#$&* course mth158 august 3 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * The inverse proportionality to the square root gives us y = k / sqrt(x). y = 4 when x = 9 gives us 4 = k / sqrt(9) or 4 = k / 3 so that k = 4 * 3 = 12. The equation is therefore y = 12 / sqrt(x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.5.12 / 2.7.12 (was 2.6.10). z directly with sum of cube of x and square of y; z=1 and x=2 and y=3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: z = k (x^3 + y^2). 1 = k ( 2^3 + 3^2) 17 k = 1 k = 1/17. Prop = z = (x^3 + y^2) / 17. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The proportionality is z = k (x^3 + y^2). If x = 2, y = 3 and z = 1 we have 1 = k ( 2^3 + 3^2) or 17 k = 1 so that k = 1/17. The proportionality is therefore z = (x^3 + y^2) / 17. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.5.20 / 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = 2 pi / sqrt(32) sqrt(L). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The equation is T = k sqrt(L), with k = 2 pi / sqrt(32). So we have T = 2 pi / sqrt(32) * sqrt(L). ** **** What equation relates period and length? **** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.5.42 / 2.7.42 (was 2.7.34 (was 2.6.30). Resistance dir with lgth inversely with sq of diam. 432 ft, 4 mm diam has res 1.24 ohms. **** What is the length of a wire with resistance 1.44 ohms and diameter 3 mm? Give the details of your solution. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R = k * L / D^2. 1.24 = k * 432 / 4 k = 1.24 * 4^2 / 432 = .046 R = .046 * L / D^2 let R = 1.44 and d = 3 L = R * D^2 / (.046) L = 1.44 * 3^2 / .046 = 280 approx. Wire is approximately 280 ft long confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * We have R = k * L / D^2. Substituting we obtain 1.24 = k * 432 / 4^2 so that k = 1.24 * 4^2 / 432 = .046 approx. Thus R = .046 * L / D^2. Now if R = 1.44 and d = 3 we find L as follows: First solve the equation for L to get L = R * D^2 / (.046). Then substitute to get L = 1.44 * 3^2 / .046 = 280 approx. The wire should be about 280 ft long. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 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? 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,-1,3,4) 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.1.46 / 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)?* 3.1.30. y = (3x-1)/(x+2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION WITH INSTRUCTOR COMMENTS: f(x) = 1- 1/(x+2)^2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.1.36 / 44 (was 3.1.30) Is y = (3x-1)/(x+2) the equation of a function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All real numbers except for -2 because there is an x there can be a y confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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) g(x)= (x+4) / (x (x^2-4)) g(x) = (x+4) / (x(x-2)(x+2)). The denominator is zero when x = 0, 2 or -2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.2.16 (was 3.1.54) Circle rad 2 about origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not a function confidence 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 denominator is zero when x = 0, 2 or -2. The range gives the possible y for all numbers. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The denominator is zero when x = 0, 2 or -2. The function extends to the right and to the left without breaks so the domain consists of all real numbers. The range consists of all possible y values. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 3.1.84 / 82 (was 3.1.70). f(x) =(2x - B) / (3x + 4). • If f(0) = 2 then what is the value of B? • If f(2)=1/2 what is value of B? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If f(0) = 2 then 2 = (2 * 0 - B) / (3 * 0 + 4) = -B / 4, so that B = -4 * 2 = -8 If f(2) = 1/2 then 1/2 = ((2*2)-B) / ((3*2)+4) 1/2 = (4-B) / 10 5 = 4-B 1=-B B=-1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 (if necessary): ------------------------------------------------ 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: If f(0) = 2 then 2 = (2 * 0 - B) / (3 * 0 + 4) = -B / 4, so that B = -4 * 2 = -8 If f(2) = 1/2 then 1/2 = ((2*2)-B) / ((3*2)+4) 1/2 = (4-B) / 10 5 = 4-B 1=-B B=-1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If f(0) = 2 then we have 2 = (2 * 0 - B) / (3 * 0 + 4) = -B / 4, so that B = -4 * 2 = -8. If f(2) = 1/2 then we have 1/2 = ((2*2)-B) / ((3*2)+4) 1/2 = (4-B) / 10 5 = 4-B 1=-B B=-1 ** STUDENT COMMENT I tried to write it on paper and follow how it was solved, but it is still a little confusing to me. Especially the second part where f(2) = ½ I don’t see where you plug in the ½ or where you plug in the 2? INSTRUCTOR RESPONSE f(x) = (2x - B) / (3x + 4), so f(2) = (2 * 2 - B) / (3 * 2 + 4) = (4 - B) / 10. Since f(2) = (4 - B) / 10, f(2) = 1/2 means (4 - B) / 10 = 1/2. We solve this equation for B, as in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: * 3.1.84 / 82 (was 3.1.70). f(x) =(2x - B) / (3x + 4). • If f(0) = 2 then what is the value of B? • If f(2)=1/2 what is value of B? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If f(0) = 2 then 2 = (2 * 0 - B) / (3 * 0 + 4) = -B / 4, so that B = -4 * 2 = -8 If f(2) = 1/2 then 1/2 = ((2*2)-B) / ((3*2)+4) 1/2 = (4-B) / 10 5 = 4-B 1=-B B=-1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!