course MTH 271 12/15, 1430 assignment #024024. `query 24
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14:18:06 **** 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 --> (-infinity, -2) concave downward, 2nd derivative neg. (-2,1) concave downward, 2nd derivative neg. (1,2) concave upward, 2nd derivative pos. (2,infinity)concave upward, 2nd derivative pos. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:18:13 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 ' = 5·x^4 + 20·x^3 - 80·x, which is zero when x = 0 and when x = 1.679 approx.. The second derivative is y '' = 20·x^3 + 60·x^2 - 80. Setting this equal to zero we obtain 20·x^3 + 60·x^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 Rating:ent: 3
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14:18:34 **** 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 --> 1. concave up, increasing: f'(x) pos, f''(x) pos 2. concave down, increasing: f'(x) pos, f''(x) neg 3. concave down, decreasing: f'(x) neg, f''(x) neg 4. concave up, decreasing: f'(x) neg, f''(x) pos confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:18:39 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 Rating:ent: 3
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14:26:21 **** 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 --> (1-t)(t-4)(t^2-4) -(t-1)(t-4)(t^2-4) -(t^2-5t+4)(t^2-4) (-t^2 + 5t -4)(t^2 - 4) -t^4 + 5t^3 - 4t^2 + 4t^2 -20t +16 f(x)=-t^4 + 5t^3 - 20t + 16 f'(x)=-4t^3 + 15t^2 - 20 f''(x)= -12t^2 + 30t f''(x)= -6t(2t - 5) 0=-6t t=0 2t-5=0 t=5/2 Possible points of inflection: t=0, t=5/2 but, f''(x) value must be opposite on the intervals beside the points. (-infinity,0)...f''(x): neg. (0,5/2)...f''(x): pos. (5/2, infinity)...f''(x): neg. The f''(x) values do change signs at points t=0 and t=5/2, confirming these as points of inflection. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:26:31 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 Rating:ent: 3
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14:30:47 **** 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 --> Average cost= C/x =(.002 x^3 + 20x + 500) / x Average cost'=( x(.006x^2 + 20) - (.002x^3 + 20x + 500) ) / x^2 =(.004x^3 - 500) /x^2 0=.004x^3 -500 500=.004x^3 12,500=x^3 50=x Production level to minimize cost: 50 units confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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14:31:47 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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