course mth271 r???z???????assignment #020020. `query 20
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00:24:35 `questionNumber 200000 2.7.16 dy/dx at (2,1) if x^2-y^3=3
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RESPONSE --> dy/dx= x^2-y^3=3 = 2xdy/dx-3y^2dy/dx= 0 dy/dx= 2x-3y^2dy/dx=0 = dy/dx= 2x/3y^3 and 4/3 confidence assessment: 1
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00:24:59 `questionNumber 200000 The derivative of x^2 with respect to x is 2 x. The derivative of y^3 with respect to x is 3 y^2 dy/dx. You can see this by realizing that since y is implicitly a function of x, y^3 is a composite function: inner function is y(x), outer function f(z) = z^3. So the derivative is y'(x) * 3 * f(y(x)) = dy/dx * 3 * y^3. So the derivative of the equation is 2 x - 3 y^3 dy/dx = 0, giving 3 y^2 dy/dx = 2 x so dy/dx = 2 x / ( 3 y^2). At (2,1), we have x = 2 and y = 1 so dy/dx = 2 * 2 / (3 * 1^2) = 4/3. **
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RESPONSE --> thats what i got self critique assessment: 3
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00:26:26 `questionNumber 200000 2.7.30 slope of x^2-y^3=0 at (1,1) What is the desired slope and how did you get it?
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RESPONSE --> i did 2x-3y^2dy/dx=o = dy/dx= 2x/3y^2 and plug in x=1 and y=1 and get 2/3 confidence assessment: 2
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00:27:03 `questionNumber 200000 The derivative of the equation is 2 x - 3 y^2 dy/dx = 0. Solving for dy/dx we get dy/dx = 2x / (3 y^2). At (-1,1) we have x = 1 and y = 1 so at this point dy/dx = 2 * -1 / (3 * 1^2) = -2/3. **
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RESPONSE --> i thought it said (1,1) and not (-1,1) but if it did then it would have been a -2/3 self critique assessment: 3
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00:29:05 `questionNumber 200000 2.7.36 p=`sqrt( (500-x)/(2x))
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RESPONSE --> p= (500-x)^1/2(2)^1/2 + (2x)^1/2(-1)^1/2 = confidence assessment: 0
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00:30:51 `questionNumber 200000 You could apply implicit differentiation to the present form, and that would work but it would be fairly messy. You have lots of choices for valid ways to rewrite the equation but I would recommend squaring both sides and getting rid of denominators. You get p^2 = (500-x) / (2x) so 2x p^2 = 500-x and 2x p^2 + x - 500 = 0. You want dx/dp so take the derivative with respect to p: 2x * 2p + 2 dx/dp * p^2 - dx / dp = 0 (2 p^2 - 1) dx/dp = - 4 x p dx / dp = -4 x p / (2p^2 - 1) **
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RESPONSE --> i should have gotten rid of the sqrt, would have made it easier. if you do that then you get p^2= (500-x)/2x and then bring over the 2x and get 2xp^2= 500-x and then get 2xp^2 +x-500=0 and then you take the dx/dp and get -4xp/(2p^2-1) self critique assessment: 2
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?????Z????I?M???€ assignment #021 021. `query 21 Applied Calculus I 08-04-2007
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00:34:59 `questionNumber 210000 At (3,4) you are given dx/dt as x ' = 8. Since 2x dx/dt + 2y dy/dt = 0 we have 2(3) * 8 + 2 * 4 dy/dt = 0 so dy/dt = -48/8 = -6. At (4,3) you are given dy/dt as y' = -2. So you get 2 * 4 dx/dt + 2 * 3 * -2 = 0 so 8 dx/dt - 12 = 0 and therefore 8 dx/dt = 12. Solving for dx/dt we get dx/dt = 12/8 = 3/2. **
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RESPONSE --> oops. 2xdx/dt+2ydy/dt=0 we plug in 3,4 and get a -6 and then for 4,3 we do 2*4dx/dt +2*3*-2=0 and get 3/2 self critique assessment: 2
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00:37:12 `questionNumber 210000 **** Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute?
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RESPONSE --> dV/dt= 4/3pi' r^2 and get confidence assessment: 0
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00:38:56 `questionNumber 210000 The shape is a sphere. The volume of a sphere, in terms of its radius, is V = 4/3 `pi r^3. Taking the derivative with respect to t, noting that r is the only variable, we obtain dV/dt = ( 4 `pi r^2) dr/dt You know that r increases at a rate of 2 in / min, which means that dr/dt = 2. Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx. Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. **
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RESPONSE --> i forgot to cube it instead of squaring it. then you get dV/dt= (4'pir^2)dr/dt and then plug in 2 for dr/dt and get 4pi(6^2)*2=288pi= 904 and then do the same thing for r=24 and get 14,476 self critique assessment: 2
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???????????? assignment #022 022. `query 22 Applied Calculus I 08-04-2007
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00:42:27 `questionNumber 220000 Query 3.1.4 derivative of -3x `sqrt(x+1) at x=-1, -2/3, 2`sqrt(3)/3 **** What is the derivative of the function at each of the indicated points?
