course MTH 158 ?????S???K????assignment #022
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16:08:53 3.1.20 (was 3.1.6) (-2,5),(-1,3),(3,7),(4,12)}Is the given relation a function? Why or why not? If so what are its domain and range?
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RESPONSE --> Yes, it is a function, because none of the domain repeats itself. domain = {-2, -1, 3, 4 } range = {5, 3, 7, 12 } confidence assessment: 3
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16:09:08 This relation is a function because every first element is paired with just one second element--there are no distinct ordered pairs with the same first element. the domain is ( -2,-2,3,4) the range is ( 5,3,7,12) Another way of saying that this is a function is that every element of the domain appears only once in the relation.
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RESPONSE --> ok self critique assessment: 3
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16:40:16 3.1.34 (was 3.1.20) f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h) for 1 - 1 / (x+1)^2What are you expressions for f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h)?query 3.1.30. y = (3x-1)/(x+2)
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RESPONSE --> (a) f(0) = f = 1 - 1/(0+1)^2, f = 1 - 1/1, f = 0 (b) f(1) = f = 1 - 1/(1+1)^2, f = 1 - 1/4, f = 3/4 (c) f(-1) = f = 1 - 1/(-1+1)^2, f = 1 (d) f(-x) = f = 1 - 1/(-x+1)^2, f = 1 - 1/x^2 -2x +1 (e) -f(x) = f = -1 ( 1 + 1/(x+1)^2 ), f = -1 + 1/(x+1)^2 (f) f(x + 1) = f = 1 - 1/{(x+1) +1}^2, f = 1 - 1/x^2 + 4x + 4 (g) f(2x) = f = 1 - 1/(2x + 1)^2, f = 1 - 1/4x^2 + 4x +2 (h) f(x + h) = f = 1 - 1/(x + h + 1)^2, f = 1 - 1/x^2 + h^2 + 2xh +2x + 2h + 1 confidence assessment: 2
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16:43:37 STUDENT SOLUTION WITH INSTRUCTOR COMMENTS: f(x) = 1- 1/(x+2)^2 f(0) = 1- 1/ (0+2)^2 f(0) = 1-1/4 f(0) = 3/4 f(1) = 1- 1/ (3)^2 f(1) = 1- 1/9 f(1) = 8/9 f(-1) = 1- 1/(-1+2)^2 f(-1)= 1-1 f(-1)= 0 f(-x)= 1- 1/(-x+2)^2 f(-x)= 1 -1/ (x^2-4x+4) -f(x) = -(1- 1/(x+2)^2) -f(x)= -(1 - 1/ (x^2+4x+4)) -f(x) = (1/(x^2 + 4x + 4)) - 1 ** Your answer is right but you can leave it in factored form: f(-x) = -(1 - 1/(x+2)^2) = -1 + 1 / (x+2)^2. ** f(x+1) = 1- 1/((x+1) + 2)^2. This can be expanded as follows, but the expansion is not necessary at this point: = 1- 1/ ((x+1)^2 +2(x+1) + 2(x+1)+4) = 1- 1/ ((x+1)^2 +8x+8) = 1- 1/ (x^2+2x+1+8x+8) = 1- 1/(x^2 + 10x +9) ** Good algebra, and correct, but again no need to expand the square, though it is perfectly OK to do so. ** f(2x)= 1-1/(2x+2)^2 = 1- 1/(4x^2+8x+4) ** same comment ** f(x+h)= 1- 1/((x+h)+2)^2 = 1- 1/((x+h)^2 + 4(x+h) + 4) = 1- 1/ (x^2 + 2xh + h^2+4x+4h+4) ** same comment **
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RESPONSE --> I don't understand, I thought the problem was f(x) = 1 - 1/(x + 1)... that's what was in this query. self critique assessment: 1
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16:51:33 3.1.44 (was 3.1.30) Is y = (3x-1)/(x+2) the equation of a function?
