## calculus question

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• Hi all, I have the following apparent-to-be-simple calculus question, yet I found it not as easy as it seems (maybe because I don t know calculus anymore):
Message 1 of 8 , May 9, 2005
Hi all,

I have the following apparent-to-be-simple calculus question, yet I
found it not as easy as it seems (maybe because I don't know calculus
anymore):

define f(x)=[1+c/x]^x where c is some positive constant and x>0. Can it
be proved or disproved that f(x) is increasing (for x>0)?

Thanks a lot,

Tony
• I remember from Calc I about differentiating the function and then setting it equal to zero to finding the max or critical points. You can go one step further
Message 2 of 8 , May 9, 2005
I remember from Calc I about differentiating the function and then setting it equal to zero to finding the max or critical points.

You can go one step further and differnetiate a second time to check for concavity..meaning is the function increasing/decreasing to the right or left of the max or critical points.

To make it a little easier, take the natural log of f(x) (i.e. LN{f(x)} to get the "x" out of the exponent, and then exponentiate this expression.

I get d/dx( exp[xLN{1+c/x}] ) = {x * 1/(1+c/x) * (-c/x^2) + LN{1+c/x} } * f(x).

where f(x)=(1+c/x)^x.

Now set the first derivative equal to zero and solve for x to find the max.

Differentiate a second time, set this equal to zero, and you can check concavity.

Regards,

tw813 <tw813@...> wrote:
Hi all,

I have the following apparent-to-be-simple calculus question, yet I
found it not as easy as it seems (maybe because I don't know calculus
anymore):

define f(x)=[1+c/x]^x where c is some positive constant and x>0. Can it
be proved or disproved that f(x) is increasing (for x>0)?

Thanks a lot,

Tony

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• But it doesn;t seem obvious to me how we can solve for x in the equation of setting f (x)=0 which you wrote down below. In fact, that was the part I got stuck
Message 3 of 8 , May 9, 2005
But it doesn;t seem obvious to me how we can solve for
x in the equation of setting f'(x)=0 which you wrote
down below. In fact, that was the part I got stuck at.

Tony

--- The Webmaster <maintainer_wiz@...> 撰寫:

---------------------------------
I remember from Calc I about differentiating the
function and then setting it equal to zero to finding
the max or critical points.

You can go one step further and differnetiate a second
time to check for concavity..meaning is the function
increasing/decreasing to the right or left of the max
or critical points.

To make it a little easier, take the natural log of
f(x) (i.e. LN{f(x)} to get the "x" out of the
exponent, and then exponentiate this expression.

I get d/dx( exp[xLN{1+c/x}] ) = {x * 1/(1+c/x) *
(-c/x^2) + LN{1+c/x} } * f(x).

where f(x)=(1+c/x)^x.

Now set the first derivative equal to zero and solve
for x to find the max.

Differentiate a second time, set this equal to zero,
and you can check concavity.

Regards,

tw813 <tw813@...> wrote:
Hi all,

I have the following apparent-to-be-simple calculus
question, yet I
found it not as easy as it seems (maybe because I
don't know calculus
anymore):

define f(x)=[1+c/x]^x where c is some positive
constant and x>0. Can it
be proved or disproved that f(x) is increasing (for
x>0)?

Thanks a lot,

Tony

---------------------------------

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• Here are two quick ways to check. 1) sketch out some points on the graph to see what it looks like. 2) set up d/dx in a spreadsheet or Maple and solve for x.
Message 4 of 8 , May 10, 2005
Here are two quick ways to check.

1) sketch out some points on the graph to see what it looks like.
2) set up d/dx in a spreadsheet or Maple and solve for x.

Here's something else to notice, which first struck me but I didn't immediately act on it:

Lim as x -> oo [ (1 + c/x)^x ] = e^c, which implies that f(x) is increasing for x >0.

This should be enough, I think.

Regards,

Tony Wong <tw813@...> wrote:
But it doesn;t seem obvious to me how we can solve for
x in the equation of setting f'(x)=0 which you wrote
down below. In fact, that was the part I got stuck at.

Tony

--- The Webmaster <maintainer_wiz@...> ���g:

---------------------------------
I remember from Calc I about differentiating the
function and then setting it equal to zero to finding
the max or critical points.