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RESPONSE --> not sure confidence assessment: 0
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00:44:01 `questionNumber 220000 f(x) = -3x(x + 1)^(1/2) . Using the Product Rule we get f ' (x) = (-3) ( x + 1)^(1/2) + 1/2(x +1)^(-1/2)(1)(-3x) = -3(x+1)^(1/2) - 3/2 * x (x+1)^(-1/2). You could substitute into this form and get the right answers. However it's good to see how to simplify this expression: Simplifying, we multiply the first term by (x+1)^(1/2) / (x+1)^(1/2) to obtain f ' (x) = -3(x+1) / (x+1)^(1/2) - (3 x / 2)/(x+1)^(1/2) = [-3(x+1) - 3 x / 2 ] / (x+1)^(1/2) = (-3x - 3 - 3 x / 2) / (x+1)^(1/2) = (-6x - 6 - 3x) / ( 2(x+1)^(-1/2) ) = -3 ( 3x + 2) / ( 2(x+1)^(-1/2) ) Thus f ' (x) = -3(3x + 2) / (2(x +1)^(1/2)) f ' (-1) is undefined (denominator is zero) f ' (0) = -6/2 = -3 f ' (-2/3) = 0 (critical number, numerator is zero) DER
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RESPONSE --> should use the product rule and get f ' (x) = (-3) ( x + 1)^(1/2) + 1/2(x +1)^(-1/2)(1)(-3x)= -3(x+1)^(1/2) - 3/2 * x (x+1)^(-1/2). and then simplify this and get = -3 ( 3x + 2) / ( 2(x+1)^(-1/2) ) and then plug in the numbers and get -1= undefined, 0= -3 and -2/3 = 0 and is the critical # self critique assessment: 2
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00:47:16 `questionNumber 220000 Query 3.1.8 increasing, decreasing behavior of x^2 / (x+1) On what open intervals is x^2 / (x+1) increasing, and on what intervals is it decreasing?
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RESPONSE --> get the derivative and get -2x^2/(x+1)^2 and it increases on (0,0) and decreasing at ((-1, -4) confidence assessment: 1
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00:50:22 `questionNumber 220000 We first look at where the function itself is positive, where it is negative, and where it is undefined: The numerator of the function f(x) is always positive, being a square. The denominator of f(x) goes from positive to negative at x = 1 and stays there, so the function is negative on the first two intervals, positive on the other two. To determine where the function is increasing and where decreasing we need to find the derivative and determine where it is positive and where it is negative. Using the quotient rule we get f ' (x) = (2x( x + 1) - (1)(x^2)) / (x + 1)^2 f ' (x) = (2x^2 + 2x - x^2) / (x + 1)^2 f ' (x) = (x^2 + 2x) / (x + 1)^2 f ' (x) = (x(x + 2)) / (x +1)^2 0 = (x(x + 2)) / (x + 1)^2 . The derivative is zero at x = 0 and x = -2, and undefined at x = -1. So the function can change sign only at -2, -1 or 0. The function therefore has the same sign on each of the intervals (-infinity, -2), (-2, -1), (-1, 0), (0 infinity). The derivative goes from positive to negative at -2 and from negative to positive at 0. So the function is decreasing on (-2, 0) and increasing otherwise. So the function is negative and increasing on (-infinity, -2), then negative and decreasing on (-2, -1), where it decreases asymptotic to the line x = 1. Then the function is positive and decreasing on (-1,0), and positive and increasing on (0, infinity). Note that x = -2 and x = 0 are extrema, with the function reaching a relative maximum of -4 at x = -2 and a relative minimum of 0 at x = 0. **
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RESPONSE --> I used the quotient rule and messed it up at -(1)(x^2) part i brought down the 2 and shouldnt have. so we would get 0 = (x(x + 2)) / (x + 1)^2 and d is 0 when x=0,-2 and is undefined at x=-1 so the intervals are (-infinity, -2), (-2, -1), (-1, 0), (0 infinity)The derivative goes from positive to negative at -2 and from negative to positive at 0. So the function is decreasing on (-2, 0) and increasing otherwiseo the function is negative and increasing on (-infinity, -2), then negative and decreasing on (-2, -1), where it decreases asymptotic to the line x = 1.Then the function is positive and decreasing on (-1,0), and positive and increasing on (0, infinity). self critique assessment: 2
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00:52:31 `questionNumber 220000 ** Query 3.1.30 critical numbers, increasing and decreasing intervals for f(x) = x / (x+1) What are the critical numbers for the given function, and on what intervals is it increasing, and on what intervals is it decreasing?
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RESPONSE --> i used the quotient rule and got (x+1)(1)-(x)(1)/(x+1)^2= x+1-x/(x+1)^2= 1/(x+1)^2 so x=-1 self critique assessment: 1
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00:53:29 `questionNumber 220000 Critical numbers are the numbers where the derivative is zero, and the points where the function is discontinuous. f( x ) = x / (x +1) . Using the quotient rule we get: f ' (x) = (1(x + 1) - (1)(x)) / (x + 1)^2 f ' (x) = 1 / (x + 1)^2 f ' (x) = 1 / (x + 1)^2. We can set this expression equal to zero: 1 / (x+1)^2 = 0. However we will get no solutions, since the numerator is never zero. The only possible critical numbers are those where the function is discontinous. This occurs only at x = 1, which makes the denominator zero. The function is discontinuous at x = 1. So x = 1 is a critical number. The derivative, being the reciprocal of a perfect square, is positive at numbers less than 1 and is positive at numbers greater than 1. The derivative in undefined, and approaches infinity, as x -> -1. On the interval (-infinity, 1), the derivative is positive and the function is therefore increasing. On the interval (1, infinity), the derivative is positive and the function is therefore again increasing. **
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RESPONSE --> the critical # should be 1 and not -1.On the interval (-infinity, 1), the derivative is positive and the function is therefore increasing. On the interval (1, infinity), the derivative is positive and the function is therefore again increasing self critique assessment: 2
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?~V????????????? assignment #023 023. `query 23 Applied Calculus I 08-04-2007
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00:56:19 `questionNumber 230000 3.2.10 all relative extrema of x^4 - 32x + 4 Give the x and y coordinates of all the relative extrema of x^4 - 32x + 4.
