#$&* course MTH 271 Feb. 15, 1:45pm 004. `query 4
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Given Solution: `a You have to find the average rate of change between clock times t and t + `dt: ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt = [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt = [ 2 a t `dt + a `dt^2 + b `dt ] / `dt = 2 a t + b t + a `dt. Now if `dt shrinks to a very small value the ave rate of change approaches y ' = 2 a t + b. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When we know the value of y’, then we will know the values of m and d; c is an arbitrary constant. The linear rate-of-depth-change function implies the separate quadratic depth function y=1/2mt^2 + dt + c. The rate-of-depth-change function gives us the necessary information to determine the difference between two clock times, but the absolute depth at clock time requires c, and can be found as long as we know both depth and time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Student Solution: If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent to the given function y ' (t)=mt+b . Since 2at+b=mt+b for all possible values of t the parameter b is the same in both equations, which means that the coefficients 2a and m must be equal also. So if 2a=m then a=m/2. The depth function must therefore be y(t)=(1/2)mt^2+bt+c. c is not specified by this analysis, so at this point c is regarded as an arbitrary constant. c depends only on when we start our clock and the position from which the depth is being measured. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q Explain why, given only the rate-of-depth-change function y' and a time interval, we can determine only the change in depth and not the actual depth at any time, whereas if we know the depth function y we can determine the rate-of-depth-change function y' for any time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With the rate function, there is only enough information to find the average rate within a certain time by averaging. Without the linear function, which gives us the starting and ending points, there is no way to determine the actual average depth at any time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Given the rate function y' we can find an approximate average rate over a given time interval by averaging initial and final rates. Unless the rate function is linear this estimate will not be equal to the average rate (there are rare exceptions for some functions over specific intervals, but even for these functions the statement holds over almost all intervals). Multiplying the average rate by the time interval we obtain the change in depth, but unless we know the original depth we have nothing to which to add the change in depth. So if all we know is the rate function, have no way to find the actual depth at any clock time. ** STUDENT COMMENT: Not sure I understand this I read you solution and it makes more sense. INSTRUCTOR RESPONSE: Here's an analogy: If I tell you that I drove down the interstate at 50 mph for 1/2 hour, then at 70 mph for 1 hour, could you tell me at what milepost I ended? You couldn't because I didn't tell you where I started. But you could tell me how far I traveled. In other words, if I give you the rate information and time intervals, you can tell me how far I went. But if that's the only information I give you, you can't tell me where I started or ended. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q In terms of the depth model explain the processes of differentiation and integration. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Finding the y'(t) function from the y(t) function is differentiation. Integration is when we use the differentiated function, and find depth changes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Rate of depth change can be found from depth data. This is equivalent to differentiation. Given rate-of-change information it is possible to find depth changes but not actual depth. This is equivalent to integration. To find actual depths from rate of depth change would require knowledge of at least one actual depth at a known clock time. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Growth rate: 0.1 Growth factor: 1.10 It will take 7.27 years for the principle to double. The principle will first reach $300 at 4.254 years. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t. This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis. For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22. Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course). $400 is double the initial $200. We need to find how long it takes to achieve this. Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error. To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error. The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `qAt what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ½ t=20 value is ~$673. The doubling time is about 12.73 years. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The t = 20 value is $200 * 1.1^20 = $1340, approx. Half the t = 20 value is therefore $1340/2 = $670 approx.. By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx.. For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45). At 12.75 we get 674.20 so it would probably be about 12.72 This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. ** Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `qquery #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1(1.10)^t @ 10% interest 10% 20% 30% 40% 1st year: 1.1 1.2 1.3 1.4 2: 1.21 1.44 1.69 1.96 3: 1.331 1.728 2.197 2.744 4: 1.4641 2.0736 2.8561 3.8416 Doubling time: ~7.28 ~3.81 ~2.65 ~2.06 Confidence Rating: 3
