.1

4

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You appear to have calculated the values of the product function rather than using the shapes of the two graphs to predict its shape.

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Given Solution:

`aSTUDENT RESPONSE: Where g(x) is - and f(x) is + graph will be -. where g(x) =0 graph will be at 0where both

are + graph will be positive and rise more steeply.

y=2^x asymptote negative x axis y intercept (0,1) y = .5x linear graph passing through (0,0) rising 1 unit for run of 2 units

INSTRUCTOR COMMENT:

The function y = f(x) = 2^x is positive for any value of x (since any power of 2 is positive).

The function y = g(x) = .5 x is negative when x < 0, positive when x > 0 and zero when x = 0.

When x < 0, then, f(x) is negative and g(x) is positive, so the product function f(x) g(x) must be negative.

When x > 0, both functions are positive so the product function is positive.

Since g(x) = 0 when x = 0, the product f(x) * g(x) will be 0 at x = 0.

For x > 0 the exponential rise of the one graph and the continuing rise of the other imply that the graph will rise more and more rapidly, without bound, for large positive x.

For x < 0, as we have seen, one function is positive and the other is negative so the graph will be below the x axis.

For large negative values of x, one graph approaches 0 while the other keeps increasing in magnitude; it's not immediately clear which function 'wins'.

However the exponential always 'beats' a fixed power so the graph will be asymptotic to the negative x axis.

The graph will reach a minimum somewhere to the left of the x axis, before curving back toward the x axis and becoming asymptotic. **

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Self-critique (if necessary):

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Question: `q problem 7 range(depth) = 2.9 `sqrt(depth) and depth(t) = t^2 - 40 t + 400.

At what times is depth 0. How did you show that the vertex of the graph of depth vs. time coincides with these zeros?

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Your solution:

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My solutions were 43.5, 29,14.5

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There are only two times when the depth is zero. It's not clear which two correspond to this condition, or what the meaning is of the third, though I suspect you've included the t value of the vertex. However the vertex of this graph, as you can easily verify, does not lie on the t aix and is therefore not a zero of the function.

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In any case you don't explain how you got your values, and as shown in the given solution the depth function has no zeros

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Given Solution:

`a** The depth function is quadratic. Its vertex occurs at t = - b / (2 a) = - (-40) / (2 * 1) = 20.

Its zeros can be found either by factoring or by the quadratic formula.

t^2 - 40 t + 400 factors into (t - 20)(t - 20), so its only zero is at t = 20. This point (20, 0) happens to be the vertex,

and the graph opens upward, so the graph never goes below the x axis.

STUDENT QUESTION: When I simplified range(depth(t) = 2.9*'sqrt(t^2 - 40t +400) I got 2.9t - 58 which gives a

Negative range so I reversed it and got the correct results, what have I done wrong?

`arange(depth(t) = 2.9*'sqrt(t^2 - 40t +400)

= 2.9*'sqrt( (t-20)^2) )

= 2.9( t - 20)

= 2.9t - 58

I know it should be -2.9 + 58 I just don't understand how to get there. Thanks

INSTRUCTOR RESPONSE TO QUESTION: This is a great question.

Let's first consider another question:

What is sqrt( (-5) ^ 2)?

`sqrt( (-5)^2 ) isn't -5, it's 5, since `sqrt(25) = 5. This shows that you have to be careful about possible negative values of t - 20.

This is equivalent to saying that `sqrt( (-5)^2 ) = | -5 | = 5.

We conclude that sqrt(x^2) is not always equal to x. As in the above case, sqrt(x^2) can be equal to -x.

We can settle the entire question by observing that

• sqrt(x^2) = | x |.

Applying this to the present problem:

`sqrt( (t-20) ^2 ) cannot be negative.

Using the observations made above, we see that `sqrt( (t-20)^2 ) = | t - 20 |.

Thus the composite sqrt( (t - 20)^2 ) is equal to | t - 20 |.

• If t > 20 then | t - 20 | = t - 20.

• If t < 20 then t - 20 is negative so that | t - 20 | = -(t - 20) = 20 - t.

STUDENT QUESTION

I see how it's possible to get a positive value from solving the quadratic and a negative value if

the quadratic factor, but it doesn't entirely make sense to me how that is possible or which form is truly correct? Dealing

with a range in this circumstance, the positive value of course makes sense, but out of context, how would we know?

