course Mth 271 ????t??????w^+????assignment #009
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16:50:10 **** Query problem 1.4.06 diff quotient for x^2-x+1 **** What is the simplified form of the difference quotient for x^2-x+1?
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RESPONSE --> I dont think this problem matches my edition of the textbook. confidence assessment: 0
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16:54:33 The difference quotient would be [ f(x+`dx) - f(x) ] / `dx = [ (x+`dx)^2 - (x+`dx) + 1 - (x^2 - x + 1) ] / `dx. Expanding the squared term, etc., this is [ x^2 + 2 x `dx + `dx^2 - x - `dx + 1 - x^2 + x - 1 ] / `dx, which simplifies further to }[ 2 x `dx - `dx + `dx^2 ] / `dx, then dividing by the `dx we get 2 x - 1 + `dx. For x = 2 this simplifies to 2 * 2 - 1 + `dx = 3 + `dx. **
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RESPONSE --> I see that the question was asking me to find [ f(x+`dx) - f(x) ] / `dx for the given function f(x) = x^2 - x + 1. self critique assessment: 2
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17:16:59 1.4.40 (was 1.4.34 f+g, f*g, f/g, f(g), g(f) for f=x/(x+1) and g=x^3 the requested functions and the domain and range of each.
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RESPONSE --> a. f + g = x/(x+1) + x^3 = x/(x+1) + (x^4 + x^3)/(x+1) = (x^4 + x^3 + x)/(x+1) domain: all real numbers except -1 range: all real numbers b. f * g = (x/(x+1)) * x^3 = x^4 / (x+1) domain: all real numbers except -1 range: all real numbers c. f / g = (x/(x+1))/x^3 = (x/(x+1)) * 1/x^3 = x/((x^3)(x+1)) = x/((x^2)(x+1) domain: all real numbers except 0 and -1 range: all real numbers d. f(g) = x^3/(x^3 + 1) domain: all real numbers except -1 range: all real numbers e. g(f) = (x/(x+1))^3 = x^3/(x^3 + 3x^2 + 3x + 1) domain: all real numbers except -1 range: all real numbers confidence assessment: 2
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17:17:58 (f+g)(x) = x / (x + 1) + x^3 = (x^4 + x^3 + x) / (x + 1). Domain: x can be any real number except -1. (f * g)(x) = x^3 * x / (x+1) = x^4 / (x+1). Domain: x can be any real number except -1. (f / g)(x) = [ x / (x+1) ] / x^3 = 1 / [x^2(x+1)] = 1 / (x^3 + x), Domain: x can be any real number except -1 or 0 f(g(x)) = g(x) / (g(x) + 1) = x^3 / (x^3 + 1). Domain: x can be any real number except -1 g(f(x)) = (f(x))^3 = (x / (x+1) )^3 = x^3 / (x+1)^3. Domain: x can be any real number except -1 **
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RESPONSE --> self critique assessment: 3
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17:28:41 1.4.66 (was 1.4.60 graphs of |x|+3, -.5|x|, |x-2|, |x+1|-1, 2|x| from graph of |x|
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RESPONSE --> a. |x|+3. The graph is the same except all points have been raised by 3. b. (-1/2)|x| The graph is flipped upside down and the slope is half as steep. c. |x-2| The graph has been shifted 2 units to the right. d. |x+1| - 1 The graph has been shifted 1 unit down and 1 unit to the left. e. 2|x| The graph has twice the slope. confidence assessment: 2
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17:29:55 The graph of y = | x | exists in quadrants 1 and 2 and has a 'v' shape with the point of the v at the origin. It follows that: The graph of | x |+3, which is shifted 3 units vertically from that of | x |, has a 'v' shape with the point of the v at (0,3). The graph of -.5 | x |, which is stretched by factor -.5 relative to that of | x |, has an inverted 'v' shape with the point of the v at (0,0), with the 'v' extending downward and having half the (negative) slope of the graph of | x |. The graph of | x-2 |, which is shifted 2 units horizontally from that of |x |, has a 'v' shape with the point of the v at (2, 0). The graph of | x+1 |-1, which is shifted -1 unit vertically and -1 unit horizontally from that of | x |, has a 'v' shape with the point of the v at (-1, -1). The graph of 2 |x |, which is stretched by factor 2 relative to that of | x |, has a 'v' shape with the point of the v at (0,0), with the 'v' extending upward and having double slope of the graph of | x |. |x-2| shifts by +2 units because x has to be 2 greater to give you the same results for |x-2| as you got for |x|. This also makes sense because if you make a table of y vs. x you find that the y values for |x| must be shifted +2 units in the positive direction to get the y values for |x-2|; this occurs for the same reason given above For y = |x+1| - 1 the leftward 1-unit shift is because you need to use a lesser value of x to get the same thing for |x+1| that you got for |x|. The vertical -1 is because subtracting 1 shifts y downward by 1 unit **
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RESPONSE --> self critique assessment: 3
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17:39:51 1.4.71 (was 1.4.64 find x(p) from p(x) = 14.75/(1+.01x)
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RESPONSE --> p(x) = 14.75/(1+.01x) cross multiply 14.75 = p(1+.01x) 14.75 = p + .01*x*p 14.75 - p = .01*x*p 100(14.75 - p) = x*p (100(14.75 - p))/p = x x(p) = (100(14.75 - p))/p x(10) = (100(14.75 - 10)) / 10 x(10) = (100 * 4.75) / 10 x(10) = 10 * 4.75 = 47.5 which rounds up to 48 units sold confidence assessment: 3
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17:41:15 p = 14.75 / (1 + .01 x). Multiply both sides by 1 + .01 x to get (1 + .01 x) * p = 14.75. Divide both sides by p to get 1 + .01 x = 14.75 / p. Subtract 1 from both sides to get 1 x = 14.75 / p - 1. Multiply both sides by 100 to get = 1475 / p - 100. Put the right-hand side over common denominator p: = (1475 - 100 p) / p. If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **
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RESPONSE --> self critique assessment: 3
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17:42:22 What is the x as a function of p, and how many units are sold when the price is $10?
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RESPONSE --> x(p) = (100(14.75 - p)) / p x(10) = 47.5 which rounds to 48 units confidence assessment: 3
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17:42:36 If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **
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RESPONSE --> self critique assessment: 3
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