course MTH 163
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21:18:26 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> F(-2)=-2^3 F(-A)=-A^3 F(X-4)=X-4)^3 F(X)-4=X^3-4
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21:18:31 ** 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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21:20:19 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> F(2)=2^2 F(-A)=2^-A F(X+3)=2^(X+3) F(X)+3=2^X+3
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21:20:27 ** 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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21:20:30 13:06:29
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21:22:21 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> DON'T GET AS CONFUSED AND YOU ACTUALLY GIVE A VARIABLE A FUNCTION OF HOW TO SOLVE AND WHAT TO SOLVE FOR.
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21:22:24 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example...
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21:22:33 '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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21:26:35 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> v(0)=1000(1.07)^0 V(2)=1000(1.07)^2 V(T+3)=1000(1.07)^T+3
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21:26:39 ** 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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21:26:42 13:14:30
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21:27:09 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> I DID NOT GET ANYTHING FOR THIS ONE
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21:27:13 ** 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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21:30:10 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 --> IF Y=F(X) THEN F(2)=80 WOULD GIVE POINTS (80,2) F(5)=40 WOULD GIVE POINTS (40,5) F(10)=25 WOULD GIVE POINTS (25,10)
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21:30:19 ** 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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21:31:09 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> 3.5
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21:31:14 **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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21:31:32 what is your estimate of the value f(7)?
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RESPONSE --> 35
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21:31:36 ** 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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21:31:50 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> 10
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21:31:56 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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21:31:58 13:29:11
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21:32:46 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 --> 5
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21:32:52 ** 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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21:33:25 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> Y(T)
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21:33:37 ** 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 --> THE QUESTION WAS NOT COMPLETE
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21:33:51 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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21:34:03 ** 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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21:34:27 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 --> f(X)=34-47
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21:34:33 ** 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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21:34:46 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> 10
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21:34:49 ** 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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21:34:57 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 --> 25
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21:34:59 ** 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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21:35:06 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 --> 50.6
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21:35:09 ** 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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21:36:16 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 --> TOOK EACH POINT PUT THEM INTO A Y=AX2+B+C FUNCTION OR QUADRATIC FUNCTION AND USED THE ELIMINATION PROCESS. i ALSO FOUND THE VERTEX AND MY FUNDAMENTAL POINTS
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21:36:19 ** 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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21:36:29 What 3 data point did you use as a basis for your model?
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21:36:33 ** 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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21:36:35 14:17:00
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21:36:47 What was your function model?
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RESPONSE --> UPWARD PARABOLA
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21:36:49 ** 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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21:36:52 14:19:45
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21:36:59 What is the average deviation for your model?
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RESPONSE --> DID NOT GET ONE
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21:37:02 14:23:26
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21:37:04 ** 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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21:37:31 How close is your model to the curve you sketched earlier?
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RESPONSE --> NOT VERY CLOSE AT ALL I THINK THAT I NEED TO GO BACK AND FIND MY ERROR
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21:37:33 14:25:12
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21:37:35 ** 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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21:37:38 14:25:13
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21:37:41 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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21:37:43 14:27:22
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21:37:45 ** 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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