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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
1x 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 11
3!
n!
2!
n1 !
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! j0 exp n1 c
n1 ! and this equation says that
ec
.
e 11 1 1 1
3!
n!
2!
n1 !
b. Deduce that if n is a positive integer then
n 1
j! 0e
j0 e
.
n1 ! The desired inequality follows at once from the fact that if c 1 and n is a positive integer then
e
ec
n1 !
n1 !
c. By putting n 2 in the latter inequality, prove that e 3.
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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.
 Fall '08
 STAFF
 Math, Calculus

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