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Unformatted text preview: Math 191 FINAL EXAM SOLUTIONS Fall 2000 1. (a) Using lHopitals rule, the limit is 1. (b) The only critical point is x = e 1 . ( f ( 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 Newtons method is x 1 = 1 . 01 = 1 . 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 2tan 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 theabove the trapezoids are larger than those that are below....
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 Fall '07
 BERMAN
 Critical Point

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