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.

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Question: *  6.3.40 / 7th edition 5.3.38. Explain how you used transformations to graph f(x) = 2^(x+2)

The function y = 2 ^ ( x+2) is a transformation of the basic function y = 2^x, with x replaced by x - 2. So the graph moves 2 units in the x direction.

Your solution:

Confidence Assessment:

Given Solution:

The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). The graph of y = 2^(x+2), being shifted in only the x direction, is also asymptotic to the negative x axis and passes through the points (0 - 2, 1) = (2, 1) and (1 - 2, 2) = (3, 2). The graph also increases at a rapidly increasing rate.

All the points of the graph of y = 2^(x + 2) lie 2 units to the left of points on the graph of y = 2^x.

To understand why the graph shifts to the left, and why we used the points (0 - 2, 1) = (2, 1) and (1 - 2, 2) = (3, 2) as a basis for the graph, consider the tables for y = 2^x and y = 2^(x + 2).  The tables are given below:

x	y = 2^x		x	y = 2^(x+2)
-3	1/8		-3	1/2
-2	1/4		-2	1
-1	1/2		-1	2
0	1		0	4
1	2		1	8
2	4		2	16
3	8		3	32

Observe that the y values 1/2, 1, 2, 4 and 8 in the y = 2^(x+2) column also occur in the y = 2^x column, but for different values of x:

• The values of x for the y = 2^(x+2) function, corresponding to y values 1/2, 1, 2 and 4, are 2 units less than for the y = 2^x function.
• This occurs because the exponent x + 2 is 2 units greater than the exponent x, so that x + 2 is always 2 units 'ahead' of the value of x.  Thus y = 2^(x + 2) reaches its values 'earlier' than y = 2^x (for example y = 2^(x + 1) reaches the value y = 8 when x = 1, whereas y = 2^x doesn't reach y = 8 until x = 3). 
• This causes the y = 2^(x + 2) graph to be shifted 2 units to the left, relative to the graph of y = 2^x.

The figure below depicts the graphs of the two functions:

Self-critique (if necessary):

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Question: *  Extra Problem / 7th edition 5.3.42. Transformations to graph f(x) = 1 – 3 * 2^x

Your solution:

Confidence Assessment:

Given Solution:

* * 1 - 3 * 2^x is obtained from y = 2^x by first vertically stretching the graph by factor -3, then by shifting this graph 1 unit in the vertical direction.

The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2).

-3 * 2^x will approach zero as 2^x approaches zero, so the graph of y = -3 * 2^x will remain asymptotic to the negative x axis. The points (0, 1) and (1, 2) will be transformed to (0, -3* 1) = (0, -3) and (1, -3 * 2) = (1, -6). So the graph will decrease from its asymptote just below the x axis through the points (0, -3) and (1, -6), decreasing at a rapidly decreasing rate.

To get the graph of 1 - 3 * 2^x the graph of -3 * 2^x will be vertically shifted 1 unit. This will raise the horizontal asymptote from the x axis (which is the line y = 0) to the line y = 1, and will also raise every other point by 1 unit. The points (0, -3) and (1, -6) will be transformed to (0, -3 + 1) = (0, -2) and (1, -6 + 5) = (1, -5), decreasing from its asymptote just below the line y = 1 through the points (0, -2) and (1, -5), decreasing at a rapidly decreasing rate.

If you don't understand the above, then do as follows, without looking up at the solution given so far:

• Plot the basic points (0, 1) and (1, 2) of the y = 2^x function.

• Multiply your y values by 3 to get the basic points of the y = 3 * 2^x function, plot your points and sketch the graph (this is your 'vertical stretch'; you should see that it moves your original points 3 times as far from the x axis as before.  The same thing happens to all the points of the original y = 2^x graph--they all move 3 times as far from the x axis.)

• Multiply your y values by -1.  This gives you the basic points of the y = - 3 * 2^x function, plot your points and sketch the graph..

• Add 1 to your y values, plot your points and sketch your graph.  This gives you the basic points of the y = 1 -3 * 2^x function.  (It should be clear that this 'shifts' the points of your y = - 3 * 2^x graph 1 unit in the vertical direction).

Having done this, look again at the given solution.

You might also consider the following table:

x	y = 2^x	y = 3 * 2^x	y = -3 * 2^x	y = 1 -3 * 2^x
0	1	3	-3	-2
1	2	6	-6	-5

It should be clear how this table demonstrates the process described above (get y values of basic function, multiply by 3, multiply by -1, add 1), and how you transform the 'basic points' from (0, 1) and (1, 2) to get (0, 3) and (1, 6), then (0, -3) and (1, -6) and finally (0, -2) and (1, -5).

You should identify these points on the graph depicted below, and having idenfied the basic points you should be able to identify which function is which.

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Question: *  Extra Problem / 7th edition 5.3.60 Solve (1/2)^(1-x) = 4.

Your solution:

Confidence Assessment:

Given Solution:

The reliable way to solve a problem of this nature is to recognize that the variable x occurs in an exponent, so we can 'get at it' by taking logs of both sides:

(1/2)^(1-x) = 4.  Taking the natural log of both sides, using the laws of logarithms, we get

ln( (1/2)^(1 - x) = ln(4)) so that

(1 - x) ln(1/2) = ln(4) and

(1 - x) = ln(4) / ln(1/2).  A calculator will reveal that ln(4) / ln(1/2) = -2, so that

1 - x = -2.  We easily solve for x, obtaining

x = 1.

It is also possible to reason this problem out directly, and in this case our reasoning leads us to an exact solution:

We first recognize one fact: 

• 4 is an integer power of 2, and 1/2 is an integer power of 2, so 4 must also be an integer power of 1/2.

Since 4 = 2^2, and since 1/2 = 2^-1, we can recognize that 4 = 2^-2, and reason as follows:

• (1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 4, so we know that the exponennt 1 - x must be -2

• If 1 - x = -2, it follows that -x = -2 - 1 = -3 and x = 3.

Self-critique (if necessary):

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Question: *  6.3.104 / 6.3.98 / 7th edition 5.3.78 A(n) = A0 e^(-.35 n), area of wound after n days. What is the area after 3 days and what is the area after 10 days? Init area is 100 mm^2.

Your solution:

Confidence Assessment:

Given Solution:

* * If the initial area is 100 mm^2, then when n = 0 we have A(0) = A0 e^(-.35 * 0) = 100 mm^2.

Since e^0 = 1 this tells us that A0 = 100 mm^2.

So the function is A(n) = 100 mm^2 * e^(-.35 n).

To get the area after 3 days we evaluate the function for n = 3, obtainind A(3) = 100 mm^2 * e^(-.35 * 3) = 100 mm^3 * .35 = 35 mm^2 approx..

After to days we find that the area is A(10) = 100 mm^2 * e^(-.35 * 10) 100 mm^3 * .0302 = 3.02 mm^2 approx..

Self-critique (if necessary):

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Question: *  6.3.109 / 6.3.104 / 7th edition 5.3.84. Poisson probability 4^x e^-4 / x!, probability that x people will arrive in the next minute. What is the probability that 5 will arrive in the next minute, and what is the probability that 8 will arrive in the next minute?

Your solution:

Confidence Assessment:

Given Solution:

The probability that 5 will arrive in the next minute is P(5) = 4^5 * e^-4 / (5 !) = 1024 * .0183 / (5 * 4 * 3 * 2 * 1) = 1024 * .0183 / 120 = .152 approx..

The probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-4 / (8 !) = .030 approx..