course Mth 163 Any advice on getting a better understanding on creating formulas? I can solve given formulas but I struggled on the questions asked to create one. dϺe~`ܘǍgxassignment #004 004. `query 4 Precalculus I 08-05-2008
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21:38:21 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> where f(x)=x^3 f(-2)=-2^3=-8 F(-a)=-a^3 f(x-4)=(x-4)^3 =(x-4)(x-4)(x-4) =[x^2-4x-4x+16](x-4) =(x^2-8x+16)(x-4) =x^2(x-4)+-8x(x-4)+16(x-4) =x^3-4x^2+-8x^2+32x+16x-64 =x^3-12x^2+48x-64 F(x)-4=x^3-4 confidence assessment: 2
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21:40:46 ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. **
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RESPONSE --> self critique assessment: 3
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21:42:50 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> f(x)=2^x f(2)=2^2=4 f(-a)=2^-a f(x+3)=2^(x=3) f(x)+3= 2^x+3 confidence assessment: 3
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21:43:16 ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **
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RESPONSE --> self critique assessment: 3
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21:44:41 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> By using meaningful names one is able to keep straight the information being used to solve the problem. You do not have to keep straight in your head what each letter means the information is right there. confidence assessment: 3
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21:44:58 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.**
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RESPONSE --> self critique assessment: 2
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21:48:25 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> Value(0)=$1000(1.07)^0 =$1000(1) =$1000 Value(2)=$1000(1.07)^2 =$1000(1.1449) =$1144.9 value(t+3)=$1000(1.07)^(t+3) value(t+3)/value(t)= $1000 (1.07)^(t+3)/t confidence assessment: 3
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21:50:19 ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. **
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RESPONSE --> self critique assessment: 3
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22:03:06 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> illumination (distance)=50/distance^2 =[50/distance^2]/[50/(2*distance)^2 =distance^2/(2*distance)^2 confidence assessment: 2
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22:06:03 ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. **
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RESPONSE --> I did not simplify correctly but I understand now. self critique assessment: 2
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22:28:38 query #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph.
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RESPONSE --> I am confused at this point. What formula do I substitute x in? confidence assessment: 0
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22:35:11 ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. **
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RESPONSE --> I am feeling very non intelligent right now. using the information from the student answer I realized I was making this very difficult. I used graph paper and each square as 5. I plotted the three points and created a slight curve between the three points. self critique assessment: 2
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22:36:29 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> according to my graph I estimate the value of x to be 2.5 confidence assessment: 2
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22:37:15 **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3**
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RESPONSE --> need to work on my estimation skills self critique assessment: 2
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22:37:56 what is your estimate of the value f(7)?
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RESPONSE --> 35 confidence assessment: 2
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22:38:16 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. **
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RESPONSE --> self critique assessment: 2
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22:40:10 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> 5 confidence assessment: 1
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22:40:31 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> self critique assessment: 3
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22:44:07 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
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RESPONSE --> if x is estimated at 2.5 for f(x)=70 and 8 for f(x)=30 then the difference is 5.5 confidence assessment: 3
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22:44:37 ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. **
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RESPONSE --> self critique assessment: 3
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22:49:01 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> y=T(3) confidence assessment: 1
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22:50:44 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 **
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RESPONSE --> got it self critique assessment: 2
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23:00:57 What equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30?
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RESPONSE --> y=temperature=T(t) temperature given as 150 150(t) 80(t)-30(t)=time confidence assessment: 0
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23:03:44 ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] **
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RESPONSE --> I still am struggling with this one. self critique assessment: 2
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23:17:42 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
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RESPONSE --> I can estimate using my graph, but I am unsure of the correct formula to create. confidence assessment: 0
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23:24:12 ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' **
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RESPONSE --> okay I was actually trying to solve for an answer instead of just writing the formula that I would use to find the answer. How to I stop myself from making this so difficult? self critique assessment: 1
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23:30:20 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> if depth = f(x) then for I would slove for the depth at each time value depth1= f(23) depth2=f(34) then to find the difference I would subtract Depth 2 - depth 1= depth change confidence assessment: 1
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23:31:46 ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. **
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RESPONSE --> ugggh! just when I think I get it I don't. I can solve given formulas but can't seem to create one from given information. self critique assessment: 1
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23:32:10 On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds?
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RESPONSE --> confidence assessment: 0
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23:33:16 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. **
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RESPONSE --> ? I am lost self critique assessment: 1
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23:34:10 On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds?
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RESPONSE --> confidence assessment: 0
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23:34:46 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. **
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RESPONSE --> was this the same as the question before? self critique assessment: 1
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23:46:45 query. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph.
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RESPONSE --> using graph paper I used each square as a value of 5. I plotted the given points and then as directed I drew a curve that did not touch but came close to the given points. confidence assessment: 1
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23:47:11 ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). **
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RESPONSE --> self critique assessment: 2
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00:07:21 What 3 data point did you use as a basis for your model?
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RESPONSE --> I used the following 3 points (25,70) (40,50) (75,35) confidence assessment: 3
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00:07:31 ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.**
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RESPONSE --> self critique assessment: 2
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00:08:46 What was your function model?
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RESPONSE --> y=.1038x^2+ -11.08x+282.125 confidence assessment: 3
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00:09:00 ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089x^2 - 1.4992x + 98.8544. **
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RESPONSE --> self critique assessment: 3
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00:33:22 What is the average deviation for your model?
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RESPONSE --> average deviation was 63.744 confidence assessment: 3
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00:34:01 ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.**
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RESPONSE --> self critique assessment: 2
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00:34:44 How close is your model to the curve you sketched earlier?
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RESPONSE --> not very close at all confidence assessment: 2
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00:35:16 ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.**
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RESPONSE --> self critique assessment: 2
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00:36:00 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I was surprised at the difficulty I had in the questions regarding creatign formulas. self critique assessment: 2
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00:36:19 ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation **
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RESPONSE --> self critique assessment:
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