QUERY 16

course MTH 271

10/23,2100

assignment #016016. `query 16

Applied Calculus I

10-23-2009

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21:08:34

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?

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RESPONSE -->

f(x) = (x + 1) / (x - 1)

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

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

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

f ' (x)= 2

confidence rating:

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21:09:16

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?

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RESPONSE -->

confidence rating:

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

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21:12:10

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. **

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RESPONSE -->

self critique rating: 3

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21:18:27

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?

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RESPONSE -->

h(t)=(t+2)/(t^2+5t+6)

( (t^2 + 5t + 6)(1) - (t + 2)(2t + 5) ) / (t^2 + 5t + 6)^2

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

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

-(t + 2)(t + 2) / (t + 2)^2 (t + 3)^2

-1/(t + 3)^2

confidence rating:

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

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21:18:38

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**

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RESPONSE -->

self critique rating: 3

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21:23:09

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

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RESPONSE -->

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

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

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

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

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

2x=0

x=0, y=3 (0,3)

confidence rating:

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21:25:13

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. **

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RESPONSE -->

My mistake was in omitting the zeros obtained with the x^2.

self critique rating: 3

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21:26:35

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

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RESPONSE -->

With a horizontal tangent, the point of tangency would be atop a peak or at the bottom of it.

confidence rating:

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

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21:26:47

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. **

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RESPONSE -->

self critique rating: 3

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21:37:25

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?

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RESPONSE -->

P= (t + 1750) / (50t + 100)

P'(t)=( 1(50t + 100) - 50(t + 1750) ) / (50t + 100)^2

P'(t)= (50t + 100 - 50t - 87500) / (50t + 100)^2

P'(t)=-87400/(50t + 100)^2

P'(1)= -87400/(150)^2

P'(1)= -3.88

P'(10)= -87400/(600)^2

P'(10)= -0.243

confidence rating:

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21:37:38

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. **

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RESPONSE -->

self critique rating: 3

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You left one problem blank, but I assume you understood it with no trouble.

Keep up the great work.

rating:

QUERY 16

course MTH 271

10/23,2100

assignment #016016. `query 16

Applied Calculus I

10-23-2009

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

21:08:34

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 -->

f(x) = (x + 1) / (x - 1)

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

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

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

f ' (x)= 2

confidence rating:

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

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

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

21:09:16

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 rating:

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

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

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

21:12:10

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. **

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RESPONSE -->

self critique rating: 3

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21:18:27

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 -->

h(t)=(t+2)/(t^2+5t+6)

( (t^2 + 5t + 6)(1) - (t + 2)(2t + 5) ) / (t^2 + 5t + 6)^2

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

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

-(t + 2)(t + 2) / (t + 2)^2 (t + 3)^2

-1/(t + 3)^2

confidence rating:

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

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

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

21:18:38

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**

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RESPONSE -->

self critique rating: 3

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21:23:09

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

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

RESPONSE -->

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

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

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

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

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

2x=0

x=0, y=3 (0,3)

confidence rating:

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

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

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

21:25:13

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 -->

My mistake was in omitting the zeros obtained with the x^2.

self critique rating: 3

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21:26:35

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

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

RESPONSE -->

With a horizontal tangent, the point of tangency would be atop a peak or at the bottom of it.

confidence rating:

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

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

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

21:26:47

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 -->

self critique rating: 3

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21:37:25

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 -->

P= (t + 1750) / (50t + 100)

P'(t)=( 1(50t + 100) - 50(t + 1750) ) / (50t + 100)^2

P'(t)= (50t + 100 - 50t - 87500) / (50t + 100)^2

P'(t)=-87400/(50t + 100)^2

P'(1)= -87400/(150)^2

P'(1)= -3.88

P'(10)= -87400/(600)^2

P'(10)= -0.243

confidence rating:

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

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

21:37:38

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. **

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RESPONSE -->

self critique rating: 3

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I believe this is a repeat. In any case it looks good.

rating: