course Mth 271 016. `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 Rating: I took the derivative and then substituted the value of x into the function. ********************************************* Question: `q 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: I used the quotient rule: f/g = f(x) = fg fg/g^2 f(x) = (t + 2)/(t^2 + 5t + 6) f(x) = [1(t^2 + 5t + 6) (t + 2)(2t + 5)]/(t^2 + 5t + 6)^2 f(x) = (t^2 + 5t + 6 2t^2 5t 4t 10)/[(t 3)(t 2)]^2 f(x) = (-t^2 -4t 4)/[(t-3)(t-2)]^2 f(x) = -(t 2)^2/[(t 3)(t 2)]^2 f(x) = -1/(t-3)^2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I feel very confident about this problem.
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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. ------------------------------------------------ Self-critique Rating: I used the quotient rule and took the derivative. ********************************************* Question: `q 2.4.48 (was 2.4.44) What are the points of horizontal tangency for(x^4+3)/(x^2+1)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(x) = (x^4+3)/(x^2 + 1) f/g = (fg fg)/g^2 f(x) = [4x^3(x^2 +1) (2x(x^4 + 3))]/(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 f(x) = 2x(x^4 + 4x^2 3)/(x^2 + 1)^2 2x(x^4 + 4x^2 3) = 0 2x = 0 x = 0 (x^4 + 4x^2 3) = 0 = (x^2 + 3)(x^2 1) (x^2 + 3) = 0 x^2 = -3 x = isqrt(3) (x^2 1) = 0 x^2 = 1 x = +-1 horizontal tangents are (-1, 0, 1) confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I feel very confident about this problem.
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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 Rating: I used the quotient rule to take the derivative and then solved the numerator for zero. ********************************************* Question: `qWhat would the graph of the function look like at and near a point where it has a horizontal tangent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would be an asymptote. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I feel confident about my solution.
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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 Rating: I dont quite understand your answer, would this be the same as an asymptote?
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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. ** (MOSTLY CORRECT) STUDENT SOLUTION TO NON-ASSIGNED PROBLEM WITH INSTRUCTOR COMMENTS IN BOLDFACE: f(x) = [x^3 + 3x + 2] / (x^2 - 1) f ' (x) = [(x^2 - 1)(x^3 + 3x + 2)' - (x^3 + 3x + 2) (x^2 - 1) ' ] / (x^2 -1)^2 Since the next step is correct I imagine this is just a typo on your part, but I wanted to be sure to point it out, just in case. f'(x) = (x^2 - 1)(3x^2 + 3) - (x^3 + 3x + 2) (2x] / (x^2 -1)^2 f'(x) = (3x^4 - 3x^2 + 3x^2 - 3) - ( 2x^4 +6x^2 + 4x)] / (x^2-1)^2 f'(x) = x^4 - 6x^2 -4x - 3 ] / (x^2 - 1)^2 f'(x) = x^4 - 6x^2 - 4x - 3 ] / x^4 - 2x^2 + 1 Except for signs of grouping this is the correct derivative. With signs of grouping it is written (x^4 - 6 x^2 - 4 x - 3)/(x^2 - 1)^2 or (x^4 - 6 x^2 - 4 x - 3)/(x^4 - 2 x^2 + 1). f'(x) = -3 - 4x - 3 f'(x) = -4x - 6 These last two steps are not correct. You can divide out common factors of the numerator and denominator but you can't divide out terms. For example (4 + 7) / (4 + 3) is not 7/3. (4 + 7) / (4 + 3) = 11 / 7. The 4's don't divide out. Neither do the x^4 terms, or the x^2 factors of the -6x^2 and 2x^2 terms of your (correct-up-to-that-point) expression. You did a very good job taking the derivative. That algebra error isn't one you should be making at this stage, so be sure to take note, but it doesn't detract from the fact that you've done a nice job applying the quotient rule. ------------------------------------------------ Self-critique Rating: I used the same procedure as the preceding problems. " "