#$&* course Mth 158 3/17/12 Around 12:30 PM 022. `* 22
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENTS: f(x) = 1- 1/(x+2)^2 f(0) = 1- 1/ (0+2)^2 f(0) = 1-1/4 f(0) = 3/4 f(1) = 1- 1/ (3)^2 f(1) = 1- 1/9 f(1) = 8/9 f(-1) = 1- 1/(-1+2)^2 f(-1)= 1-1 f(-1)= 0 f(-x)= 1- 1/(-x+2)^2 f(-x)= 1 -1/ (x^2-4x+4) -f(x) = -(1- 1/(x+2)^2) -f(x)= -(1 - 1/ (x^2+4x+4)) -f(x) = -(1/(x^2 + 4x + 4)) - 1 ** Your answer is right but you can leave it in factored form: f(-x) = -(1 - 1/(x+2)^2) = -1 + 1 / (x+2)^2. ** f(x+1) = 1- 1/((x+1) + 2)^2. This can be expanded as follows, but the expansion is not necessary at this point: = 1- 1/ ((x+1)^2 +2(x+1) + 2(x+1)+4) = 1- 1/ ((x+1)^2 +8x+8) = 1- 1/ (x^2+2x+1+8x+8) = 1- 1/(x^2 + 10x +9) ** Good algebra, and correct, but again no need to expand the square, though it is perfectly OK to do so. ** f(2x)= 1-1/(2x+2)^2 = 1- 1/(4x^2+8x+4) ** same comment ** f(x+h)= 1- 1/((x+h)+2)^2 = 1- 1/((x+h)^2 + 4(x+h) + 4) = 1- 1/ (x^2 + 2xh + h^2+4x+4h+4) ** same comment ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Is my expression for -f(x) correct???? I can't tell.
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Given Solution: ** This is a function. Any value of x will give you one single value of y, and all real numbers x except -2 are in the domain. So for all x in the domain this is a function. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 3.1.54 (was 3.1.40). G(x) = (x+4)/(x^3-4x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Domain = {x | x ≠ 0, x ≠ 2} confidence rating #$&*: 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}. ** 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 didn't have x does not equal -2 in the domain. I see what I did wrong. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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): Yes it is a function because it passes the vertical-line test. Domain = (-infinity, infinity) Range = (-infinity, infinity) y-int = (0 , 1) There is no x-int on the part of the graph I can see. No symmetry. Self-critique: I see why the range is {y | y>0}. The graph looked like it was still angling down very very slightly to me so I figured it would pass through all the entire range of y. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 3.2.16 (was 3.1.54) Circle rad 2 about origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph is not a function because vertical lines can pass through it at more than one point. x-int = (2 , 0) and (-2 , 0) y-int = (0 , -2) and (0 , 2) It has symmetry with respect to the x-axis, y-axis, and origin. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Using the vertical line test we see that every vertical line lying between x = -2 and x = 2, not inclusive of x = -2 and x =2, intersects the graph in two points. So this is not a function STUDENT COMMENT I made the mistake by including the x intercepts, when the vertical line test is only the y intercepts and it has only 2 points. INSTRUCTOR RESPONSE You don't necessarily have to use the y intercepts. Any x value between -2 and 2, not including -2 or 2, defines a vertical line which intecepts the circle at two points. That is, for -2 < x < 2, the vertical line through (x, 0) passes through the circle at two points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 3.2.22 (was 3.1.60). Downward hyperbola vertex (1,2 5).Does the given graph depict a function? Explain how you determined whether or not the graph depicts a function.If the graph does depict a function then what are the domain and range of this function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This graph is a function because vertical lines can only pass through the graph at one point. Domain = (-infinity, infinity) Range = [5 , -infinity) y-int = (0 , 4) x-int = (-1 , 0) and (2 , 0) No symmetry. confidence rating #$&*: 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 don't see how the range consists of all possible y values if the vertex of the curve is at (1/2 , 5) and the opening of the curve is facing downwards. Like an upside down V. Wouldn't the range of y values stop at 5 since the graph doesn't go any higher than that????? ------------------------------------------------ Self-critique 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), sof(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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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 H(1) = 20 - 13(1)^2 = 20 - 13 = 7 H(1.1) = 20 - 13(1.1)^2 = 20 - 15.73 = 4.27 H(1.2) = 20 - 13(1.2)^2 = 20 - 18.72 = 1.28 H(x) = 20 - 13x^2 = 15 = -13x^2 = -5 = x^2 = 5/13 = x = sqrt(5/13) H(x) = 20 - 13x^2 = 10 = -13x^2 = -10 = x^2 = 10/13 = x = sqrt(10/13) H(x) = 20 - 13x^2 = 5 = -13x^2 = -15 = x^2 = 15/13 = x = sqrt(15/13) H(x) = 20 - 13x^2 = 0 = -13x^2 = -20 = x^2 = 20/13 = x = sqrt(20/13) 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): I need to actually complete the equation and get the decimal numbers of all the measurements and time. ------------------------------------------------ Self-critique Rating: 3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!