course MTH 158 7/26/09 1:00 A.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.
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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): Self-critique Rating: ********************************************* Question: * 3.1.36 / 44 (was 3.1.30) Is y = (3x-1)/(x+2) the equation of a function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is a function Domain {x|x not= -2} confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: (x^3 - 4x) = x(x^2 -4)=x(x-2)(x+2) The domain of g is{x|x is not= -2,2,0} confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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}. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * 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? 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I had no idea on this problem.....Pos incr exp?
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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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: This is a function Not sure how to find the domain and range confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 just assuming a hyperbola was a Parabola and its mirrored image opening upward and downward.??
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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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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(x) = 20-13x^2 = 20 - 13 = 7 m H(1.1) = 20 - 13(1.1)^2 = 20 - 15.73 = 4.27 m H(1.2)= 20 - 13(1.2)^2 = 20 - 18.72 = 1.28m 15m = 20-13x^2 15-20 = -13x^2 -13x^2 = -5 divide by -13 on both sides x^2= 5/13 square root both sides x=+/- .62 .62 seconds 20-13x^2 = 10 m -13x^2 = -10 divide by - 13 both sides x^2 = 10/13 square root both sides x = +/- .88 seconds 20 - 13x^2 = 5 -13x^2 = -15 divide both sides by -13 x^2 = 15/13 square root both sides x = +/- 1.07 1.07 seconds 20-13x^2 = 0 -13x^2 = -20 divide both sides by -13 x^2 = 20/13 square root both sides x= +/- 1.24 seconds x= 1.24 seconds confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: "