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Given Solution: `a We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double. for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double. Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. ** Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `qquery #11. equation for doubling time YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1 + r)^ t = 2, Confidence Rating: 3
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Given Solution: `a the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore P0 * (1+r)^t = 2 P0. Note that this simplifies to (1 + r)^ t = 2, and that this result depends only on the interest rate, not on the initial amount P0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1.08)^dt = 2 Dt= ~ 9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get 1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2]. This can be written as 1.08^2 * 1.08^doublingtime = 2 * 1.08^2. Dividing both sides by 1.08^2 we obtain 1.08^doublingtime = 2. We can then use trial and error to find the doubling time that works. We get something like 9 years. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q Desribe how on your graph how you obtained an estimate of the doubling time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By following the doubled money point over to the graphed line, then down to its point on the x-axis, the double point is easily found. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIn this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis. The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q#17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)? Your solution: Q(t) = 550mg(.89)^t Confidence Rating: 3
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Given Solution: `a Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or Q(t)=550(.89)^t ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `qHow much antibiotic is present at 3:00 p.m.? Your solution: 307.1233 mg confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 3:00 p.m. is 5 hours after the initial time so at that time there will be Q(5) = 550 mg * .89^5 = 307.123mg in the blood ** Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: Describe your graph and explain how it was used to estimate half-life. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph I drew decreases on the y-axis as it increases on the x-axis. It is a steep downward curve. By finding the point on the y-axis which corresponds to half of the starting amount: 275mg, then going down to the x-axis to find its co-coordinating point, I found the approximate half-life. Confidence Rating: 3
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Given Solution: `a Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point. The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down. The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `qWhat is the equation to find the half-life? What is the most simplified form of this equation? Your solution: ½ = (.89)^dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Q(doublingTime) = 1/2 Q(0)or 550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have .89^doublingTime = .5. We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q#19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0. For what values of t did Q(t) lie between .005 Q0 and .01 Q0? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was unsure of how to complete this problem until I read the solution. I was unable to see, for some reason, what exactly was being asked in the problem. I understand now, and am able to see how the problem is worked out. I believe I was confusing myself and over-thinking the problem, trying to solve for Q0, rather than time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0. Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0. Solving Q(t) = .05 Q0 we rewrite this as Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get 1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -31.4 approx. Solving Q(t) = .1 Q0 we rewrite this as Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get 1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -24.2 approx. (The solution for .005 Q0 is about -55.6, for .01 is about -48.3 For this solution any value between about t = -48.3 and t = -55.6 will work). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was unsure of how to complete this problem until I read the solution. I was unable to see, for some reason, what exactly was being asked in the problem. I understand now, and am able to see how the problem is worked out. I believe I was confusing myself and over-thinking the problem, trying to solve for Q0, rather than time. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qexplain why the negative t axis is a horizontal asymptote for this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As the time value gets larger, the 1.1^t value will get smaller, gradually approaching zero. