open qa 92

#$&*

course Mth 271

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 + 2xdx + dx^2 -x -dx +1 - x^2 +x -1]/dx = [2xdx + dx^2 -dx]/dx = [dx(2x + dx -1)]/dx = 2x + dx -1

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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):

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Self-critique Rating:

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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:

F + g = x/(x+1) + x^3 = [x + x^3(x+1)]/(x+1) = [x^4 + x^3 +x]/(x+1)

Domain:

X + 1 =/ 0

X =/-1

(-infinity, -1) U (-1, infinity)

Range = all real numbers

[x/(x+1)]/x^3 = x/x^4 +3

Domain = all real numbers

Range = all real numbers

f*g = x/(x+1) * x^3 = x^4/(x+1)

Domain:

(-infinity, -1) U (-1, infinity)

Range: All real numbers

f(g) = x^3/(x^3 +1)

Domain: (-infinity, -1) U (-1, infinity)

Range: all real numbers

g(f) = x^3/(x+1)^3

Domain:

(-infinity, -1) U (-1, infinity)

Range: all real numbers

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:

|x| + 3 - moves up 3 points on the y axis.

-1/2|x| - will be vertically compressed by the factor of ½ and will reflect about the x axis

|x-2| = will shift horizontally to the right, 2 units

|x+1| -1shifts one unit to the left and moves vertically down 1 unit

2|x| = is vertically stretched by the factor of 2.

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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):

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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 + 0.01x)p = 14.75

p + 0.01 xp = 14.75

0.01xp = 14.75 - p

X = (14.75 - p) / 0.01p

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):

Is it necessary to multiply by 100??? Because my answer is basically correct minus that step, at least from what I can see.

Btw, I CAN NOT find this problem in the book!!! But the one in book is similar I guess.

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Self-critique Rating:

@& The expression

(14.75 - p) / 0.01p

should be written

(14.75 - p) / (0.01p)

to avoid any ambiguity regarding order of operations.

However your intent is clear.

That expression can be used in an obvious manner to answer the question about the value of x when p = 10.

However the expression

(14.75 - p) / (0.01p)

is not in simplified form.

The simplified form could be

100 * (14.75 - p) / p,

or the expression could be expanded to give

1475 / p - 100

which is correct as written but could be written

(1475 / p) - 100

to avoid a possible misreading of the order of operations.

*@

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 = (14.75 - 10)/0.01 * 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 **

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