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RESPONSE --> the d'= 4x^3-32. two critical #s would be 0, -8 confidence assessment: 1
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00:58:38 `questionNumber 230000 The procedure is to find the critical numbers, where the derivative is zero, since at a 'peak' or a 'valley' the function levels off and the derivative is for that one instant zero. The derivative of this function is 4 x^3 - 32. 4 x^3 - 32 = 4 ( x^3 - 8) = 4 ( x-2)^3 has a zero at x = 2. This is the only value for which the derivative is zero and hence the only critical point. For x < 2, x - 2 is negative and hence (x-2)^3 is negative. For x > 2, x-2 is positive and hence (x-2)^3 is positive. So the derivative changes from negative to positive at this zero. This means that the function goes from decreasing to increasing at x = -2, so x = -2 is a relative minimum of x^4 - 32x + 4. The value of the function at the relative minimum is -44. That is the function has its minimum at (2, -44). **
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RESPONSE --> i forgot to do this 4 x^3 - 32 = 4 ( x^3 - 8) = 4 ( x-2)^3 has a zero at x = 2. to find the critical point. for x<2, x-2 is negative so (x-2)^3 is negative. x>2 and x-2 is + so (x-2)^3 is +. so d changes from - to + at this zero relative min. is (2,-44) self critique assessment: 2
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00:59:29 `questionNumber 230000 3.2.30 abs extrema of 4(1+1/x+1/x^2) on [-4,5] What are the absolute extrema of the given function on the interval?
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RESPONSE --> not sure confidence assessment: 0
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01:01:20 `questionNumber 230000 the derivative of the function is -4/x^2 - 8 / x^3. Multiplying through by the common denominator x^3 we see that -4/x^2 - 8 / x^3 = 0 when x^3( -4/x^2 - 8 / x^3 ) = 0, x not 0. This simplified to -4 x - 8 = 0, which occurs when x = -2. At x = -2 we have y = 4 ( 1 + 1 / (-2) + 1 / (-2)^2 ) = 4 ( 1 - .5 + .25) = 4(.75) = 3. Thus (-2, 3) is a critical point. Since large negative x yields a negative derivative the derivative for all x < -2 is negative, and since as x -> 0 from 'below' the derivative approaches +infinity the derivative between x=-2 and x = 0 is positive. Thus the derivative goes from negative to positive at x = 2, and the point is a relative minimum. A second-derivative test could also be used to show that the point is a relative minimum. We also need to test the endpoints of the interval for absolute extrema. Testing the endpoints -4 and 5 yields 4(1+1/(-4)+1/(-4)^2) = 3.25 and 4(1+1/5+1/25) = 4(1.24) = 4.96 at x = 5. However these values aren't necessarily the absolute extrema. Recall that the derivative approaches infinity at x = 0. This reminds us to check the graph for vertical asymptotes, and we find that x = 0 is a vertical asymptote of the function. Since as x -> 0 the 1 / x^2 terms dominates, the vertical asymptote will approach positive infinity on both sides of zero, and there is no absolute max; rather the function approaches positive infinity. However the min at (-2, 3), being lower than either endpoint, is the global min for this function. **
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RESPONSE --> find the d'= -4/x^2 - 8 / x^3. and then get a common denominator and then get x^3( -4/x^2 - 8 / x^3 ) = 0, x not 0. This simplified to -4 x - 8 = 0, which occurs when x = -2. At x = -2 we have y = 4 ( 1 + 1 / (-2) + 1 / (-2)^2 ) = 4 ( 1 - .5 + .25) = 4(.75) = 3. so (-2,3) is critical point. test the endpts. and get -4 and 5 yields 4(1+1/(-4)+1/(-4)^2) = 3.25 and 4(1+1/5+1/25) = 4(1.24) = 4.96 at x = 5. self critique assessment: 2
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01:02:07 `questionNumber 230000 3.2.44 demand x inversely proportional to cube of price p>1; price $10/unit -> demand 8 units; init cost $100, cost per unit $4. Price for maximum profit?
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RESPONSE --> not sure on this one confidence assessment: 1
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01:06:11 `questionNumber 230000 If x is inversely prop to the cube of price, with x = 8 when p =10, then we have: x = k/p^3. Substituting and solving for k: 8 = k / 10^3 8 = k / 1000 k = 8000 So x = 8000/ p^3. We want to maximize profit in terms of x. Profit is revenue - cost and revenue is price * demand = x * p. The demand function is found by solving for p in terms of x: p^3 = 8000/x^3 p = 20/ x^(1/3) The revenue function is therefore R = xp = x (20/ x^(1/3) = 20 x ^ (2/3). The cost function is characterized by init cost $100 and cost per item = $4 so we have C = 100 + 4x The profit function is therefore P = profit = revenue - cost =20x ^(2/3) - 100 - 4x. We want to maximize this function, so we find its critical values: P ' = 40/ 3x^(1/3) - 4 Setting P' = 0 we get 0 = 40/ 3x^(1/3) - 4 4 = 40/ 3x^(1/3) 3x^(1/3) = 40/4 3x^(1/3) = 10 x^(1/3) = 10/3 x = 37.037 units For x < 37.037 we have P ' positive and for x > 37.037 we have P ' negative. So the derivative goes from positive to negative, making x = 37.037 a relative maximum. At the endpoint x = 0 the profit is negative, and as x -> infinity the profit function is dominated by the -4x and becomes negative. At x = 37.037 we find that profit = 20* 37.037^(2/3) - 100 - 4 x profit = -26, approx. This is greater than the endpoint value at x = 0 so this is the maximum profit. This is negative, so we're going to lose money. The graph of the profit function starts at profit -100, peaks at profit -26 when about 37 items are sold, then decreases again. Alternative solution, with demand expressed and maximized in terms of price p: Demand is inversely proportional to cube of price so x = k / p^3. When p = 10, x is 8 so 8 = k / 10^3 and k = 8 * 10^3 = 8000. So the function is x = 8000 / p^3. The cost function is $100 + $4 * x, so the profit is profit = revenue - cost = price * demand - cost = p * 8000 / p^3 - ( 100 + 4 x) = 8000 / p^2 - 100 - 4 ( 8000 / p^3) = -100 + 8000 / p^2 - 32000 / p^3. We maximize this function by finding the derivative -16000 / p^3 + 96000 / p^4 and setting it equal to zero. We obtain -16000 / p^3 + 96000 / p^4 = 0 or -16000 p + 96000 = 0 so p = (96000 / 16000) = 6. For large p the derivative is negative, so the derivative is going from positive to negative and this is a relative max.. We also have to check the endpoint where p = 1. At this price the profit would be -23,900, so the function does have a maximum at p = 6. Note that the above solution in terms of p then gives demand x = 8000 / p^3 = 8000 / 6^3 = 37 approx, which is consistent with the solution we got in terms of x. The revenue would be 6 * 37 = 222, approx.. Cost would be 100 + 4 * 37 = 248 approx, and the profit would be $222-$248=-$26. That is, we're going to lose money, but better to lose the $26 than the $23,900 **
