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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 $x>0$.

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  • $\begingroup$ Did you compute it's derivative? $\endgroup$
    – tst
    Nov 25, 2012 at 2:34
  • $\begingroup$ $x$^$X$ 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. $\endgroup$
    – Abhinav Anand
    Nov 25, 2012 at 2:51
  • $\begingroup$ 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. $\endgroup$
    – Abhinav Anand
    Nov 25, 2012 at 3:10
  • $\begingroup$ This function is not defined for $x\in[-1,0]$, since you cannot neatly raise negative numbers to noninteger powers. So your domain is $(-\infty,-1)\cup(0,\infty)$. $\endgroup$
    – 2'5 9'2
    Nov 25, 2012 at 3:45
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    $\begingroup$ Did you check the answers in the link JavaMan posted? $\endgroup$
    – user856
    Nov 25, 2012 at 6:27

2 Answers 2

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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 ).

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  • $\begingroup$ 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 $x$. $\endgroup$
    – Abhinav Anand
    Nov 25, 2012 at 16:46
  • $\begingroup$ @AbhinavAnand: In this case $g(x)=(1+x)\ln(1+1/x)$ which is a decreasing function. $\endgroup$
    – Mhenni Benghorbal
    Nov 25, 2012 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.

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    $\begingroup$ This is probably circular reasoning. To know the derivative of $e^x$, you first need to deduce some properties of the number $e$ itself. $\endgroup$
    – JavaMan
    Nov 25, 2012 at 2:38
  • $\begingroup$ Fair enough, but its not clear what tools are available for approaching this problem. I'm just suggesting a possible approach. $\endgroup$
    – ackshooairy
    Nov 25, 2012 at 2:41
  • $\begingroup$ 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. $\endgroup$
    – Abhinav Anand
    Nov 25, 2012 at 2:42
  • $\begingroup$ 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. $\endgroup$
    – tst
    Nov 25, 2012 at 4:35
  • $\begingroup$ What would zero derivative do help in solving the problem? $\endgroup$
    – Abhinav Anand
    Nov 25, 2012 at 4:51

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