query 17

course mth 271

šÒþèî½s„àÞÚ‘‹ŸÂ–€üJ}]¯assignment #017

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.

017. `query 17

Applied Calculus I

11-18-2007

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

10:16:20

2.5.2 inner, outer fns for (x^2-3x+3)^3

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

RESPONSE -->

g(x)=x^2-2x+3

g(x)=z^3

confidence assessment: 3

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

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

10:16:30

The first function you evaluate is x^2 - 3x + 3.

You then cube this result.

So the breakdown to get f(g(x)) form is

f(z) = z^3

g(x) = x^2 - 3x + 3. **

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

RESPONSE -->

yup

self critique assessment: 3

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

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

10:17:01

2.5.8 inner, outer fns for (x+1)^-.5

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

RESPONSE -->

g(x)=x+1

f(x)=z^-.5

confidence assessment: 3

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

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

10:17:09

The first function you evaluate is x+1.

You then take this result to the -5 power.

So the breakdown to get f(g(x)) form is

f(z) = z^-.5

g(x) = x+1. **

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

RESPONSE -->

yup

self critique assessment: 3

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

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

10:18:33

2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule

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

RESPONSE -->

z=9t+2

n=2/3

z'=9

f'(t)=2/3(9t+2)^-1/3(9)

f'(t)=6(9t+2)^-1/3

confidence assessment: 3

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

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

10:19:01

This function is of the form u^(2/3), where u = 9 t + 2.

The derivative of f(t) = u^p is f ' (t) = p * u^(p-1) * u ', by the general power rule.

Here p = 2/3, and u ' = (9t + 2)' = 9 so we have

f ' (t) = p u^(p-1) * u ' = 2/3 * u^(2/3 - 1) * 9 = 6 * u^(-1/3), or since u = 9 t + 2

f ' (t) = 6 ( 9 t + 2)^(-1/3). **

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

RESPONSE -->

yup

self critique assessment: 3

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

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

10:20:49

2.5.32 der of f(x) = (25+x^2)^(-1/2) by gen power rule

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

RESPONSE -->

z=25+x^2

n=-1/2

z'=2x

f'(x)= -1/2(25+x2)^-3/2(2x)

f'(x)= (2x)/(-2(25+x^2)^3/2)

confidence assessment: 3

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

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

10:22:03

Using inner function u = 25+x^2 and outer function = (u)^(-1/2) we get

n u^(n-1) du/dx is -1/2 * u^(-3/2) * 2x = - x ( 25+x^2)^(-3/2). **

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

RESPONSE -->

oops forgot to simplify

self critique assessment: 3

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

"

Very good.

&#

Let me know if you have questions. &#