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RESPONSE --> Yes confidence assessment: 2
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16:51:44 ** This is a function. Any value of x will give you one single value of y, and all real numbers x except -2 are in the domain. So for all x in the domain this is a function. **
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RESPONSE --> ok self critique assessment: 3
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18:00:10 3.1.54 (was 3.1.40). G(x) = (x+4)/(x^3-4x)
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RESPONSE --> I guess the domain is all real numbers, since it doesn't really matter if x = +/- 1 or 2... confidence assessment: 1
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18:01:14 ** Starting with g(x) = (x+4) / (x^3-4x) we factor x out of the denominator to get g(x)= (x+4) / (x (x^2-4)) then we factor x^2 - 4 to get g(x) = (x+4) / (x(x-2)(x+2)). The denominator is zero when x = 0, 2 or -2. The domain is therefore all real numbers such that x does not equal {0,2,-2}. **
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RESPONSE --> How did you just factor x out of the denominator and not anything else? self critique assessment: 1
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18:28:46 3.2.12 (was 3.1.50). Pos incr exp fnDoes the given graph depict a function? Explain how you determined whether or not the graph depicts a function.If the graph does depict a function then what are the domain and range of this function?
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RESPONSE --> Yes, it is a function because it passed the vertical line test. Domain = all real numbers Range = appears to be all real numbers above and including 0 confidence assessment: 2
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18:30:43 using the vertical line test we determine this is a function, since any given vertical line intersects the graph in exactly one point. The function extends all the way to the right and to the left, and there are no breaks, so the domain consists of all real numbers. The range consists of all possible y values. The function takes all y values greater than zero so the range is {y | y>0}, expressed in interval notation as (0, infinity). The y intercept is (0,1); there is no x intercept but the negative x axis is an asymptote. This graph has no symmetery.
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RESPONSE --> Alright, I see that the range does not include 0 now. I didn't know that we were to answer the questions on intercept and symmetery since the query did not ask them. self critique assessment: 2
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18:32:02 3.2.16 (was 3.1.54) Circle rad 2 about origin.
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RESPONSE --> No, it is not a function because it does not pass the vertical line test. confidence assessment: 3
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18:32:10 Using the vertical line test we see that every vertical line lying between x = -2 and x = 2, not inclusive of x = -2 and x =2, intersects the graph in two points. So this is not a function
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RESPONSE --> ok self critique assessment: 3
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18:36:19 3.2.22 (was 3.1.60). Downward hyperbola vertex (1,2 5).Does the given graph depict a function? Explain how you determined whether or not the graph depicts a function.If the graph does depict a function then what are the domain and range of this function?
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RESPONSE --> Yes, it is a function because it passes the vertical-line test. Domain is all real numbers Range is (-infinity, 5] Y-intercept is at 4, X intercepts are at -1 and 2 There is no symmetry confidence assessment: 2
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18:37:04 Every vertical line intersects the graph at exacty one point so the graph depicts a function. The function extends to the right and to the left without breaks so the domain consists of all real numbers. The range consists of all possible y values.
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RESPONSE --> Was my answer concerning the range still correct, though?
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19:07:55 3.1.82 (was 3.1.70). f(x) =(2x - B) / (3x + 4). If f(0) = 2 then what is the value of B?
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RESPONSE --> Substitute f(x) with f(0).... (2(0) - B) / (3(0) + 4 -B / 4 B would have to equal -8 in order for f(0) to equal 2 confidence assessment: 2
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19:10:06 If f(0) = 2 then we have 2 = (2 * 0 - B) / (3 * 0 + 4) = -B / 4, so that B = -4 * 2 = -8. if f(2)=1/2 what is value of B? If f(2) = 1/2 then we have 1/2 = ((2*2)-B) / ((3*2)+4) 1/2 = (4-B) / 10 5 = 4-B 1=-B B=-1 **
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RESPONSE --> I didn't know it was asking both If f(0) = 2 and if f(2)=1/2. I had solved for if f(2)=1/2 first, then saw that the query asked for, so I changed my answer accordingly. I did, however, in my original problem (f(2)=1/2) come up with the answer B = -1 self critique assessment: 2
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19:23:49 3.1.90 (was 3.1.80). H(x) = 20 - 13 x^2 (falling rock on Jupiter)What are the heights of the rock at 1, 1.1, 1.2 seconds?When is the rock at each altitude: 15 m, 10 m, 5 m.When does the rock strike the ground?
