Assignment 34 53

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course Mth 158

4/25/13 Around 9:00PM

034. *   34 

 

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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)

 

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Your solution:

 Step 1: Factored form = ((x + 4)(x - 3))/((x - 2)(x + 2)), Domain {x | x ≠ 2, x ≠ -2}

 Step 2: It is in lowest form.

 Step 3: xint = (-4 , 0) and (3 , 0); yint = (0 , 3); At (-4 , 0) the graph acts like -7/12(x + 4) so a line with a negative slope. At (3 , 0) the graph acts like 7/5(x - 3) so a line with a positive slope.

 Step 4: The vertical asymptotes are x = -2 and x = 2

 Step 5: The rational functions numerator and denominator have the same degree it will be horizontal asymptote: x^2/x^2 = 1; y = 1; To see if this line is intercepted:

 (x^2 + x - 12) / (x^2 - 4) = 1

 x^2 + x - 12 = x^2 - 4

 x = 8

 Step 6: The function is divided into the intervals: (-infinity, -4) (-4 , -2) (-2 , 2) (2 , 3) (3 , infinity)

 Step 7: At interval (-infinity, -3), R(-5) = 8/21, above x-axis. At interval (-4 , -2), R(-3) = -6/5, below x-axis. At interval (-2 , 2), R(-1) = 4 and R(1) = 10/3, above x-axis. At interval (2 , 3), R(2*1/2) = -13/9, below x-axis. At interval (3 , infinity), R(4) = 2/3, above x- axis.

 Step 8: Graphed the function. From its intercept (-4 , 0) it goes left above the x-axis, increasing but not reaching y = 1; it also goes right, below the x-axis, decreasing but not reaching x = -2. From its y intercept (0 , 3) it looks like a parabola opening up, passing through the points (-1 , 4) and (1 , 10/3), increasing but not reaching x = 2 or -2. From its intercept (3 , 0) it goes left below the x-axis, decreasing but not reaching x = 2; it also goes right, above the x-axis, increasing as it passes through the asymptote intercept of y =1 at (8 , 1).

 

 

confidence rating #$&*: 2

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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)

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Step 1: ((x + 3)(x - 4)) / (x + 1); Domain {x | x ≠ -1}

Step 2: It is in its lowest form

Step 3: xint: (-3 , 0), (4 , 0); At (-3 , 0) the graph acts like 7/2(x + 3) so a line with a positive slope. At (4 , 0) the graph acts like 7/5(x - 4) so a line with a positive slope. yint: (0 , -12)

Step 4: Vertical asymptote: x = -1

Step 5: The function has an oblique asymptote of y = x - 2; Determining if the are any intercepts along the asymptote:

(x^2 - x - 12) / (x + 1) = x - 2

x^2 - x - 12 ≠ x^2 - x -2; No intercepts on the oblique asymptote.

Step 6: Intervals: (-infinity , -3) (-3 , -1) (-1 , 4) (4 , infinity)

Step 7: Negative at (-infinity , -3), positive at (-3 , -1), negative at (-1 , 4), and positive at (4 , infinity).

Step 8: The graph increases from -infinity (with negative values, following the curving with the oblique asymptote) to x = -3 where it continues to increase towards the vertical asymptote x = -1 (with positive values). That happens over the intervals (-infinity , -3) and (-3 , -1). The graph also ascends from near the vertical asymptote (with negative values) up to x = -4 where it continues to increase, curving with the oblique asymptote. This happens over the intervals (-1 , 4) and (4 , infinity).

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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.

 

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

 

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Your solution:

I'm not sure what to do with this function; I am unsure how to go about it when it is not in the form of (something)/(something)

 

 

confidence rating #$&*: 0

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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.

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I think I understand. Could you tell me what effect having a rational function not in the form x/x might have on the graph? If it has an effect????

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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.

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I also am unsure how to do this one.

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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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Self-critique (if necessary):

I understand this; I simply didn't wrap my mind around it.

 

 

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&#Your work looks good. Let me know if you have any questions. &#