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RESPONSE --> ok, we use x=k/p^3 and get 8=k/1000 so k=8000 so x=8000/p^3 and then we p^3=8000/x^3 and get p=20/x^(1/3) the revenue= R=xp so we plug that in and get 20x^(2/3) and C=100+4x and then P=R-C so we do 20x^(2/3)-100-4x = 40/3x^(1/3)-4 and get 0=40/3x^1/3-4= 4=40/3x^(1/3) = 37.037units and is relative max. then we do 20*73.037^(2/3)-100-4x and get -26 and this is the max profit self critique assessment: 2
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?????K?>?j??B????assignment #024 024. `query 24 Applied Calculus I 08-04-2007
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01:11:07 `questionNumber 240000 **** 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 --> not sure confidence assessment: 0
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01:14:08 `questionNumber 240000 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?^4 + 20?^3 - 80?, which is zero when x = 0 and when x = 1.679 approx.. The second derivative is y '' = 20?^3 + 60?^2 - 80. Setting this equal to zero we obtain 20?^3 + 60?^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 --> y = x^5 + 5x^4 - 40x^2 theny' = 5 x^4 + 20 x^3 - 80 x so y'' = 20 x^3 + 60 x^2 - 80. the second d= y'' = (x-1)(x+2)^2. and when u plug in 0 u get x=1,-2 and the only change is at -1 (infinity,-1) and positive on (1,infinity)y ' = 5?^4 + 20?^3 - 80? and when x=0 u get 1.679 and then on the d''= 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 pt x=0 is max. and the critical pt at 1.679 is min. self critique assessment: 2
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01:14:25 `questionNumber 240000 **** 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 --> no idea confidence assessment: 0
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01:15:11 `questionNumber 240000 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 --> First derivative is positive for an increasing function, negative for a decreasing function. So we have for the first derivative: 19. positive20. positive21. negative22. negative Second derivative is positive if concave up, negative if concave down. So we have for the second derivative: 19. positive20. negative21. negative22. positive self critique assessment: 2
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01:20:04 `questionNumber 240000 **** 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 --> find the d' and then d'' and then factor and then you will be able to solve for pts of inflection confidence assessment: 1
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01:22:23 `questionNumber 240000 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 --> 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 so we have (infinity,0) and y'' is - and so (0,5/2) will be + and t=0 is inflection pt. as well as t=5/2 self critique assessment: 2
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01:23:31 `questionNumber 240000 **** 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 --> find the derivative and get C= .006x^2+20 confidence assessment: 1
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01:25:25 `questionNumber 240000 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 --> average cost per unit = (.002x^3 + 20 x + 500)/x = .002 x^2 + 20 + 500/x. the d'= .004 x - (500/ x^2) and the critical #s happen when d'=0 so 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) d''= .004 + (1000 / x^3) self critique assessment: 2
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{??????O????? assignment #025 025. `query 25 Applied Calculus I 08-04-2007
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01:27:16 `questionNumber 250000 3.4.6 find two positive numbers such that the product is 192 and a sum of the first plus three times the second is a minimum What are the two desired numbers and how did you find them?
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RESPONSE --> no idea confidence assessment: 0
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01:28:36 `questionNumber 250000 First set up the primary equation S=x+3y (y being the 2nd number) and the secondary equation xy=192. So S = x + 3(192/x). We now maximize the function by finding critical points (points where the derivative is zero) and testing to see whether each gives a max, a min, or neither. S ' = 1 - 576 / x^2, which is zero when x = sqrt(576) = 24 (or -24, but the problem asks for positive numbers). For this value of x we get y = 192 / x = 192 / 24 = 8. So the numbers are x = 24 and y = 8. }Note that x = 24 does result in a min by the first derivative test, since S ' is negative for x < 24 and positive for x > 24. **
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RESPONSE --> so we use S=x+3y and xy=192 so we have S=x+3(192/x)then S ' = 1 - 576 / x^2, which is zero when x = sqrt(576) = 24 so x is y = 192 / x = 192 / 24 = 8. self critique assessment: 2
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01:29:09 `questionNumber 250000 3.4.21 80 apple trees yield and average of 400 per tree; each additional tree decreases the yield by 32 apples per tree. Maximize yield.
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RESPONSE --> S=400x-32 confidence assessment: 0
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01:29:24 `questionNumber 250000 How many trees should be planted and what will be the maximum yield?
