open qa 171

#$&*

course Mth 271

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

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Your solution:

Inner function is x^2 -3x +3

Outer function is u^3

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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. **

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

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Your solution:

Inner function is g(x) = x+1

Outer function is [g(x)]^-.5

confidence rating #$&*:

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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. **

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Self-critique (if necessary):

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

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Your solution:

2/3(9t+2) ^-1/3 * 9 = [18(9t + 2)^-1/3]/3 = 18/3cuberoot(9t +2)

confidence Rating:3

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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). **

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Self-critique (if necessary):

*** I assume that it is OK to write the solution the way I did it. I am not sure what is correct way to type 3rd root…. But I hope the idea is right.

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Self-critique Rating:

@& I can interpret what you mean, so it's no problem.

However by the order of operations your expression

(9t+2) ^-1/3

means 'raise (9t + 2) to the -1 power, then divide by 3'.

To express this as you intent you would simply group the -1/3 into parentheses

(9t+2) ^(-1/3)*@

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

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Your solution:

-1/2 (25 + x^2) ^ -3/2 * (2x) = [-2x (25 + x^2) ^ -3/2]/2 =

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

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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). **

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Self-critique (if necessary):

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

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Your solution:

-1/2 (25 + x^2) ^ -3/2 * (2x) = [-2x (25 + x^2) ^ -3/2]/2 =

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

confidence rating #$&*:

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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). **

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Self-critique (if necessary):

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#*&!

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