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calculus question

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  • tw813
    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
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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):

      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
    • The Webmaster
      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
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        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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      • Tony Wong
        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
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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 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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        • The Webmaster
          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
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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.

            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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          • OVIDIU NEGRILA
            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
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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 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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            • mfradon
              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
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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 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
                >
                >
                >
                >
                >
                >
                >
                >
                > ---------------------------------
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                >
                > 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!
                > Terms of Service.
                >
                >
                >
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                > Yahoo! Mail - Find what you need with new enhanced
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                >
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                >
                >
                >
                >
                >
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              • Tony Wong
                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
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                  I end up working out the same as what you suggested.

                  Thanks Myriam

                  --- mfradon <Myriam.Fradon@...> 說:


                  ---------------------------------
                  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
                  >
                  >
                  >
                  >
                  >
                  >
                  >
                  >
                  > ---------------------------------
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                • OVIDIU NEGRILA
                  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
                  • 0 Attachment
                    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
                    >
                    >
                    >
                    >
                    >
                    >
                    >
                    >
                    > ---------------------------------
                    > Yahoo! Groups Links
                    >
                    > 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!
                    > Terms of Service.
                    >
                    >
                    >
                    > ---------------------------------
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                    > Yahoo! Mail - Find what you need with new enhanced
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                    > [Non-text portions of this message have been removed]
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                    >
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