course Mth 271 ÅÇYzчGJwwwɔassignment #009
......!!!!!!!!...................................
10:37:42 **** 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?
......!!!!!!!!...................................
RESPONSE --> [ f(x+`dx) - f(x) ] / `dx = [ (x+`dx)^2 - (x+`dx) + 1 - (x^2 - x + 1) ] / `dx [ x^2 + 2 x `dx + `dx^2 - x - `dx + 1 - x^2 + x - 1 ] /`dx [ 2 x `dx - `dx + `dx^2 ] / `dx 2 x - 1 + `dx. x=2 (2(2) - 1) + `dx = 3 + `dx confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:37:54 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. **
......!!!!!!!!...................................
RESPONSE --> yup self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:43:11 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.
......!!!!!!!!...................................
RESPONSE --> f(x)=x/(x+1) g(x)=x^3 a) (x+x^4+x^3)/(x+1) x cannot equal -1 b) x^4/(x+1) x cannot equal -1 c) x/(x^4+1) x cannot equal -1 d) x^3/(x^3+1) x cannot equal -1 e) (x)^3/ (x+1) x cannot equal -1 confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:45:16 (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 **
......!!!!!!!!...................................
RESPONSE --> forgot to cube the denominator on part e self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:51:05 1.4.66 (was 1.4.60 graphs of |x|+3, -.5|x|, |x-2|, |x+1|-1, 2|x| from graph of |x|
......!!!!!!!!...................................
RESPONSE --> y=abs(x) is centered at (0,0) y=abs(x)+3 graph moves up 3 and is centered at the point (0,3) y=-.5abs(x) is centered at (0,0) but is negative, and its points are exactly one half the points of y=abs(x) y=abs(x-2) graph moves two places to the right and is centered at (2,0) y=abs(x+1)-1 is centered at (0,0) but is moved to the left and down one y=2abs(x) is a narrower graph centered at (0,0) confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:52:01 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 **
......!!!!!!!!...................................
RESPONSE --> yup self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:52:44 1.4.71 (was 1.4.64 find x(p) from p(x) = 14.75/(1+.01x)
......!!!!!!!!...................................
RESPONSE --> x=1475/p -100 confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:54:28 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 **
......!!!!!!!!...................................
RESPONSE --> yup self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:55:58 What is the x as a function of p, and how many units are sold when the price is $10?
......!!!!!!!!...................................
RESPONSE --> if x= 1475/p -100, then x= 1475/10 -100 then x=47.5 units confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:56:11 If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **
......!!!!!!!!...................................
RESPONSE --> yup self critique assessment: 3
.................................................
"