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `q#22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )? Your solution: Y = 12( 2.718^ -.5x) - Y = 12(2.718^-0.5)^x - Y= 12(0.607)^x B = 0.607 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 ********************************************* Question: `qwhat is b for the function y = .007 ( e^(.71 x) )? Your solution: Y = 0.007(e^(.071x)) - Y = 0.007(2.034)^x B = 2.034 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhat is b for the function y = -13 ( e^(3.9 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: B = 49.403 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qList these functions, each in the form y = A b^x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = 12(0.607)^x Y = 0.007(2.034)^x Y = -13(49.403)^x confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q0.4.44 (was 0.4.40 find all real zeros of x^2+5x+6 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X^2 + 5x + 6 -> {-5 [+-] sqrt(5^2 - 4*1*6}/ 2 -> -5/2 [+-] [sqrt(1)/2] -> -6/2, -4/2 -> (-3, -2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a We can factor this equation to get (x+3)(x+2)=0. (x+3)(x+2) is zero if x+3 = 0 or if x + 2 = 0. We solve these equations to get x = -3 and x = -2, which are our two solutions to the equation. COMMON ERROR AND COMMENT: Solutions are x = 2 and x = 3. INSTRUCTOR COMMENT: This error generally comes after factoring the equation into (x+2)(x+3) = 0, which is satisfied for x = -2 and for x = -3. The error could easily be spotted by substituting x = 2 or x = 3 into this equation; we can see quickly that neither gives us the correct solution. It is very important to get into the habit of checking solutions. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qExplain how these zeros would appear on the graph of this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The y coordinates for each of these x-values would simply be named as zero. The graph will cross the x-axis at each of these points. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 0.4.50 (was 0.4.46 x^4-625=0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X^4 - 625 = 0 -> x^4 = 625 -> 4rt(x^4) = 4rt(625) -> x = 5, -5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Common solution: x^4 - 625 = 0. Add 625 to both sides: x^4 = 625. Take fourth root of both sides, recalling that the fourth power is `blind' to the sign of the number: x = +-625^(1/4) = +-[ (625)^(1/2) ] ^(1/2) = +- 25^(1/2) = +-5. This is a good and appropriate solution. It's also important to understand how to use factoring, which applies to a broader range of equations, so be sure you understand the following: We factor the equation to get (x^2 + 25) ( x^2 - 25) = 0, then factor the second factor to get (x^2 + 25)(x - 5)(x + 5) = 0. Our solutions are therefore x^2 + 25 = 0, x - 5 = 0 and x + 5 = 0. The first has no solution and the solution to the second two are x = 5 and x = -5. The solutions to the equation are x = 5 and x = -5. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is >1000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -200x^2 + 2000x - 3800 > 1000 -> -200x^2 + 2000x - 4800 >0 -> divide each side by 200 -> -1x^2 + 10x - 24 = 0 -> using the quadratic formula, I get x = 6,4. The interval is (4, 6). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You have to solve the inequality 1000<-200x^2+2000x-3800, which rearranges to 0<-200x^2+2000x-4800. This could be solved using the quadratic formula, but it's easier if we simplify it first, and it turns out that it's pretty easy to factor when we do: -200x^2+2000x-4800 = 0 divided on both sides by -200 gives x^2 - 10 x + 24 = 0, which factors into (x-6)(x-4)=0 and has solutions x=4 and x=6. So -200x^2+2000x-4800 changes signs at x = 4 and x = 6, and only at these values (gotta go thru 0 to change signs). At x = 0 the expression is negative. Therefore from x = -infinity to 4 the expression is negative, from x=4 to x=6 it is positive and from x=6 to infinity it's negative. Thus your answer would be the interval (4,6). COMMON ERROR: x = 4 and x = 6. INSTRUCTOR COMMENT: You need to use these values, which are the zeros of the function, to split the domain of the function into the intervals 4 < x < 6, (-infinity, 4) and (6, infinity). Each interval is tested to see whether the inequality holds over that interval. COMMON ERROR: Solution 4 > x > 6. INSTRUCTOR COMMENT: Look carefully at your inequalities. 4 > x > 6 means that 4 > x AND x > 6. There is no value of x that is both less than 4 and greater than 6. There are of course a lot of x values that are greater than 4 and less than 6. The inequality should have been written 4 < x < 6, which is equivalent to the interval (4, 6). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is >1000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -200x^2 + 2000x - 3800 > 1000 -> -200x^2 + 2000x - 4800 >0 -> divide each side by 200 -> -1x^2 + 10x - 24 = 0 -> using the quadratic formula, I get x = 6,4. The interval is (4, 6). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a You have to solve the inequality 1000<-200x^2+2000x-3800, which rearranges to 0<-200x^2+2000x-4800. This could be solved using the quadratic formula, but it's easier if we simplify it first, and it turns out that it's pretty easy to factor when we do: -200x^2+2000x-4800 = 0 divided on both sides by -200 gives x^2 - 10 x + 24 = 0, which factors into (x-6)(x-4)=0 and has solutions x=4 and x=6. So -200x^2+2000x-4800 changes signs at x = 4 and x = 6, and only at these values (gotta go thru 0 to change signs). At x = 0 the expression is negative. Therefore from x = -infinity to 4 the expression is negative, from x=4 to x=6 it is positive and from x=6 to infinity it's negative. Thus your answer would be the interval (4,6). COMMON ERROR: x = 4 and x = 6. INSTRUCTOR COMMENT: You need to use these values, which are the zeros of the function, to split the domain of the function into the intervals 4 < x < 6, (-infinity, 4) and (6, infinity). Each interval is tested to see whether the inequality holds over that interval. COMMON ERROR: Solution 4 > x > 6. INSTRUCTOR COMMENT: Look carefully at your inequalities. 4 > x > 6 means that 4 > x AND x > 6. There is no value of x that is both less than 4 and greater than 6. There are of course a lot of x values that are greater than 4 and less than 6. The inequality should have been written 4 < x < 6, which is equivalent to the interval (4, 6). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: 3 " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!