course MTH 158 This is over section 4.4 only. I stopped the query when it got to section 5.2.
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20:01:56 4.4.18. Analyze the graph of y = (x^2 + x – 12) / (x^2 – 4)
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RESPONSE --> 1) Domain: x cannot = -2 or 2 2.) x-int. = -4,3 y-int = R(0) = 3 3.) R(-x) = (x^2 - x - 12)/(x^2 - 4) does not = R(x) -R(x) = (-x^2 - x +12) / (x^2 - 4) does not = r(-x) no symmetry 4.) V.A. x = 2 and x = -2 5) H.A. y = 1 because m = 2 and n =2 when you solve R(x) = 1 you find x = 8 so H.A. intercepts graph at (8,1) 6) creating table with intervals: (-inf, -4) (-4,-2) (-2, 2) (2,3) (3,inf) # -5 -3 0 5/2 4 value 8/21 -1.2 3 -1.4 2/3 lies: above below above below above point (-5,8/21) (-3, -1.2) (0,3) (5/2, -1.4) (4,2/3)
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20:03:56 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.
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RESPONSE --> right
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20:11:47 4.3.30. Analyze the graph of y = (x^2 - x – 12) / (x + 1)
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RESPONSE --> factors to (x-4)(x+3)/(x+1) 1) domain {x|x cannot = -1} 2.) x-int. x = 4 x= -3 for y-int solve R(x) = 0 = 12 3.) no symmetry 4.) VA x = -1 5.) n = m+1 solve by long division to find oblique asymptote: (x^2 - x -12)/(x+1) = x -1 6.) zeros of num. are 4 and -3 zero of denom is -1 creating table with intervals: (-inf, -3) (-3,-1) (-1,4) (4, inf) # -4 -2 0 5 value -8/3 6 -12 4/3 lies: below above below above point (-4,-8/3) (-2,6) (0,12) (5,4/3)
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20:13:11 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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RESPONSE --> ok
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20:20:05 4.3.42. Analyze the graph of y = 2 x^2 + 9 / x.
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RESPONSE --> (2x^3 + 9 ) / x 1) domain : x cannot = 0 2)for x-int solve 2x^3 + 9 = 0 x = -1.65 no y-int because x cannot = 0 3) no symmetry 4) V.A. is x = 0 5) H.A. or Oblique? none because n>m+1 6) dividing up on the intervals: (-inf, -1.65) (-1.65, 0) (0, inf) # -2 -1 1 value 3.5 -7 11 lies: above below above point (-2,3.5) (-1,-7) (1,11)
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20:22:06 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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RESPONSE --> ok
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20:28:29 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 --> 100 is the volume so 100=pi*r^2*h A(r) = (pi*r^2*h) / 100 I am unsure if you can solve this because you don't have a height. I will do it this way nonetheless. A(3) = (pi*3^2 * h)/100 A(4) = (pi*4^2 * h)/100 A(5) = (pi*5^2 * h)/100
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20:39:17 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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RESPONSE --> Ok, I see how you got the equation for height. I guess I wasn't thinking in terms of surface area.
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20:39:30 035. Query 35
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RESPONSE --> ??
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