#$&* course MTH 271 3.31.1211:45 AM 016. `query 16
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Given Solution: `a f ' (x) = [ (x+1)'(x-1) - (x+1)(x-1)'] / (x-1)^2 = [ (x-1) - (x+1) ] / (x-1)^2 = -2 / (x-1)^2. When x = 2 we get f ' (x) = f ' (2) = -2 / (2-1)^2 = -2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 2 g 7th edition 2.4.34 (was 2.4.30) der of (t+2)/(t^2+5t+6) What is the derivative of the given function and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (t+2) /(t^2 + 5t +6) à [(t + 2)’ *(t^2 + 5t + 6 ) - (t + 2) *( t^2 + 5t + 6)’]/ (t^2 + 5t +6)^2 à [(1)*(t^2 + 5t + 6) - (t+ 2)*(2t + 5)]/ (t^2 + 5t + 6 )^2 à (t^2 + 5t + 6 -2t^2 - 9t - 10)/ (t^2 + 5t + 6 )^2 à -t^2 - 4t - 4/ ((t^2 + 5t + 6 )^2 à - (t+2)^2 / [ (t + 2) ( t + 3) ]^2 à - 1 / (t + 3)^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 4b 7th edition 2.4.48 (was 2.4.44) What are the points of horizontal tangency for(x^4+3)/(x^2+1)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x^4 + 3)/(x^2 + 1) à [(x^4 +3)’ *(x^2 + 1) - (x^4 + 3)*(x^2 +1)]/ (x^2 + 1)^2 à [(4x^3)(x^2 + 1) - (x^4 + 3)(2x)]/ (x^2 +1)^2 à (4x^5 + 4x^3 - 2x^5 - 5x)/ (x^2 + 1)^2 à 2x^5 + 4x^3 - 5x/ (x^2 + 1)^2 à 2x(x^2 +3)(x^2 - 1)]/ (x^2 +1)^2. The points of horizontal tangency 0, -1, and 1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a the derivative is ( f ' g - g ' f) / g^2 = (4 x^3 ( x^2 + 1) - (2x) (x^4 + 3) ) / (x^2+1)^2 = [ 4 x^5 + 4 x^3 - 8 x^5 - 6 x ] / (x^2 + 1)^2 = -(2 x^5 + 4x^3 - 6x ) / (x^2 + 1)^2 = 2x (x^4 + 2 x^2 - 3) / (x^2+1)^2. The tangent line is horizontal when the derivative is zero. The derivative is zero when the numerator is zero. The numerator is 2x ( x^4 + 2 x^2 - 3), which factors to give 2x ( x^2 + 3) ( x^2 - 1). 2x ( x^2 + 3) ( x^2 - 1) = 0 when 2x = 0, x^2 + 3 = 0 and x^2 - 1 = 0. }2x = 0 when x = 0; x^2 + 3 cannot equal zero; and x^2 - 1 = 0 when x = 1 or x = -1. Thus the function has a horizontal tangent when x = -1, 0 or 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat would the graph of the function look like at and near a point where it has a horizontal tangent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph goes fully horizontal at a point of horizontal tangent. This is generally small window in the graph where there is no change in the y value for a moment. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 7 7th edition2.4.58 (was 2.4.54) 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = (t + 1750)/[50*(t + 2)] The rate of change function is the derivative of this equation. P = [50*(t + 2) - 50*(t + 1750)]/ [50(t + 2)]^2 à 50t + 100 - 50t - 87500/ 2500(t + 2)^2 à -874/ 25(t+2)^2 When t = 1, P = -3.88 When t = 10, P = - 0.243 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!