You can go one step further and differnetiate a second
time to check for concavity..meaning is the function
increasing/decreasing to the right or left of the max
or critical points.

To make it a little easier, take the natural log of
f(x) (i.e. LN{f(x)} to get the "x" out of the
exponent, and then exponentiate this expression.

I get d/dx( exp[xLN{1+c/x}] ) = {x * 1/(1+c/x) *
(-c/x^2) + LN{1+c/x} } * f(x).

where f(x)=(1+c/x)^x.

Now set the first derivative equal to zero and solve
for x to find the max.

Differentiate a second time, set this equal to zero,
and you can check concavity.

Regards,

tw813 <tw813@...> wrote:
Hi all,

I have the following apparent-to-be-simple calculus
question, yet I
found it not as easy as it seems (maybe because I
don't know calculus
anymore):

define f(x)=[1+c/x]^x where c is some positive
constant and x>0. Can it
be proved or disproved that f(x) is increasing (for
x>0)?

Thanks a lot,

Tony

---------------------------------

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• we have the function f(x)=[1+c/x]^x and we want to check if is incresing or not for x 0 if we prove that f (x) 0 for every x 0 then everything is ok we can
Message 5 of 8 , May 11, 2005
we have the function f(x)=[1+c/x]^x and we want to check if is incresing or not for x>0
if we prove that f'(x)>0 for every x>0 then everything is ok

we can use the equality x^y=exp{yln(x)}

so f(x)=[1+c/x]^x =exp{xln(1+c/x)} (we can use the logarithm to see that this is true)

so we start to differentiate f, and we get

f'(x)=exp{xln(1+c/x)} * {xln(1+c/x)} ' =exp{xln(1+c/x)} *{ln(1+c/x) + x* [1/(1+c/x)]*(-c/x^x)}

=exp{xln(1+c/x)} *{ln(1+c/x) -c/(c+x)}

the exponetial part is obvously positive, so it remains to check if the second part of f'(x)

ln(1+c/x)-c/(c+x) is positive or not for x>0

I tried to sketch the graphs of ln(1+c/x) and of c/(c+x)

if x -> 0 then ln (1+c/x) -> oo and c/(c+x) - > 1 so ln(1+c/x)-c/(c+x) >0

if x - > oo then ln(1+c/x) -> 0+ and c/(c+x) - > 0+ , and it remains to prove is bigger ln(1+c/x) or c/(c+x) when x - > oo

I tried using the l'Hopital rule for the ratio betwen ln(1+c/x) and c/(c+x) and I got that the limit is greater than 1 when x -> oo , so it follows that ln(1+c/x)>c/(c+x)

so ln(1+c/x)-c/(c+x)>0 for every x>0 then f'(x)>0 it follows that f is incresing

if I did something wrong please let me know

Regards

The Webmaster <maintainer_wiz@...> wrote:
Here are two quick ways to check.

1) sketch out some points on the graph to see what it looks like.
2) set up d/dx in a spreadsheet or Maple and solve for x.

Here's something else to notice, which first struck me but I didn't immediately act on it:

Lim as x -> oo [ (1 + c/x)^x ] = e^c, which implies that f(x) is increasing for x >0.

This should be enough, I think.

Regards,

Tony Wong wrote:
But it doesn;t seem obvious to me how we can solve for
x in the equation of setting f'(x)=0 which you wrote
down below. In fact, that was the part I got stuck at.

Tony

--- The Webmaster ���g:

---------------------------------
I remember from Calc I about differentiating the
function and then setting it equal to zero to finding
the max or critical points.

You can go one step further and differnetiate a second
time to check for concavity..meaning is the function
increasing/decreasing to the right or left of the max
or critical points.

To make it a little easier, take the natural log of
f(x) (i.e. LN{f(x)} to get the "x" out of the
exponent, and then exponentiate this expression.

I get d/dx( exp[xLN{1+c/x}] ) = {x * 1/(1+c/x) *
(-c/x^2) + LN{1+c/x} } * f(x).

where f(x)=(1+c/x)^x.

Now set the first derivative equal to zero and solve
for x to find the max.

Differentiate a second time, set this equal to zero,
and you can check concavity.

Regards,

tw813 wrote:
Hi all,

I have the following apparent-to-be-simple calculus
question, yet I
found it not as easy as it seems (maybe because I
don't know calculus
anymore):

define f(x)=[1+c/x]^x where c is some positive
constant and x>0. Can it
be proved or disproved that f(x) is increasing (for
x>0)?