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RESPONSE --> no idea confidence assessment: 0
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01:30:51 `questionNumber 250000 If we let x stand for the number of trees added to the 80 then the yield per tree is 400 - 32 x, and there would be 80 + x trees. The total yield would therefore be total yield = number of trees * yield per tree = (80 + x) * ( 400 - 32 x) = -32 x^2 + -2160 x + 32000, which is maximized when x = -34 approx.; this indicates -34 trees in addition to the 80, or 46 trees total. Another approach is to assume that only the additional trees experience the decrease. However it doesn't make sense for the yield decrease to apply only to the added trees and not to the original 400. If you're gonna crowd the orchard every tree should suffer. In any case, if we make the unrealistic assumption that the original 80 trees maintain their 400-apple-per-tree yield, and that the x additional trees each have a yield of 32 x below the 400, we have x added trees each producing 400 - 32 x apples, so we produce x (400 - 32 x) = 400 x - 32 x^2 additional apples. We therefore maximize the expression y = 400 x - 32 x^2. We obtain y ' = 400 - 64 x, which is 0 when -64 x + 400 = 0 or x = 6.25. Since y ' is positive for x < 6.25 and negative for x > 6.25 we see that 6.25 will be our maximizing value. We can't plant 6.25 trees, so the actual maximum must occur for either 6 or 7 trees. We easily see that the max occurs for 6 additional trees. So according to this interpretation we plant 86 trees. **
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RESPONSE --> oh i messed up the equation, so it should be 400-32x and 80+x total yield = (80+x)*(400-32x)=-32x^2-2160x+32000 which is maximized when x = -34 approx.; this indicates -34 trees in addition to the 80, or 46 trees total self critique assessment: 2
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01:30:56 `questionNumber 250000 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none self critique assessment: 3
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^?????Y??I????assignment #026 026. `query 26 Applied Calculus I 08-04-2007
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01:34:40 `questionNumber 260000 **** Query 3.5.12 find the price per unit p for maximum profit P if C = 35x+500, p=50-.1`sqrt(x) **** What price per unit produces the maximum profit?
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RESPONSE --> P= R-C and R=xp so x(50-.1sqrtx)- 35x+500 = confidence assessment: 0
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01:36:58 `questionNumber 260000 Revenus is price * number sold: R = xp. Since p = 50 - .1 sqrt(x) we have R = x(50 - .1 `sqrt (x)) = 50x - .1x^(3/2) Price is revenue - cost: P = R - C = 50x - .1 x^(3/2) - 35x - 500. Simplifying: P = 15x - .1x^(3/2) - 500 Derivative of profit P is P ' = 15 -.15 x^(1/2). Derivative is zero when 15 - .15 x^(1/2) = 0; solving we get x = 10,000. 2d derivative is .075 x^-(1/2), which is negative, implying that x = 10000 gives a max. When x = 10,000 we get price p = 50 - .1 sqrt(x) = 50 - .1 * sqrt(10,000) = 40. Price is $40. **
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RESPONSE --> so we get P = 15x - .1x^(3/2) - 500 and p'= 15-.15x^(1/2) and make that equal to 0 and we get 10,000 and the P''= .075x^-(1/2) and gives us x=10000 (max) then we plug in 10000 into p=50-.1sqrt(x)= 50-.1*sqrt(10000)=40 self critique assessment: 2
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01:37:50 `questionNumber 260000 ** Query 3.5.22 amount deposited proportional to square of interest rate; bank can reinvest at 12%. What interest rate maximizes the bank's profit? **** What is the desired interest rate?
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RESPONSE --> .12 self critique assessment: 0
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01:40:06 `questionNumber 260000 According to my note here amount deposited A is proportional to the square of interest rate r so A = k r^2 for some proportionality constant k. The interest paid at rate r on amount A is A * r. The bank can reinvest at 12% so it gets return A * .12. The bank therefore nets .12 * A - r * A = (.12 - r) * A. Since A = k r^2 the bank nets profit P = (.12 - r) * (k r^2) = k * (.12 r^2 - r^3). We maximize this expression with respect to r: dP/dr = k * (.24 r - 3 r^2). dP/dr = 0 when .24 r - 3 r^2 = 0, when 3 r ( .08 - r) = 0, i.e., when r = 0 or r = .08. The second derivative is -6 r + .24, which is negative for r > .06. This shows that the critical point at r = .08 is a maximum. The max profit is thus P = (.12 * .08 - .08^3) * k = (.096 - .0016) k = .080 k. In order to find the optimal interest rate it is not necessary to find the proportionality constant k. However if the proportionality constant was known we could find the max profit. **
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RESPONSE --> A=kr^2 so A*.12 and then .12*A-r*A=(.2-r)*A so P = (.12 - r) * (k r^2) = k * (.12 r^2 - r^3) so dP/dr = k * (.24 r - 3 r^2). and r=0 and .08 the d''= -6r+.24 and r=.08 for that as the max. The max profit is P = (.12 * .08 - .08^3) * k = (.096 - .0016) k = .080 k self critique assessment: 2
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???w????j????~??? assignment #027 027. `query 27 Applied Calculus I 08-04-2007
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01:41:50 `questionNumber 270000 Query 3.6.16 lim {x -> 2-} (1/(x+2)); graph shown. **** What is the desired the limit?
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RESPONSE --> not sure confidence assessment: 0
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01:42:31 `questionNumber 270000 As you approach the vertical line x = -2 from the left (i.e., x -> -2-) y values drop asymptotically into unbounded negative values. So the limit is -infinity. **
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RESPONSE --> vertical line is at x=-2 and the y value falls into negative values so the limit is -infinity self critique assessment: 2
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01:42:46 `questionNumber 270000 Query 3.6.24 lim{x->infinity}( (5x^3+1) / (10x^3 - 3x^2 + 7) ) **** What is the desired limit and how did you obtain it?
