query9

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course Mth 271

10/11 10p

uestion: `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?YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

(x^2+2dx^2+dx^2+1-x^2+x-1)dx

2dx^2+dx^2+x/dx= [x(3dx +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):

Where are the notes for this? I used what was in the book on pages 81-82 and that did not match the solution.

why is (x+dx)(x+dx) not x^2 + 2dx^2+dx^2 ?

are the points supposed to be x and f(x) = x^2-x+1 ?

Self-critique Rating:0

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You've had notes on difference quotients in previous assignment, and they are covered in the text material.

By the distributive law

(x + `dx)^2

= (x + `dx) * (x + `dx)

= x ( x + `dx) + `dx * (x + `dx)

= x^2 + x `dx + x `dx + `dx^2

= x^2 + 2 x `dx + `dx^2.

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Your expression

2dx^2+dx^2+x/dx

is therefore not correct, but if it was you would need to put parentheses around the entire numerator. As it is, only the x would be divided by `dx. With the required parentheses the expression would be

(2dx^2+dx^2+x)/dx

To repeat, this isn't correct, but you made an error in simplifying it:

x(3dx +1) = 3 x dx + 1,

whereas

2dx^2+dx^2+x = 3 dx^2 + x.

You can't factor an x out of that expression, since 3 dx^2 does not contain x as a factor.

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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

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the requested functions and the domain and range of each.

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Your solution:

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

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

f(g) = f(x^3)

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

onfidence Rating:2

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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):

I didn’t state the domain but I understand that the domain has to be such that the denominator can’t = 0, and I answered what was asked for in the last 2.

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Self-critique Rating:2

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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:

y=|x| (1,1)(-1,1)(2,2)(-2,2)

y=|x|+3 (1.4)(-1,4)(2,5)(-2,5)

y=-.5|x| (0,0)(1, -.5)(-1, -.5)

y=|x-2| (0,2)(1,1)(-1,3)

y=|x+1|-1 (0,0)(2,2)(-1,-1)(-2,0)

y=2|x| (0,0)(1,2)(-1,2)

all graphs form 2 straight lines reflected over the y-axis, except y=|x-1|-1 which is reflected across y=-1

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):n/a

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Self-critique Rating:3

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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:

14.75/(1+.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):

I understand what you did but I did not know what to do. Again, I can’t find the notes on this. They were not in the Lesson 8 or 9 notes.

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Self-critique Rating:0

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This is an exercise in algebra. You can't expect to find specific notes on every algebraic manipulation you might encounter.

You have an equation in p and x, and you need to solve it for x.

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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:

f(x) = x/(10p)

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):Is there any way you could make it clearer when a question is related to a previous question ? This is not at all clear, especially when trying to work through the problems with no notes.

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Self-critique Rating:0

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It should be clear on an assigned problem when I'm breaking it down to make in manageable.

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#