1873_solutions

# Large log x log x log x 1 log x log x 1 log x 1 log

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Unformatted text preview: at  is a nonzero real number and that S x R 1  x  0 . a. Prove that if for all x g x  x 1  x log 1  x S then g x  0 for every nonzero number x S. Solution: We begin by observing that if x S then g x   log 1  x . 1 1 We look first at the case   0. In this case we have S   , Ý . Whenever   x  0 we have 1  x  1 which gives us g x  0. Furthermore, whenever x  0 we have 1  x  1 which gives us 1 g x  0. Therefore, since g is strictly increasing on the interval  , 0 and strictly decreasing on the interval 0, Ý , the function g must have a maximum value only at 0 . Since g 0  0 we deduce that g x  0 for all x S 0. 1 Ý,  . Whenever x  0 we have 1 1  x  1 which gives us g x  0. Furthermore, whenever 0  x   we have 1  x  1 which gives us g x  0. Thus g is strictly increasing on the interval Ý, 0 and strictly descreasing on the 1 interval 0,  and so, once again, g has its maximum only at 0. Now we look at the case   0. In this case we have S  b. Prove that if log 1  x x for every nonzero number x S and if f 0   then the function f is continuous and strictly decreasing on S. Deduce that the inequality f x   holds for every positive number x S. fx  Solution: For every number x S 0 we have log 1  x gx  fx 1  x x 2 x2 and so it follows from part a that f x  0 for every nonzero number x S. The fact that f is continuous at the number 0 follows from the observation that log 1  x  . lim x x0 x 1x Now, to see why f must be strictly decreasing on S, look first at the case in which   0. Since f has a 1 1 negative derivative everywhere in the interval  , 0 and is continuous on the interval  , 0 we 1 know that f is strictly decreasing on  , 0 . In a similar way we know that f is strictly decreasing on the interval 0, Ý . Therefore f is strictly decreasing on S. Now fill in the details why f must be strictly decreasing on S when   0. Finally, since f 0  , the fact that f x   whenever x is a positive member of S follows at once from the fact that f is strictly descreasing. c. Prove that the inequality 1  x holds for every positive number x S. 258 1/x  e Solution: This fact follows at once from the fact that 1  x whenever x 8. S 1/x log 1  x x  exp  exp f x 0 a. By applying the higher order mean value theorem to the function exp on the interval 0, 1 , show that if n is a positive integer then there must exist a number c 0, 1 such that ec 1  1  1  . e  11 3! n! 2! n1 ! The higher order mean value theorem tells us that if n is a positive integer then there is a number c between 0 and 1 such that n exp 1  j exp 0 j! j0  exp n1 c n1 ! and this equation says that ec . e  11 1  1  1  3! n! 2! n1 ! b. Deduce that if n is a positive integer then n 1 j! 0e j0 e . n1 ! The desired inequality follows at once from the fact that if c  1 and n is a positive integer then e ec  n1 ! n1 ! c. By putting n  2 in the latter inequality, prove that e  3. From the fa...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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