Thanks a lot,

Tony

---------------------------------

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• Please be careful : the fact that f has a positive limit for x - 0 and for x - oo doesn t mean f is positive on ]0,oo[. There are some nasty U-shaped
Message 6 of 8 , May 11, 2005
Please be careful : the fact that f' has a positive limit for x -> 0
and for x -> oo doesn't mean f' is positive on ]0,oo[. There are some
nasty U-shaped functions which are positive near 0 and oo and negative
on some x>0.

But for f(x)=[1+c/x]^x, there is no such problem. One can write it
as f(x)=exp(g(x)) and since exp is increasing, we juste have to prove
that g(x)=xln(1+c/x) is increasing on ]0,oo[.
We know that g'(x)=ln(1+c/x)-c/(c+x) and g''(x)=-c^2/(x(c+x)^2).
g'' is negative thus g' is decreasing, thus g'(x) is greater than
the limit of g' for x -> oo, which is 0. This means that g' is
positive and g is increasing.

Regards,

Myriam

--- In probability@yahoogroups.com, OVIDIU NEGRILA <diddy1512@y...> wrote:
> we have the function f(x)=[1+c/x]^x and we want to check if is
incresing or not for x>0
> if we prove that f'(x)>0 for every x>0 then everything is ok
>
> we can use the equality x^y=exp{yln(x)}
>
> so f(x)=[1+c/x]^x =exp{xln(1+c/x)} (we can use the logarithm to see
that this is true)
>
> so we start to differentiate f, and we get
>
> f'(x)=exp{xln(1+c/x)} * {xln(1+c/x)} ' =exp{xln(1+c/x)} *{ln(1+c/x)
+ x* [1/(1+c/x)]*(-c/x^x)}
>
> =exp{xln(1+c/x)} *{ln(1+c/x) -c/(c+x)}
>
> the exponetial part is obvously positive, so it remains to check if
the second part of f'(x)
>
> ln(1+c/x)-c/(c+x) is positive or not for x>0
>
> I tried to sketch the graphs of ln(1+c/x) and of c/(c+x)
>
>
> if x -> 0 then ln (1+c/x) -> oo and c/(c+x) - > 1 so
ln(1+c/x)-c/(c+x) >0
>
> if x - > oo then ln(1+c/x) -> 0+ and c/(c+x) - > 0+ , and it remains
to prove is bigger ln(1+c/x) or c/(c+x) when x - > oo
>
> I tried using the l'Hopital rule for the ratio betwen ln(1+c/x) and
c/(c+x) and I got that the limit is greater than 1 when x -> oo , so
it follows that ln(1+c/x)>c/(c+x)
>
> so ln(1+c/x)-c/(c+x)>0 for every x>0 then f'(x)>0 it follows that f
is incresing
>
> if I did something wrong please let me know
>
> Regards
>
> The Webmaster <maintainer_wiz@y...> wrote:
> Here are two quick ways to check.
>
> 1) sketch out some points on the graph to see what it looks like.
> 2) set up d/dx in a spreadsheet or Maple and solve for x.
>
> Here's something else to notice, which first struck me but I didn't
immediately act on it:
>
> Lim as x -> oo [ (1 + c/x)^x ] = e^c, which implies that f(x) is
increasing for x >0.
>
> This should be enough, I think.
>
> Regards,
>
>
>
> Tony Wong wrote:
> But it doesn;t seem obvious to me how we can solve for
> x in the equation of setting f'(x)=0 which you wrote
> down below. In fact, that was the part I got stuck at.
>
> Tony
>
>
> --- The Webmaster ¼¶¼g:
>
> ---------------------------------
> I remember from Calc I about differentiating the
> function and then setting it equal to zero to finding
> the max or critical points.
>
> You can go one step further and differnetiate a second
> time to check for concavity..meaning is the function
> increasing/decreasing to the right or left of the max
> or critical points.
>
> To make it a little easier, take the natural log of
> f(x) (i.e. LN{f(x)} to get the "x" out of the
> exponent, and then exponentiate this expression.
>
> I get d/dx( exp[xLN{1+c/x}] ) = {x * 1/(1+c/x) *
> (-c/x^2) + LN{1+c/x} } * f(x).
>
> where f(x)=(1+c/x)^x.
>
> Now set the first derivative equal to zero and solve
> for x to find the max.
>
> Differentiate a second time, set this equal to zero,
> and you can check concavity.
>
> Regards,
>
>
> tw813 wrote:
> Hi all,
>
> I have the following apparent-to-be-simple calculus
> question, yet I
> found it not as easy as it seems (maybe because I
> don't know calculus
> anymore):
>
> define f(x)=[1+c/x]^x where c is some positive
> constant and x>0. Can it
> be proved or disproved that f(x) is increasing (for
> x>0)?
>
> Thanks a lot,
>
> Tony
>
>
>
>
>
>
>
>
> ---------------------------------
>
> To visit your group on the web, go to:
> http://groups.yahoo.com/group/probability/
>
> To unsubscribe from this group, send an email to:
> probability-unsubscribe@yahoogroups.com
>
> Your use of Yahoo! Groups is subject to the Yahoo!
>
>
>
> ---------------------------------
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> Yahoo! Mail - Find what you need with new enhanced
>
> [Non-text portions of this message have been removed]
>
>
>
>
>
> ---------------------------------
>
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>
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>
>
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> ¤U¸ü Yahoo! Messenger
> http://messenger.yahoo.com.hk/
>
>
>
>
> ---------------------------------
>
> To visit your group on the web, go to:
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>
> To unsubscribe from this group, send an email to:
> probability-unsubscribe@yahoogroups.com
>
>
>
>
> ---------------------------------
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> Read only the mail you want - Yahoo! Mail SpamGuard.
>
> [Non-text portions of this message have been removed]
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> ---------------------------------
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• I end up working out the same as what you suggested. Thanks Myriam ... Please be careful : the fact that f has a positive limit for x - 0 and for x - oo
Message 7 of 8 , May 11, 2005
I end up working out the same as what you suggested.