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RESPONSE --> not sure confidence assessment: 0
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01:43:40 `questionNumber 270000 You can compare the leading terms of both numerator and denominator, which are both x^3 terms. The limit is therefore just the ratio of leading coefficiets: 5 / 10 = 1/2. A more rigorous algebraic treatment is sometimes called for if you are asked for a proof. See below: lim (x -> infinity) (5x^3 + 1)/(10x^3 - 3x^2 +7) = lim(x -> infinity) (5 x^3) / (10 x^3) = 5/10 or 1/2. You find this limit using the horizontal asymptote rules. Since the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is equal to a/b with a being the leading coefficient of the numerator and b being the leading coefficient of the denominator. You can also rearrange the expression by dividing numerator and denominator both by x^3 to get lim(x -> infinity) ( 5 + 1 / x^3) / ( 10 - 3 / x + 7 / x^3). Since the limits of 1/x^3, -3/x and 7 / x^3 are all zero you end up with just 5/10 = 1/2. **
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RESPONSE --> lim (x -> infinity) (5x^3 + 1)/(10x^3 - 3x^2 +7) = lim(x -> infinity) (5 x^3) / (10 x^3) = 5/10 or 1/2 and that is where the asymptote is at 1/2 self critique assessment: 2
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01:45:15 `questionNumber 270000 ** Query 3.6.50 sketch f(x) = (x-2) / (x^2-4x+3). **** Describe your graph, including a description of all intercepts, extrema, asymptotes and concavity.
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RESPONSE --> its an L shape downward at x=1 and a curvy line at x=2 and another L shape upward at x=3 self critique assessment: 1
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01:48:22 `questionNumber 270000 The function has zeros when the numerator is 0, at x = 2; only one x intercept which occurs at x = 2. The numerator is negative on (-inf,2) and positive on (2, inf). The y intercept is at x = 0; we get (0, -2/3). Vertical asymptotes when denominator is zero, which occurs at x = 1 and at x = 3. On (-inf,1), (1, 3) and (3, inf) the denominator is respectively positive, negative and positive. The function is always of one sign between zeros and asymptotes, i.e., on (-inf,1), (1, 2), (2, 3) and (3, inf). Putting together what we know about signs of the numerator and denominator we see that the respective signs on these intervals are negative, positive, negative and positive. For large pos and neg x values f(x) -> 0 so + and - x axis is an asymptote. First derivative is - (x^2 - 4? + 5)/(x^2 - 4? + 3)^2 and 2d derivative is 2?x - 2)?x^2 - 4? + 7)/(x^2 - 4? + 3)^3. The function is critical when 1st derivative is 0; numerator of 1st derivative is never 0 (quad formula, discriminant is negative) so there are no critical points. 2d derivative: x^2 - 4 x + 7 is always positive (discriminant negative, expression positive for x=0). Numerator zero only when x = 2. The denominator is zero at x=3, x=1. The 2d derivative is therefore of one sign on (-inf,1), on (1, 2), on (2,3) and on (3, inf). Substitution shows -, +, -, + on these respective intervals and concavity is therefore down, up, down and up. So the function starts with a horizontal asymptote below the negative x axis, remains negative and concave down to a vertical asymptote at x=1. It then becomes positive with upward concavity, descending from a vertical asymptote at x = 1 to a zero at x = 2, then becoming negative with downward concavity as it approaches a negative asymptote at x = 3. To the right of x = 3 the graph descends from a positive vertical asymptote to a horizontal asymptote at the positive x axis, remaining positive and concave up. **
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RESPONSE --> the concavity is up n down n up n down n up, The y intercept is at x = 0; we get (0, -2/3) Vertical asymptotes when denominator is zero, which occurs at x = 1 and at x = 3. On (-infinity,1), (1, 3) and (3, infinity) the denominator is respectively positive, negative and positive.For large positive and negative x values f(x) -> 0 so + and - x axis is an asymptote. d'= - (x^2 - 4? + 5)/(x^2 - 4? + 3)^2 and d''=2?x - 2)?x^2 - 4? + 7)/(x^2 - 4? + 3)^3. The denominator is zero at x=3, x=1 self critique assessment: 2
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01:50:10 `questionNumber 270000 SOLUTION TO PROBLEM #43:
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RESPONSE --> g(x)= x^2/ x^2-16 and the horizontal asymptote is at x=1 self critique assessment: 1
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01:51:40 `questionNumber 270000 The graph of x y^2 = 4 is not defined for either x = 0 or y = 0. The function has horizontal and vertical asymptotes at the axes. The graph of x y^2 = 4 is not defined for negative x because y^2 cannot be a negative number. However for any x value y can be positive or negative. So the first-quadrant graph is also reflected into the fourth quadrant and both are part of the graph of the relation. You therefore have the graphs of both y = 2 / sqrt(x) and y = -2 / sqrt(x). The graph of the first-quadrant function will be decreasing since its derivative is negative, and will as you say be asymptotic to the x axis. The graph of the fourth-quadrant function is increasing and also asymptotic to the x axis. **
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RESPONSE --> guess i did the wrong problem, xy^2=4 and it has both horizontal and vertical asymptotes. You have the graphs of both y = 2 / sqrt(x) and y = -2 / sqrt(x) self critique assessment: 1
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g???e???e???????assignment #028 028. `query 28 Applied Calculus I 08-04-2007
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01:56:23 `questionNumber 280000 Query 3.7.12 sketch y = -x^3+3x^2+9x-2 **** Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.
