assign 35

course mth 158

Since I was snowed in at my job and had to stay all night here, it gave me a lot of time to work on my math! This chapter was much easier to understand than the previous one...at least until I got to the end where the word problems were.

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.

035. * 35

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Question: * 6.2.18 / 7th edition 5.2.18. Horiz line test, looks like log.

What did the horizontal line test tell you for this function?

There is no horizontal line that passes through this graph more than once. The function is strictly increasing, taking each y value only once. The function is therefore one-to-one on its domain.

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Self-critique (if necessary):

I think the given solution is part of the question on this one. I agree with the solution, to look at the function for 6.2.18 there is only one y value at a time when the line passes through the function. Therefore, we conclude it is one-to-one on its domain.

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Self-critique Rating: 3

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Question: * 6.2.18 / 7th edition 5.2.20. Horiz line test, looks like inverted parabola or hyperbola.

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Your solution:

I believe the question number may be 6.2.20 on this one because of the description in the question. When I look at the function in number 20 and do a horizontal line test, the line passes through more than one y points at a time therefore the function is not one-to-one.

confidence rating: 3

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Given Solution:

* * What did the horizontal line test tell you for this function?

For every horizontal below the 'peak' of this graph the graph will intersect the horizontal line in two points. This function is not one-to-one on the domain depicted here.

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Self-critique (if necessary):

No discrepancies.

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Self-critique Rating:

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Question: * 6.2.43 / 7th edition 5.2.28 looks like cubic thru origin, (1,1), (-1,-1), sketch inverse.

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Your solution:

I believe the points on the question stated here and the one in the book are slightly different. The points from the book on 6.2.43 are (-1,-1), (2,1)

(I am sorry if I keep correcting these. I absolutely do not mean to offend you by saying that. I just felt that you would want to know if there were any discrepancies between the book and the question posted. I know that I personally could not stand someone correcting me all the time, and it’s hard to get a feel for someone’s personality and intent through words on a page, so I wanted to make that clear.)

With that said, to sketch the inverse of the points (1,1), (-1,-1) the graph would look exactly the same, To sketch the inverse of the points (-1, -1), (2, 1) the graph would have the points (-1, -1) and (1, 2).

confidence rating: 3

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Given Solution:

* * Describe your sketch of the inverse function.

The graph of the function passes through (0, 0), (1,1), and (-1,-1). The inverse function will reverse these coordinates, which will give the same three points.

Between x = -1 and x = 1 the graph of the original function is closer to the x axis than to the y axis, and is horizontal at the origin. The graph of the inverse function will therefore be closer to the y axis than to the x axis for y values between -1 and 1, and will be vertical at the origin.

For x < 1 and for x > 1 the graph lies closer to the y axis than to the x axis. The graph of the inverse function will therefore lie closer to the x axis than to the y axis for y < 1 and for y > 1.

In the first quadrant the function is increasing at an increasing rate. The inverse function will therefore be increasing at a decreasing rate in the first quadrant.

In the third quadrant the function is increasing at a decreasing rate. The inverse function will therefore be increasing at an increasing rate in the third quadrant.

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Self-critique (if necessary):

I am a little confused on the description of where the graph lies closer to the y axis and the x axis. If the sketch of the function passes through (0,0) (1,1) (-1,-1) and the inverse of the function reverses the coordinates, but passes through the same points, wouldn’t it just look like a straight line or a linear function? If that is true, then how would the graph of the original function be closer to the x axis between x = -1 and x = 1 and closer to the y axis for y values between -1, and 1? I’m not clear on how that would be.

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Self-critique Rating: 3

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Question: * 6.2.32 / 7th edition 5.2.32 f = 2x + 6 inv to g = 1 / 2 * x – 3.

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Your solution:

I’m sorry!! Really, I hate to say this, but the problem given here is 6.2.34.

f(x) = 2x + 6 g(x) = 1/2x - 3

f(g(x)) = 1/2x – 3 = f(1/2x – 3) = 2(1/2x – 3) + 6 = x – 6 + 6 = x

g(f(x)) = 2x + 6 = g(2x + 6) = ˝(2x + 6) – 3 = x + 3 – 3 = x

confidence rating: 3

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Given Solution:

* * Show that the functions f(x) and g(x) are indeed inverses.

f(g(x)) = 2 g(x) + 6 = 2 ( 1 / 2 * x - 3) + 6 = x - 6 + 6 = x.

g(f(x)) = 1 / 2 * f(x) - 3 = 1/2 ( 2 x + 6) - 3 = x + 3 - 3 = x.

Since f(g(x)) = g(f(x)) = x, the two functions are inverse.