INSTRUCTOR RESPONSE

It is of course possible that the quadratic function would have been negative for some values of t, in which case the square root of the quadratic would not have been defined for those values of t. Those values of t would therefore have to be omitted from the domain of the composite function.

There are two things that restrict the domain of a composite:

• The function f(g(t)) is undefined if g(t) is undefined--i.e., if t is not in the domain of g, then it can't be in the domain of f(g(t)).

• I addition the function f(g(t)) is undefined if z = g(t) is not in the domain of the function f(z).

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Self-critique (if necessary):

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I keep forgetting I use the quadratic factor on basically anything to find zeros

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Question: `q The depth function is the product of two linear factors. What are these factors?

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Your solution:

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I think I’ve done this wrong

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Given Solution:

The quadratic function factors as depth(t) = (t-20)(t-20).

This function consists of two identical linear factors, each equal to t - 20.

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Self-critique (if necessary):

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Did this come from including the zeros??? Maybe that is why I did not get linear factors

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It you factor t2 - 40 t + 400 you get (t - 20) * (t - 20).

The quadratic formula tells you that there are no zeros (negative under square root) and the vertex is at t = 20 (since b / (2 a) = 20).

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Question: `qFor t = 5, 10 and 15, what are the ranges of the stream?

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Your solution:

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My ranges were 43.5, 29,14.5

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Given Solution:

`a** depth(t) = t^2 - 40 t + 400 = (t-20)^2 so

depth(5) = (5-20)^2 = (-15)^2 = 225

depth(10) = (10-20)^2 = (-10)^2 = 100

depth(15) = (15-20)^2 = (-5)^2 = 25.

It follows that the ranges of the stream are

range(depth(5)) = 2.9 sqrt(225) = 43.5

range(depth(10) = 2.9 sqrt(100) = 29 and

range(depth(15) = 2.9 sqrt(25) = 14.5. **

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Self-critique (if necessary):

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Question: `qWhat is the composite function range(depth(t))? Show that its simplified form is a linear function of t.

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Your solution:

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. range(depth(t))=2.6 sqrt(t^2-40t+400)

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Right.

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. t^2 -40t+400

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This is the original depth function, not the simplified form of the above.

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Given Solution:

`a** range(depth(t) ) = 2.9 sqrt(depth(t)) = 2.9 sqrt(t^2 - 40 t + 400) = 2.9 sqrt( (t - 20)^2 ) = 2.9 | t - 20 |. **

This function isn't actually linear. Its graph consists of two linear rays, the first of negative slope ending at t = 20 and the second of positive slope starting at t = 20. The graph forms a 'v' shape.

In reality if the quadratic function represents the depth of water in a leaking container, the useful domain of the function ends when the container is empty (i.e., when the depth reaches its 'low point' corresponding to the vertex of the parabola), and the composite function would represent the stream range only up to this point. So the 'real-world' range(t) function would in fact be linear.**

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Question: `qproblem 8 Illumination(r) = 40 / r^2; distance = 400 - .04 t^2.

What is the composite function illumination(distance(t))?

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Your solution:

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40/(400-.04(t^2))

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This would be 40 / r, not 40 / r^2.

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Given Solution:

`a** Illumination(r) = 40 / r^2 so

Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2. **

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Question: `qGive the illumination at t = 25, t = 50 and t = 75.

At what average rate is illumination changing during the time interval from t = 25 to t = 50, and from t = 50 to t = 75?

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Your solution:

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.026

.096

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@&

You haven't shown how you got these results.

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Given Solution:

`a** illumination(distance(t)) = 40 / (400 - .04 t^2)^2 so

illumination(distance(25)) = 40 / (400 - .04 * 25^2)^2 = .000284

illumination(distance(50)) = 40 / (400 - .04 * 50^2)^2 = .000444

illumination(distance(75)) = 40 / (400 - .04 * 75^2)^2 = .001306.

from 25 to 50 change is .000444 - .000284 = .000160 so ave rate is .000160 / 25 = .0000064

from 50 to 75 change is .001306 - .000444 = .00086 so ave rate is .00086 / 25 = .000034 approx. **

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Self-critique (if necessary):

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I did not square twice as you did

illumination(distance(25)) = 40 / (400 - .04 * 25^2)^2 = .000284

illumination(distance(50)) = 40 / (400 - .04 * 50^2)^2 = .000444

illumination(distance(75)) = 40 / (400 - .04 * 75^2)^2 = .001306.

that may have been why did not get the same answer I did not combined my squares

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Question: `q problem 10 gradeAverage = -.5 + t / 10. t(Q) = 50 (1 - e ^ (-.02 (Q - 70) ) ).

