query 16

course mth 271

sx???E?d????a??assignment #016016. `query 16

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

Applied Calculus I

11-15-2007

......!!!!!!!!...................................

09:21:13

11-15-2007 09:21:13

2.4.12 der of f(x) = (x+1)/(x-1) at (2,3)

What is the derivative of f(x) at the given point?

......!!!!!!!!...................................

NOTES -------> f'(2)= -2

.......................................................!!!!!!!!...................................

09:21:14

2.4.12 der of f(x) = (x+1)/(x-1) at (2,3)

What is the derivative of f(x) at the given point?

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

.................................................

......!!!!!!!!...................................

09:21:30

f ' (x) = [ (x+1)'(x-1) - (x+1)(x-1)'] / (x-1)^2 =

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

-2 / (x-1)^2.

When x = 2 we get f ' (x) = f ' (2) = -2 / (2-1)^2 = -2. **

......!!!!!!!!...................................

RESPONSE -->

yup

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:22:55

2.4.30 der of (t+2)/(t^2+5t+6)

What is the derivative of the given function and how did you get it?

......!!!!!!!!...................................

RESPONSE -->

using the quotient rule, i get (x^4-6x^2-4x-3)/((x^2-1)2)

confidence assessment: 2

.................................................

......!!!!!!!!...................................

09:24:22

we get (f ' g - g ' f) / g^2 = [ 1 ( t^2 + 5 t + 6) - (2t + 5)(t + 2) ] / (t^2 + 5 t + 6 )^2 =

[ t^2 + 5 t + 6 - ( 2t^2 + 9 t + 10) ] / (t^2 + 5 t + 6)^2 =

(-t^2 - 4 t - 4) / (t^2 + 5 t + 6)^2 ) =

- (t+2)^2 / [ (t + 2) ( t + 3) ]^2 =

- 1 / (t + 3)^2.

DER**

......!!!!!!!!...................................

RESPONSE -->

this is not problem #30

problem 30 is (x^3+3x+2)/(x^2-1)

self critique assessment: 3

Good solution.

.................................................

......!!!!!!!!...................................

09:28:01

2.4.48 What are the points of horizontal tangency for(x^4+3)/(x^2+1)?

......!!!!!!!!...................................

RESPONSE -->

the deriv is (-4x^6+2x^5-4x^3+6x)/((x^2+1)^2)

i dont know how to solve this for x

confidence assessment: 1

The keys are:

Horizontal tangency implies 0 derivative.

The numerator must be 0 for the fraction to be 0.

To find the zeros of a polynomial, do everything you can to factor it first.

.................................................

......!!!!!!!!...................................

09:32:28

the derivative is

( f ' g - g ' f) / g^2 =

(4 x^3 ( x^2 + 1) - (2x) (x^4 + 3) ) / (x^2+1)^2 =

[ 4 x^5 + 4 x^3 - 8 x^5 - 6 x ] / (x^2 + 1)^2 =

-(2 x^5 + 4x^3 - 6x ) / (x^2 + 1)^2 =

2x (x^4 + 2 x^2 - 3) / (x^2+1)^2.

The tangent line is horizontal when the derivative is zero. The derivative is zero when the numerator is zero.

The numerator is 2x ( x^4 + 2 x^2 - 3), which factors to give 2x ( x^2 + 3) ( x^2 - 1).

2x ( x^2 + 3) ( x^2 - 1) = 0 when 2x = 0, x^2 + 3 = 0 and x^2 - 1 = 0.

}2x = 0 when x = 0;

x^2 + 3 cannot equal zero; and

x^2 - 1 = 0 when x = 1 or x = -1.

Thus the function has a horizontal tangent when x = -1, 0 or 1. **

......!!!!!!!!...................................

RESPONSE -->

ok

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:33:02

What would the graph of the function look like at and near a point where it has a horizontal tangent?

......!!!!!!!!...................................

RESPONSE -->

the graph reaches that point but never crosses it

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:33:17

At or near a point of horizontal tangency the graph would become at least for an instant horizontal. This could occur at a peak (like a hilltop, which is level at the very top point) or a valley (level at the very bottom). It could also occur if an increasing function levels off for an instant then keeps on increasing; or if a decreasing function levels off for and instant then keeps decreasing. **

......!!!!!!!!...................................

RESPONSE -->

yup

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:34:45

2.4.58 defective parts P = (t+1750)/[50(t+2)] t days after employment

What is the rate of change of P after 1 day, and after 10 days?

......!!!!!!!!...................................

RESPONSE -->

f'(1)= -3.88

f'(10)= -.243

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:35:01

It doesn't look like you evaluated the rate of change function to get your result.

You have to use the rate of change function to find the rate of change. The rate of change function is the derivative.

The derivative is

( f ' g - g ' f) / g^2 =

( 1 * 50(t+2) - 50(t + 1750) ) / ( 50(t+2) ) ^ 2 =

-50 (1748) / ( 2500 ( t^2)^2 ) =

- 874 / ( 25 ( t + 2) ^ 2 ).

Evaluating when t = 1 and t = 10 we get -3.88 and -.243. **

......!!!!!!!!...................................

RESPONSE -->

yup

self critique assessment: 3

.................................................

Good work overall. Try to include a little more detail in your solutions, so I can spot the trouble in case you have an error.