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Possible Duplicate:
How to prove $(1+1/x)^x$ is increasing when $x>0$?

$$f(x)=(1+1/x)^x$$ Where $x>0$

I am in search to find a proof that the function $f(x)$ is always increasing in its any real number domain. As the above function always increasing a slight variation in the form of function will change the outcome in opposite way.That is when we change the exponent $x$ by ($1+x$) of the above function and letting all the expression on the right hand side intact, this new function will always be decreasing for real domain $x>0$.

Community
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Abhinav Anand
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  • Did you compute it's derivative? – tst Nov 25 '12 at 02:34
  • $x$^$X$ can be decided,then only if slight changes can be made with$(X+1)$^$x$ things begin to be look complicated,so i thought there is something intriguing to play with it. – Abhinav Anand Nov 25 '12 at 02:51
  • for $x$<0 answer is obvious.it is always increasing.when the exponent is $(x+1)$ this problem i found on net when could not solve the first one,but could not understand their proofs so posted here. – Abhinav Anand Nov 25 '12 at 03:10
  • This function is not defined for $x\in[-1,0]$, since you cannot neatly raise negative numbers to noninteger powers. So your domain is $(-\infty,-1)\cup(0,\infty)$. – 2'5 9'2 Nov 25 '12 at 03:45
  • my dear fellow,what kind of mathematics are you talking about,the function on which you commented is only not defined for $X$=-1.next time first be so sure. – Abhinav Anand Nov 25 '12 at 04:16
  • Which function is not defined only for $x=-1$. Also read the FAQ on how to write properly, you should use Latex commands to write mathematics. – tst Nov 25 '12 at 04:40
  • I am sorry about my writing style so much so that please dont feel it,otherwise i cant forget myself.The function of which domain you suggested me actually not quit so.I can only say that check it again.I very recently joint this awesome thing,so forgive those writing mistakes too. – Abhinav Anand Nov 25 '12 at 04:56
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    Did you check the answers in the link JavaMan posted? –  Nov 25 '12 at 06:27

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Here is an idea, it is based on the fact that the composition of two increasing functions is an increasing function. Let

$$ f(x) = \left(1+\frac{1}{x}\right)^x \,. $$

Now, consider the function

$$ g(x) = x\ln\left(1+\frac{1}{x}\right) $$

and prove that it is an increasing function. Then note that, $ f(x) = e^{g(x)} $ is a composition of two increasing functions ( since $e^x$ is an increasing function ).

Mhenni Benghorbal
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  • then the computed answer is that f($x$) is an increasing function but if i replace exponent of f($x$) ,$x$ by ($1+x$),then the function will be a decreasing function but it should be increasing too as our discussion ,i found this on net as i was inquiring on internet for f($x$) with exponent $x$. – Abhinav Anand Nov 25 '12 at 16:46
  • @AbhinavAnand: In this case $g(x)=(1+x)\ln(1+1/x)$ which is a decreasing function. – Mhenni Benghorbal Nov 25 '12 at 18:54
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Compute the derivative. Are you familiar with first derivative test? The derivative will tell you the slope of the function at any given point, you can use this information to tell you about how the function behaves.

For example, positive slope means an increasing function.

ackshooairy
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    This is probably circular reasoning. To know the derivative of $e^x$, you first need to deduce some properties of the number $e$ itself. – JavaMan Nov 25 '12 at 02:38
  • Fair enough, but its not clear what tools are available for approaching this problem. I'm just suggesting a possible approach. – ackshooairy Nov 25 '12 at 02:41
  • i have computed the first derivative trough taking log of the function but the derivative is even more complicated to decide whether the derivative function is always positive or whatever. – Abhinav Anand Nov 25 '12 at 02:42
  • If you compute the derivative you will see that is is zero when $\log(1+\frac 1x)=\frac{1}{1+x}$. Now you have to think when this does happen. – tst Nov 25 '12 at 04:35
  • What would zero derivative do help in solving the problem? – Abhinav Anand Nov 25 '12 at 04:51
  • @AbhinavAnand, how familiar are you with the subject? Do you know the first derivative test? – tst Nov 25 '12 at 11:51
  • I am extremely sorry that I overlooked your modest suggestion. The equation you mentioned and i arrived at after is a transcendental equation but still I reached at the proof through this.Don`t you have an email id.I have a wish to discuss this. – Abhinav Anand Nov 26 '12 at 15:56