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating:

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Question: * 6.2.52 / 7th edition 5.2.44. inv of x^3 + 1; domain range etc..

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Your solution:

I am not sure about this one. From reading the book, I know you take the inverse of x by switching roles with y so if y = x^3 + 1 you would have x = y^3 + 1. Then solve for y you get y = (x-1)^3

Good, but that would be y = (x - 1)^(1/3).

The domain of x = y^3 + 1 is the set of all real numbers and the range is the set of all real numbers so the range of the inverse = the domain of x = y^3 + 1 and the domain of the inverse = the range of x = y^3 + 1

confidence rating: 0

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Given Solution:

* * Give the inverse of the given function and the other requested information.

The function is y = x^3 + 1. This function is defined for all real-number values of x and its range consists of all real numbers.

If we switch the roles of x and y we get x = y^3 + 1. Solving for y we get

y = (x - 1)^(1/3).

This is the inverse function. We can take the 1/3 power of any real number, positive or negative, so the domain of the inverse function is all real numbers. Any real-number value of y can be obtained by using an appropriate value of x. So both the domain and range of the inverse function consist of all real numbers.

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Self-critique (if necessary):

I think I did it right, but I’m not sure.

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Self-critique Rating: 0

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Question: * 6.2.64 / 7th edition 5.2.56. inv of f(x) = (3x+1)/(-x). Domain and using inv fn range of f.

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Your solution:

f(x) = (3x+1) / (-x) Substituting x for y you get

x = (3y+1) / (-y) then solve for y D:{All real numbers, except y cannot equal 0}

I don’t know how to solve for y here.

confidence rating:

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Given Solution:

* * What is the domain of f? What is the inverse function? What does the inverse function tell you about the range of f?

f(x) is defined for all x except x = 0, since division by 0 is not defined.

If we switch x and y in the expression y = (3x + 1) / (-x) we get x = (3y + 1) / (-y). To solve for y we first multiply by -y, noting that this excludes y = 0 since multiplication of both sides by 0 would change the solution set. We get

-x y = 3y + 1. Subtracting 3 y from both sides we get

-x y - 3 y = 1. Factoring y out of the left-hand side we get

(-x - 3) y = 1, and dividing both sides by (-x - 3), which excludes x = -3, we get

y = -1 / (x + 3).

The domain of this function is the set of all real numbers except 3. Since the domain of the inverse function is the range of the original function, the range of the original function consists of all real numbers except 3.

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Self-critique (if necessary):

How did you get –xy on the left and 3y+1 on the right? Then when you subtracted 3y from both sides how did it become (-x-3) y = 1?

I've modified the solution to answer those questions:

If we switch x and y in the expression y = (3x + 1) / (-x) we get

x = (3y + 1) / (-y).

To solve for y we first multiply by -y, noting that this excludes y = 0 since multiplication of both sides by 0 would change the solution set. We get

-y * x = -y * (3 y + 1) / (-y). The left-hand side simplifies to - x y and the right-hand side to 3 y + 1, giving us

 

-x y = 3y + 1. Subtracting 3 y from both sides we get

 

-x y - 3 y = 1.

Factoring y out of the left-hand side (which becomes (-x - 3) * y; if you multiply this out you see that it is the same as -x y - 3 y) we get

 

(-x - 3) y = 1, and dividing both sides by (-x - 3), which excludes x = -3, we get

 

y = -1 / (x + 3).

 

The domain of this function is the set of all real numbers except 3. Since the domain of the inverse function is the range of the original function, the range of the original function consists of all real numbers except 3.

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Self-critique Rating: 2

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Question: * 6.2.94 / 7th edition 5.2.74. T(L) = 2 pi sqrt ( L / 32.2). Find L(T).

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Your solution:

I don’t even know how to begin this one. I tried to think of it as a composite function, but couldn’t even do that.

confidence rating: 0

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Given Solution:

* * We solve T = 2 pi sqrt( L / 32.2) for L. First squaring both sides we obtain

T^2 = 4 pi^2 * L / 32.2. Multiplying both sides by 32.2 / ( 4 pi^2) we get

L = T^2 * 32.2 / (4 pi^2).

So our function L(T) is

L(T) = 32.2 T^2 / (4 pi^2).

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Self-critique (if necessary):

I tried following along by writing it on paper to see how you solved this, but I still don’t really understand it.

Starting with the equation T = 2 pi sqrt( L / 32.2) we solve for L. We square both sides, then multiply both sides by 32.2 / (4 pi^2).

There was a typo in the solution, which might have been the source of confusion. Hopefully my correction above will help you make more sense of the problem.

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Self-critique Rating: 0

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