course Mth 271 q}JުdN¿assignment #024
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19:23:21 **** 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 y' = 5x^4 + 20x^2 - 80x y'' = 20x^3 + 40x - 80 y''= 20(x^3 + 40X - 4) X^3 + 2X - 4 = 0 (x +4)()x-1) = 0 Critical numbers = -4, 1 Interval = -infinity
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19:27:14 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 --> I'm confused on how we can divide my x-1 after finding the solution to setting the derivative to zero?
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19:28:43 **** 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) First:positive Second:positve 24)First:positve Second:positive 25)First:negative Second:negative 26)First:negative Second:negative confidence assessment: 2
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19:29:44 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 --> I only missed 2 and now realize why I missed them. self critique assessment: 2
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19:32:48 **** 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) = t - t^2 -4 + 4t(t^2 - 4) f(t) = t^3 - t^4 - 4t^2 + 4t^3 - 4t + 4t^2 + 16 - 16t f(t) = -t^4 + 5t^3 - 20t - 16 f'(t) = -4t^3 + 15t^2 - 20 f''(t) = -12t^2 + 30t t(-12t + 30) = 0 Critical: 0 and 2.5 (0, 0) and (2 1/2, 5 1/16) confidence assessment: 1
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19:33:49 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 --> I just interpreted my results wrong, but got the process right. self critique assessment: 2
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19:34:29 **** 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 --> C = .002x^3 + 20x + 500 C' = .006x^2 + 20 C''= .012x confidence assessment: 1
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19:36:32 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 didn't think enough on this problem. I forgot that ave. cost was C/x. Otherwise, I knew that the second derivitave would give the right answer. self critique assessment: 2
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