If the student's mental health quotient is an average 100, then what grade average should the student expect?

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Your solution:

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. t(Q)= 50(1-e^(-.02(100-70))=22.6

When t=110 t(Q)=27.5

When t=120 t(Q)=31.6

When t=130 t(Q)=34.9

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Given Solution:

`a** gradeAverage = -.5 + t / 10 = -.5 + 50 ( 1 - e^(-.02 (Q - 70) ) ) / 10 = -.5 + 5 ( 1 - e^(-.02 (Q - 70) ) )

So

gradeAverage(t(100)) = -.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = -.5 + 5( 1 - .5488 ) = -.5 + 5 ( .4522 ) = -.5 + 2.26 =

1.76.

gradeAverage(t(110)) = -.5 + 5 ( 1 - e^(-.02 ( 110 - 70) ) = -.5 + 5( 1 - .4493 ) = -.5 + 5 ( .5517 ) = -.5 + 2.76 =

2.26.

gradeAverage(t(120)) = -.5 + 5 ( 1 - e^(-.02 ( 120 - 70) ) = -.5 + 5( 1 - .3678 ) = -.5 + 5 ( .6322 ) = -.5 + 3.16 =

2.66.

gradeAverage(t(130)) = -.5 + 5 ( 1 - e^(-.02 ( 130 - 70) ) = -.5 + 5( 1 - .3012 ) = -.5 + 5 ( .6988 ) = -.5 + 3.49 =

2.99.

As Q gets larger and larger Q - 70 will get larger and larger, so -.02 ( Q - 70) will be a negative number with increasing

magnitude; its magnitude increases without limit.

It follows that e^(-.02 ( Q - 70) ) = will consist of e raised to a negative number whose magnitude increases without limit.

As the magnitude of the negative exponent increases the result will be closer and closer to zero.

So -.5 + 5 ( 1 - e^(-.02 ( Q - 70) ) ) will approach -.5 + 5 ( 1 - 0) = -.5 + 5 = 4.5.

Side note: For Q = 100, 200 and 300 we would have grade averages 1.755941819, 4.128632108, 4.449740821. To

get a 4-point Q would have to be close to 200. Pretty tough course

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Self-critique (if necessary):

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I keep forgetting to plug in the grade ave for Q

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When you get to the end of your answer, you need to reread the question and be sure you've answered it.

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Question: `qWhat grade averages would be expected for mental health quotients of 110, 120 and 130?

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Your solution:

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. t(Q)=50(1-e^(-.02(-.5+100/10)-70)

t(Q)=50(1-e^(-.02(-.5+110/10)-70)

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Given Solution:

`a110...2.2534, 120...2.66, 130...2.99

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Self-critique (if necessary):

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I already knew my answers were wrong because I get getting these numbers that are no where close to a GPA scale what is wrong with my equations???

. t(Q)=50(1-e^(-.02(-.5+100/10)-70)

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t(Q) = 50 (1 - e ^ (-.02 (Q - 70) ) ).

This result is then plugged into the grade average function

-.5 + t / 10.

For Q = 100, you get

t(100) = 50 (1 - e ^ (-.02 (100 - 70) ) ).

You then take the result and use it to evaluate

-.5 + t / 10.

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t(Q)=50(1-e^(-.02(-.5+110/10)-70)

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Question: `qWhat is the upper limit on the expected grade average that can be achieved by this student?

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Your solution:

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The upper limit on a student with 100 would be a 2.26, my hypothesis is that a student with Q 100 cannot exceed the lowest of Q 110

“gradeAverage(t(100)) = -.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = -.5 + 5( 1 - .5488 ) = -.5 + 5 ( .4522 ) = -.5 + 2.26 =

1.76.

