course Mth 271 I accidently skipped a problem and I had to go back into the program to do it. I have been studying for my chapter 2 test and I will take next week. QcƲRɗԟassignment #024
......!!!!!!!!...................................
18:29:21 **** 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?
......!!!!!!!!...................................
RESPONSE --> Using differentiation I got the following: y'=5x^4+20x^3-80x y''= 20x^3 + 60x^2 - 80 = 0 I divided through by 20 to get: x^3 + 3x^2 - 4 = 0 (x-1)(x+2)^2 x=1 and x=-2 (- inf, -2) (1, infin) The function concaves downward at (-inf, -1) and the function concaves upward at (-2,1) confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:31:09 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
......!!!!!!!!...................................
RESPONSE --> I did this mostly correct. self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:32:56 **** 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.
......!!!!!!!!...................................
RESPONSE --> 23) f'>0 f''>0 24) f'>0 f''<0 25) f'<0 f''<0 26) f'<0 f''>0 confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:35:01 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 **
......!!!!!!!!...................................
RESPONSE --> My answers were for problems 23-26 self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:35:05 **** 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.
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
_oL⣻VǕbz assignment #002 002. `Query 2 Applied Calculus I 04-14-2007 ξS؎Mzo߿ assignment #024 024. `query 24 Applied Calculus I 04-14-2007
......!!!!!!!!...................................
18:40:00 **** 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?
......!!!!!!!!...................................
RESPONSE --> I began by dividing the function by x : 0.002x^2 + 20 + 500/x = 0 .004x - 500/x^2 = 0 I multiplied thru by x^2 and got .004x^3 - 500=0 I divided thru by .004 and got 3x = 125,000 divide through by 3 and get x = 41,666.66 confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:41:40 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. **
......!!!!!!!!...................................
RESPONSE --> I got a little off track on this one. I have copied your response and will review how you did the problem further. self critique assessment: 2
.................................................
"