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.
027. `query 27
Question: `qQuery problem 2a 7th edition 3.6.16 lim {x -> 2-} (1/(x+2)); graph shown. **** What is the desired the limit?
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`a As you approach the vertical line x = -2 from the left (i.e., x -> -2-) y values drop asymptotically into unbounded negative values. So the limit is -infinity. **
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Question: `qQuery problem 3e 7th edition 3.6.24 lim{x->infinity}( (5x^3+1) / (10x^3 - 3x^2 + 7) ) **** What is the desired limit and how did you obtain it?
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`a You can compare the leading terms of both numerator and denominator, which are both x^3 terms. The limit is therefore just the ratio of leading coefficiets: 5 / 10 = 1/2.
A more rigorous algebraic treatment is sometimes called for if you are asked for a proof. See below:
lim (x -> infinity) (5x^3 + 1)/(10x^3 - 3x^2 +7) = lim(x -> infinity) (5 x^3) / (10 x^3) = 5/10 or 1/2.
You find this limit using the horizontal asymptote rules. Since the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is equal to a/b with a being the leading coefficient of the numerator and b being the leading coefficient of the denominator.
You can also rearrange the expression by dividing numerator and denominator both by x^3 to get
lim(x -> infinity) ( 5 + 1 / x^3) / ( 10 - 3 / x + 7 / x^3). Since the limits of 1/x^3, -3/x and 7 / x^3 are all zero you end up with just 5/10 = 1/2. **
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`a** Query 3.6.52 (was 3.6.50) sketch f(x) = (x-2) / (x^2-4x+3). **** Describe your graph, including a description of all intercepts, extrema, asymptotes and concavity.
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`a The function has zeros when the numerator is 0, at x = 2; only one x intercept which occurs at x = 2. The numerator is negative on (-inf,2) and positive on (2, inf).
The y intercept is at x = 0; we get (0, -2/3).
Vertical asymptotes when denominator is zero, which occurs at x = 1 and at x = 3. On (-inf,1), (1, 3) and (3, inf) the denominator is respectively positive, negative and positive.
The function is always of one sign between zeros and asymptotes, i.e., on (-inf,1), (1, 2), (2, 3) and (3, inf). Putting together what we know about signs of the numerator and denominator we see that the respective signs on these intervals are negative, positive, negative and positive.
For large pos and neg x values f(x) -> 0 so + and - x axis is an asymptote.
First derivative is
- (x^2 - 4•x + 5)/(x^2 - 4•x + 3)^2
and 2d derivative is
2•(x - 2)•(x^2 - 4•x + 7)/(x^2 - 4•x + 3)^3.
The function is critical when 1st derivative is 0; numerator of 1st derivative is never 0 (quad formula, discriminant is negative) so there are no critical points.
2d derivative:
x^2 - 4 x + 7 is always positive (discriminant negative, expression positive for x=0). Numerator zero only when x = 2.
The denominator is zero at x=3, x=1.
The 2d derivative is therefore of one sign on (-inf,1), on (1, 2), on (2,3) and on (3, inf).
Substitution shows -, +, -, + on these respective intervals and concavity is therefore down, up, down and up.
So the function starts with a horizontal asymptote below the negative x axis, remains negative and concave down to a vertical asymptote at x=1. It then becomes positive with upward concavity, descending from a vertical asymptote at x = 1 to a zero at x = 2, then becoming negative with downward concavity as it approaches a negative asymptote at x = 3. To the right of x = 3 the graph descends from a positive vertical asymptote to a horizontal asymptote at the positive x axis, remaining positive and concave up. **
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`aSOLUTION TO PROBLEM #43:
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`a The graph of x y^2 = 4 is not defined for either x = 0 or y = 0. The function has horizontal and vertical asymptotes at the axes.
The graph of x y^2 = 4 is not defined for negative x because y^2 cannot be a negative number.
However for any x value y can be positive or negative. So the first-quadrant graph is also reflected into the fourth quadrant and both are part of the graph of the relation.
You therefore have the graphs of both y = 2 / sqrt(x) and y = -2 / sqrt(x).
The graph of the first-quadrant function will be decreasing since its derivative is negative, and will as you say be asymptotic to the x axis.
The graph of the fourth-quadrant function is increasing and also asymptotic to the x axis. **
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