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RESPONSE --> (a) the height of the rock when x = 1 is 7 meters when x = 1.1 the height is 4.27 meters when x = 1.2 the height is 1.28 meters (b) the time that the rock is at the altitude 15 m is approximately .62 seconds when at altitude 10 m, the rock has been falling for approximately .88 seconds when at altitude 5 m, the rock has been falling for approximately 1.07 seconds (c) the rock will strike the ground at approximately 1.24 seconds confidence assessment: 2
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19:24:31 GOOD STUDENT SOLUTION: The height at t = 1 is H(1) = 20-13 H(1) = 7m The height at t = 1.1 is H(1.1)= 20-13(1.1)^2 = 20-13(1.21) = 20-15.73 H(1.1)= 4.27m. The height at t = 1.2 is H(1.2)= 20 - 13*(1.2)^2 = 20- 13 *(1.44) = 20-18.72 H(1.2) = 1.28m. The rock is at altitude 15 m when H(x) = 15: 15=20-13x^2 -5=-13x^2 5/13= x^2 x= +- .62 .62sec. The rock is at altitude 10 m when H(x) = 10: 10=20-13x^2 -10=-13x^2 10/13 = x^2 x= +-.88 .88sec. The rock is at 5 meter heigh when H(x) = 5: 5=20-13x^2 -15 = -13x^2 15/13=x^2 x= +- 1.07 1.07sec. To find when the rock strikes the ground let y = 0 and we get 0= 20-13x^2. Adding -20 to both sides we have -20=-13x^2. Multiplying both sides by -1/13 we get 20/13=x^2. Taking the square root of both sides we obtain the approximate value of x: x=+-1.24 We conclude that x = 1.24sec. when the rock strikes the ground **
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RESPONSE --> ok self critique assessment: 3
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19:27:23 3.1.90. Sales S vs. advertising expenditures A. 335 339 337 343 341 350 351 vs. 20, 22, 22.5, 24, 24, 27, 28.3 in thousands of dollars.Does the given table describe a function? Why or why not?What two points on your straight line did you pick and what is the resulting equation?What is the meaning of the slope of this line?Give your equation as a function and give the domain of the function.What is the predicted sales if the expenditures is $25,000?
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RESPONSE --> This problem is nowhere in my book that I can find, and since the way it is described on the query is not adequate, I don't see how I can solve it at this point... I need to see how it is set up in the book... confidence assessment: 0
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19:27:51 The table does not describe a function because ordered pairs that have the same first element and a different second element. Specifically 24,000 is paired with both 343,000 and 341,000. I picked the points (20000,335000) (27000,350000). INSTRUCTOR COMMENT: These are data points, not points on the best-fit straight line graph. You should have sketched your line then picked two points on the line, and the line will almost never pass through data points. STUDENT SOLUTION CONTINUED: The slope between these points is slope = (350000-335000)/(27000-20000) = 15000/7000 = 15/7 = 2.143 approximately. Our equation, using this slope and the first chosen point, is therefore y-335000=2.143(x-20000) y- 335 = 2.143x-42857.143 y= 2.143x+29214.857 equation of the line Expressed as a function we have f(x) = 2.143x+292142.857. Predicted sales for expenditure $25000 will be f(25000) = 2.143(25000) + 292142.857 = 53575 + 292142.857 = 345717.857 We therefore have predicted sales f(25000)= $345,717.86 INSTRUCTOR COMMENT: Excellent solution, except for the fact that you used data points and not points on the best-fit line.
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RESPONSE --> Again, I couldn't see the problem, so I didn't know what to do. self critique assessment: 0
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