assign 33

course mth 158

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.

033. * 33

* * * 5.2.20 / 7th edition 4.3.20. Find the domain of G(x)= (x-3) / (x^4+1).

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Your solution:

G(x) = x – 3/ x^4 + 1

D: {All real numbers} There are no limitations

confidence rating: 3

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Given Solution:

The denominator is x^4 + 1, which could be zero only if x^4 = -1. However x^4 being an even power of x, it cannot be negative. So the denominator cannot be zero and there are no vertical asymptotes.

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating: 3

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Question: * 5.2.43 / 7th edition 4.3.43. Find the vertical and horizontal asymptotes, if any, of H(x)= (x^4+2x^2+1) / (x^2-x+1).

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Your solution:

I think 5.2.43 is a different problem, but I will solve the one given here to see if I can get it right by the given solution.

H(x)= (x^4+2x^2+1) / (x^2-x+1)

factors into (x^2 + 1)^2 / (x – 1 )^2

The zeros of the denominator are x= 1 so the vertical asymptote is when x = 1

The degree of the numerator is larger than that of the denominator so we use long division.

(Here is where I have such a hard time with this. I don’t understand how to do long division like this. I have gone back to the review chapters and I still don’t understand how to do it.)

When I tried doing the long division, I came up with an answer like:

x^2 – 2 + 2x^2 – 1/x^2 – x + 1

If that is the correct way, which I don’t think it is, how would you find the horizontal asymptote from that?

confidence rating: 0

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Given Solution:

The function (x^4+2x^2+1) / (x^2-x+1) factors into

(x^2 + 1)^2 / (x-1)^2.

The denominator is zero if x = 1. The numerator is not zero when x = 1 so there is a vertical asymptote at x = 1.

The degree of the numerator is greater than that of the denominator so the function has no horizontal asymptotes.

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Self-critique (if necessary):

So, if the degree of the numerator is greater than that of the denominator, is this saying it will always be no horizontal asymptotes? Where does the long division come in?

Long division shows the function to which the graph is asymptotic.

The correct quotient is x^2 + x + 2 + (x - 1) / (x^2 - x + 1). As x gets large, the fraction (x - 1) / (x^2 - x + 1) approaches 0. As a result, the graph of the function approaches that of x^2 + x + 2, whose graph is a parabola with vertex at (-1/2, 7/4), opening upward.

You don't actually need to know how to deal with this particular situation, but you might need to know about 'slant asymptotes'. The principle is the same:

Slant asymptotes occur when the degree of the numerator exceeds that of the denominator by 1. For example if the question had been about the function

f(x) = (x^3+2x^2+1) / (x^2-x+1)

the long division would have given you x + 3 + (2x - 2) / (x^2 - x + 1). Again for large x the fraction would approach zero, this time leaving you with the linear function y = x + 3 (the line with slope 1 and y intercept 3). Whatever else the graph does, as you move to the right or to the left the graph eventually approaches this line.

See the query, which I have edited to include some graphs.

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Self-critique Rating: 2

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Question: * 5.2.50 / 7th edition 4.3.50. Find the vertical and horizontal asymptotes, if any, of R(x)= (6x^2+x+12) / (3x^2-5x-2).

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Your solution:

R(x)= (6x^2+x+12) / (3x^2-5x-2)

Denominator factors into (x-2)(3x + 1)

The zeros are x = 2 and x = -1/3 which is the vertical asymptote

Since the degree of the numerator is the same as that of the denominator, you can use the leading terms 6/3 = 2 to get the horizontal asymptote

confidence rating: 3

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Given Solution:

* * The expression R(x)= (6x^2+x+12) / (3x^2-5x-2) factors as

(6•x^2 + x + 12)/((x - 2)•(3•x + 1)).

The denominator is zero when x = 2 and when x = -1/3. The numerator is not zero at either of these x values so there are vertical asymptotes at x = 2 and x = -1/3.

The degree of the numerator is the same as that of the denominator so the leading terms yield horizontal asymptote

y = 6 x^2 / (3 x^2) = 2.

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating: 3

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You're doing well with this. See my notes and let me know if you have additional questions.