Assignment 17

#$&*

course Mth 271

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

017. `query 17

*********************************************

Question: `q 1b 7th edition 2.5.2 inner, outer fns for (x^2-3x+3)^3

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The first function you evaluate is x^2 - 3x + 3. This function is evaluated and then ^3.

f(g(x))

f(z) = z^3

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

`a 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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

Question: `q 1e 7th edition 2.5.8 inner, outer fns for (x+1)^-.5

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

First evaluate x+1 g(x)

You then take this result to the -5 power = z^-.5

f(g(x))

f(z) = z^-.5

g(x) = x+1

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

`a 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. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

Question: `q 2b 7th edition 2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

u^(2/3)

u = 9 t + 2

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

p = 2/3, and u ' = (9t + 2)' = 9

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

= 6 ( 9 t + 2)^(-1/3)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

`a 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). **

STUDENT COMMENT: This explanation loses me. Is there any way you could explain it different. I am having a hard time following exactly where I messed up.

INSTRUCTOR RESPONSE: The power function rule (x^n) ' = n x ^ (n - 1) could be written just as well using u as the variable; it would read (u^n) ' = n u ^ (n - 1), where the ' now means the derivative with respect to u.

Thus (u^(2/3)) ' = 2/3 u^(2/3 - 1) = 2/3 u^(-1/3).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

Question: `q 2c 7th edition 2.5.28 der of f(x) = (25+x^2)^(-1/2) by gen power rule

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Decompose to inner function u = 25+x^2 and outer function = (u)^(-1/2)

n u^(n-1) du/dx is -1/2 * u^(-3/2) * 2x

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

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

Given Solution:

`a 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). **

STUDENT QUESTION

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

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

f(x) = (-12.5 - 1/2x^2) (2x)

Is what I had for my answer. I still don't quite understand where I am messing up. Although, this section I am having trouble with a little bit.

INSTRUCTOR RESPONSE

you appear to occasionally overlook the power function rule

in this case you have

(u^(-1/2)) ' = -1/2 u^(-1/2 - 1) = -1/2 u^(-3/2)

So your expression

f(x) = -1/2 (25 + x^2) (25 + x^2)' should have read

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

That appears to be the only step you're missing, so it's clear you are on the right track. I'm confident you'll clear this up, but be sure to let me know if not.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

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