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Find the minimum value of $(\frac{x^n}{n} + \frac{1}{x})$ for $n \ge 4$.

One possible approach could be by first writing $$ \left(\frac{x^n}{n} + \frac{1} {x}\right) = \left( \frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx} + \text{upto n terms}\right) $$ then using the property $\mathrm{AM} \ge \mathrm{GM}$, we would get $$\left(\frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx}+ \text{upto }n \text{ terms}\right) = \left( \frac{x^n}{n} + \frac{1}{x} \right) \ge \frac{n+1}{n}$$ But this does not hold good for negative $x$; I am inquisitive to know how to approach for that case?

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Approaching $x=0$ from the left gives a vertical asymptote going to negative infinity, implying there will never be a global minimum. I take it you're looking for local minima then? Are you familiar with the method of using derivatives to find local extrema? –  anon Aug 1 '11 at 23:30
    
yes,I am familiar of using the derivatives for finding the maxima and minima,however I got mired while using it here :/ –  Quixotic Aug 1 '11 at 23:41
    
You shouldn't have come across any problems. Can you tell me what the derivative of $x^n/n+1/x$ is? –  anon Aug 1 '11 at 23:44
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Ehhh... my fault,I somehow assumed as the minimum point as the minimum value :/,thanks a lot,however this A.M,G.M approach seems a bit more intuitive to me :) –  Quixotic Aug 1 '11 at 23:59
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The AM-GM approach is indeed nice. But the calculus deals just as easily, for example, with the minimum value of $$\frac{x^{\sqrt{2}}}{\sqrt{2}}+\frac{1}{x}$$ for positive $x$. –  André Nicolas Aug 2 '11 at 0:39
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1 Answer 1

up vote 2 down vote accepted

For $x \lt 0$ there is no minimum. As $x$ gets very close to $0$, the $\frac{1}{x}$ term gets very large and negative, faster than the $\frac{x^n}{n}$ can get positive (assuming $n$ is even-if $n$ is odd this term is negative, too).

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Thanks,I should have taken the limit first,one more doubt,how $n \ge 4$ is justified here?I mean what is the problem with $n=3$? –  Quixotic Aug 1 '11 at 23:37
    
@FoolForMath: There is no problem. The argument is the same no matter what natural number $n$ is. I encourage you to test and see for yourself using WolframAlpha. –  anon Aug 1 '11 at 23:42
    
@ anon:Yes,I noticed that but I was just wondering why the problem setter mentioned it explicitly. –  Quixotic Aug 1 '11 at 23:45
    
@FoolForMath: Doesn't look like there's any meaningful reason to set $n\ge 4$. Also, you might want to repair your spacebar key. :) –  anon Aug 1 '11 at 23:48
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It's a joke! You're supposed to put spaces after punctuation. No biggie. –  anon Aug 1 '11 at 23:58
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