#$&* course Mth 271 9/26 1030pI have literally worked on this every day since Sunday. I don't understand a lot of it, but did what I could so I would not get further behind. The notes I need for this must be on the CDs because I could not find much elsewhere.
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Given Solution: `a You have to find the average rate of change between clock times t and t + `dt: ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt = [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt = [ 2 a t `dt + a `dt^2 + b `dt ] / `dt = 2 a t + b t + a `dt. Now if `dt shrinks to a very small value the ave rate of change approaches y ' = 2 a t + b. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: We can determine only the change in depth and not the actual depth because the rate-of-change function does not include a constant. The depth function includes a constant so the actual depth can be determined. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Given the rate function y' we can find an approximate average rate over a given time interval by averaging initial and final rates. Unless the rate function is linear this estimate will not be equal to the average rate (there are rare exceptions for some functions over specific intervals, but even for these functions the statement holds over almost all intervals). Multiplying the average rate by the time interval we obtain the change in depth, but unless we know the original depth we have nothing to which to add the change in depth. So if all we know is the rate function, have no way to find the actual depth at any clock time. ** STUDENT COMMENT: Not sure I understand this I read you solution and it makes more sense. INSTRUCTOR RESPONSE: Here's an analogy: If I tell you that I drove down the interstate at 50 mph for 1/2 hour, then at 70 mph for 1 hour, could you tell me at what milepost I ended? You couldn't because I didn't tell you where I started. But you could tell me how far I traveled. In other words, if I give you the rate information and time intervals, you can tell me how far I went. But if that's the only information I give you, you can't tell me where I started or ended. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think what I meant is the same thing because the constant would tell where you started. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q In terms of the depth model explain the processes of differentiation and integration. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Differentiation is the process of finding the rate of change function from the depth function. Integration is the opposite of differentiation: using the rate of change function to find the depth function. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I think I really need some practice problems to be sure I understand it. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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 $300? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Growth rate= .10 Growth factor= 1.10 Year 1 200*1.10=220 Year 2 220*1.10=242 Year 3 242*1.10= 266.20 Year 4 266.20*1.10 = 292.80 Year 5 292.80*1.10 = 322.102 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I didn’t see in the original question that I needed to determine when the amount reaches $600. ------------------------------------------------ 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: Assuming that the principal referred to is $200 from the previous problem, P(t)=200(1.10)^20= 1345.50 1345.50/2 = 672.75 200(1.10)^12= 627.69 200(1.10)^13 = 690.45 so, time t that the principal is equal to ½ the t=20 value occurs somewhere near the end of year 12. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $627.69 and at t = 13 you get 690.45). At 12.75 we get 674.20 so it would probably be about 12.72 This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):n/a ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qquery #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:The curve increases at a rate that increases as the rate increases. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not see in the original question that I was supposed to relate information about the doubling rate. ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery #11. equation for doubling time YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P(t0 + doubling time)= 2P(t0) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):I wrote down what I found in the notes but I see that that is not right. I will correct my notes.
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Given Solution: `adividing 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): I really don’t understand why you couldn’t just use trial and error to find the doubling time to begin with. It seems like a lot of work to go through just to use trial and error anyway. I really need more practice with this.
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Given Solution: `aIn this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis. The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t really understand the given solution. How do you know how far over to go?
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Given Solution: `a Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or Q(t)=550(.89)^t ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I had a basic concept of what to do, but I really needed an example and couldn’tfind one. Self-critique Rating: 2
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Given Solution: `a 3:00 p.m. is 5 hours after the initial time so at that time there will be Q(5) = 550 mg * .89^5 = 307.123mg in the blood ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):n/a ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qDescribe your graph and explain how it was used to estimate half-life. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Vertical axis: antibiotic in mg. Horizontal axis: time in hours. Subtract 11% of the remaining amount each hour. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I am still not sure about estimating using the graph. It makes more sense to me to just calculate it.
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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):I’m not sure why doubling time was used. It seems like half the doubling time would just be time *1 and not half-life. I tried to find a formula for half-life and that is what I used.
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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 substituted 12 numbers just to get the values I used. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `qexplain why the negative t axis is a horizontal asymptote for this function. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
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Given Solution: `a 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):
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Given Solution: `a 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I rounded incorrectly. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `qwhat is b for the function y = -13 ( e^(3.9 x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: b= 49.4 approximately confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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= .007(^2.03x) y= -13(^49.4x) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wrote b in the wrong place ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q0.4.44 (was 0.4.40 find all real zeros of x^2+5x+6 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x+2)(x+3) x=-2 and x=-3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:Zeros are the x-intercepts. This graph is a parabola that intercepts the x-axis at -2 and -3. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 0.4.50 (was 0.4.46 x^4-625=0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x=5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I tend to forget the +/- ------------------------------------------------ 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: x is greater than 24 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): I set the problem up right but then this is what I did: 4800= 200x(-1x+10) 200x = 4800 and -1x+10 = 4800 which would not work because a negative x would not work in this situation. ------------------------------------------------ Self-critique Rating:"