Assignment 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:

Denominator can not be zero, x cannot be negative, no vertical asymptotes

confidence rating:

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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):

Self-critique Rating:

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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:

First factor : (x^2 + 1)^2 / (x-1)^2

Vertical asymptote of x=1

No horizontal asymptotes

confidence rating:

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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):

Self-critique Rating:

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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:

vertical asymptotes at x = 2 and x = -1/3

horizontal asymptote y = 6 x^2 / (3 x^2) = 2

confidence rating:

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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):

Self-critique Rating:

"

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