asst7-f10-solns

asst7-f10-solns - MATH 137, FALL 2010 Assignment #7 sample...

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Unformatted text preview: MATH 137, FALL 2010 Assignment #7 sample solution 1. We note, first, that if b 0 (so- b 0) and a 6 = 0, then lim x axe- bx 2 = and lim x - axe- bx 2 = depending, respectively, on whether or not a > 0 or a < 0. [In- deed, given > 0 ( =big), if | x | > / | a | then | axe bx 2 | = | ax | e bx 2 > e b ( / | a | ) 2 > , and sign of the latter is determined by the sign of ax .] Hence in neither of these cases does f ( x ) = axe bx 2 even have an abso- lute maximum. Of course if a 0, then it is impossible for f (2) = 1. Thus we assume a > 0 and b > 0. We note that (i) f is a product of compositions of elementary functions, hence dif- ferentiable (hence continuous) on R ; and (ii) lim x f ( x ) = lim x ax e bx 2 = 0, by LHospitals rule; indeed lim x ax lim x e bx 2 = (indeterminate form) lim x f ( x ) = lim x d dx ax d dx e bx 2 = lim x a e bx 2 2 bx = 0 . By (ii) and (i), it is possible that the maximum may occur at x = 2. By (i), we have that if the maximum occurs at x = 2, then f (2) = 0. Hence, since f ( x ) = d dx axe- bx 2 = ae- bx 2 + axe- bx 2 (- 2 bx ) = a (1- 2 bx 2 ) e- bx 2 we solve 0 = f (2) = a (1- 2 b 2 2 ) e- b 2 2 to see that 0 = 1- 2 b 2 2 = 1- 8 b , so it must be the case that b = 1...
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This note was uploaded on 01/18/2012 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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asst7-f10-solns - MATH 137, FALL 2010 Assignment #7 sample...

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