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RESPONSE --> the graph goes down and then up and then down again. confidence assessment: 0
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02:01:12 `questionNumber 280000 First we find the zeros: You can find the zero at x = -2 by inspection (i.e., try a few simple values of x and see if you 'hit' one). Knowing that x = -2 is a zero, you know that (x - (-2) ) = x + 2 is a factor of the expression. You can find the other factor by long division of x + 2 into -x^3 + 3 x^2 + 9x - 2. You get quotient x^2 - 5 x + 1. Thus -x^3 + 2 x^2 + 9 x - 2 = - (x + 2)?x^2 - 5? + 1). This expression is zero when x+2 = 0 or when x^2 - 5 x + 1 = 0. The first equation we already know gives us x = -2. The second is solved by the quadratic formula. We get x = [ -(-5) +- `sqrt( (-5)^2 - 4 * 1 * 1 ) ] / (2 * 1) = [ 5 +- `sqrt(21) ] / 2. Simplifying we get approximate x values .21 and 4.7. Then we find maxima and minima using 1st and 2d derivative: The derivative is -3x^2+6x + 9, which gives the equation -3x^2+6x + 9 = 0 for critical points. Dividing thru by -3 we get x^2 - 2x - 3 = 0 or(x-3)(x+1) = 0 so x = 3 or x = -1. Second derivative is -6x + 6, which is negative when x = 3 and positive when x = -1. At x = 3 we have a maximum. Evaluating y = -x^3+3x^2+9x-2 at x = 3 we get y = 25. The negative second derivative indicates that (3,25) is a maximum. At x = -1 we have a minimum. Evaluating y = -x^3+3x^2+9x-2 at x = -1 we get y = -7. The positive second derivative indicates that this (-1,-7) is a minimum. Finally we analyze 2d derivative for concavity and pts of inflection: The second derivative -6x + 6 is zero when x = 1; at this point the derivative, which is linear in x, changes from positive to negative. Thus the x = 1 point (1, 9) is a point of inflection. The derivative is positive and the function therefore concave up on (-infinity, 1). The derivative is negative and the function therefore concave down on (1, infinity). The function is defined for all x so there are no vertical asymptotes. As | x | -> infinity the magnitude of the function -> infinity so there are no horizontal asymptotes. **
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RESPONSE --> a zero is at x=-2 and then we do some long division and get a quotient of x^2-5x+1 so -x^3 + 2 x^2 + 9 x - 2 = - (x + 2)?x^2 - 5? + 1) and is zero when x+2-0 or when x^2-5x+1=0. the 2nd equation has to be solved with the quadratic formula and we get x = [ -(-5) +- `sqrt( (-5)^2 - 4 * 1 * 1 ) ] / (2 * 1) = [ 5 +- `sqrt(21) ] / 2. and the x = .21 and 4.7 the d' = -3x^2+6x + 9 and make that equal to 0 and we solve for x=3 or -1. d''=-6x + 6 and is - at x=3 and + when x=-1 and at x=3 is our max. and y=25 so (3,25) and at -1 is min. and y=-7 so (-1,-7) is min. and x=1 (1,9) is a pt of inflection there are no vertical asymptotes self critique assessment: 2
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02:02:19 `questionNumber 280000 Query 3.7.34 sketch (x^2+1)/(x^2-1) Note: The problem in the text might be (x^2+1)/(x^2-2). If so the solution given below can be easily adapted to that function. Describe your graph, including a description of any intercepts, extrema, asymptotes and concavity.
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RESPONSE --> a horizontal asymptote at y=1 confidence assessment: 0
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02:05:05 `questionNumber 280000 First we look for zeros and intercepts: The numerator is never zero, being the sum of the positive number 1 and the nonnegative quantity x^2. So the function has no zeros, i.e., never crosses the x axis. When x=0 we have y = (0^2+1)/(0^2-1) = -1 so the y intercept is (0,-1). Next we analyze the derivative to see if we can find relative maxima and minima: The derivative is - 4x/(x^2 - 1)^2, which has its only zero when x = 0. So (0, -1) is the only critical point. The second derivative is 4(3x^2 + 1)/(x^2 - 1)^3, which is negative when x = 0. So the critical point gives us a maximum. We analyze the second derivative to determine concavity: The second derivative 4(3x^2 + 1)/(x^2 - 1)^3 has a numerator which is always positive, since x^2 is always positive. The denominator is negative where x^2 - 1 < 0, which occurs between x = -1 and x = 1. So the 2d derivative is positive on (-infinity, -1) and on (1, infinity), where the graph will be concave up, and negative on (-1, 1), where the graph will be concave down. Now we look for vertical and horizontal asymptotes: The denominator of (x^2+1)/(x^2-1) is zero and the numerator isn't when x = +1 and also when x = -1. So we have vertical asymptotes at x = +1 and at x = -1. }For very large positive x or for very large negative x the +1 in the numerator and the -1 in the denominator are both insignificant and the function value is very close to x^2 / x^2 = 1. The function approaches its horizontal asymptote y = 1 for both large positive x and large negative x. We finally determine where the function is positive and where negative: For x < -1 the function cannot change sign, since it is continuous and has no zeros. Testing any x < -1 gives us a positive value. On this interval the function is therefore positive. The same is true for x > 1. For -1 < x < 1 the same argument shows that the function is negative**
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RESPONSE --> x=0 so we get (0,-1). d'= - 4x/(x^2 - 1)^2 and (0,-1) is the only critical pt and the d''= 4(3x^2 + 1)/(x^2 - 1)^3 and the d'' is + at (-infinity,-1), (1, infinity) and it concaves up too, and is neg. on (-1,1) and it will be concave down. we have vertical asymptotes at x=+-1 self critique assessment: 2
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????H???U?J}????assignment #029 029. `query 29 Applied Calculus I 08-04-2007
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02:08:15 `questionNumber 290000 Query 3.8.6 differential of cube root of (6x^2) **** Give the differential of the expression and explain how you obtained it.
