#$&* course MTH 271 3/4 9:30 009. `query 9*********************************************
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Given Solution: `a The difference quotient would be [ f(x+`dx) - f(x) ] / `dx = [ (x+`dx)^2 - (x+`dx) + 1 - (x^2 - x + 1) ] / `dx. Expanding the squared term, etc., this is [ x^2 + 2 x `dx + `dx^2 - x - `dx + 1 - x^2 + x - 1 ] / `dx, which simplifies further to }[ 2 x `dx - `dx + `dx^2 ] / `dx, then dividing by the `dx we get 2 x - 1 + `dx. For x = 2 this simplifies to 2 * 2 - 1 + `dx = 3 + `dx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Originally, I forgot to carry the second half of the first part of the quotient across after squaring the (x + cx). After figuring out this algebra mistake, everything made perfect sense. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q1.4.40 (38)(was 1.4.34 f+g, f*g, f/g, f(g), g(f) for f=x/(x+1) and g=x^3 the requested functions and the domain and range of each. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For f+g: (x/x+1) + x^3 → x + (x^3(x+1))/ x+1 → x^4 + x^3 + x/ x+1 For f*g: (x/x+1) * x^3 → x^4/ (x+1) For f/g: x/(x + 1) / x^3 → x/x+1 * 1/x^3 → x/x^3 * (x+1) → x/ x^4 + x^3 → 1/ x^3 + x^2 For f(g): x^3/(x+1)^3 For g(f): (x/ x+1)^3 → x^3/ (x+1)^3 The domain for all of these expressions is that x can equal anything other than -1. For f/g, x cannot equal 0 either. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a (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^2), 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 1.4.66 (was 1.4.60 graphs of |x|+3, -.5|x|, |x-2|, |x+1|-1, 2|x| from graph of |x| YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: |x|+3: Same shape, raised 3 units. -.5|x|: Has a more “squashed” shape; it’s slope increases more gradually. |x-2|: This is moved to the right two units. |x+1|-1: Moved left one unit, and down one unit. 2|x|: This graph is the same, however it has a more dramatic slope. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q1.4.71 (was 1.4.64 find x(p) from p(x) = 14.75/(1+.01x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To obtain an inverse, you just switch values and solve: y = 14.75/ (1 +.01x) → x = 14.75/ (1 + 0.01y) → (1475 - 100p) /p = x(p) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat is the x as a function of p, and how many units are sold when the price is $10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When p= 10, then (1475 - 100p) /p = x(p) → (1475 - 100*10)/ 10 → 475/ 10 → 47.5 units. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!