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final_fa00_solutions

final_fa00_solutions - Math 191 FINAL EXAM SOLUTIONS Fall...

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Math 191 FINAL EXAM SOLUTIONS Fall 2000 1. (a) Using l’Hopital’s rule, the limit is 1. (b) The only critical point is x = e - 1 . ( f 0 ( x ) = (1 + ln x ) e x ln x ) (c) The absolute minimum occurs when x = e - 1 where the value is e - 1 /e . There is no absolute maximum since lim x →∞ = . (d) The linearization of f at x = 1 is the function l ( x ) = 1 + ( x - 1) = x . (e) The approximation from Newton’s method is x 1 = 1 . 01 = 1 - - 0 . 01 1 . 2. Separate variables and use partial fractions: Z dt y 2 - 4 = Z sin xdx to obtain 1 4 ln( y - 2 y + 2 ) = - cos x + C Substitute the initial condition to compute C = 1 + 1 4 ln 1 5 . 3. (a) 1 4 ( 1 2 + 2 4 5 + 2 + 2 4 5 + 1 2 ) = 31 20 (b) This problem is more difficult than intended. The approximation is an under- estimate of 2 tan - 1 (1) 1 . 57. The function 1 1+ t 2 is concave downwards on the interval ( - 1 / 3 , 1 3) and concave upwards for | t | > 1 3). The parts that are above the trapezoids are larger than those that are below. If one evaluates the difference between f ( ± 1 / 4) and f ( ± 3 / 4) and the points on the secants, then at ± 1 /
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