Thanks Myriam

---------------------------------
Please be careful : the fact that f' has a positive
limit for x -> 0
and for x -> oo doesn't mean f' is positive on ]0,oo[.
There are some
nasty U-shaped functions which are positive near 0 and
oo and negative
on some x>0.

But for f(x)=[1+c/x]^x, there is no such problem. One
can write it
as f(x)=exp(g(x)) and since exp is increasing, we
juste have to prove
that g(x)=xln(1+c/x) is increasing on ]0,oo[.
We know that g'(x)=ln(1+c/x)-c/(c+x) and
g''(x)=-c^2/(x(c+x)^2).
g'' is negative thus g' is decreasing, thus g'(x) is
greater than
the limit of g' for x -> oo, which is 0. This means
that g' is
positive and g is increasing.

Regards,

Myriam

--- In probability@yahoogroups.com, OVIDIU NEGRILA
<diddy1512@y...> wrote:
> we have the function f(x)=[1+c/x]^x and we want to
check if is
incresing or not for x>0
> if we prove that f'(x)>0 for every x>0 then
everything is ok
>
> we can use the equality x^y=exp{yln(x)}
>
> so f(x)=[1+c/x]^x =exp{xln(1+c/x)} (we can use the
logarithm to see
that this is true)
>
> so we start to differentiate f, and we get
>
> f'(x)=exp{xln(1+c/x)} * {xln(1+c/x)} '
=exp{xln(1+c/x)} *{ln(1+c/x)
+ x* [1/(1+c/x)]*(-c/x^x)}
>
> =exp{xln(1+c/x)} *{ln(1+c/x) -c/(c+x)}
>
> the exponetial part is obvously positive, so it
remains to check if
the second part of f'(x)
>
> ln(1+c/x)-c/(c+x) is positive or not for x>0
>
> I tried to sketch the graphs of ln(1+c/x) and of
c/(c+x)
>
>
> if x -> 0 then ln (1+c/x) -> oo and c/(c+x) - > 1 so

