These sections were somewhat harder than earlier chapter 4. I am going to reread the sections and examine everything carefully, but I feel like I am on Mars when I try to do this.
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13:57:16 4.4.18. Analyze the graph of y = (x^2 + x 12) / (x^2 4)
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RESPONSE --> The domain is all real numbers. The equation factors into (x-3)(x+4)/ (x-2)(x+2). The zeros of numerator are x=3, x=-4. There are vertical asymptotes at x=2, x=-2 Because the degree of numerator= degree of denom, there is a horizontal asymptote at y=1. This function goes back and forth positive and negative at every zero. it is negative (-4,-2), positive (-2,2), negative (2,3) and positive (3, infinity). I am not sure how to describe this graph as far as the asymptotes, step 7 of the process.
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13:58:10 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.......!!!!!!!!...................................
RESPONSE --> Okay, this is more complex as far as explanations go, but looking at the problem, I do understand now more of what you were looking for. I will try this again.
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14:07:07 4.3.30. Analyze the graph of y = (x^2 - x 12) / (x + 1)
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RESPONSE --> The domain of this is all real # not equal to -1 (I think). This factors into (x-4)(x+3)/x+1. There are zeros at x=-3, x=4. There's no horizontal asymptote because the degree of the numerator is higher than the degree of the denominator. This function is undefined. The vertical asymptote is x=-1. This function alternates positive and negative, changes signs at every zero and at the vert. asymptote. At (-infinity, -3), the function is negative. It is postive (-3,-1), negative ( -1,4) and positive (4, infinity) It passes through the x axis at the zeros 4 and -3. Facing the vertical asymptote from the left side, it is in the (-3,-1) interval. The function is positive approaching the asymptote this way. On the right side of the asymptote is in the interval (-1,4). The function is negative on this side.
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14:08:09 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.......!!!!!!!!...................................
RESPONSE --> Okay, I need to give longer explanation. this is so involved and complicated! The rest of chapter 4 was easier. I will work on this more.
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14:19:39 4.3.42. Analyze the graph of y = 2 x^2 + 9 / x.
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RESPONSE --> The domain of this function is all real numbers not equal to zero. I can't factor this. I set the whole expression equal to zero. I get x^3=-9/2, so I take that and put it to the 1/3 power to get x by itself. This is about -1.65. There's no horizontal asymptote because the degree of the numerator is higher than the degree of the denominator. The denominator puts the vertical asymptote on the y axis (x=0) This function alternates between positive and negative (-infinity, -1.65)(-1.65,0)(0,infinity). It is positive (-infinity, -1.65) negative (-1.65,0) and positive (0, infinity) This looks like two parabolas on my graphing utility. It approaches the vertical asymptote on the left in negative y and curves up and descends back down toward the y axis. To the right of the vert. asymptote, it descends in a parabolic shape in positive values of y and curves back up and away from the y axis toward positive infinity. I think the values for scale on my utility may not be set up correctly for this.
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14:20:17 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.......!!!!!!!!...................................
RESPONSE --> okay, I need some work here. I have the basics of this, I think, but my powers of description and calculator use may be bad.
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14:21:49 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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RESPONSE --> I am having trouble setting this up
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14:22:29 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.......!!!!!!!!...................................
RESPONSE -->
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14:22:34 035. Query 35
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RESPONSE -->
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14:24:11 5.2.18. Horiz line test, looks like log.
What did the horizontal line test tell you for this function?......!!!!!!!!...................................
RESPONSE --> The horizontal line test on this graph has the line passing through the graph only once. The domain is one to one.
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14:24:19 There is no horizontal line that passes through this graph more than once. The function is strictly increasing, taking each y value only once. The function is therefore one-to-one on its domain.
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RESPONSE --> ok
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14:25:20 5.2.20. Horiz line test, looks like inverted parabola or hyperbola.
What did the horizontal line test tell you for this function?......!!!!!!!!...................................
RESPONSE --> The horizontal line passes through this graph at more than one point, so the domain is not one to one.
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14:25:27 For every horizontal below the 'peak' of this graph the graph will intersect the horizontal line in two points. This function is not one-to-one on the domain depicted here.
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RESPONSE --> ok
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14:29:06 5.2.28 looks like cubic thru origin, (1,1), (-1,-1), sketch inverse.
Describe your sketch of the inverse function.......!!!!!!!!...................................
RESPONSE --> The inverse function reverses the original points in their individual pairs, so the points are still the same. The inverse will be closer to the y axis than the x axis for x=-1 to x=1 (the opposite of the original).
