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Question: *  5.3.18 / 7th edition 4.4.18. Analyze the graph of y = (x^2 + x – 12) / (x^2 – 4)

Your solution:

Confidence Assessment:

Given Solution:

* * The factored form of the function is

y = (x – 3) ( x + 4) / [(x – 2) ( x + 2)].

As x -> infinity the function is dominated by the highest-power terms in numerator and denominator, and the value approaches y = x^2 / x^2 = 1. The same occurs as x -> -infinity. So the graph has a horizontal asymptote at y = 1.

The function has zeros where the numerator has zeros, at x = 3 and x = -4.

The function is undefined and approaches vertical asymptotes when the denominator is zero, which occurs at x = 2 and x = -2.

Since every factor is linear the function will change sign at every zero and vertical asymptote. So the function will alternate between positive and negative on the intervals (-infinity, -4), (-4, -2), (-2, 2), (2, 3) and (3, infinity).

For large negative x, as we have seen, the function is positive (it approaches y = +1 as x -> -infinity). So on the interval (-infinity, -4) the function will be positive.

Alternating between positive and negative, the function is negative on (-4, -2), positive on (-2, 2), negative on (2, 3) and positive on (3, infinity). It passes through the x axis at x = -4 and at x = 3.

We can use these facts to determine the nature of the vertical asymptotes.

As we approach x = -2 from the left we are in the interval (-4, -2) so function values will be negative, and we approach the asymptotes through negative values, descending toward the asymptote. To the right of x = -2 we are in the interval (-2, 2) so function values are positive, and the asymptote to the right of x = -2 descends from positive values.

As we approach x = 2 from the left we are in the interval (-2, 2) so function values will be positive, and we approach the asymptotes through positive values, rising toward the asymptote. On the interval (-2, 2), then, the values of the function descend from a positive asymptote at the left and ascend toward a positive asymptote on the right. It does this without passing through the x axis, since there are no zeros in the interval (-2, 2), and therefore remains above the x axis on this interval.

To the right of x = 2 we are in the interval (2, 3) so function values are negative, and the asymptote to the right of x = 2 ascends from negative values.

At x = 3 we have a zero so the graph passes through the x axis from negative to positive, and thereafter remains positive while approaching y = 1 as a horizontal asymptote.

5.3.30 / 7th edition 4.4.30. Analyze the graph of y = (x^2 - x – 12) / (x + 1)

The factored form of the function is

y = (x – 4) ( x + 3) / (x + 1).

As x -> infinity the function is dominated by the highest-power terms in numerator and denominator, and therefore approaches y = x^2 / x = x. So the graph is asymptotic to the line y = x at both left and right.

The function has zeros where the numerator has zeros, at x = -3 and x = 4.

The function is undefined and approaches vertical asymptotes when the denominator is zero, which occurs at x = -1.

Since every factor is linear the function will change sign at every zero and vertical asymptote. So the function will alternate between positive and negative on the intervals (-infinity, -3), (-3, -1), (-1, 4) and (4, infinity).

For large negative x, the function is close to y = x, which is negative. So on the interval (-infinity, -3) the function will be negative.

Alternating between positive and negative, the function is positive on (-3, -1), negative on (-1, 4) and positive on (4, infinity). It passes through the x axis at x = 4 and at x = -3.

We can use these facts to determine the nature of the vertical asymptote.

As we approach x = -1 from the left we are in the interval (-3, -1) so function values will be positive, and we approach the asymptotes through positive values, ascending toward the asymptote. To the right of x = -1 we are in the interval (-1, 4) so function values are negative, and the asymptote to the right of x = -1 ascends from negative values.

The function passes through the x axis at x = 4, and then approaches the line y = x as an asymptote, remaining positive from x = 4 on.

*&$*&$

Question:  5.3.40 / 5.3.42 / 7th edition 4.4.42. Analyze the graph of y = 2 x^2 + 9 / x.

Your solution:

Confidence Assessment:

Given Solution:

The denominator x indicates a vertical asymptote at x = 0, i.e., at the y axis.

The function has zeros when 2 x^2 + 9 / x = 0 . Multiplying both sides by x we get

2 x^3 + 9 = 0 so that

x^3 = -9/2 and

x = -(9/2)^(1/3) = -1.65 approx..

The function therefore alternates between positive and negative on the intervals (-infinity, -1.65), (-1.65, 0) and (0, infinity).

For large positive or negative values if x the term 9 / x is nearly zero and the term 2 x^2 dominates, so the graph is asymptotic to the y = 2 x^2 parabola. This function is positive for both large positive and large negative values of x.

So the function is positive on (-infinity, -1.65), negative on (-1.65, 0) and positive on (0, infinity).

Approaching the vertical asymptote from the left the function therefore approaches through negative y values, descending toward its vertical asymptote at the y axis.

To the right of the vertical asymptote the function is positive, so it descends from its vertical asymptote.

From left to right, therefore, the function starts close to the parabola y = 2 x^2, eventually curving away from this graph toward its zero at x = -1.65 and passing through the x axis at this point, then descending toward the y axis as a vertical asymptote.

To the right of the y axis the graph descends from the y axis before turning back upward to become asymptotic to the graph of the parabola y = 2 x^2.

5.3.56 / 4.4.56. Steel drum volume 100 ft^3, right circular cylinder. Find amount of material as a function of r and give amounts for r = 3, 4, 5 ft. Graph and indicate the min.

If the radius of the cylinder is r then the area of its circular base is pi r^2. The volume of the drum is area of base * height = 100, so that

pi r^2 * height = 100 and

height = 100 / (pi r^2).

The surface area is the sum of the surface areas of the bases, which is 2 pi r^2, and the surface area of the sides, which is circumference * height = 2 pi r * height = 2 pi r * (100 / ( pi r^2 )) = 200 / r. So the total surface area is

Surface Area = 2 pi r^2 + 200 / r.

For r = 3 we get

2 pi * 3^2 + 200 / 3 = 123.2.

Similarly for r = 4 and r = 5 we get areas 150.5 and 197.1.

Analysis of the function tells us that the graph descends from the positive vertical axis as an asymptote, reaches a minimum then begins ascending toward the 2 pi r^2 parabola, to which it is asymptotic. There must therefore be a minimum in there somewhere. Our areas 123.2, 150.5 and 197.1 are increasing, so the minimum lies either to the left of r = 3 or between r = 3 and r = 4.

Evaluating the function half a unit to the left and right of r = 3 gives us values 119.2699081, 134.1118771 at r = 2.5 and r = 3.5. We conclude that the minimum lies to the left of r = 3.

Evaluating at r = 2.6 and r = 2.4 we get areas 119.3974095 and 119.5244807, both greater than the 119.27 we got at r = 2.5. So our minimum will lie close to r = 2.5.

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