ln(1+c/x)-c/(c+x) >0
>
> if x - > oo then ln(1+c/x) -> 0+ and c/(c+x) - > 0+
, and it remains
to prove is bigger ln(1+c/x) or c/(c+x) when x - > oo
>
> I tried using the l'Hopital rule for the ratio
betwen ln(1+c/x) and
c/(c+x) and I got that the limit is greater than 1
when x -> oo , so
it follows that ln(1+c/x)>c/(c+x)
>
> so ln(1+c/x)-c/(c+x)>0 for every x>0 then f'(x)>0 it
follows that f
is incresing
>
> if I did something wrong please let me know
>
> Regards
>
> The Webmaster <maintainer_wiz@y...> wrote:
> Here are two quick ways to check.
>
> 1) sketch out some points on the graph to see what
it looks like.
> 2) set up d/dx in a spreadsheet or Maple and solve
for x.
>
> Here's something else to notice, which first struck
me but I didn't
immediately act on it:
>
> Lim as x -> oo [ (1 + c/x)^x ] = e^c, which implies
that f(x) is
increasing for x >0.
>
> This should be enough, I think.
>
> Regards,
>
>
>
> Tony Wong wrote:
> But it doesn;t seem obvious to me how we can solve
for
> x in the equation of setting f'(x)=0 which you wrote
> down below. In fact, that was the part I got stuck
at.
>
> Tony
>
>
> --- The Webmaster 撰寫:
>
> ---------------------------------
> I remember from Calc I about differentiating the
> function and then setting it equal to zero to
finding
> the max or critical points.
>
> You can go one step further and differnetiate a
second
> time to check for concavity..meaning is the function
> increasing/decreasing to the right or left of the
max
> or critical points.
>
> To make it a little easier, take the natural log of
> f(x) (i.e. LN{f(x)} to get the "x" out of the
> exponent, and then exponentiate this expression.
>
> I get d/dx( exp[xLN{1+c/x}] ) = {x * 1/(1+c/x) *
> (-c/x^2) + LN{1+c/x} } * f(x).
>
> where f(x)=(1+c/x)^x.
>
> Now set the first derivative equal to zero and solve
> for x to find the max.
>
> Differentiate a second time, set this equal to zero,
> and you can check concavity.
>
> Regards,
>
>
> tw813 wrote:
> Hi all,
>
> I have the following apparent-to-be-simple calculus
> question, yet I
> found it not as easy as it seems (maybe because I
> don't know calculus
> anymore):
>
> define f(x)=[1+c/x]^x where c is some positive
> constant and x>0. Can it
> be proved or disproved that f(x) is increasing (for
> x>0)?
>
> Thanks a lot,
>
> Tony
>
>
>
>
>
>
>
>
> ---------------------------------
>
> To visit your group on the web, go to:
> http://groups.yahoo.com/group/probability/
>
> To unsubscribe from this group, send an email to:
> probability-unsubscribe@yahoogroups.com
>
> Your use of Yahoo! Groups is subject to the Yahoo!
>
>
>
> ---------------------------------
> Do you Yahoo!?
> Yahoo! Mail - Find what you need with new enhanced
>
> [Non-text portions of this message have been
removed]
>
>
>
>
>
> ---------------------------------
>
> To visit your group on the web, go to:
> http://groups.yahoo.com/group/probability/
>
> To unsubscribe from this group, send an email to:
> probability-unsubscribe@yahoogroups.com
>
> Your use of Yahoo! Groups is subject to the Yahoo!
>
>
> __________________________________
> 想即時收到新 email 通知？
> 下載 Yahoo! Messenger
> http://messenger.yahoo.com.hk/
>
>
>
>
> ---------------------------------
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• your remark is very correct, but I didn t stated my assumptions as being of general validity my ideea was to prove that f (x) is positive for every x 0 , not
Message 8 of 8 , May 11, 2005
your remark is very correct, but I didn't stated my assumptions as being of general validity

my ideea was to prove that f'(x) is positive for every x>0 , not only when x->0 and x->oo;

the exponetial part of the derivative is positive , so it remained to check for the second part

first of all I sketched the graph of the two functions, namely ln(1+c/x) and c/(c+x) for the case

x>0 and c is a positive constant , and both of them are positive , the problem was to prove that

ln(1+c/x)>c/(c+x) everywhere for x>0, c positive constant which is true

Please be careful : the fact that f' has a positive limit for x -> 0
and for x -> oo doesn't mean f' is positive on ]0,oo[. There are some
nasty U-shaped functions which are positive near 0 and oo and negative
on some x>0.