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14:29:11 The graph of the function passes through (0, 0), (1,1), and (-1,-1). The inverse function will reverse these coordinates, which will give the same three points.
Between x = -1 and x = 1 the graph of the original function is closer to the x axis than to the y axis, and is horizontal at the origin. The graph of the inverse function will therefore be closer to the y axis than to the x axis for y values between -1 and 1, and will be vertical at the origin. For x < 1 and for x > 1 the graph lies closer to the y axis than to the x axis. The graph of the inverse function will therefore lie closer to the x axis than to the y axis for y < 1 and for y > 1. In the first quadrant the function is increasing at an increasing rate. The inverse function will therefore be increasing at a decreasing rate in the first quadrant. In the third quadrant the function is increasing at a decreasing rate. The inverse function will therefore be increasing at an increasing rate in the third quadrant.......!!!!!!!!...................................
RESPONSE --> ok
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14:31:32 5.2.32 f = 2x + 6 inv to g = 1 / 2 * x 3.
Show that the functions f(x) and g(x) are indeed inverses.......!!!!!!!!...................................
RESPONSE --> f(g(x))= 2 (1/2*x-3)+6= x-6+6=x g(f(x))= 1/2(2x+6)-3= x+3-3=x.
These are equal and therefore are inverse..................................................
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14:31:37 f(g(x)) = 2 g(x) + 6 = 2 ( 1 / 2 * x - 3) + 6 = x - 6 + 6 = x.
g(f(x)) = 1 / 2 * f(x) - 3 = 1/2 ( 2 x + 6) - 3 = x + 3 - 3 = x. Since f(g(x)) = g(f(x)) = x, the two functions are inverse.......!!!!!!!!...................................
RESPONSE --> ok
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14:42:57 5.2.44. inv of x^3 + 1; domain range etc..
Give the inverse of the given function and the other requested information.......!!!!!!!!...................................
RESPONSE --> y=x^3+1, inverse is x=y^3+1 y=(x-1)^1/3 The domain and range of the original function and the inverse function are the same: all real numbers. The graph of the original function passes through the point (0,1) and looks like the graph of x^3. The graph of the inverse looks like a linear function passing through y at -0.33
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14:43:17 The function is y = x^3 + 1. This function is defined for all real-number values of x and its range consists of all real numbers.
If we switch the roles of x and y we get x = y^3 + 1. Solving for y we get y = (x - 1)^(1/3). This is the inverse function. We can take the 1/3 power of any real number, positive or negative, so the domain of the inverse function is all real numbers. Any real-number value of y can be obtained by using an appropriate value of x. So both the domain and range of the inverse function consist of all real numbers.......!!!!!!!!...................................
RESPONSE --> I went a little farther and attempted to describe the graph, otherwise, ok
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14:46:49 f(x) is defined for all x except x = 0, since division by 0 is not defined.
If we switch x and y in the expression y = (3x + 1) / (-x) we get x = (3y + 1) / (-y). To solve for y we first multiply by -y, noting that this excludes y = 0 since multiplication of both sides by 0 would change the solution set. We get -x y = 3y + 1. Subtracting 3 y from both sides we get -x y - 3 y = 1. Factoring y out of the left-hand side we get (-x - 3) y = 1, and dividing both sides by (-x - 3), which excludes x = -3, we get y = -1 / (x + 3). The domain of this function is the set of all real numbers except 3. Since the domain of the inverse function is the range of the original function, the range of the original function consists of all real numbers except 3.......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. Sorry, I hit enter before I meant to. The domain of the original function is all real number except 0, the range is all y values except 0. For the inverse, I end up with y=-1/(x+3). Y cannot equal 0 and the domain and range cannot equal 3. I have the idea here, I think.................................................
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14:49:42 5.2.74. T(L) = 2 pi sqrt ( L / g). Find L(T).
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RESPONSE --> I think here I would first square everything. So T^2=4 pi^2(L)/g. I will switch L and T, and invert the equation= L(T)=T^2*g/4(pi)^2
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14:50:12 We solve T = 2 pi sqrt( L / g) for L. First squaring both sides we obtain
T^2 = 4 pi^2 * L / g. Multiplying both sides by 6 / ( 4 pi^2) we get L = T^2 g / (4 pi^2). So our function L(T) is L(T) = T^2 g / (4 pi^2).......!!!!!!!!...................................
RESPONSE --> More reading and paying attention pays off- much easier than the steel drum question in the last assignment!
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14:50:40 11-20-2005 14:50:40 036. Query 36
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NOTES ------->
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