Assignment 16 Query

course Mth 271

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18:36:05

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

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

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

f'(2) = -2

confidence assessment: 3

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18:36:07

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 assessment: 3

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18:36:21

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

h'(t) = t^2-5t+6-2t^2-9t-10/(t^2+5t+6)^2

h'(t) = -t^2-4t-4/(t^2+5t+6)^2

h'(t) = -(t+2)^2/[(t+2)(t+3)]^2 = -1/(t+3)^2

confidence assessment: 3

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18:36:25

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 assessment: 3

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18:36:36

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

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RESPONSE -->

f'(x) = x^2 + 1(4x^3) - x^4+3(2x)/(x^2+1)^2

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

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

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

numerator = 0 when x = 0, 1 or -1

horizontal tangency is when x = -1, 0 or 1

confidence assessment: 3

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18:36:39

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 -->

self critique assessment: 3

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18:36:46

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 -->

confidence assessment: 3

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18:36:49

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 assessment: 3

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18:37:00

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) = 50t+100(1) - t+1750(50)/(50t+100)^2

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

p'(1) = -3.88

p'(10) = -.243

confidence assessment: 3

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18:37:03

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 assessment: 3

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