But for f(x)=[1+c/x]^x, there is no such problem. One can write it
as f(x)=exp(g(x)) and since exp is increasing, we juste have to prove
that g(x)=xln(1+c/x) is increasing on ]0,oo[.
We know that g'(x)=ln(1+c/x)-c/(c+x) and g''(x)=-c^2/(x(c+x)^2).
g'' is negative thus g' is decreasing, thus g'(x) is greater than
the limit of g' for x -> oo, which is 0. This means that g' is
positive and g is increasing.

Regards,

Myriam

--- In probability@yahoogroups.com, OVIDIU NEGRILA <diddy1512@y...> wrote:
> we have the function f(x)=[1+c/x]^x and we want to check if is
incresing or not for x>0
> if we prove that f'(x)>0 for every x>0 then everything is ok
>
> we can use the equality x^y=exp{yln(x)}
>
> so f(x)=[1+c/x]^x =exp{xln(1+c/x)} (we can use the logarithm to see
that this is true)
>
> so we start to differentiate f, and we get
>
> f'(x)=exp{xln(1+c/x)} * {xln(1+c/x)} ' =exp{xln(1+c/x)} *{ln(1+c/x)
+ x* [1/(1+c/x)]*(-c/x^x)}
>
> =exp{xln(1+c/x)} *{ln(1+c/x) -c/(c+x)}
>
> the exponetial part is obvously positive, so it remains to check if
the second part of f'(x)
>
> ln(1+c/x)-c/(c+x) is positive or not for x>0
>
> I tried to sketch the graphs of ln(1+c/x) and of c/(c+x)
>
>
> if x -> 0 then ln (1+c/x) -> oo and c/(c+x) - > 1 so
ln(1+c/x)-c/(c+x) >0
>
> if x - > oo then ln(1+c/x) -> 0+ and c/(c+x) - > 0+ , and it remains
to prove is bigger ln(1+c/x) or c/(c+x) when x - > oo
>
> I tried using the l'Hopital rule for the ratio betwen ln(1+c/x) and
c/(c+x) and I got that the limit is greater than 1 when x -> oo , so
it follows that ln(1+c/x)>c/(c+x)
>
> so ln(1+c/x)-c/(c+x)>0 for every x>0 then f'(x)>0 it follows that f
is incresing
>
> if I did something wrong please let me know
>
> Regards
>
> The Webmaster <maintainer_wiz@y...> wrote:
> Here are two quick ways to check.
>
> 1) sketch out some points on the graph to see what it looks like.
> 2) set up d/dx in a spreadsheet or Maple and solve for x.
>
> Here's something else to notice, which first struck me but I didn't
immediately act on it:
>
> Lim as x -> oo [ (1 + c/x)^x ] = e^c, which implies that f(x) is
increasing for x >0.
>
> This should be enough, I think.
>
> Regards,
>
>
>
> Tony Wong wrote:
> But it doesn;t seem obvious to me how we can solve for
> x in the equation of setting f'(x)=0 which you wrote
> down below. In fact, that was the part I got stuck at.
>
> Tony
>
>
> --- The Webmaster ���g:
>
> ---------------------------------
> I remember from Calc I about differentiating the
> function and then setting it equal to zero to finding
> the max or critical points.
>
> You can go one step further and differnetiate a second
> time to check for concavity..meaning is the function
> increasing/decreasing to the right or left of the max
> or critical points.
>
> To make it a little easier, take the natural log of
> f(x) (i.e. LN{f(x)} to get the "x" out of the
> exponent, and then exponentiate this expression.
>
> I get d/dx( exp[xLN{1+c/x}] ) = {x * 1/(1+c/x) *
> (-c/x^2) + LN{1+c/x} } * f(x).
>
> where f(x)=(1+c/x)^x.
>
> Now set the first derivative equal to zero and solve
> for x to find the max.
>
> Differentiate a second time, set this equal to zero,
> and you can check concavity.
>
> Regards,
>
>
> tw813 wrote:
> Hi all,
>
> I have the following apparent-to-be-simple calculus
> question, yet I
> found it not as easy as it seems (maybe because I
> don't know calculus
> anymore):
>
> define f(x)=[1+c/x]^x where c is some positive
> constant and x>0. Can it
> be proved or disproved that f(x) is increasing (for
> x>0)?
>
> Thanks a lot,
>
> Tony
>
>
>
>
>
>
>
>
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