#$&* course Mth 158 4/25/13 Around 9:00PM 034. * 34
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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) ######## 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). ######### 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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. ####### 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???? ####### 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. ###### I also am unsure how to do this one. ##### 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand this; I simply didn't wrap my mind around it. ------------------------------------------------ Self-critique Rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!