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The following is totally unacceptable as documentation:

-.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = -.5 + 5( 1 - .5488 ) = -.5 + 5 ( .4522 ) = -.5 + 2.26 =

1.76.

It contains many false statements. For example this statement would imply that

-.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = 1.76

when

-.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = .5488.

You appear to be quoting calculator steps, which do not constitute an equal series of algebraic expressions and cannot serve as the basis for algebraic manipulation.

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Nevertheless, you do need to use a calculator to evaluate this, and those calculator steps would work and have given you the right answer. So you have decoded this series of steps correctly.

Just don't put them like that in writing.

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gradeAverage(t(110)) = -.5 + 5 ( 1 - e^(-.02 ( 110 - 70) ) = -.5 + 5( 1 - .4493 ) = -.5 + 5 ( .5517 ) = -.5 + 2.76 =

2.26.” - using your equations from previous answer because mine are wrong

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Given Solution:

`a** If Q is very, very large, e^-( .02(Q-70) ) would have a negative exponent with a very large magnitude and so would

be very close to 0.

• In this case 50 ( 1-e^(-.02 (Q-70)) would be close to 50(1-0) = 50.

Then the grade average would be -.5 + 50 / 10 = -.5 + 5 = 4.5 .

DER [0.5488116360, 0.4493289641, 0.3678794411, 0.3011942119][1.755941819, 2.253355179, 2.660602794,

2.994028940]

STUDENT COMMENT:

I was able to determine the cap from trial and error, but I don't think I would've recognized it just from """"reading"""" the equation. I find it really interesting that no matter what number is plugged in higher than 1760 (approx) the result is always the same.

INSTRUCTOR RESPONSE:

An exponential function approaches its asymptote quickly, just as in the opposite direction it moves quickly away from its asymptote.

It's useful to remember how and why an exponential function approaches zero as its exponent becomes increasingly negative.

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Self-critique (if necessary):

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Question: `qWhat is the composite function gradeAverage( t(Q) )?

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Your solution:

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t(Q)=50(1-e^(-.02(-.5+Q/10)-70)

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Given Solution:

`a-.5+(50(1-e^(-.02(Q-70))/10

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Self-critique (if necessary):

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This explains so much.

You plug in the t(Q) into the t of the grade average, which totally makes sense I just assumed you stuck the first function into the second …oops not doing that again haha!

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OK. You see it.

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Question: `qWhat do you get when you evaluate your composite function at t = 100, 110, 120 and 130?

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Your solution:

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1.8

2.3

2.7

3.0

(these are approximations)

Just plugging Q in for -.5+(50(1-e^(-.02(Q-70))/10

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confidence rating #$&*:

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Given Solution:

You should get the same values you got before for these Q values. For example an approximate calculation for t =

130 is

-.5 + 50(1-e^(-.02(130-70) ) / 10 =

-.5 + 50 (1-e^-1.2) / 10 =

-.5 + 50 (1 - .3) / 10 =

-.5 + 35/10 =

-.5 + 3.5 =

3, approx., pretty close to your 2.99.

Evaluating the composite function more accurately, we get

gradeAverage(100) = -.5 + ( 50 (1 - e ^ (-.02 ((100) - 70) ) ) ) / 10 = 1.76

gradeAverage(110) = -.5 + ( 50 (1 - e ^ (-.02 ((110) - 70) ) ) ) / 10 = 2.25

gradeAverage(120) = -.5 + ( 50 (1 - e ^ (-.02 ((120) - 70) ) ) ) / 10 = 2.66

gradeAverage(130) = -.5 + ( 50 (1 - e ^ (-.02 ((130) - 70) ) ) ) / 10 = 2.99

These values agree, as they must, with the values calculated previously.

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Self-critique (if necessary):

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Self-critique rating:OK

STUDENT COMMENT: Composite functions are fun. It's almost exactly the same as functions encountered in

computer programming so I find it quite fascinating, helpful, and complementary to programming. I think this correlation

helps me to better understand much of the precalculus I've seen thus far. Also, as I mentioned in one of the assignment's

problems, I am really fascinated by the gradeAverage(t(Q)) function in that it has an upper limit that can't be breached no

matter what number is plugged in for Q. It's a little disorienting when you're used to getting different output for different

input. I don't quite comprehend how the function causes this to happen, though.

