course Mth 271 ??w??~??|?{??assignment #016016. `query 16
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12:53:28 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) = [(x+1) ' (x-1) - (x-1) ' (x+1)] / (x-1)^2 f'(x) = [x - 1 - (x+1)] / (x-1)^2 f'(x) = -2 / (x-1)^2 At the point (2, 3) f'(2) = -2 / (2 - 1)^2 = -2 / 1^2 = -2 confidence assessment: 3
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12:53:43 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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13:05:56 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) h'(t) = [(t+2)' (t^2+5t+6) - (t^2+5t+6)' (t+2)] / (t^2+5t+6)^2 h'(t) = [(t^2+5t+6) - (2t+5)(t+2)] / (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 + 4t + 4) / (t^2 + 5t + 6)^2 confidence assessment: 3
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13:07:44 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 --> I see that both the numerator and the denominator had (t+2)^2 as a factor, and I could have factored them out to make my fraction simpler. self critique assessment: 2
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13:25:36 2.4.48 What are the points of horizontal tangency for(x^4+3)/(x^2+1)?
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RESPONSE --> I took the derivative and got (2x^5 + 4x^3 - 6x) / (x^2 + 1). I dont know how to find the points of horizontal tangency though. confidence assessment: 0
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13:30:47 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 --> I understand. I see that a tangent line will be horizontal when the slope is zero. I also see that the slope is zero when the numerator is zero. I could have then factored my numerator and set each factor equal to zero to find the points with a horizontal tangent line. self critique assessment: 2
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13:32:05 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 --> At a point where the tangent is horizontal, it is either the highest or lowest point on a curve. Near the point, the graph is leveling out until it becomes flat. confidence assessment: 2
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13:32:23 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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13:52:26 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)/[50(t+2)] P' = {(t+1750)' [50(t+2)] - [50(t+2)]' (t+1750)} / [50(t+2)]^2 P' = [50(t+2) - (t+1750)] / [50(t+2)]^2 P' = (49t - 1650) / [50(t+2)]^2 After day 1 P'(1) = [49(1) - 1650] / [50(1+2)]^2 P'(1) = -1601 / 150^2 = -1601 / 22,500 = approx -.071% After day 10 P'(10) = [49(10) - 1650] / [50(10+2)]^2 p'(10) = -1160 / 360,000 = approx. -.003% confidence assessment: 2
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13:56:35 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 --> I messed up when I took the derivative of g. I fogot the constant multiple rule. self critique assessment: 2
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