#$&* course Mth 158 10/27/11 at 3:30 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
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Given Solution: * * 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): the equation that I was given in the question was 1 - 1 / (x+1)^2. The equation that was worked in the Given Solution was 1 - 1 / (x+2)^2. I think something is not right about the question that I was given. I see that I gave to much information. ------------------------------------------------ Self-critique Rating:
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.1.54 (was 3.1.40). G(x) = (x+4)/(x^3-4x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: G(x) = (x + 4) / (x^3 - 4x) is a function. The domain is {x| x does not = 0} confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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}. ** STUDENT QUESTION: Well, I went about it the long way and plugged in the numbers until I found what would make the denominator 0. I still have trouble factoring. I don’t see how factoring x out of the denominator helped to come up with the solution. Wouldn’t you still have to guess at x? INSTRUCTOR RESPONSE Once you've factored out x, the other factor is x^2 - 4. This is the difference of two squares and factors accordingly as (x-2) (x+2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I may have been supposed to know what to do with the equation. But I was given no instruction in the question of what to do. I see that I need to look at the equation in the text book also. ------------------------------------------------ Self-critique Rating:
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Given Solution: 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 STUDENT COMMENT I made the mistake by including the x intercepts, when the vertical line test is only the y intercepts and it has only 2 points. INSTRUCTOR RESPONSE You don't necessarily have to use the y intercepts. Any x value between -2 and 2, not including -2 or 2, defines a vertical line which intecepts the circle at two points. That is, for -2 < x < 2, the vertical line through (x, 0) passes through the circle at two points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph does depict a line. I found this by using the vertical line test. The domain = {x| x >= -3} The range = {y| y >= 0} confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Every vertical line intersects the graph at exacty one point so the graph depicts a function. The function extends to the right and to the left without breaks so the domain consists of all real numbers. The range consists of all possible y values. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was looking at the rang graph. ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.1.84 / 82 (was 3.1.70). f(x) =(2x - B) / (3x + 4). If f(0) = 2 then what is the value of B? If f(2)=1/2 what is value of B? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2(0) - B) / (3(0) + 4) = 2 -B / 4 = 2 -B = 2(4) B = -8 (2(2) - B) / (3(2) + 4) = 1/2 (4 - B) / (10) = ½ 4 - B = ½(10) 4 - B = 5 - B = 1 B = -1 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 then we have 1/2 = ((2*2)-B) / ((3*2)+4) 1/2 = (4-B) / 10 5 = 4-B 1=-B B=-1 ** STUDENT COMMENT I tried to write it on paper and follow how it was solved, but it is still a little confusing to me. Especially the second part where f(2) = ½ I don’t see where you plug in the ½ or where you plug in the 2? INSTRUCTOR RESPONSE f(x) = (2x - B) / (3x + 4), so f(2) = (2 * 2 - B) / (3 * 2 + 4) = (4 - B) / 10. Since f(2) = (4 - B) / 10, f(2) = 1/2 means (4 - B) / 10 = 1/2. We solve this equation for B, as in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.1.94 / 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: H(1) = 20 - 13(1)^2 20 - 13 = 7 The height of the rock would be 7 ft. H(1.1) = 20 - 13(1.1)^2 20 - 13(1.21) 20 - 15.73 = 4.27 The height of the rock would be 4.27 ft. H(1.2) = 20 - 13(1.2)^2 20 - 13(1.44) 20 - 18.72 = 1.28 15 = 20 - 13(x)^2 -5 = -13(x)^2 0.38 = x^2 x = 0.62 The rock is 15m off the ground after 0.62 sec. 10 = 20 - 13(x)^2 -10 = -13(x)^2 0.77 = x^2 x = 0.88 The rock is 10m off the ground after 0.88 sec. 5 = 20 - 13(x)^2 -15 = -13(x)^2 1.15 = x^2 x = 1.07 The rock is 5m off the ground after 1.07 sec. 0 = 20 - 13(x)^2 -20 = -13(x)^2 1.54 = x^2 x = 1.24 The rock would strike the ground after 1.24 sec. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!