#$&* course MTH 271 4/212:08 024. `query 24
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Given Solution: `a 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 STUDENT QUESTION: I'm confused on how we can divide my x-1 after finding the solution to setting the derivative to zero? INSTRUCTOR RESPONSE: We learn in precalculus that a polynomial factors into linear factors and irreducible quadratic factors (this is the Fundamental Theorem of Algebra). (x - z) is a linear factor if, and only if, the polynomial is zero when x = z. So if the polynomial is zero when x = z, when it is divided by (x - z) the result will be a polynomial. Furthermore the zeros of the quotient polynomial will also be zeros of the original polynomial. In the present case x = 1 is a zero of the polynomial, so if we divide the polynomial by x - 1 we will get another polynomial. If a polynomial function is zero when x = 1, as in the present case, then dividing it by x - 1 yields a polynomial whose zeros are also zeros of the original polynomial. The equation is 20 x^3 + 60 x^2 - 80 = 0, which we reduce to the equivalent equation x^3 + 3 x^2 - 4 = 0. We use a little trial and error to find that x = 1 is a solution, then we divide by x - 1 to get x^2 + 4 x + 4 = 0. This is a quadratic equation and could be solved using the quadratic formula, which always works; however it is easy to factor so the above solution indicates this method. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q **** Query problem 3 7th edition 3.3.23-26 (was 19-22) 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1st derivative: 2nd derivative: 1) positive positive 2) positive negative 3) negative negative 4) negative positive confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a First derivative is positive for an increasing function, negative for a decreasing function. So we have for the first derivative (numbers refer to problem numbers from 7th edition): top left (#19): positive top right (#20). positive bottom left (#22). negative bottom right (#23). negative Second derivative is positive if concave up, negative if concave down. So we have for the second derivative: top left (#19): positive top right (#20). negative bottom left (#22). negative bottom right (#23). positive &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q **** Query problem 4 c 7th edition 3.3.34 (was 3.3.30) 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(t) = (1 - t)(t - 4)( t^2 - 4) → -t^4 + 5t^3 - 16 f’(t) = -4t^3 + 15t^2 - 20 f’’(t) = -12t^2 + 30t The zeroes for the second derivative are at 0 and 2.5. For the interval (-infinity, 0), y’’ is negative. 0 is the inflection point. For the interval (0, 2.5), y’’ is positive. 2.5 is the inflection point. Which leaves y’’ negative on the interval (2.5, infinity). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q **** Query problem 6 7th edition 3.3.54 (was 3.3.50) 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The equation for the cost per unit is cost, C, divided by x number of units. The equation is: Pl = C/x → Pl = (0.002x^3 + 20x + 500)/ x → 0.002x^2 + 20 + 500*(x^-1) Pl’ = 0.004x - 500x^(-2) When this is equated to zero, the critical point will be x: 0 = 0.004x - 500x^(-2) → 0.004x = 500/x^2 → 0.004(x^3) = 500 → x^3 = 125000 → x = 50, which is the critical point. Using the second degree test on the second derivative: Pl’’ = 0.004 + (1000/x^3) This function is concave upwards and decreasing, so this is a relative minimum. This is the production level to minimize the average cost. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: