#$&* course Mth 158 7/24/12 at 10:51 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
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Given Solution: * * What did the horizontal line test tell you for this function? For every horizontal below the 'peak' of this graph the graph will intersect the horizontal line in two points. This function is not one-to-one on the domain depicted here. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.45 / 6.2.43 / 7th edition 5.2.28 looks like cubic thru origin, (1,1), (-1,-1), sketch inverse. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It passes through (0,0), (1,1) and (-1,-1). The inverse function will give you the same 3 points. It is increasing in the 1st quadrant and decreasing in the inverse, which is in the 3rd quadrant. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Describe your sketch of the inverse function. The graph of the function passes through (0, 0), (1,1), and (-1,-1). The inverse function will reverse these coordinates, which will give the same three points. Between x = -1 and x = 1 the graph of the original function is closer to the x axis than to the y axis, and is horizontal at the origin. The graph of the inverse function will therefore be closer to the y axis than to the x axis for y values between -1 and 1, and will be vertical at the origin. For x < 1 and for x > 1 the graph lies closer to the y axis than to the x axis. The graph of the inverse function will therefore lie closer to the x axis than to the y axis for y < 1 and for y > 1. In the first quadrant the function is increasing at an increasing rate. The inverse function will therefore be increasing at a decreasing rate in the first quadrant. In the third quadrant the function is increasing at a decreasing rate. The inverse function will therefore be increasing at an increasing rate in the third quadrant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.36 / 6.2.32 / 7th edition 5.2.32 f = 2x + 6 inv to g = 1 / 2 * x - 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You have f(g(x)) = 2g(x) + 6. You plug in g, which gives you: 2(1/2x - 3) + 6. This will equal: x - 6 + 6 = x. You have g(f(x)) = 1/2x - 3. You plug in f(x), which gives you: ½(2x + 6) - 3 = x + 3 - 3 = x. So since you have x for both answers, the functions are inverse. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Show that the functions f(x) and g(x) are indeed inverses. f(g(x)) = 2 g(x) + 6 = 2 ( 1 / 2 * x - 3) + 6 = x - 6 + 6 = x. g(f(x)) = 1 / 2 * f(x) - 3 = 1/2 ( 2 x + 6) - 3 = x + 3 - 3 = x. Since f(g(x)) = g(f(x)) = x, the two functions are inverse. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.54 / 6.2.52 / 7th edition 5.2.44. inv of x^3 + 1; domain range etc.. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You have f(x) = x^3 + 1. The domain and range are all real numbers. If we solve for y, we get (x-1)^1/3. This is an inverse function. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Give the inverse of the given function and the other requested information. The function is y = x^3 + 1. This function is defined for all real-number values of x and its range consists of all real numbers. If we switch the roles of x and y we get x = y^3 + 1. Solving for y we get y = (x - 1)^(1/3). This is the inverse function. We can take the 1/3 power of any real number, positive or negative, so the domain of the inverse function is all real numbers. Any real-number value of y can be obtained by using an appropriate value of x. So both the domain and range of the inverse function consist of all real numbers. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.66 / 6.2.64 / 7th edition 5.2.56. inv of f(x) = (3x+1)/(-x). Domain and using inv fn range of f. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have (3x +1)/(-x). If we substitute y for x, we have f(x) = (3y + 1)/(-y). To solve for y, we multiply by -y and get: -xy = 3y + 1. You subtract 3y and get -xy - 3y = 1. You factor y out to get y(-x - 3) = 1. We divide by (-x - 3) to get y = -1/x + 3. The domain and range will be all real numbers except for 3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * What is the domain of f? What is the inverse function? What does the inverse function tell you about the range of f? f(x) is defined for all x except x = 0, since division by 0 is not defined. If we switch x and y in the expression y = (3x + 1) / (-x) we get x = (3y + 1) / (-y). To solve for y we first multiply by -y, noting that this excludes y = 0 since multiplication of both sides by 0 would change the solution set. We get -y * x = -y * (3 y + 1) / (-y). The left-hand side simplifies to - x y and the right-hand side to 3 y + 1, giving us -x y = 3y + 1. Subtracting 3 y from both sides we get -x y - 3 y = 1. Factoring y out of the left-hand side (which becomes (-x - 3) * y; if you multiply this out you see that it is the same as -x y - 3 y) we get (-x - 3) y = 1, and dividing both sides by (-x - 3), which excludes x = -3, we get y = -1 / (x + 3). The domain of this function is the set of all real numbers except 3. Since the domain of the inverse function is the range of the original function, the range of the original function consists of all real numbers except 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 6.2.96 / 6.2.94 / 7th edition 5.2.74. T(L) = 2 pi sqrt ( L / 32.2). Find L(T). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have T(L) = 2pi sqrt(L/32.2). We are finding L. We will square both sides to get: T^2 = 4 pi^2 * L /32.2. We multiply both sides by 32.2 and divide by 4 pi^2. You have L= T^2 * 32.2/4pi^2. You get L(T) = 32.2T^2/4 pi^2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * We solve T = 2 pi sqrt( L / 32.2) for L. First squaring both sides we obtain T^2 = 4 pi^2 * L / 32.2. Multiplying both sides by 32.2 / ( 4 pi^2) we get L = T^2 * 32.2 / (4 pi^2). So our function L(T) is L(T) = 32.2 T^2 / (4 pi^2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!