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RESPONSE --> d'= 12x= 12(1)(.01)=.12 confidence assessment: 1
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02:08:57 `questionNumber 290000 dy is the differential; `dy means 'delta-y' and is the exact change. y = (6x^2)^(1/3) y' = dy/dx = 1/3(6x^2)^(-2/3)(12x) y' = dy/dx = 4x(6x^2)^(-2/3) y' = dy/dx = 4x / (6x^2)^(2/3). So dy = (4x / (6x^2)^(2/3)) dx **
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RESPONSE --> y = (6x^2)^(1/3) y' = dy/dx = 1/3(6x^2)^(-2/3)(12x) y' = dy/dx = 4x(6x^2)^(-2/3) y' = dy/dx = 4x / (6x^2)^(2/3) = dy = (4x / (6x^2)^(2/3)) dx self critique assessment: 2
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02:09:47 `questionNumber 290000 ** Query 3.8.12 compare dy and `dy for y = 1 - 2x^2, x=0, `dx = -.1 **** What is the differential estimate dy for the given function and interval, and what is the actual change `dy?
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RESPONSE --> (1-2x^2) self critique assessment: 0
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02:10:31 `questionNumber 290000 y ' = dy /dx = - 4 x so dy = -4x dx. The differential estimate is dy = -4 * 0 * (-.1) = 0. Actual change is y(x + `dx) - y(x) = y(0 - .1) - y(0) = (1 - 2 ( -.1)^2) - (1-0) = -.02. **
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RESPONSE --> y ' = dy /dx = - 4 x so dy = -4x dx. then dy = -4 * 0 * (-.1) = 0 and we get y(x + `dx) - y(x) = y(0 - .1) - y(0) = (1 - 2 ( -.1)^2) - (1-0) = -.02 self critique assessment: 2
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02:10:47 `questionNumber 290000 Query Extra Problem: Give the equation of the tangent line to y= 2 * x^(1/3) - 1 at (8,3); tan line prediction and actual fn value at `dx = -.01 and .01. **** What is the equation of the tangent line and how did you obtain it?
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RESPONSE --> no idea confidence assessment: 0
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02:12:40 `questionNumber 290000 f(x) = 2x^(1/3) - 1 f' (x) = 2/3x ^(-2/3) f ' (8) = 2/3(8)^(-2/3) f ' (8) = 1/6 y - 3 = 1/6(x - 8) y - 3 = 1/6x - 8/6 y = 1/6x + 10/6 y = (1/6)x + (5/3) after simplification. Using `dx = .01 we get x + `dx = 8.01. The tangent-line approximation is thus y = 1/6 * 8.01 + 5/3 = 3.001666666. The actual function value is 2 * 8.01^(1/3) - 1 = 3.001665972. The difference is .0000007, approx. A similar difference is found approximating the function for `dx = -.01, i.e., at 7.99. We see that for this short interval `dx the tangent-line approximation is very good, predicting the change in the y value (approximately .001666) accurate to 4 significant figures. COMMON ERROR: Students often round off to 3.0017, or even 3.002, which doesn't show any discrepancy between the tangent-line approximation and the accurate value. Taken to enough decimal places the values of the function and the tangent-line approximation do not coincide; this must be the case because the function isn't linear, whereas the tangent line approximation is. You should in general use enough significant figures to show the difference between two approximations. The difference should be no greater than -.02 * .01^2 = -.000002 (based on a Taylor's Theorem estimate I did in my head so don't hold me responsible for its accuracy, and you aren't responsible for Taylor's Theorem at this point of the course); the discrepancy might therefore appear in the 6th decimal place, almost certainly not later than the 7th. **
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RESPONSE --> f(x) = 2x^(1/3) - 1 f' (x) = 2/3x ^(-2/3) f ' (8) = 2/3(8)^(-2/3) f ' (8) = 1/6 then y - 3 = 1/6(x - 8) y - 3 = 1/6x - 8/6 y = 1/6x + 10/6 and then we get y = (1/6)x + (5/3) we use dx=.01 and get x+'dx=8.01 and the tangent line is y = 1/6 * 8.01 + 5/3 = 3.001666666 The actual function value is 2 * 8.01^(1/3) - 1 = 3.001665972. The difference is .0000007, approx. self critique assessment: 2
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02:13:26 `questionNumber 290000 **** Query 3.8.42 drug concentration C = 3t / (27+t^3) **** When t changes from t = 1 to t = 1.5 what is the approximate change in C, as calculated by a differential approximation? Explain your work.
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RESPONSE --> C= 3t/ (27+t^3) confidence assessment: 0
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02:14:46 `questionNumber 290000 By the Quotient Rule we have C' = ((3) (27 + t^3) - (3t^2)(3t))/ (27 + t^3)^2, or C' = (81 - 6t^3) / (27 + t^3)^2. The differential is therefore dC =( (81 - 6t^3) / (27 + t^3)^2) dx. Evaluating for t = 1 and `dt = .5 we get dC = ((81 - 6(1)^3) / (27 + (1)^3)^2) * (.5) dC = (75 / 784) (.5) dC = .0478 mg/ml **
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RESPONSE --> so the Quotient Rule we have C' = ((3) (27 + t^3) - (3t^2)(3t))/ (27 + t^3)^2, or C' = (81 - 6t^3) / (27 + t^3)^2 then we get dC=( (81 - 6t^3) / (27 + t^3)^2) dx and then solve for t=1 and 'dt=.5 and we get dC = ((81 - 6(1)^3) / (27 + (1)^3)^2) * (.5) =dC = (75 / 784) (.5) = dC = .0478 self critique assessment: 2
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