Query 9

course Mth 271

ÒŸÖ’šxWœ¡ÍÒÑ©•ôºµ´³Ï·–Zâassignment #009

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Applied Calculus I

07-05-2006

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10:18:11

**** 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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RESPONSE -->

2x-1

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10:19:00

** 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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RESPONSE -->

I understand how to do this but I do it a different way...is this ok....I was taught differently in high school and it stuck...

Show me the details of how you got the difference quotient. It should be 2 x - 1 + `ds, not just 2 x - 1. If you take the limit as `dx -> 0, you get 2 x - 1, but the `dx is definitely part of the difference quotient.

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10:25:21

1.4.38 (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.

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RESPONSE -->

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

x^3 (x+1)= x^4 +x^3

x^4+x^3+x/ (x+1)

Domain expect -1

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

= x^4/ x+1

Domain not -1

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

1/ x^3+ x^2

Domain not 0

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

Domain not -1

G(fx)= (x/x+1) ^3

Domain not -1

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10:27:04

** (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 **

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RESPONSE -->

for f/g ......How come that it should be (x^3 + x) ....I thought that it would be (x^3 + x^2).....also why is the domain not 0 or -1....I just thought it would be -1......thanks

It should be x^3 + x^2. However the denominator is still zero when x = 0 or x = -1. You have to exclude every value for which the function g is zero.

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10:35:11

1.4.64 (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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RESPONSE -->

F(x)= abs(x) is just an upward v centered at 0,0

f(x)= abs(x)+ 3 this doesnt do anything except change the center (0,3) its a shifter ....vertically up 3

f(x)= -.5 abs(x) it just flip flops it and makes the v 1/2 closer to the x axis.

f(x)= abs(x-2) this is the horizontal shifter so then it changes the center to (2,0) it moves over to the right because we do the opposite of whats inside

f(x)= (x+1) -1this moves the center to the left 1 and shiftes it vertically -1 units so the center is not (-1,-1)

y= 2 abs(x) this moves the v twice as close to the x axis becuase 2 is a strecher

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10:35:54

** 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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RESPONSE -->

You seem to have expalined it better than me ...but all in all I think I did ok and I understand it

Your explanation was good.

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10:40:24

1.4.69 (was 1.4.64 find x(p) from p(x) = 14.75/(1+.01x)

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p= 14.75/ 1+ 0.01x

p(1+ 0,01x)= 14.75

p+ .01xp= 14.75

.01xp= 14.75-p

xp= 1475- p/.01

x= (1475 - (p/.01)) /p

x= 1475/p - 100

x= 47.5

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10:40:42

** 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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RESPONSE -->

I understhand this ...this one was a tuffy

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10:41:15

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

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I already answered this in the previous question 47.5

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10:41:19

** If p = 10 then x = (1475 - 100 p) / p = (1475 - 100 * 10) / 10 = 475 / 10 = 47.5 **

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RESPONSE -->

ok

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Very good work. See my notes and let me know if you have questions.