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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

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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

1 Answer 1

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

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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 $g'(x)=$ , wolframalpha.com/input/… –  math110 Apr 17 '13 at 9:33
    
Oh,Thank you,That's nice solution! –  math110 Apr 17 '13 at 9:42

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