course Mth 271 ҠYkRassignment #024
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13:04:06 **** Query 3.3.8 picture of y = x^5 + 5x^4 - 40x^2; zeros around 0, 2.3, crit around 0, 1.8 **** On which intervals is the function concave upward and of which concave downward? **** On which intervals is the second derivative negative, on which positive?
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RESPONSE --> y = x^5 + 5x^4 - 40x^2 I need to take the second derivative. y' = 5x^4 + 20x^3 - 80x y"" = 20x^3 +60x^2 - 80 I tried setting it equal to zero to find the points of inflection and I got 20(x^3 + 3x^2 - 4) = 0 I don't know how to take x^3 + 3x^2 - 4 and find the zeros. confidence assessment: 1
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13:06:56 A function is concave upward when the second derivative is positive, concave downward when the second derivative is negative. If y = x^5 + 5x^4 - 40x^2 then y' = 5 x^4 + 20 x^3 - 80 x so y'' = 20 x^3 + 60 x^2 - 80. y'' = 0 when 20 x^3 + 60 x^2 - 80 = 0, or dividing this equation by 20 when x^3 + 3 x^2 - 4 = 0. We easily find the solution x = 1 by trial and error, just substituting simple integers. Then we can divide by x-1 to get x^2 + 4 x + 4, which factors to give us (x+2)^2. Thus y'' = (x-1)(x+2)^2. So y'' = 0 when x = 1 or x = -2. y'' can therefore change signs only at x = 1 or at x = -2. However the nature of the zero at x = -2 is parabolic so y'' doesn't change sign at this point. The only sign change is at x = -1. }For large negative x we have y'' < 0, so y'' < 0 on (infinity, -1), and is positive on (1, infinity). The first derivative is y ' = 5x^4 + 20x^3 - 80x, which is zero when x = 0 and when x = 1.679 approx.. The second derivative is y '' = 20x^3 + 60x^2 - 80. Setting this equal to zero we obtain 20x^3 + 60x^2 - 80 = 0 with solutions x = -2 and x = 1. The second derivative is negative for x < -2, again negative for -2 < x < 1 and positive for x > 1, so the function is concave down for all x < 1 except for the x = -2 point, and concave up for all x > 1. The critical point at x = 0 therefore yields a maximum and the critical point at 1.679 yields a minimum. DER
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RESPONSE --> self critique assessment: 3
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13:14:05 **** Query 3.3.23-26 concave up increasing, concave down increasing, concave down decreasing, concave up increasing **** In order list the sign of the first derivatives of the functions represented by the four graphs, and the same for the second derivatives.
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RESPONSE --> 23. concave up decreasing f' is positive f"" is negative 24. concave down increasing f' is negative f"" is negative 25. concave down decreasing f' is negative f"" is positive 26. concave up increasing f' is positive f"" is positive confidence assessment: 1
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13:15:42 First derivative is positive for an increasing function, negative for a decreasing function. So we have for the first derivative: 19. positive 20. positive 21. negative 22. negative Second derivative is positive if concave up, negative if concave down. So we have for the second derivative: 19. positive 20. negative 21. negative 22. positive **
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RESPONSE --> self critique assessment: 3
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13:28:30 **** Query 3.3.34 points of inflection for (1-t)(t-4)(t^2-4) **** List the points of inflection of the graph of the given function and explain how you obtained each.
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RESPONSE --> f(t) = (1-t)(t-4)(t^2-4) To find the points of inflection, I need to take the second derivative. I thought it would be easier to differentiate if I multiplied the factors together first. f(t) = (t - t^2 - 4 + 4t)(t^2 - 4) f(t) = (-t^2 + 5t - 4)(t^2 - 4) f(t) = -t^4 + 5t^3 - 4t^2 + 4t^2 - 20t + 16 f(t) = -t^4 + 5t^3 - 20t + 16 Now I take the first derivative f'(t) = -4t^3 + 15t^2 - 20 Next, I take the second derivative f""(t) = -12t^2 + 30t Now, I set the expression equal to zero -12t^2 + 30t = 0 t(-12t^2 + 30t) = 0 -12t + 30 = 0 12t = 30 t = 5/2 So the points of inflection are t = 0 and t = 5/2 confidence assessment: 3
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13:29:27 A point of inflection is a point where the concavity changes. Since the sign of the second derivative determines concavity, a point of inflection is a point where the sign of the second derivative changes. For a continuous function (the case here since we have a polynomial) the sign of the second derivative changes when the second derivative passes through 0. The function is f = -t^4+5t^3-20t-16 Derivative is f ' = -4t^3+15t^2-20 so f '' = -12t^2+30t, which is a quadratic function with zeros at t=0 or 5/2 . The graph of y'' vs. x is therefore a parabola. For large negative x we have y'' negative, since the leading term is -30 t^2. So on (infinity, 0) we see that y'' will be negative. y'' changes sign at x=0 so on (0, 5/2) we see that y'' is positive, and t =0 is an inflection point. y'' again changes sign at x=5/2 so on (5/2, infinity) we see that y'' is negative and t = 5/2 is an inflection point. COMMON ERROR:Common error by student: f'(t)=(-1)(1)(2t) f''(t)=-2 (- infinity, + infinity) the point of inflection doesn't exist INSTRUCTOR CORRECTION: Your derivative was based on the incorrect idea that (f g ) ' = f ' * g '. Be sure you understand how how fell into this error. You would need to apply the product rule to this function twice. The easier alternative is to multiply it out and take the derivative of the resulting polynomial. **
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RESPONSE --> self critique assessment: 3
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13:55:58 **** Query 3.3.54 production level to minimize average cost per unit for cost function C = .002 x^3 + 20 x + 500 **** What is the production level to minimize the average cost and how did you obtain it?
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RESPONSE --> I am really having a hard time with this problem. I tried taking the derivative and using the second derivative test to see which critical point was the minimum. I also though that maybe I needed to divide both sides of the function by x to get the function for the average cost. I havent been able to get anywhere using either approach. confidence assessment: 0
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14:00:38 Ave cost per unit is cost / # of units = C / x, so average cost per unit = (.002x^3 + 20 x + 500)/x = .002 x^2 + 20 + 500/x. derivative of average cost per unit = .004 x - (500/ x^2) Critical numbers occur when derivative is 0: 0 = .004 x - (500/x^2) 500 / x^2 = .004 x so 500 = .004 x^3 and x^3 = 500 / .004 = 125,000. x = 50 (critical number) The second derivative is .004 + (1000 / x^3). For all x > 0 the second derivative is therefore positive. So for x > 0 the graph is concave up, and this shows that the critical point at x = 50 is a minimum. Alternatively you could show that .004 x - (500 / x^2) changes sign at x = 50. COMMON ERROR: Students commonly make the error of minimizing only the given function. Note that you aren't supposed to minimize the cost, but the cost per unit. This is C / x = .002 x^2 + 20 + 500/x. (C/x)' = .004 x - 500 / x^2, which is zero when .004 x - 500 / x^2 = 0; multiplying by x^2 we get .004 x^3 - 500 = 0 so x^3 = 500 / .004 = 62500. Either a first- or second-degree test shows this to be a minimum. **
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RESPONSE --> I was on the right track when I divided both sides by x to get the average cost. I messed up when I took the derivative of that function. I made a dumb mental error and got the derivative of average cost per unit = .004x - 1/x^2 instead of .004x - (500/x^2). If I hadnt made this mistake, I think I would have figured the problem out. self critique assessment: 2
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