Query 9

course Mth 271

10/06 12:03am

009. `query 9*********************************************

Question: `q **** 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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Your solution:

[(X+DX)^2-(X+DX)+1-X^2-x+1]/DX

X^2+2X(DX)+DX^2-X-DX+1-X^2+X-1 (cancel opposite terms)

(2X(DX)-DX+DX^2)/DX

2X-1+DX

confidence rating:

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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. **

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Self-critique (if necessary):

OK

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Self-critique Rating:

OK

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Question: `q1.4.40 (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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Your solution:

x/(x+1)+x^3 x cannot equal -1 because you can’t divide by 0

(x/(x+1))*(x^3/1) = x^4/(x+1) x can’t equal -1 “ “

(x/(x+1))/(x^3/1) = 1/x^2(x+1) = 1/x^3 + x^2 ; x can’t equal -1 and 0 “ “

x^3/(x^3+1) x can’t equal -1 “ “

(x/(x+1)^3 = x^3/(x+1)^3 x can’t equal -1

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 **

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Self-critique (if necessary):

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Self-critique Rating:

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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|

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Your solution:

Abs(x)+3: V shaped graph with a y intercept at 3

-.5Abs(x): expanded V shape with its tip at (0, 0)

Abs(x-2): V shape with the tip 2 points right of the origin

Abs(x+1)-1: V shape with tip 1 left and 1 down from the origin

2Abs(x): Narrowed V shape with the tip at the origin

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 **

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Self-critique (if necessary):

OK

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Self-critique Rating:

OK

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Question: `q1.4.71 (was 1.4.64 find x(p) from p(x) = 14.75/(1+.01x)

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Your solution:

(1+.01x)p=14.75

14.75/p-100= x

14.75-100p/p= x

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 **

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Self-critique (if necessary):

OK

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

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Self-critique Rating:

OK

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Question: `qWhat is the x as a function of p, and how many units are sold when the price is $10?

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Your solution:

X = (1475-1000)/10

X= 475/10

X = 47.5

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 **

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Self-critique (if necessary):

OK

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Self-critique Rating:

OK

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&#Your work looks good. See my notes. Let me know if you have any questions. &#