Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x\ge0$. Find maximum $$f(x)=\dfrac{e^{\frac{2x}{x+1}}-1}{x}$$

I think this maximum is $2$, I hope this problem have some nice solution,Thank you

share|cite|improve this question
What have you done? Why do you think that the answer is 2? – Nikita Evseev Apr 17 '13 at 8:50
because I find this $\lim_{x\to 0}f(x)=2$,if we have prove $f'(x)<0,x\ge 0$, that $f(x)\le 2$,this problem maybe have other nice methods – math110 Apr 17 '13 at 8:52
No maximum found when $x\geq 0$. – Takasima Senko Apr 17 '13 at 9:11
up vote 1 down vote accepted

$$f=\dfrac{e^{\frac{2x}{x+1}}-1}{x}$$ $$f'=\dfrac{{\Bigg(\frac{2x}{(1+x)^2}\cdot e^\left(\frac{2x}{1+x}\right)}\Bigg)-\Big(e^\left(\frac{2x}{1+x}\right) -1\Big)}{x^2}$$ $$f'=-\dfrac{(1+x^2)\cdot e^\left(\frac{2x}{1+x}\right) -(1+x)^2}{x^2(1+x)^2}$$ $$f'=\dfrac{-(1+x^2)\bigg(e^\frac{2x}{1+x}-1\bigg)+2x}{x^2(1+x)^2}\le0 \forall x\ge0$$

So, $f$ decreases for all $x\ge0$ because $f'\le0$, see this plot of $f'$ numerator.

and $$\lim_{x\to 0}f=2;\ \text{ :use L-Hospitals' Rule}$$

And at $x=-1$ function goes undefined and $f=-\infty$ And after which graph is as shown.

It rises till x=0 and then again steeps down. enter image description here

share|cite|improve this answer
Thank you,@exploringnet,but f' you have some wrong. – math110 Apr 17 '13 at 9:27
@exploringnet,why $f'\le 0$,have nice solve? – math110 Apr 17 '13 at 9:31
mean:why $e^{\frac{2x}{x+1}}\ge\dfrac{(1+x)^2}{x^2+1}$?have nice solve? I think we can use ugly methods:$g(x)=(x^2+1)e^{\frac{2x}{x+1}}-(x+1)^2$, then we have $g'(x)=$ ,… – math110 Apr 17 '13 at 9:33
Oh,Thank you,That's nice solution! – math110 Apr 17 '13 at 9:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.