INSTRUCTOR RESPONSE: We can go back to the statements

'If Q is very, very large, e^-( .02(Q-70) ) would have a negative exponent with a very large magnitude and so would

be very close to 0.

In this case 50 ( 1-e^(-.02 (Q-70)) would be close to 50(1-0) = 50.

Then the grade average would be -.5 + 50 / 10 = -.5 + 5 = 4.5 .'

The key is that e^-( .02(Q-70) ) essentially 'disappears' for large values of Q, so that ( 1-e^(-.02 (Q-70)) will be pretty much equal to 1 and 50 ( 1-e^(-.02 (Q-70)) pretty much equal to 50.

Plugging 50 into the grade average function we get -.5 + 50 / 10 = -.5 + 5 = 4.5, which is therefore the limiting value of the composite function.

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: It is possible to use a fair amount of algebra on this problem, but it isn't necessary to use any. Basic graphing skills and an understanding of polynomials are sufficient and should be used in preference to algebraic manipulations, that latter of which may be used (reluctantly) if necessary.

Sketch the graphs of f(x) = (x - 2)^2 - 1 and g(x) = -(x + 2)^2 + 1. It is recommended that you graph each of these functions using a sequence of transformations applied to basic points.

Use these two graphs to sketch the graph of the product function f(x) * g(x).

How did you determine the zeros of the product function?

How would (or, hopefully, did) graphing by transformations reveal the zeros without the need for algebra?

How did you determine the intervals on which the product function was positive, and those on which it was negative?

There exists an interval on which the product function is closer to the x axis that the f(x) function, and on the same side of the axis as this function. Can you determine this interval?

Optional challenging question: Based on the answers you have just given you can determine the factored form of the product function, without the need to actually multiply the two expressions. How could this be done, and what is your result?

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Your solution:

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How did you determine the zeros of the product function?

By looking at where points or the graph passed the x axis

How would (or, hopefully, did) graphing by transformations reveal the zeros without the need for algebra?

You can see where the points and graphs intersect and touch the x axis (as well as y axis and other graphs)

How did you determine the intervals on which the product function was positive, and those on which it was negative?

By observing the positive y axis because the negative values for this graph took place in the second quadrant on the left side (nothing went under the x axis )

There exists an interval on which the product function is closer to the x axis that the f(x) function, and on the same side of the axis as this function. Can you determine this interval?

The interval of the f(x) function which is closest to the x axis has the coordinates (1,0), 0

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confidence rating #$&*:

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Question: Power P is proportional to the square of speed v, which is in turn a linear function of time.

Specifically, P = 4 v^2 and v = 5 t + 12, where v is in meters / second when t is in seconds, and P is in watts when v is in meters / second.

What is the velocity when t = 4 seconds, and what therefore is the power at that instant?

Write P as a function of t, and verify that when 4 is substituted for t in your expression the result agrees with your preceding answer.

Optional challenge questions:

Verify that the rate of change of v with respect to t is 5, no matter what the value of t.

When t = 4, v = 32, as you probably know already.

If v = 32, then the rate at which P is changing with respect to v is 256. Verify this by finding the average rate of change of P with respect to v, between v = 31 and 33.

Now you know, at the instant t = 4, the rate at which v changes with respect to t, and the rate at which P changes with respect to v. How do you expect these rates would be combined to determine the rate at which P changes with respect to t?

Test your conjecture by determining the average rate of change of P with respect to t, between t = 3 and t = 5.

What would be the units of the following rates of change, and how are they related to one another?

• The rate of change of v with respect to t.

• The rate of change of P with respect to v.

• The rate of change of P with respect to t.

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Your solution:

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What is the velocity when t = 4 seconds, and what therefore is the power at that instant?

4096 watts

Write P as a function of t, and verify that when 4 is substituted for t in your expression the result agrees with your preceding answer.

Composite fn

P=4(5t+12)^2

When t=4, P=4(5(4)+12)^2

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Question: Add comments on any surprises or insights you experienced as a result of this assignment.

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Your response:

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

&#Your work looks good. See my notes. Let me know if you have any questions. &#