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I have $f(x) = \displaystyle\frac{2e^{2x}}{x^2}$, and I 'm looking for minimums/maximums of the function. Now for some reason, I differentiate and only find $x=1$, while in the solution page it says $x=-1$ is also a minimum. Can anyone show me the RIGHT way to differentiate this, step by step? Do i

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Is it possible that the function should have $e^{x^2}$ as numerator? Then it would be symmetric with critical points $x\pm1$. –  Simon Markett Feb 17 '13 at 16:38
    
YOU ARE RIGHT i mistook it for e^2x thinking the square operation is on e^x. Thanks! –  Or Cyngiser Feb 17 '13 at 16:57

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up vote 1 down vote accepted

$$f(x) = \frac{2e^{2x}}{x^2}$$

Using the quotient rule and simplifying: $$f'(x) = \frac{4e^{2x}x^2 - 4x e^{2x}}{x^4} = \dfrac{4xe^{2x}(x - 1)}{x^4}$$

To find critical points (in particular, maximum and minimums), set $f'(x) = 0$: $$f'(x) = 0 \implies 4xe^{2x}(x - 1) = 0 \implies x= 1$$

So your solution is correct, and the answer key must have made a typo!

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Or: are you okay with this? Does this check out with your differentiation result? If the given function is as you wrote, then your solution to it is correct. –  amWhy Feb 17 '13 at 16:57
    
I see in the comments that you Is it possible that the function should have $e^{x^2}$ instead of $e^{2x} \neq e^{x^2}?$ Then it would be symmetric with critical points $x±1$. Let me know if you can obtain the desired result, after differentiating the correct function. –  amWhy Feb 17 '13 at 17:01
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When I saw the title I thought that it was your icon name. Why am....+ –  Babak S. Feb 17 '13 at 19:50

What you did is correct, $x=1$ is the only minimum (only critical point, even) of your function. Your solution page seems to be wrong.

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But if $x$ can assume negative values too, there is no global minimum value of $\frac{2e^x}{x^2}$,right? –  lab bhattacharjee Feb 17 '13 at 16:41

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