QUERY 17

course MTH 271

10/26,1600

assignment #017017. `query 17

Applied Calculus I

10-26-2009

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15:29:52

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

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RESPONSE -->

Inner: z=x^2 - 3x + 3

Outer: y=z^3

confidence rating: 3

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15:30:08

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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RESPONSE -->

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Self-critique Rating:ent: 3

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15:30:54

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

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RESPONSE -->

Inner: z=(x + 1)

Outer: y=z^(-1/2)

confidence rating: 3

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15:31:04

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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RESPONSE -->

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Self-critique Rating:ent: 3

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15:32:46

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

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RESPONSE -->

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

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

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

confidence rating: 2

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15:33:44

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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RESPONSE -->

I took my solution one step further by placing the (9t + 2)^(-1/3) as the denominator.

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Self-critique Rating:ent: 3

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15:37:08

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

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RESPONSE -->

nu^(n-1)du/dx

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

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

confidence rating: 2

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15:38:37

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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RESPONSE -->

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Self-critique Rating:ent: 3

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&#Very good work. Let me know if you have questions. &#